quantum tops

Quantum Mechanics of Rotating Molecules

W Happer

November 6 2013

1 Introduction

It is hard to find a discussion of rotating bodies in either elementary or advanced quantummechanics texts Perhaps the best available is the chapter on polyatomic molecules in thetext Quantum Mechanics by Landau and Lifschitz[1] These notes outline the basic quantummechanics of rotating bodies a key part of the physics of polyatomic molecules deformednuclei and some other parts of physics Since the notes were written in a hurry over themid-term break they undoubtedly contain many errors and unclear parts Feedback on howto improve the notes either in person or by e-mail would be welcome I will post updatedversions of these notes every day or two as we discuss rotating molecules over the next twoweek

Rigid rotors The simplest rotating system is a ldquorigid rotatorrdquo A classical example is aninfinitely thin straight rod that spins freely in space much like a twirling baton that hasbeen tossed by a band majorette A quantum mechanical example would be a rotating H2

molecule One can specify the orientation of the internuclear axis of the molecule with acolatitude angle θ and an azimuthal angle φ The rotational Hamiltonian of the molecule is

H = 2

2I perpL middot L (1)

where L is the angular momentum operator and I perp is the moment of inertia for rotationsaround any axis through the center of mass and perpendicular to the internuclear axis Theenergy eigenfunctions and quantized energies are

ψ jm = Y jm and E j = 2

2I perp j( j + 1) (2)

where Y jm = Y jm(θ φ) denotes a spherical harmonic The angular momentum quantumnumbers j = 0 1 2 and m = l l minus 1 minusl are given by the eigenvalue equations

L middot Lψ jm = j( j + 1)ψ jm

L3ψ jm = mψ jm (3)

Here L3 = x3 middotL is the projection of the angular momentum operator along the z axis x3 of afixed laboratory coordinate system The energy levels E j of (2) are (2 j +1)ndashfold degeneratesince all of the wave functions ψ jj ψ jjminus1 ψ jminus j have the same energies The azimuthalquantum number m has no effect on the energy

1

Symmetric tops The next simplest rotating system is the symmetric top A classicalexample is an elongated American football rotating freely in space with no air resistanceA quantum mechanical example is a molecule like methyl chloride ClCH3 with a three-foldrotational symmetry axis through the Cl and C nuclei Symmetric tops have three nonzeromoments of inertia two of which are identical Three parameters often taken to be theEuler angles αβγ are needed to specify the orientation of a symmetric top The first figureof the Wikipedia article on ldquoEuler anglesrdquo gives a good sketch of αβγ We can think of the symmetry axis as being defined by an aximuthal angle φ = α and by a colatitude angleθ = β The amount that the top has been rotated about its symmetry axis is given by thethird Eulers angle γ The rotational Hamiltonian of a symmetric-top molecule is

H = 2

2I perpL middot L +

2

2

983080

1

I ∥minus 1

I perp

983081

L23 (4)

Here L3 = x3 middot L is the projection of the angular momentum operator along the symmetryaxis x3 of the molecule The energy eigenfunctions and quantized energies are

ψ jml =

991770

2 j + 1

8π2 D jlowast

ml and E jl = 2

2I perp j( j + 1) +

2

2

983080

1

I ∥minus 1

I perp

983081

l2 (5)

Here D jlowastmk is the complex conjugate of the Wigner D-Function D j

mk = D jmk(α β γ ) The co-

efficientradic

(2 j + 1)(8π2) is for normalization A very through discussion of the D-Functionscan be found in Quantum Theory of Angular Momentum by Varshalovich Morskalev andKhersonskii [2] We will have more to say about the D-Functions later in these notes Themoment of inertia around the symmetry axis of the symmetric top is I ∥ and the two equalmoments about independent principal axes perpendicular to the symmetry axis are I perp If I perp gt I ∥ (eg ClCH3) the top is said to be prolate If I perp lt I ∥ (eg SO3) the top is said tobe oblate A symmetric top with I perp = I ∥ (eg CH4) is said to be a spherical top

The angular momentum quantum numbers j = 0 1 2 m = j j minus 1 minus j l = j j minus 1 minusl are given by the eigenvalue equations

L middot Lψ jml = j( j + 1)ψ jml

L3ψ jml = mψ jml

L3ψ jml = lψ jml (6)

The energy levels E jl of (5) are (2 j + 1)ndashfold degenerate since all of the wave functionsψ jj lψ jjminus1l ψ jminus jl have the same energies The quantum number l is the projectionof the angular momentum on the symmetry axis From (5) we see that E jl = E jminusl sounless l = 0 the energy states with specified values of j and m will have an additional two-fold degeneracy for plusmnl Various small perturbations always lift the degeneracy in plusmnl andproduce a closely-spaced energy doublet The characteristic doublet structure of symmetric-top energy levels plays an significant role in many physical processes

In the limit that the molecule is linear of nearly linear we will have I ∥ ≪ I perp From (5) wesee that as I ∥ rarr 0 E jl rarr 2l2(2I perp rarr infin if l = 0 States with quantum number l = 0 willhave so much energy that they are almost never excited for molecules in thermal equilibrium

2

Then the only states of (5) that occur are those with l = 0 for which (5) becomes

ψ jm0(α β γ ) =

991770

2 j + 1

8π2 D jlowast

m0 = 1radic

2πY jm(β α) and E jl =

2

2I perp j( j + 1) (7)

The states and energies (7) are the same as those of rigid rotor given by (2) aside fromthe factor of 1

radic 2π multiplying the spherical harmonic which ensures that the wave func-

tion is normalized for integrating over dα sin βdβdγ instead of sin βdβdα For the identityD jlowast

m0

radic

(2 j + 1)(4π) = Y jm see 417 (1) of Varshalovich et al[2]

Asymmetric tops The most complicated rotating system is the asymmetric top Aclassical example might be an Australian boomerang rotating freeing in space with no airresistance A quantum mechanical example would be a water molecule H2O for which thethree principal moments of inertia I 1 I 2 and I 3 are all different The molecular orientationcan be still specified by Euler angles αβγ just as for the symmetric top with the orientationof one of the principal axes say x3 defined by an azimuthal angle φ = α and by a colatitudeangle θ = β The amount that the top has been rotated about x3 is given by the thirdEulers angle γ The rotational Hamiltonian of an asymmetric-top molecule is

H = 2

2

1048616

L21

I 1+

L22

I 2+

L23

I 3

1048617

(8)

Here Lk = xk middot L for k = 1 2 3 is the projection of the angular momentum operator alongthe kth principle axis xk of the molecule The energy eigenfunctions are superpositions of symmetric-top eigenfunctions

ψ jmn =

991770

2 j + 1

8π2

sum

l

D jlowastmlvln with

sum

l

|vln|2 = 1 (9)

For asymmetric tops there is no general formula for the energies analogous to those of (2)and (5) The expansion coefficients vln are the eigenvectors of the Hamiltonian (8)

The angular momentum quantum numbers j = 0 1 2 m = j j minus 1 minus j and theenergy E jn of the wave function (9) are given by the eigenvalue equations

L middot Lψ jmn = j( j + 1)ψ jmn

L3ψ jmn = mψ jmn

Hψ jmn = E jnψ jmn (10)

The energy levels E jn of (10) are (2 j + 1)ndashfold degenerate since all of the wave functionsψ jjnψ jjminus1n ψ jminus jn have the same energies

2 Rotations

Let a molecule contain ν atoms with masses M 1 M 2 M ν We suppose that the centerof mass of the molecules is at rest at the origin of a fixed laboratory coordinates system

3

and that the atomic constituents are displaced from the center of mass by the distancesr1 r2 rν Then the masses and position vectors satisfy the constraint

sum

micro

M micrormicro = 0 (11)

We introduce Cartesian basis vectors along the axes of the laboratory coordinate system

x = x1 y = x2 and z = x3 (12)

We can write the position vector of the microth atom in the molecule as

rmicro =3sum

j=1

x jr jmicro equiv x jr jmicro where r jmicro = x j middot rmicro (13)

In (13) and elsewhere we will often use the Einstein convention that repeated Cartesianindices of vector or tensor elements are to be summed over their three possible values Thebasis vectors (12) form a right-handed orthonormal coordinate system such that

xi middot x j = δ ij and xi times x j = ϵijkxk (14)

The antisymmetric unit tensor ϵijk is defined by

ϵijk =

1 if ijk = 123 231 312minus1 if ijk = 213 132 321

0 otherwise(15)

We see from (14) that antisymmetric unit tensor can be written as

(xi times x j) middot xk = xi middot (x j times xk) = ϵijk (16)

We will often use (16) to write the cross product of two vectors A = xiAi and B = x jB j as

A times B = AiB jxkϵijk (17)

On the right side of (17) it is understood that the repeated indices i j and k are summedover

Rigid molecules We assume that the center of mass of the molecule at rest and that themolecules are rigidly attached to each other so they can only move if the entire moleculerotates with instantaneous angular velocity Ω The velocity of the microth atom must be

vmicro = Ω times rmicro (18)

Perturbation theory can be used to account for small vibrational contributions to the ve-locities when the constituent atoms of the molecule move with respect to each other Theangular velocity Ω can be written in terms of the basis vectors (12) as

Ω = Ω jx j (19)

4

We can use (18) to calculate the classical angular momentum L (in units of ) of the moleculeabout its center of mass

L =sum

micro

rmicro times M microvmicro =sum

micro

M micrormicro times (Ω times rmicro)

=sum

micro

M micrormicro times (Ωir jmicroxk)ϵijk =sum

micro

M microrsmicroΩir jmicroϵijkϵsktxt

=sum

micro

M micro(r jmicroΩir jmicroxi minus rimicroΩir jmicrox j) (20)

Here we used the simple identity that can be verified by inspection

ϵijkϵuvk = (xi times x j) middot (xu times xv) = δ uvij = δ iuδ jv minus δ ivδ ju (21)

We often write (20) as the formally simpler expression

L = I middotΩ (22)

The moment-of-inertia dyadic is

I =sum

micro

M micro(r jmicror jmicroxixi minus r jmicrorimicrox jxi) =sum

micro

M micro(r2microUminus rmicrormicro) (23)

We will use ldquomath boldfacerdquo fonts to denote dyadics ldquodirect productsrdquo of two vectors asopposed to dot-products or a cross-products An example is the unit dyadic used in (23)

U = xixi = x1x1 + x2x2 + x3x3 (24)

The matrix elements of the inertial dyadic between laboratory basis vectors are the elementsof the inertial tensor

I ij = xi middot I middot x j =sum

micro

M micro(r2microδ ij minus rimicror jmicro) (25)

We see that the inertial tensor (25) is real and symmetric I ij = I ji It therefore has threereal eigenvalues I k for k = 1 2 3 and three orthogonal eigenvectors which we denote by xkAside from normalization the eigenvectors are defined by the eigenvalue equation

I middot xk = I kxk (26)

The doubled index on the right of (26) is not summed over We will call the eigenvectorsxk ldquobody-fixed basis vectorsrdquo and we will enumerate them so they have orthonormalityproperties completely analogous to those of (14) and (24) that is

xi middot x j = δ ij xi times x j = ϵijk xk and xixi = U (27)

The eigenvalues I k of the inertial tensor are the ldquoprincipal momentsrdquo From(26) (27) and(23) we see that

I k = xk middot I middot xk =sum

micro

M micro(r2micro minus |xk middot rmicro|2) =sum

microj=k

M micro|x j middot rmicro|2 ge 0 (28)

5

According to (28) the principal moments cannot be negative although for linear moleculeslike CO2 one of the moments can be nearly zero so the molecule can be well approximatedas a rigid rotor For the remainder of this discussion we will assume that all three principalmoments are positive and of comparable sizes This is always true for nonlinear moleculesconsisting of three or more atoms

Multiplying both sides of (26) on the right by xk summing over the repeated index kand using the completeness property of (27) xkxk = U we find that the inertial dyadic canbe written as

I =sum

k

I kxkxk (29)

Since we assume that none of the principal moments I k is zero we see by inspection of (29)that the inverse inertial dyadic is

Iminus1 =sum

k

xkxk

I k with I middot Iminus1 = Iminus1 middot I = U (30)

If we multiply both sides of (22) on the left by Iminus1middot and recall that Iminus1middot is symmetric we find

Ω = Iminus1 middot L = L middot Iminus1 (31)

Using (18) we find that he kinetic energy H of the rotating molecule is

H =sum

micro

1

2M microvmicro middot vmicro =

sum

micro

1

2M micro(Ω times rmicro) middot (Ω times rmicro)

= 1

2

sum

micro

M microΩir jmicroΩurvmicroϵijkϵuvk = 1

2

sum

micro

M microΩir jmicroΩurvmicroδ uvij

= 1

2

sum

micro

M micro(Ωir jmicroΩir jmicro minus Ωir jmicroΩ jrimicro) = 1

2ΩiI ijΩ j (32)

From (32) and (31) we see that we can write the energy as

H = 1

2Ω middot I middotΩ =

2

2 L middot Iminus1 middot L (33)

The Hamiltonian (33) is often taken to be the starting point for discussing rotations of rigidbodies in both classical and quantum mechanics The Hamiltonians (4) and (8) come from(33) with the angular momentum L interpreted to be a quantum mechanical operator

21 Euler angles

The orientation of a rigid body is conveniently described with the Euler angles αβγ whichparameterize the relative orientation of laboratory-fixed basis vectors xk and body-fixed basisvectors x j We can use the unit dyadic (24) to write the body-fixed basis vectors x j in termsof the laboratory-fixed basis vectors xk

x j = U middot x j = xkX kj (34)

6

The elements of the direction-cosine matrix are

X kj = xk middot x j (35)

The direction-cosine dyadic is defined to be

X = xkx jX kj = x jx j (36)

The direction-cosine dyadic rotates laboratory-fixed unit vector to body-fixed unit vectors

x j = X middot x j (37)

We will write X as the product of three successive rotations

X = Cprimeprime middot Bprime middot A (38)

The dyadic A rotates the laboratory basis vectors x j by an angle α about the axis x3 intothe primed basis vectors

xprime j = A middot x j = xkAkj (39)

The elements of the 3times 3 matrix A are

Akj = xk middot A middot x j with A =

cosα minus sinα 0sinα cosα 0

0 0 1

(40)

In analogy to (36) we can write the rotation dyadic A as

A = xkx jAkj = xprime jx j (41)

The rotation dyadic A is unitary and the elements Akj are real so that

Aminus1 = x jxkAminus1 jk = Adagger = x jxkAkj = x jxprime

j or Aminus1 jk = Akj (42)

The middle dyadic Bprime of (38) rotates the primed basis vectors xprime j by an angle β about

the axis xprime2 (the ldquoline of nodesrdquo) into the double-primed basis vectors

xprimeprime j = Bprime middot xprime

j = xprimekBkj (43)

The elements of the 3times 3 matrix B are

Bkj = xprimek middot Bprime middot xprime

j or B =

cos β 0 sinβ 0 1 0

minus sin β 0 cosβ

(44)

In analogy to (36) we can write the rotation dyadic Bprime as

Bprime = xprimekxprime

jBkj = xprimeprime j xprime

j (45)

In analogy to (42) we can write

Bprimeminus1 = xprime

jxprimekBminus1

jk = Bprimedagger = xprime

jxprimekBkj = xprime

jxprimeprime j or Bminus1

jk = Bkj (46)

7

The leftmost dyadic Cprimeprime of (38) rotates the double-primed basis vectors x primeprime j by an angle γ

about the axis xprimeprime3 into the body-fixed basis vectors

x j = Cprimeprime middot xprimeprime j = xprimeprime

kC kj (47)

The elements of the 3times 3 matrix C are

C kj = xprimeprimek middot Cprimeprime middot xprimeprime

j or C =

cos γ minus sin γ 0sin γ cos γ 0

0 0 1

(48)

In analogy to (36) we can write the rotation dyadic Cprimeprime as

Cprimeprime = xprimeprimekxprimeprime

jC kj = x jxprimeprime j (49)


Cprimeprimeminus1 = xprimeprime

j xprimeprimekC minus1 jk = C

primeprimedagger = xprimeprime j xprimeprime

kC kj = xprimeprime j x j or C minus1 jk = C kj (50)

We can use (39) and (45) to write

Bprime = A middot x jxk middot Aminus1B jk

= A middot B middot Aminus1 (51)

where we noted from (42) that xprimek = xk middot Aminus1 and we defined the unprimed rotation dyadic

B asB = x jxkB jk (52)

Proceeding in a similar manner and using (51) we find

Cprimeprime = Bprime middot xprime jxprime

k middot Bprimeminus1C jk = A middot B middot Aminus1 middot xprime

jxprimek middot A middot Bminus1 middot Aminus1C jk

= A middot B middot x jxk middot Bminus1 middot Aminus1C jk = A middot B middot C middot Bminus1 middot Aminus1 (53)

where we noted from (41) and (42) that Aminus1 middotxprime j = x j and xprime

j middotA = x j and where we definedthe unprimed rotation dyadic C as

C = x jxkC jk (54)

Using (51) and (53) we find that we can write the rotation dyadic (38) as

X = Cprimeprime middot Bprime middot A = A middot B middot C (55)

orX = ABC (56)

Inserting the 3times3 matrices for A B and C from (40) (44) and (48) into (56) we find matrixdirection cosines X kj = xk middot x j between laboratory-fixed axes xk and body-fixed axes x j

X =

cosα cos β cos γ minus sinα sin γ minus cosα cos β sin γ minus sinα cos γ cosα sin β sinα cosβ cos γ + cosα sin γ minus sinα cos β sin γ + cosα cos γ sinα sin β

minus sin β cos γ sin β sin γ cosβ

(57)

We can parameterize every possible rotation with Euler angles in the range

0 le α lt 2π 0 le β lt π 0 le γ lt 2π (58)

8

22 Spin matrices

If we replace the finite rotation angle α by an infinitesimal angle |δα| ltlt 1 we can expandof (40) as a power series to terms of order δα to find the infitesimal rotation

A = U minus iδαS 3 (59)

Here the unit matrix is

U =

1 0 00 1 00 0 1

(60)

Here S 3 is one of the three spin matrices

S 1 = 1

i

0 0 00 0 10 minus1 0

S 2 = 1

i

0 0 minus10 0 01 0 0

S 3 = 1

i

0 1 0minus1 0 0

0 0 0

(61)

One can readily see that the elements of the spin matrices are given by

(S i) jk = ϵijk

i (62)

The elements of the antisymmetric unit tensor were given by (15) Squaring the spin matricesof (61) we find

S 21 =

0 0 00 1 00 0 1

S 22 =

1 0 00 0 00 0 1

S 23 =

1 0 00 1 00 0 0

(63)

Summing the squares we find

S iS i = 2U = s(s + 1)U (64)

It is also straightforward to demonstrate that the matrices (61) satisfy the commutationrelations

[S i S j] = iϵijkS k (65)

We conclude that the three laboratory-fixed basis vectors xk act like the three independentspin components of a particle of spin quantum number s = 1

We can use (62) (63) and the 3 times 3 unit matrix U with matrix elements U jk = δ jk asbasis matrices to write the matrix (39) for a finite rotation angle α as

A = U minus S 23 minus iS 3 sin α + S 23 cos α = eminusiαS 3 (66)

As indicated in (66) we can also write A = eminusiαsect3 by expanding the exponential in a powerseries ex = 1 + x + x22 + x33 + middot middot middot with x = minusiαS 3 and noting that for i = 1 2 3

S i = S 3i = S 5i = S 7i = middot middot middotS 2i = S 4i = S 6i = S 8i = middot middot middot (67)

9

There is no sum on repeated indices i in (67) In like manner we can show that

C = U minus S 23 minus iS 3 sin γ + S 23 cos γ = eminusiγS 3

B = U minus S 22 minus iS 2 sin β + S 22 cos β = eminusiβS 2 (68)

Equivalent dyadic spin matrices are readily constructed for example

S3 = x jxk(S 3) jk Sprime

2 = xprime jxprime

k(S 2) jk Sprimeprime

3 = xprimeprime j xprimeprime

k(S 3) jk (69)

The rotation dyadics of (42) (44) and (48) for the individual Euler angles become

Amiddot = exp(minusiαS3middot) or A = Uminus S3 middot S3 minus iS3 sin α + S3 middot S3 cos α (70)

Bprimemiddot = exp(minusiβ Sprime2middot) or Bprime = Uminus Sprime2 middot Sprime2 minus iSprime2 sin β + Sprime2 middot Sprime2 cos β (71)

Cprimeprimemiddot = exp(minusiγ Sprimeprime3middot) or Cprimeprime = Uminus Sprimeprime3 middot Sprimeprime3 minus iSprimeprime3 sin γ + Sprimeprime3 middot Sprimeprime3 cos γ (72)

Here it is understood that for any dyadic operator M constructed from 3 times 3 matrices inanalogy to (69) the exponential function is

exp(Mmiddot) = U middot +M middot + 1

2(Mmiddot)2 +

1

3(Mmiddot)3 + middot middot middot (73)

The full rotation dyadic (55) becomes

Xmiddot = exp(minusiγ Sprimeprime3middot) exp(minusiβ Sprime2middot)exp(minusiαS3middot)= exp(minusiαS3middot) exp(minusiβ S2middot) exp(minusiγ S3middot) (74)

23 Rotational velocity

We can write the angular velocity Ω of (18) in terms of the rates of change of Euler anglesand their axes of rotation as

Ω = αx3 + β xprime2 + γ xprimeprime

3 (75)


xprime2 = xkAk2 = minusx1 sinα + x2 cosα (76)

Noting that xprimeprime3 = x3 we use (36) and (57) to write

xprimeprime3 = xkX k3 = x1 cosα sin β + x2 sinα sin β + x3 cosβ (77)

Substituting (76) and (77) into (75) we find that the rotational velocity can be written as

Ω = Ω jx j (78)

where the coefficients Ω j of the laboratory-fixed unit vectors x j can be described by thematrix equation

Ω1

Ω2

Ω3

=

0 minus sinα cosα sin β 0 cosα sinα sin β 1 0 cosβ

α

β γ

(79)

10

The inverse of (79) is

α

β γ

= 1

sin β

minus cosα cos β minus sinα cos β sin β minus sinα sin β cosα sin β 0

cosα sinα 0

Ω1

Ω2

Ω3

(80)

To find the projections Ω j of the rotational velocity on the body-fixed bases we use (47)and (48) with the fact that C minus1 = C dagger to write

xprime2 = xprimeprime

2 = x jC minus1 j2 = C 2 jx j = x1 sin γ + x2 cos γ (81)

and we use (36) and (57) with the fact that X minus1 = X dagger to write

x3 = X 3 jx j = minusx1 sin β cos γ + x2 sin β sin γ + x3 cos β (82)


Ω = Ω jx j (83)

where the coefficients Ω j of the body-fixed unit vectors x j can be described by the matrixequation

Ω1

Ω2

Ω3

=

minus sin β cos γ sin γ 0sin β sin γ cos γ 0

cos β 0 1

α

β γ

(84)


α

β γ

= 1

sin β

minus cos γ sin γ 0sin β sin γ sin β cos γ 0

cos β cos γ minus cos β sin γ sin β

Ω1

Ω2

Ω3

(85)

24 Angular Momentum Operators

Our discussion of rigid bodies so far has been entirely classical We can use (84) with (33)to write the Hamiltonian in terms of the coordinates αβγ and their derivatives α β γ andsolve for the classical motion with Lagrangian or Hamiltonian classical mechanics But inquantum mechanics the coordinate derivaties α β γ cannot be taken as independent variablesRather the dependence of the rigid-body orientation on time t is defined a wave function

ψ = ψ(α β γ t) (86)

The probabilityat time t of finding the orientation of the rigid body in the angle-space vol-ume element dα sin βdβ dγ centered on specified by Euler angles αβγ is |ψ(α β γ t)|2dα sin βdβ dγ The evolution of the wave function is described by the time-dependent Schroedinger equation

i part

parttψ = Hψ =

2

2 L middot Iminus1 middot Lψ (87)

11

We use the classical Hamiltonian (33) but with the angular momentum L interpreted as anoperator on functions of the Euler angles

Setting aside rigid bodies briefly to discuss the simpler case of a point particle withthe wave function ψ(r t) we recall that we can define wave functions that have subject toinfinitesimal displacements in δt time δ r space or δ φ in angular orientation by

T δtψ(t + δt) = ψ(t) or T δtψ(t) = ψ(tminus δt) (88)

T δrψ(r + δ r) = ψ(r) or T δrψ(r) = ψ(rminus δ r) (89)

T δφψ(r + δ φtimes r) = ψ(r) or T δφψ(r) = ψ(rminus δ φtimes r) (90)

The values of the displaced wave functions at the displaced values of the coordinates arethe same as the values of the original wave functions at the at the original values of thecoordinates From (88)ndash(90) we see that

T δtψ =

983080

1minus δt part

partt

983081

ψ = (1 + iδtH )ψ (91)

T δrψ =

983080

1minus δ r middotnabla983081

ψ = (1minus iδ r middot p )ψ (92)

T δφψ = (1minus [δ φtimes r] middotnabla)ψ = (1minus iδ φ middot L)ψ (93)

The familiar expression L = minusir timesnabla for the orbital angular momentum (in units of ) of aparticle moving in three dimensions comes from (93)

Suppose we rotate the wave function (86) through the infinitesimal angle δ φ As a resultof the rotation an orientation that was originally specified by the Euler angles αβγ willbe specified by the slightly incremented Euler angles α + δα β + δβ and γ + δγ We willassume that the angular momentum operator L for a rigid body is determined by expressionsanalogous to (90) and (93) so that

T δφψ(α β γ ) = ψ(αminus δαβ minus δβ γ minus δγ ) = (1minus iδ φ middot L)ψ(α β γ ) (94)

From (94) we see that the orbital angular momentum operator for a rigid-body wave functionmust be defined by

δ φ middot L = 1

i

983080

δα part

partα + δβ

part

partβ + δγ

part

partγ

983081

(95)

To find differential operators to represent the projections L j = x j middotL on the right of (95) welet δα = αδt δβ = βδt and δγ = γδt we use (80) to express α β and γ in terms of Ω j Onthe left of (95) we set δ φ middotL = δtΩ jL j Equating coefficients of Ω j we find the projections of the orbital angular momentum operators on the laboratory-fixed basis vectors

L1 = 1

i sin β

983080

minus cosα cosβ part

partα minus sinα sin β

part

partβ + cosα

part

partγ

983081

(96)

L2 = 1

i sin β

983080

minus sinα cos β part

partα + cosα sin β

part

partβ + sin α

part

partγ

983081

(97)

L3 = 1

i

part

partα (98)

12

From (96) and (97) we find

L1 plusmn iL2 = eplusmniα

i sinβ

983080

minus cos β part

partα plusmn i sin β

part

partβ +

part

partγ

983081

(99)

To find differential operators to represent the projections L j = x j middotL on the right of (95) welet δα = αδt δβ = βδt and δγ = γδt we use (85) to express α β and γ in terms of Ω j Onthe left of (95) we set δ φ middotL = δtΩ jL j Equating coefficients of Ω j we find the projections of the orbital angular momentum operators on the body-fixed basis vectors

L1 = 1

i sin β

983080

minus cos γ part

partα + sinβ sin γ

part

partβ + cosβ cos γ

part

partγ

983081

(100)

L2 = 1

i sin β

983080

sin γ part

partα + sin β cos γ

part

partβ minus cos β sin γ

part

partγ

983081

(101)

L3 = 1

i

part

partγ (102)


L1 plusmn iL2 = e∓iγ

i sin β

983080

minus part


part

partβ + cosβ

part

partγ

983081

(103)

25 Commutators of Lk and Lk

Not surprisingly the projections L j = x j middot L of L on laboratory axes x j satisfy the usualcommutation relations

[L j Lk] = iϵ jkvLv and [L j LkLk] = 0 (104)

One can verify that the differential operators (96) ndash (98) satisfy the commutation relations(104)

People are often surprised to find that the commutation relations for the projectionsLk = xk middot L on the body-fixed axes x j have the ldquowrong signrdquo

[L j Lk] = minusiϵ jkvLv and [L j LkLk] = 0 (105)

Other interesting properties of the angular momentum operators are

[L j Lk] = 0 (106)

that is any laboratory-fixed projection L j commutes with any body-fixed projection Lk Wealso find that

L jL j = L jL j = L middot L (107)

The identities (105) ndash (107) can be proved in a straightforward but tedious way with thedifferential-operator expressions (100)ndash(102) and (96)ndash(98) We outline a less tedious proof here

13

Rotating the body-fixed basis vectors xk by an additional infinitesimal angle δ φ willtransform them to xk + δ φtimes xk and we see from (35) that the value of the direction-cosinematrix at the incremented Euler angles will be

X jk(α + δαβ + δβ γ + δγ ) = x j middot (xk + δ φtimes xk) = (1 + iδ φ middot L)X jk (108)

In writing (108) we set ψ = X jk in () and used () for E minus1 From (108) we find

δ φ middot LX jk = iδ φ middot (x j times xk) (109)

Setting δ φ = xrδφ in (109) and recalling that xk = xsX sk we can equate coefficients of δφto find

LrX jk = iϵrjsX sk = [Lr X jk ] (110)

Setting δ φ = xrδφ in (109) are recalling that x j = X jsxs we can equate coefficients of δφ tofind

LrX jk = minusiϵrksX js = [Lr X jk ] (111)

According to (110) Lr changes the left subscript of X jk and according to (111) Lr changesthe right subscript

Writing the body-fixed projections as

L j = x j middot L = x j middot xkLk = X kjLk (112)

we find the commutation relations

[L j Lk] = (X rjLrX skLs minusX skLsX rjLr)

= (X rjX skLrLs minusX skX rjLsLr)

+i (ϵrslX rjX lkLs minus ϵsrlX skX ljLr)

= X rjX sk[Lr Ls] + i (ϵrslX rjX lkLs minus ϵsrlX skX ljLr)

= i (X rjX skϵrslLl + ϵrslX rjX lkLs minus ϵsrlX skX ljLr)

= i (ϵ jkvX lvLl minus ϵ jkvX svLs minus ϵ jkvX rvLr)

= minusiϵ jkvLv (113)

In the proof (113) we made use (110) (104) and of the identity

ϵrstX riX sj = ϵijkX tk (114)

which follows fromxi times x j = ϵijk xk with xi = xrX ri etc (115)

Any laboratory-fixed component L j of the angular momentum operator commutes withany body-fixed component Lk since

[L j Lk] = L jX rkLr minusX rkLrL j

= iϵ jrvX vkLr + X rkL jLr minusX rkLrL j

= iϵ jrvX vkLr + iX rkϵ jrvLv

= iϵ jrvX vkLr + iX rvϵ jvrLr

= 0 (116)

14

The sum of the squares of the three projections L j is equal to the sum of the squares of the three projections L j

L middot L = L jL j

= X jrLrX jsLs

= X jrX jsLrLs minus iX jrϵrsvX jvLs

= δ rsLrLs minus iϵrsrLs

= LrLr (117)

3 Spherical Basis

We define complex spherical basis vectors ξmicro with azimuthal quantum numbers micro = 1 0minus1in the laboratory-fixed system as

ξ1 = x1 + ix2

minusradic

2 ξ0 = x3 ξminus1 =

x1 minus ix2radic 2

(118)

or in the body-fixed system as

ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(119)

We can write (118) and (119) as

ξmicro = xiN imicro or ξmicro = x jN jmicro where N = 1radic

2

minus1 0 1minusi 0 minusi

0radic

2 0

(120)

One can verify that the transformation matrix N is unitary

N dagger = N minus1 (121)

so that the inverse of (120) is

x j = ξmicroN daggermicroj or x j = ξmicroN daggermicroj where N dagger = 1radic

2

minus1 i 0

0 0radic

21 i 0

(122)

The spherical basis vectors are orthonormal and complete in the sense that

ξmicro middot ξlowastν = δ microν and ξmicroξlowastmicro = U

or

ξmicro middot ξlowastν = δ microν and ξmicroξlowastmicro = U (123)

We can write the unit dyadic as

U = ξmicroξlowastmicro = xiN imicroξ

lowastmicro = x jx j = ξmicroN daggermicrojx j

or


lowastmicro = x jx j = ξmicroN daggermicrojx j (124)

15

From (120) (40) (42) and (48) we find

N daggerAN =

eminusiα 0 00 1 00 0 eminusiα

N daggerBN =

1+cosβ2 minus sinβradic

2

1minuscosβ2

sinβradic 2 cosβ minus sinβradic

21minuscosβ

2sinβradic

2

1+cosβ2

N daggerCN =

eminusiγ 0 00 1 00 0 eminusiγ

(125)

orN daggerANN daggerBN N daggerCN = N daggerABCN = N daggerXN = D1(α β γ ) (126)

where the Wigner D-function D1(αβγ ) denotes the 3 times 3 matrix

D1(α β γ ) =

1+cosβ2 eminusi(α+γ ) minus sinβradic

2 eminusiα 1minuscosβ

2 eminusi(αminusγ )

sinβradic 2 eminusiγ cos β minus sinβradic

2 eiγ

1minuscosβ2 ei(αminusγ ) sinβradic

2 eiα 1+cosβ

2 ei(α+γ )

(127)

The row index micro and column index ν of the matrix element D1microν take on the values micro =

1 0minus1 and ν = 1 0minus1 Using (124) and (126) we can write the rotation dyadic X of (51)as

X = U middot X middot U = ξmicroN daggermicroiX ijN iν ξlowastν

= ξmicroD1microν ξ

lowastν (128)

31 Wigner D-functions

It is convenient to describe the wave functions of polyatomic molecules as superpositionsof Wigner D-functions The most important properties of the D-functions follow from thefact that are irreducible representations of the rotation group of three-dimensional spaceTo take advantage of this connection of D-functions to rotations a special case of whichalready encountered in (128) we consider a quantum-mechanical particle with integer spinquantum number j We can describe the spin either in the laboratory or the body-fixed basisdiscussed above with spin operator J = xiJ i = xi

J iThe components of the spin operatorsreferred to laboratory and body-fixed basis vectors are (2 j + 1)times (2 j + 1) matrices with theusual commutation relations

[J i J j] = iϵijkJ k and [ J i J j] = iϵijk J k (129)

There is no anomalous sign for the commutator [ J i J j] as there was for the commutator[Li L j] for the orbital angular momentum operators of a rigid body in (104) because unlike

16

the orbital angular momentum operators of the rigid body L j or L j the spin operators J iand J i do not operate on the Euler angles αβγ

We introduce spin basis functions | jm⟩ in the laboratory system with the familiar prop-erties

J middot J| jm⟩ = j( j + 1)| jm⟩J 3| jm⟩ = m| jm⟩J plusmn| jm⟩ =

radic

( j ∓m)( j plusmn m + 1)| j m plusmn 1⟩ (130)

The D-function[2] can be defined as

D jml(α β γ ) = ⟨ jm|R| jl⟩ (131)

where the rotation operator is

R = eminusiαJ zeminusiβJ yeminusiγJ z (132)

The D-functions can be written as

D jml(α β γ ) = eminusimαd jml(β )eminusilγ where d jml(β ) = ⟨ jm|eminusiβJ y | jl⟩ (133)

Completeness of the D-Functions The D-functions D jlowastml form a complete set for ex-

panding functions of the Euler angles αβγ As shown by Varshalovich 410(5) [2] theyhave the orthogonality relations

int 2π

0

dα

int π

0

sin βdβ

int 2π

0

dγD jlowastmlD

jprime

mprimelprime = 8π2

[ j] δ jj primeδ mmprime (134)

Unless otherwise noted in (134) and in subsequent expressions D jml = D j

ml(α β γ )

32 The Products LmicroD jml and LmicroD j

ml

Let a D-Function D jml(αβγ ) be defined by (131) and consider a subsequent rotation T δφ =

1 minus iδ φ middot J by the infinitesimal angle δ φ If applied after the rotation R the infinitesimalrotation T δφ would produce a net rotation T δφR = (1minus iδ φ middotJ)R parameterized by the Eulerangles α + δαβ + δβ and γ + δγ Taking ψ = D j

ml in (94) we find

D jml(α + δαβ + δβ γ + δγ ) = ⟨ jm|(1minus iδ φ middot J)R| jl⟩ = (1 + iδ φ middot L)D j

ml (135)

or⟨ jm|JR| jl⟩ = minusLD j

ml (136)

Multiplication by L j Multiplying both sides of (136) by x3 we find

⟨ jm|J 3R| jl⟩ = mD jml = minusL3D j

ml (137)

17

Noting from Varshalovich 44(4) [2] that

D jlowastml = (minus1)mminuslD j

minusmminusl (138)

and setting m rarr minusm and l rarrminusl we find that (137) gives

L3D jlowastml = mD jlowast

ml (139)

Multiplying both sides of (136) by xplusmn = x1 plusmn ix2 we find

⟨ jm|J plusmnR| jl⟩ = ⟨ jm|J dagger∓R| jl⟩ =radic

( j plusmn m)( j ∓m + 1)D jm∓1l = minusLplusmnD j

ml (140)

Setting m rarr minusm and l rarrminusl in (139) and using (137) we find

LplusmnD jlowastml =

radic

( j ∓m)( j plusmn m + 1)D jlowastmplusmn1l (141)

In summary the operators L3 and Lplusmn have no effect on the superscript j or the secondsubscript l of the D-Function D j

ml As shown in (141) the raisinglowering operator Lplusmnraises and lowers the first subscript m in by the same amount and with the same coefficientsas for an angular momentum eigenfunction | jm⟩ As shown in (138) multiplying D j

ml by theoperator L3 is equivalent to multiplication by m as though L3 were multiplying an angularmomentum eigenfunction | jm⟩

Multiplication by L j We rotate the spin functions | jm⟩ through the Euler angles αβγ with the rotation operator R of (132) to get rotated spin functions

| jl = R| jl⟩ or ⟨ jm| = jm|R (142)

The rotated spin operators J j already mentioned in (129) can be written in two alternateways as

J j = RJ jRminus1 = x j middot J (143)

In analogy to (131) we have

J middot J| jl = j( j + 1)| jl

J 3| jl = l| jl

J plusmn| jl =radic

( j ∓ l)( j plusmn l + 1)| j l plusmn 1 (144)

Using (142) we can write (131) as

D jml = ⟨ jm|R| jl⟩ = jm|R| jl (145)

We can also use (142) to write (136) as

jm|RJ| jl = minusLD jml (146)

Multiplying (146) on the left by x3 and using (145) and (144) we find

jm|R J 3| jl = lD jml = minusL3D j

ml (147)

18

Making the substitution m rarr minusm and l rarrminusl in (147) and using (137) we find

L3D jlowastml = lD jlowast

ml (148)

Multiplying (146) on the left by x+ = x1 plusmn x2 and using (145) and (144) we find

jm|R J plusmn| jl =radic

( j ∓ l)( j plusmn l + 1)D jmlplusmn1 = minusLplusmnD j

ml (149)



radic

( j plusmn l)( j ∓ l minus 1)D jlowastml∓1 (150)

In summary the operators L3 and Lplusmn have no effect on the superscript j or the first subscriptm of the D-Function D j

ml As shown in (150) the operators Lplusmn lower and raise the secondsubscript l the opposite of what one might expect As shown in (148) multiplying D j

ml bythe operator L3 is equivalent to multiplication by l

Multiplication by L middot L Writing 2L middot L = L+Lminus + LminusL+ + 2L2z and using (138) and

(141) we findL middot LD jlowast

ml = j ( j + 1)D jlowastml (151)

We find the same expression (151) if we write 2L middot L = L+Lminus + LminusL+ + 2L2z and use (148)

and (150)

33 About-face rotations

In addition to the angular momentum quantum numbers j and m the energy eigenstatesfor rotating rigid bodies have additional quantum numbers that are helpful in classifying thestates and in determining which radiative transitions are allowed The most important of these symmetries are connected with how the eigenfunctions are affected by ldquoabout facerdquorotations around body-fixed axes xk An about-face rotation about the body-fixed axis xk

can be described by the dyadics

Q1 = +x1x1 minus x2x2 minus x3x3 = Uminus 2S1 middot S1

Q2 = minusx1x1 + x2x2 minus x3x3 = Uminus 2S2 middot S2

Q2 = minusx1x1 minus x2x2 + x3x3 = Uminus 2S3 middot S3 (152)

We see that the effect of the about-face rotations on body-fixed unit vectors is

Qk middot x j =

983163

x j if j = kminusx j if j = k

(153)

Only two of the about-face rotations are independent since one can verify from 152) that

Q1 = Q2 middot Q3 = Q3 middot Q2


Q3 = Q1 middot Q2 = Q2 middot Q1 (154)

19

We can use (70) ndash (74) to write

Q1middot = exp(minusiπS1middot) (155)

Q2middot = exp(minusiπS2middot) = exp(minusiγ Sprimeprime3middot)exp(minusiπSprimeprime2middot)exp(iγ Sprimeprime3middot) (156)

Q3middot = exp(minusiπS3middot) = exp(minusiπSprimeprime3middot) (157)

If the rotation X of (55) is parameterized by the Euler angles αβ γ the rotation Ql middotX willbe parameterized by different Euler angles αkβ k γ k Since two about-face rotation aboutthe same axis restore the body-fixed unit vectors xk to their original values we must have(αk)k = α (β k)k = β and (γ k)k = γ We can define a wave function Qkψ(α β γ ) that hasbeen subject to an about-face transformation as

Qkψ(αkβ k γ k) = ψ(α β γ ) or Qkψ(α β γ ) = ψ(αkβ k γ k) (158)

Suppose that ψ is an eigenvalue of Qk with eigenvalue σk so that

Qkψ = σkψ (159)

Multiplying both sides of (159) by Qk and noting that Q2k = 1 we find that we must have

σ2k = 1 so the possible eigenvalues of the about-face rotation are

σk = plusmn1 (160)

Taking k = 3 we find from (55) and (157)

Q3 middot Xmiddot = exp(minusi[π + γ ]Sprimeprime3middot) exp(minusiβ Sprime2middot) exp(minusiαS3middot) (161)

from which we see that the about-face Euler angles are

α3

β 3γ 3

=

αβ γ + π

(162)

Taking k = 2 and noting that Sprimeprime2 = Sprime2 we can use (55) and (156) to find

Q2 middot Xmiddot = exp(minusiπS2middot) exp(minusiγ Sprimeprime3middot)exp(minusiβ Sprime2middot) exp(minusiαS3middot)= exp(minusiγ Sprimeprime3middot) exp(minusiπSprimeprime2middot) exp(minusiβ Sprime2middot)exp(minusiαS3middot)= exp(minusiγ Sprimeprime3middot) exp(minusi[β + π]Sprime2middot) exp(minusiαS3middot)= exp(minusi[π minus γ ]Sprimeprime3middot)exp(minusi[π minus β ]Sprime2middot) exp(minusi[α + π]S3middot) (163)

If the angle β is in the required range 0 le β lt π the angle β + π that occurs in the thirdline of (163) will exceed the limit and must be replaced by a rotation of β 2 = π minus β aboutthe axis minusx1 which is parameterized (modulo 2π) by the Euler angle α2 = α + π In orderto have the axes x2 continue to point in the initial direction as it must for an about facerotation about x2 the third Euler angle must be γ 2 = π minus γ In summary the about-faceEuler angles are

α2

β 2γ 2

=

α + ππ minus β π minus γ

(164)

20

Because Q1 = Q2 middot Q3 we must have

α1

β 1γ 1

=

(α3)2(β 3)2(γ 3)2

=

α + ππ minus β minusγ

(165)

Using the about-face angles (162) (164) and (165) in (158) with ψ = D jlowastml we find

Q1D jlowastml = (minus1) jD jlowast

mminusl

Q2D jlowastml = (minus1) j+lD jlowast

mminusl

Q3D jlowastml = (minus1)lD jlowast

ml (166)

Here we used the identity (see formula 44 (1) of reference[2])

d jml(π minus β ) = (minus1) j+md jmminusl(β ) (167)

The about-face quantum number of D jlowastml are therefore

σ1 = (minus1) j

σ2 = (minus1) j+l

σ3 = (minus1)l (168)

34 Rotations of L j

In analogy to () if we rotate the operator L1 by an angle φ around the axis x3 we find arotated version Y 1 = Y 1(φ) of the operator which we denote by

Y 1 = eminusiφL3L1eiφL3 (169)

Differentiating Y 1 with respect to φ and using the commutation relation [L3 L1] = minusiL2

from (104) we findd

dφY 1 = minusY 2 (170)

whereY 2 = eminusiφL3L2eiφL3 (171)

Differentiating (171) again we find in analogy to (170)

d

dφY 2 = Y 1 (172)

Differentiating (170) once again and using (172) we find the simple differential equation forY 1

d2

dφ2Y 1 = minusY 1 (173)

21

which has the general solution

Y 1 = A cosφ + B sinφ and Y 2 = minus d

dφY 1 = A sinφminus B cosφ (174)

Setting φ = 0 in (169) (171) and equating to the corresponding values from (174) we find

A = L1 and B = minusL2 (175)

and therefore rotated versions of L1 and L2 are

Y 1 = eminusiφL3L1eiφL3 = L1 cosφminus L2 sinφ

Y 2 = eminusiφL3L2eiφL3 = L1 sin φ + L2 cos φ (176)

For the special case of an about-face rotation by φ = π we find from (176)

Q3L1 Qdagger

3 = minusL1

Q3L2 Qdagger

3 = minusL2

Q3L3 Qdagger

3 = L3

(177)

There are identities analogous to (177) for about turns around the axes x1 and x2

4 Rotational Energies

In molecular spectroscopy for asymmetric-top molecules it is customary to denote threeprincipal moments of inertia I a I b and I c where the indices abc are some permutation of 123 The corresponding body fixed axes by xa xb and xc In the customary wave-numberunits of molecular spectroscopy the Hamiltonian (8)

H = H

hc = AL2

a + BL2b + C L2

c (178)

The rotational coefficients are related to the principal moments of inertia by

A =

4πcI a B =

4πcI b C =

4πcI c (179)

Following a common convention we usually will order the moments as

I a le I b le I c or A ge B ge C (180)

If we let La = L3 Lb = L1 and Lc = L2 we can write (178) as

H = BL21 + C L2

2 + AL23

= B + C

2 L middot L +

2AminusB minus C

2L23 +

C minus B

4 (L2

+ + L2minus) (181)

with Lplusmn = L1 plusmn iL2

22

41 Quantum numbers

From inspection of (178) we see that we can select four independent operators that commutewith the Hamiltonian and with each other

[L middot L H ] = 0 [L3 H ] = 0 [ C 1 H ] = 0 [ C 2 H ] = 0 (182)

We can therefore choose the molecular wave function to be the simultaneous eigenfunctionof H L middot L L3 C 1 and C 1 We can use the eigenvalues of these five operators to label thewave functions and energies and write

Hψ Eσ1σ2 jm = Eψ

Eσ1σ3 jm

L middot Lψ Eσ1σ2 jm = j( j + 1)ψ

Eσ1σ2 jm

L3ψ Eσ1σ2 jm = mψ

Eσ1σ2 jm

Q1ψ Eσ1σ2 jm = σ1ψ

Eσ1σ2 jm


Eσ1σ2 jm (183)

Parity For planar three-atom molecules like H2O and CO2 we will assume that the prin-cipal axis x1 is normal to the plane of the molecule In this case the about-face rotationQ1 is equivalent to the parity operation P which reflects the coordinates of all particlesthrough the origin of the coordinate system the center of mass in our case so

P = Q1 (184)

Then the parity quantum number p is the same as the about-face quantum number σ1

p = σ1 = plusmn1 (185)

42 Schroedinger equation

To simplify notation we will use a single index k to label symmetric top wave functions whichwe write as

|k⟩ =

991770

2 j + 1

8π2 D jlowast

mk for k = j j minus 1 minus j (186)


L3|k⟩ = k|k⟩ (187)

Lplusmn|k⟩ =radic

( j plusmn k)( j ∓ k + 1)|k ∓ 1⟩ (188)

L middot L|k⟩ = j( j + 1)|k⟩ (189)

We have omitted the quantum numbers j and m from the labels of the symmetric top basisvectors (186) since the operators L3 and Lplusmn from which the asymmetric top Hamiltonian(183) is constructed cannot change the quantum numbers j or m when applied to one of the basis states (186)

23

We write the asymmetric top wave function |ψn⟩ as a superposition of the basis states(186)

|ψn⟩ =sum

k

|k⟩vkn (190)

Then the Shroedinger equation becomes

( H minus E n)|ψn⟩ = 0 orsum

k

( H ik minus E nδ ik)vkn = 0 (191)

The matrix elements Hamiltonian between the symmetric-top basis functions (186) are

H ik = ⟨i|H |k⟩ = 2 j + 1

8π2

int

D jmiHD jlowast

mkdα sin βdβdγ (192)

Using (187)ndash(189) we find that the only non-zero matrix elements of the Hamiltonian (181)are

⟨k|H |k⟩ = 1

2

983131

(B + C ) j( j + 1) + (2Aminus B minus C )k2983133

(193)

⟨k + 2|H |k⟩ = (C minusB)

4

radic

( j minus k)( j minus k minus 1)( j + k + 1)( j + k + 2) (194)

⟨k minus 2|H |k⟩ = (C minusB)

4

radic

( j + k)( j + k minus 1)( j minus k + 1)( j minus k + 2) (195)

Except for small rotational quantum numbers j where explicit expressions for eigenfunctionsand energies can be found it is necessary to numerically diagonalize the (2 j + 1) times (2 j + 1)matrix with nonzero elements given by (195 to find the eigenfunctions |ψn⟩ (the columnvectors vkn) and energies E n of (191)

43 Asymmetric top with j = 1

Some insight can be gained by considering the simple asymmetric top states for angular-momentum quantum numbers j = 1 The basis states |k⟩ of (186) become

|1⟩ =

100

harr991770

3

8π2D1lowast

m1

|0⟩ =

010

harr991770

3

8π2D1lowast

m0

|minus 1⟩ =

001

harr991770

3

8π2D1lowast

mminus1 (196)

Using the bases (196) we can represent the operators of (187)ndash(189) as the 3times 3 matrices

24

L middot L =

2 0 00 2 00 0 2

L23 =

1 0 00 0 00 0 1

L2+ =

0 0 20 0 00 0 0

L2minus =

0 0 00 0 02 0 0

(197)

Using (197) we find a 3 times 3 matrix representation of the Hamiltonian (192)

H = 1

2

2A + B + C 0 C minus B0 2B + 2C 0C minus B 0 2A + B + C

(198)

From inspection of (198) we see that the Hamiltonian has three eigenvectors |ψn⟩

|ψx⟩ = 1radic

2

101

|ψy⟩ = 1radic

2

10

minus1

|ψz⟩ =

010

(199)

with energiesE x = A + C E y = A + B E z = B + C (200)

If the three energies for j = 1 have been measured we can determine the coupling coefficientsABC from the inverse of (200)

2A = +E x + E y minus E z

2B = minusE x + E y + E z

2C = +E x minusE y + E z (201)

The about-face rotations of (152) can be represented with the matrices

Q1 =

0 0 minus10 minus1 0

minus1 0 0

Q2 =

0 0 10 minus1 01 0 0

Q3 =

minus1 0 00 1 00 0 minus1

(202)

Using the signs of the eigenvalues σ1σ2σ3 for Q1 Q2 Q3 to denote the about-face symme-tries of the eigenstates (199) we find that they have the symmetries

σ1σ2σ3 = minus+minus +minusminus minus minus+ (203)

The parity symmetries are p = σ1 = minus + minus (204)

Two of the states with j = 1 have negative parity and one has positive parity

25

44 State labels

[BEING REVISED] For asymmetric tops (but not for symmetric tops) the energies E = E nof (191) are non-degenerate so we can use the 2 j + 1 distinct but non-integer values of theenergies E together with the intgegers j and m as state labels and we can denote the states

simply as ψ E jm As discussed above for j = 1 it is not necessary to use E as a state label

since it is possible to use integer about-turn quantum numbers σ1 and σ2 together with j andm to uniquely label the states as ψ

σ1σ2 jm Since there are only 3 different energies for j = 1

the four possible values of σ1σ2 are more than enough to label the states An alternate

labeling of the states of the previous section is ψx = ψminus+ jm ψy = ψ

+minus jm and ψz = ψ

minusminus jm

For states with j ge 2 where 2 j + 1 ge 5 the four possible combinations of σ1σ2 do notprovide enough labels to replace the energy E as a quantum number and there will be twoor more independent states with the same quantum numbers jmσ1σ2 However it is possibleto uniquely label the states of an asymmetric top with quantum numbers j and m with apair of integers ka and kc instead of the rather uninformative energy quantum number E The 2 j + 1 allowed pairs kakc are exactly the number of labels required The pair labelsalso also uniquely define the state in the symmetric-top limits when there are only j + 1values of E not enough to provide 2 j + 1 distinct labels For symmetric tops all but oneof the states occur as ldquoparity doubletsrdquo with the same quantum numbers jm E but withopposite parity quantum numbers p = σ1 = plusmn1

To determine the labels ka and kc of a state ψkakc jm of an asymmetric top we imagine let-

ting the eigenfunction ψkakc jm of the Hamiltonian (178) adiabatically change as the coupling

coefficient B approaches either the smallest rotational coefficient C or the largest rotationalcoefficient A and the Hamiltonians become

H = C L middot L + (Aminus C )L2a the prolate limit A gt B = C (205)

H = AL middot Lminus (Aminus C )L2c the oblate limit C lt B = A (206)

Prolate limit For the prolate limit (205) we see that we have [L2a H ] = 0 so energy

eigenfunctions can be chosen to be eigenfunctions L2a We define the first state index ka as

the positive square root of the eigenvalue k2a of L2

a as defined by the eigenvalue equation

L2aψ

kakc jm = k2

aψkakc jm for the prolate limit A gt B = C (207)

For the prolate limit the energies Cj( j + 1) + (Aminus C )k2a with k2

a = 0 1 4 9 j2 Except

for ka = 0 the states ψkakc jm will be two-fold degenerate in the prolate limit since there are

two basis functions with the same value of k2a namely D jlowast

mplusmnka If we were to arrange the

energies E as a column vector with the highest energy at the top of the column and the

26

lowest at the bottom the corresponding values of the indices ka would be

ka =

j j j minus 1 j minus 1110

(208)

Oblate limit For the oblate limit (205) we see that we have [L2c H ] = 0 so energy

eigenfunctions can be chosen to be eigenfunctions L2c We define the second state index kc

as the positive square root of the eigenvalue k2c of L2

c as defined by the eigenvalue equation

L2cψ

kakc jm = k2

cψkakc jm for the oblate limit C lt B = A (209)

For the oblate limit the energies Aj( j + 1)minus (AminusC )k2c decrease with increasing values of

k2c = 0 1 4 9 j2 Except for kc = 0 the states ψ

kakc jm will be two-fold degenerate since

there are two independent basis functions with the same value of k2c namely D jlowast

mplusmnka If we

were to arrange the energies E as a column vector with the highest energy at the top of thecolumn and the lowest at the bottom the corresponding values of the indices kc would be

kc =

01122 j minus 1 j minus 1 j j

(210)

The allowed label pairs with labels higher in the column corresponding to higher energiesare therefore

kakc =

j 0 j 1 j minus 1 1 j minus 1 2

1 j minus 11 j0 j

(211)

27

From (211) we see thatka + kc = j or ka + kc = j + 1 (212)

In the prolate limit and for ka gt 0 we can choose two linear combination of the basis functionsD jlowast

mplusmnkathat are eigenfunction of the parity operator P = C 1

ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka

+ σ1(minus1) jD jlowastmminuska

852009

(213)

Using () () and () we see that the wave functions (213) are simultaneous eigenfunc-tions of the about-face operators C j

C jψkakc jm = σ jψ

kakc jm (214)

where the quantum numbers σ j are

σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)

If we gradually increase B from the prolate limit B = C to the oblate limit B = A thewave function (212) will adiabatically evolve into a more complicated linear combinationof D-functions but it will retain the same eigenvalue σ j of (214) In the oblate limit withB = A the symmetry axis will be x1 = xc In analogy to (214) where the symmetry axiswas x3 = xa the about-turn quantum numbers must be

σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)

Comparing (216) with (214) we see that the important quantum numbers for selection rulescan be written in terms of the pair indices ka and kc as

σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)

For the special case of ka = 0 we can write

ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)

One can readily verify that the relations (213) and (217) remain valid for (218)

References

[1] L Landau and E Lifschitz Quantum Mechanics (Nonrelativistic Theory) Addison-Wesley Reading Mass 1965

[2] D A Varshalovich A N Moskalev and V K Khersonskii Quantum Theory of Angular Momentum (World Scientific Singapore 1988)

[3] G Herzberg Infrared and Raman Spectra Van Norstrand (Princeton NJ 1945)

28

Symmetric tops The next simplest rotating system is the symmetric top A classicalexample is an elongated American football rotating freely in space with no air resistanceA quantum mechanical example is a molecule like methyl chloride ClCH3 with a three-foldrotational symmetry axis through the Cl and C nuclei Symmetric tops have three nonzeromoments of inertia two of which are identical Three parameters often taken to be theEuler angles αβγ are needed to specify the orientation of a symmetric top The first figureof the Wikipedia article on ldquoEuler anglesrdquo gives a good sketch of αβγ We can think of the symmetry axis as being defined by an aximuthal angle φ = α and by a colatitude angleθ = β The amount that the top has been rotated about its symmetry axis is given by thethird Eulers angle γ The rotational Hamiltonian of a symmetric-top molecule is

H = 2

2I perpL middot L +

2

2

983080

1

I ∥minus 1

I perp

983081

L23 (4)

Here L3 = x3 middot L is the projection of the angular momentum operator along the symmetryaxis x3 of the molecule The energy eigenfunctions and quantized energies are

ψ jml =

991770

2 j + 1

8π2 D jlowast

ml and E jl = 2

2I perp j( j + 1) +

2

2

983080

1

I ∥minus 1

I perp

983081

l2 (5)

Here D jlowastmk is the complex conjugate of the Wigner D-Function D j

mk = D jmk(α β γ ) The co-

efficientradic

(2 j + 1)(8π2) is for normalization A very through discussion of the D-Functionscan be found in Quantum Theory of Angular Momentum by Varshalovich Morskalev andKhersonskii [2] We will have more to say about the D-Functions later in these notes Themoment of inertia around the symmetry axis of the symmetric top is I ∥ and the two equalmoments about independent principal axes perpendicular to the symmetry axis are I perp If I perp gt I ∥ (eg ClCH3) the top is said to be prolate If I perp lt I ∥ (eg SO3) the top is said tobe oblate A symmetric top with I perp = I ∥ (eg CH4) is said to be a spherical top

The angular momentum quantum numbers j = 0 1 2 m = j j minus 1 minus j l = j j minus 1 minusl are given by the eigenvalue equations

L middot Lψ jml = j( j + 1)ψ jml

L3ψ jml = mψ jml

L3ψ jml = lψ jml (6)

The energy levels E jl of (5) are (2 j + 1)ndashfold degenerate since all of the wave functionsψ jj lψ jjminus1l ψ jminus jl have the same energies The quantum number l is the projectionof the angular momentum on the symmetry axis From (5) we see that E jl = E jminusl sounless l = 0 the energy states with specified values of j and m will have an additional two-fold degeneracy for plusmnl Various small perturbations always lift the degeneracy in plusmnl andproduce a closely-spaced energy doublet The characteristic doublet structure of symmetric-top energy levels plays an significant role in many physical processes

In the limit that the molecule is linear of nearly linear we will have I ∥ ≪ I perp From (5) wesee that as I ∥ rarr 0 E jl rarr 2l2(2I perp rarr infin if l = 0 States with quantum number l = 0 willhave so much energy that they are almost never excited for molecules in thermal equilibrium

2


ψ jm0(α β γ ) =

991770

2 j + 1

8π2 D jlowast

m0 = 1radic


2

2I perp j( j + 1) (7)




m0

radic



H = 2

2

1048616

L21

I 1+

L22

I 2+

L23

I 3

1048617

(8)


ψ jmn =

991770

2 j + 1

8π2

sum

l

D jlowastmlvln with

sum

l

|vln|2 = 1 (9)




L3ψ jmn = mψ jmn



2 Rotations


3


sum

micro



x = x1 y = x2 and z = x3 (12)


rmicro =3sum

j=1





ϵijk =


0 otherwise(15)









Ω = Ω jx j (19)

4


L =sum

micro


micro


=sum

micro


micro


=sum

micro





L = I middotΩ (22)


I =sum

micro


micro



U = xixi = x1x1 + x2x2 + x3x3 (24)



micro








micro


microj=k


5



I =sum

k

I kxkxk (29)


Iminus1 =sum

k

xkxk





H =sum

micro

1


sum

micro

1


= 1

2

sum

micro


2

sum

micro


= 1

2

sum

micro


2ΩiI ijΩ j (32)


H = 1


2



21 Euler angles



6














0 0 1

(40)









j = xprimekBkj (43)



j or B =



(44)




j (45)



jxprimekBminus1




jk = Bkj (46)

7




kC kj (47)



j or C =


0 0 1

(48)





















C = x jxkC jk (54)



orX = ABC (56)


X =



(57)



8

22 Spin matrices




U =

1 0 00 1 00 0 1

(60)


S 1 = 1

i

0 0 00 0 10 minus1 0

S 2 = 1

i

0 0 minus10 0 01 0 0

S 3 = 1

i

0 1 0minus1 0 0

0 0 0

(61)


(S i) jk = ϵijk

i (62)


S 21 =

0 0 00 1 00 0 1

S 22 =

1 0 00 0 00 0 1

S 23 =

1 0 00 1 00 0 0

(63)


S iS i = 2U = s(s + 1)U (64)








9






2 = xprime jxprime



k(S 3) jk (69)







2(Mmiddot)2 +

1







3 (75)






Ω = Ω jx j (78)


Ω1

Ω2

Ω3

=


α

β γ

(79)

10


α

β γ

= 1

sin β


cosα sinα 0

Ω1

Ω2

Ω3

(80)







Ω = Ω jx j (83)


Ω1

Ω2

Ω3

=


cos β 0 1

α

β γ

(84)


α

β γ

= 1

sin β



Ω1

Ω2

Ω3

(85)



ψ = ψ(α β γ t) (86)


i part

parttψ = Hψ =

2


11







T δtψ =

983080

1minus δt part

partt

983081

ψ = (1 + iδtH )ψ (91)

T δrψ =

983080








δ φ middot L = 1

i

983080

δα part

partα + δβ

part

partβ + δγ

part

partγ

983081

(95)


L1 = 1

i sin β

983080



part

partβ + cosα

part

partγ

983081

(96)

L2 = 1

i sin β

983080



part

partβ + sin α

part

partγ

983081

(97)

L3 = 1

i

part

partα (98)

12



i sinβ

983080

minus cos β part


part

partβ +

part

partγ

983081

(99)


L1 = 1

i sin β

983080

minus cos γ part


part


part

partγ

983081

(100)

L2 = 1

i sin β

983080

sin γ part


part


part

partγ

983081

(101)

L3 = 1

i

part

partγ (102)



i sin β

983080

minus part


part

partβ + cosβ

part

partγ

983081

(103)








[L j Lk] = 0 (106)




13




























= 0 (116)

14


L middot L = L jL j

= X jrLrX jsLs



= LrLr (117)

3 Spherical Basis


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(118)


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(119)



2


0radic

2 0

(120)





2

minus1 i 0

0 0radic

21 i 0

(122)



or





or



15

From (120) (40) (42) and (48) we find

N daggerAN =


N daggerBN =


2

1minuscosβ2


21minuscosβ

2sinβradic

2

1+cosβ2

N daggerCN =


(125)



D1(α β γ ) =





2 eiγ


2 eiα 1+cosβ

2 ei(α+γ )

(127)





lowastν (128)






16




radic



D jml(α β γ ) = ⟨ jm|R| jl⟩ (131)







int 2π

0

dα

int π

0

sin βdβ

int 2π

0

dγD jlowastmlD

jprime

mprimelprime = 8π2



ml(α β γ )


ml



ml in (94) we find


ml (135)


ml (136)



ml (137)

17



minusmminusl (138)



ml (139)




ml (140)



radic






| jl = R| jl⟩ or ⟨ jm| = jm|R (142)





J 3| jl = l| jl

J plusmn| jl =radic



D jml = ⟨ jm|R| jl⟩ = jm|R| jl (145)





ml (147)

18



ml (148)




ml (149)



radic









and (150)








Qk middot x j =

983163


(153)





19








Qkψ = σkψ (159)



σk = plusmn1 (160)




α3

β 3γ 3

=

αβ γ + π

(162)




α2

β 2γ 2

=


(164)

20


α1

β 1γ 1

=

(α3)2(β 3)2(γ 3)2

=


(165)



mminusl


mminusl


ml (166)




σ1 = (minus1) j

σ2 = (minus1) j+l

σ3 = (minus1)l (168)

34 Rotations of L j




from (104) we findd




d

dφY 2 = Y 1 (172)


d2

dφ2Y 1 = minusY 1 (173)

21










Q3L1 Qdagger

3 = minusL1

Q3L2 Qdagger

3 = minusL2

Q3L3 Qdagger

3 = L3

(177)




H = H

hc = AL2

a + BL2b + C L2

c (178)


A =

4πcI a B =

4πcI b C =

4πcI c (179)




H = BL21 + C L2

2 + AL23

= B + C

2 L middot L +

2AminusB minus C

2L23 +

C minus B

4 (L2

+ + L2minus) (181)


22

41 Quantum numbers





Eσ1σ3 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm (183)


P = Q1 (184)


p = σ1 = plusmn1 (185)



|k⟩ =

991770

2 j + 1

8π2 D jlowast



L3|k⟩ = k|k⟩ (187)

Lplusmn|k⟩ =radic

( j plusmn k)( j ∓ k + 1)|k ∓ 1⟩ (188)

L middot L|k⟩ = j( j + 1)|k⟩ (189)


23


|ψn⟩ =sum

k

|k⟩vkn (190)



k



H ik = ⟨i|H |k⟩ = 2 j + 1

8π2

int

D jmiHD jlowast



⟨k|H |k⟩ = 1

2

983131


(193)


4

radic



4

radic





|1⟩ =

100

harr991770

3

8π2D1lowast

m1

|0⟩ =

010

harr991770

3

8π2D1lowast

m0

|minus 1⟩ =

001

harr991770

3

8π2D1lowast

mminus1 (196)


24

L middot L =

2 0 00 2 00 0 2

L23 =

1 0 00 0 00 0 1

L2+ =

0 0 20 0 00 0 0

L2minus =

0 0 00 0 02 0 0

(197)


H = 1

2


(198)


|ψx⟩ = 1radic

2

101

|ψy⟩ = 1radic

2

10

minus1

|ψz⟩ =

010

(199)







Q1 =


minus1 0 0

Q2 =

0 0 10 minus1 01 0 0

Q3 =


(202)





25

44 State labels








minusminus jm











L2aψ

kakc jm = k2



a = 0 1 4 9 j2 Except





26


ka =


(208)





L2cψ

kakc jm = k2






mplusmnka If we


kc =


(210)


kakc =


1 j minus 11 j0 j

(211)

27




ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka


852009

(213)



kakc jm (214)


σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)


σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)


σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)


ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)


References




28


ψ jm0(α β γ ) =

991770

2 j + 1

8π2 D jlowast

m0 = 1radic


2

2I perp j( j + 1) (7)




m0

radic



H = 2

2

1048616

L21

I 1+

L22

I 2+

L23

I 3

1048617

(8)


ψ jmn =

991770

2 j + 1

8π2

sum

l

D jlowastmlvln with

sum

l

|vln|2 = 1 (9)




L3ψ jmn = mψ jmn



2 Rotations


3


sum

micro



x = x1 y = x2 and z = x3 (12)


rmicro =3sum

j=1





ϵijk =


0 otherwise(15)









Ω = Ω jx j (19)

4


L =sum

micro


micro


=sum

micro


micro


=sum

micro





L = I middotΩ (22)


I =sum

micro


micro



U = xixi = x1x1 + x2x2 + x3x3 (24)



micro








micro


microj=k


5



I =sum

k

I kxkxk (29)


Iminus1 =sum

k

xkxk





H =sum

micro

1


sum

micro

1


= 1

2

sum

micro


2

sum

micro


= 1

2

sum

micro


2ΩiI ijΩ j (32)


H = 1


2



21 Euler angles



6














0 0 1

(40)









j = xprimekBkj (43)



j or B =



(44)




j (45)



jxprimekBminus1




jk = Bkj (46)

7




kC kj (47)



j or C =


0 0 1

(48)





















C = x jxkC jk (54)



orX = ABC (56)


X =



(57)



8

22 Spin matrices




U =

1 0 00 1 00 0 1

(60)


S 1 = 1

i

0 0 00 0 10 minus1 0

S 2 = 1

i

0 0 minus10 0 01 0 0

S 3 = 1

i

0 1 0minus1 0 0

0 0 0

(61)


(S i) jk = ϵijk

i (62)


S 21 =

0 0 00 1 00 0 1

S 22 =

1 0 00 0 00 0 1

S 23 =

1 0 00 1 00 0 0

(63)


S iS i = 2U = s(s + 1)U (64)








9






2 = xprime jxprime



k(S 3) jk (69)







2(Mmiddot)2 +

1







3 (75)






Ω = Ω jx j (78)


Ω1

Ω2

Ω3

=


α

β γ

(79)

10


α

β γ

= 1

sin β


cosα sinα 0

Ω1

Ω2

Ω3

(80)







Ω = Ω jx j (83)


Ω1

Ω2

Ω3

=


cos β 0 1

α

β γ

(84)


α

β γ

= 1

sin β



Ω1

Ω2

Ω3

(85)



ψ = ψ(α β γ t) (86)


i part

parttψ = Hψ =

2


11







T δtψ =

983080

1minus δt part

partt

983081

ψ = (1 + iδtH )ψ (91)

T δrψ =

983080








δ φ middot L = 1

i

983080

δα part

partα + δβ

part

partβ + δγ

part

partγ

983081

(95)


L1 = 1

i sin β

983080



part

partβ + cosα

part

partγ

983081

(96)

L2 = 1

i sin β

983080



part

partβ + sin α

part

partγ

983081

(97)

L3 = 1

i

part

partα (98)

12



i sinβ

983080

minus cos β part


part

partβ +

part

partγ

983081

(99)


L1 = 1

i sin β

983080

minus cos γ part


part


part

partγ

983081

(100)

L2 = 1

i sin β

983080

sin γ part


part


part

partγ

983081

(101)

L3 = 1

i

part

partγ (102)



i sin β

983080

minus part


part

partβ + cosβ

part

partγ

983081

(103)








[L j Lk] = 0 (106)




13




























= 0 (116)

14


L middot L = L jL j

= X jrLrX jsLs



= LrLr (117)

3 Spherical Basis


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(118)


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(119)



2


0radic

2 0

(120)





2

minus1 i 0

0 0radic

21 i 0

(122)



or





or



15

From (120) (40) (42) and (48) we find

N daggerAN =


N daggerBN =


2

1minuscosβ2


21minuscosβ

2sinβradic

2

1+cosβ2

N daggerCN =


(125)



D1(α β γ ) =





2 eiγ


2 eiα 1+cosβ

2 ei(α+γ )

(127)





lowastν (128)






16




radic



D jml(α β γ ) = ⟨ jm|R| jl⟩ (131)







int 2π

0

dα

int π

0

sin βdβ

int 2π

0

dγD jlowastmlD

jprime

mprimelprime = 8π2



ml(α β γ )


ml



ml in (94) we find


ml (135)


ml (136)



ml (137)

17



minusmminusl (138)



ml (139)




ml (140)



radic






| jl = R| jl⟩ or ⟨ jm| = jm|R (142)





J 3| jl = l| jl

J plusmn| jl =radic



D jml = ⟨ jm|R| jl⟩ = jm|R| jl (145)





ml (147)

18



ml (148)




ml (149)



radic









and (150)








Qk middot x j =

983163


(153)





19








Qkψ = σkψ (159)



σk = plusmn1 (160)




α3

β 3γ 3

=

αβ γ + π

(162)




α2

β 2γ 2

=


(164)

20


α1

β 1γ 1

=

(α3)2(β 3)2(γ 3)2

=


(165)



mminusl


mminusl


ml (166)




σ1 = (minus1) j

σ2 = (minus1) j+l

σ3 = (minus1)l (168)

34 Rotations of L j




from (104) we findd




d

dφY 2 = Y 1 (172)


d2

dφ2Y 1 = minusY 1 (173)

21










Q3L1 Qdagger

3 = minusL1

Q3L2 Qdagger

3 = minusL2

Q3L3 Qdagger

3 = L3

(177)




H = H

hc = AL2

a + BL2b + C L2

c (178)


A =

4πcI a B =

4πcI b C =

4πcI c (179)




H = BL21 + C L2

2 + AL23

= B + C

2 L middot L +

2AminusB minus C

2L23 +

C minus B

4 (L2

+ + L2minus) (181)


22

41 Quantum numbers





Eσ1σ3 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm (183)


P = Q1 (184)


p = σ1 = plusmn1 (185)



|k⟩ =

991770

2 j + 1

8π2 D jlowast



L3|k⟩ = k|k⟩ (187)

Lplusmn|k⟩ =radic

( j plusmn k)( j ∓ k + 1)|k ∓ 1⟩ (188)

L middot L|k⟩ = j( j + 1)|k⟩ (189)


23


|ψn⟩ =sum

k

|k⟩vkn (190)



k



H ik = ⟨i|H |k⟩ = 2 j + 1

8π2

int

D jmiHD jlowast



⟨k|H |k⟩ = 1

2

983131


(193)


4

radic



4

radic





|1⟩ =

100

harr991770

3

8π2D1lowast

m1

|0⟩ =

010

harr991770

3

8π2D1lowast

m0

|minus 1⟩ =

001

harr991770

3

8π2D1lowast

mminus1 (196)


24

L middot L =

2 0 00 2 00 0 2

L23 =

1 0 00 0 00 0 1

L2+ =

0 0 20 0 00 0 0

L2minus =

0 0 00 0 02 0 0

(197)


H = 1

2


(198)


|ψx⟩ = 1radic

2

101

|ψy⟩ = 1radic

2

10

minus1

|ψz⟩ =

010

(199)







Q1 =


minus1 0 0

Q2 =

0 0 10 minus1 01 0 0

Q3 =


(202)





25

44 State labels








minusminus jm











L2aψ

kakc jm = k2



a = 0 1 4 9 j2 Except





26


ka =


(208)





L2cψ

kakc jm = k2






mplusmnka If we


kc =


(210)


kakc =


1 j minus 11 j0 j

(211)

27




ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka


852009

(213)



kakc jm (214)


σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)


σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)


σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)


ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)


References




28


sum

micro



x = x1 y = x2 and z = x3 (12)


rmicro =3sum

j=1





ϵijk =


0 otherwise(15)









Ω = Ω jx j (19)

4


L =sum

micro


micro


=sum

micro


micro


=sum

micro





L = I middotΩ (22)


I =sum

micro


micro



U = xixi = x1x1 + x2x2 + x3x3 (24)



micro








micro


microj=k


5



I =sum

k

I kxkxk (29)


Iminus1 =sum

k

xkxk





H =sum

micro

1


sum

micro

1


= 1

2

sum

micro


2

sum

micro


= 1

2

sum

micro


2ΩiI ijΩ j (32)


H = 1


2



21 Euler angles



6














0 0 1

(40)









j = xprimekBkj (43)



j or B =



(44)




j (45)



jxprimekBminus1




jk = Bkj (46)

7




kC kj (47)



j or C =


0 0 1

(48)





















C = x jxkC jk (54)



orX = ABC (56)


X =



(57)



8

22 Spin matrices




U =

1 0 00 1 00 0 1

(60)


S 1 = 1

i

0 0 00 0 10 minus1 0

S 2 = 1

i

0 0 minus10 0 01 0 0

S 3 = 1

i

0 1 0minus1 0 0

0 0 0

(61)


(S i) jk = ϵijk

i (62)


S 21 =

0 0 00 1 00 0 1

S 22 =

1 0 00 0 00 0 1

S 23 =

1 0 00 1 00 0 0

(63)


S iS i = 2U = s(s + 1)U (64)








9






2 = xprime jxprime



k(S 3) jk (69)







2(Mmiddot)2 +

1







3 (75)






Ω = Ω jx j (78)


Ω1

Ω2

Ω3

=


α

β γ

(79)

10


α

β γ

= 1

sin β


cosα sinα 0

Ω1

Ω2

Ω3

(80)







Ω = Ω jx j (83)


Ω1

Ω2

Ω3

=


cos β 0 1

α

β γ

(84)


α

β γ

= 1

sin β



Ω1

Ω2

Ω3

(85)



ψ = ψ(α β γ t) (86)


i part

parttψ = Hψ =

2


11







T δtψ =

983080

1minus δt part

partt

983081

ψ = (1 + iδtH )ψ (91)

T δrψ =

983080








δ φ middot L = 1

i

983080

δα part

partα + δβ

part

partβ + δγ

part

partγ

983081

(95)


L1 = 1

i sin β

983080



part

partβ + cosα

part

partγ

983081

(96)

L2 = 1

i sin β

983080



part

partβ + sin α

part

partγ

983081

(97)

L3 = 1

i

part

partα (98)

12



i sinβ

983080

minus cos β part


part

partβ +

part

partγ

983081

(99)


L1 = 1

i sin β

983080

minus cos γ part


part


part

partγ

983081

(100)

L2 = 1

i sin β

983080

sin γ part


part


part

partγ

983081

(101)

L3 = 1

i

part

partγ (102)



i sin β

983080

minus part


part

partβ + cosβ

part

partγ

983081

(103)








[L j Lk] = 0 (106)




13




























= 0 (116)

14


L middot L = L jL j

= X jrLrX jsLs



= LrLr (117)

3 Spherical Basis


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(118)


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(119)



2


0radic

2 0

(120)





2

minus1 i 0

0 0radic

21 i 0

(122)



or





or



15

From (120) (40) (42) and (48) we find

N daggerAN =


N daggerBN =


2

1minuscosβ2


21minuscosβ

2sinβradic

2

1+cosβ2

N daggerCN =


(125)



D1(α β γ ) =





2 eiγ


2 eiα 1+cosβ

2 ei(α+γ )

(127)





lowastν (128)






16




radic



D jml(α β γ ) = ⟨ jm|R| jl⟩ (131)







int 2π

0

dα

int π

0

sin βdβ

int 2π

0

dγD jlowastmlD

jprime

mprimelprime = 8π2



ml(α β γ )


ml



ml in (94) we find


ml (135)


ml (136)



ml (137)

17



minusmminusl (138)



ml (139)




ml (140)



radic






| jl = R| jl⟩ or ⟨ jm| = jm|R (142)





J 3| jl = l| jl

J plusmn| jl =radic



D jml = ⟨ jm|R| jl⟩ = jm|R| jl (145)





ml (147)

18



ml (148)




ml (149)



radic









and (150)








Qk middot x j =

983163


(153)





19








Qkψ = σkψ (159)



σk = plusmn1 (160)




α3

β 3γ 3

=

αβ γ + π

(162)




α2

β 2γ 2

=


(164)

20


α1

β 1γ 1

=

(α3)2(β 3)2(γ 3)2

=


(165)



mminusl


mminusl


ml (166)




σ1 = (minus1) j

σ2 = (minus1) j+l

σ3 = (minus1)l (168)

34 Rotations of L j




from (104) we findd




d

dφY 2 = Y 1 (172)


d2

dφ2Y 1 = minusY 1 (173)

21










Q3L1 Qdagger

3 = minusL1

Q3L2 Qdagger

3 = minusL2

Q3L3 Qdagger

3 = L3

(177)




H = H

hc = AL2

a + BL2b + C L2

c (178)


A =

4πcI a B =

4πcI b C =

4πcI c (179)




H = BL21 + C L2

2 + AL23

= B + C

2 L middot L +

2AminusB minus C

2L23 +

C minus B

4 (L2

+ + L2minus) (181)


22

41 Quantum numbers





Eσ1σ3 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm (183)


P = Q1 (184)


p = σ1 = plusmn1 (185)



|k⟩ =

991770

2 j + 1

8π2 D jlowast



L3|k⟩ = k|k⟩ (187)

Lplusmn|k⟩ =radic

( j plusmn k)( j ∓ k + 1)|k ∓ 1⟩ (188)

L middot L|k⟩ = j( j + 1)|k⟩ (189)


23


|ψn⟩ =sum

k

|k⟩vkn (190)



k



H ik = ⟨i|H |k⟩ = 2 j + 1

8π2

int

D jmiHD jlowast



⟨k|H |k⟩ = 1

2

983131


(193)


4

radic



4

radic





|1⟩ =

100

harr991770

3

8π2D1lowast

m1

|0⟩ =

010

harr991770

3

8π2D1lowast

m0

|minus 1⟩ =

001

harr991770

3

8π2D1lowast

mminus1 (196)


24

L middot L =

2 0 00 2 00 0 2

L23 =

1 0 00 0 00 0 1

L2+ =

0 0 20 0 00 0 0

L2minus =

0 0 00 0 02 0 0

(197)


H = 1

2


(198)


|ψx⟩ = 1radic

2

101

|ψy⟩ = 1radic

2

10

minus1

|ψz⟩ =

010

(199)







Q1 =


minus1 0 0

Q2 =

0 0 10 minus1 01 0 0

Q3 =


(202)





25

44 State labels








minusminus jm











L2aψ

kakc jm = k2



a = 0 1 4 9 j2 Except





26


ka =


(208)





L2cψ

kakc jm = k2






mplusmnka If we


kc =


(210)


kakc =


1 j minus 11 j0 j

(211)

27




ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka


852009

(213)



kakc jm (214)


σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)


σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)


σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)


ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)


References




28


L =sum

micro


micro


=sum

micro


micro


=sum

micro





L = I middotΩ (22)


I =sum

micro


micro



U = xixi = x1x1 + x2x2 + x3x3 (24)



micro








micro


microj=k


5



I =sum

k

I kxkxk (29)


Iminus1 =sum

k

xkxk





H =sum

micro

1


sum

micro

1


= 1

2

sum

micro


2

sum

micro


= 1

2

sum

micro


2ΩiI ijΩ j (32)


H = 1


2



21 Euler angles



6














0 0 1

(40)









j = xprimekBkj (43)



j or B =



(44)




j (45)



jxprimekBminus1




jk = Bkj (46)

7




kC kj (47)



j or C =


0 0 1

(48)





















C = x jxkC jk (54)



orX = ABC (56)


X =



(57)



8

22 Spin matrices




U =

1 0 00 1 00 0 1

(60)


S 1 = 1

i

0 0 00 0 10 minus1 0

S 2 = 1

i

0 0 minus10 0 01 0 0

S 3 = 1

i

0 1 0minus1 0 0

0 0 0

(61)


(S i) jk = ϵijk

i (62)


S 21 =

0 0 00 1 00 0 1

S 22 =

1 0 00 0 00 0 1

S 23 =

1 0 00 1 00 0 0

(63)


S iS i = 2U = s(s + 1)U (64)








9






2 = xprime jxprime



k(S 3) jk (69)







2(Mmiddot)2 +

1







3 (75)






Ω = Ω jx j (78)


Ω1

Ω2

Ω3

=


α

β γ

(79)

10


α

β γ

= 1

sin β


cosα sinα 0

Ω1

Ω2

Ω3

(80)







Ω = Ω jx j (83)


Ω1

Ω2

Ω3

=


cos β 0 1

α

β γ

(84)


α

β γ

= 1

sin β



Ω1

Ω2

Ω3

(85)



ψ = ψ(α β γ t) (86)


i part

parttψ = Hψ =

2


11







T δtψ =

983080

1minus δt part

partt

983081

ψ = (1 + iδtH )ψ (91)

T δrψ =

983080








δ φ middot L = 1

i

983080

δα part

partα + δβ

part

partβ + δγ

part

partγ

983081

(95)


L1 = 1

i sin β

983080



part

partβ + cosα

part

partγ

983081

(96)

L2 = 1

i sin β

983080



part

partβ + sin α

part

partγ

983081

(97)

L3 = 1

i

part

partα (98)

12



i sinβ

983080

minus cos β part


part

partβ +

part

partγ

983081

(99)


L1 = 1

i sin β

983080

minus cos γ part


part


part

partγ

983081

(100)

L2 = 1

i sin β

983080

sin γ part


part


part

partγ

983081

(101)

L3 = 1

i

part

partγ (102)



i sin β

983080

minus part


part

partβ + cosβ

part

partγ

983081

(103)








[L j Lk] = 0 (106)




13




























= 0 (116)

14


L middot L = L jL j

= X jrLrX jsLs



= LrLr (117)

3 Spherical Basis


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(118)


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(119)



2


0radic

2 0

(120)





2

minus1 i 0

0 0radic

21 i 0

(122)



or





or



15

From (120) (40) (42) and (48) we find

N daggerAN =


N daggerBN =


2

1minuscosβ2


21minuscosβ

2sinβradic

2

1+cosβ2

N daggerCN =


(125)



D1(α β γ ) =





2 eiγ


2 eiα 1+cosβ

2 ei(α+γ )

(127)





lowastν (128)






16




radic



D jml(α β γ ) = ⟨ jm|R| jl⟩ (131)







int 2π

0

dα

int π

0

sin βdβ

int 2π

0

dγD jlowastmlD

jprime

mprimelprime = 8π2



ml(α β γ )


ml



ml in (94) we find


ml (135)


ml (136)



ml (137)

17



minusmminusl (138)



ml (139)




ml (140)



radic






| jl = R| jl⟩ or ⟨ jm| = jm|R (142)





J 3| jl = l| jl

J plusmn| jl =radic



D jml = ⟨ jm|R| jl⟩ = jm|R| jl (145)





ml (147)

18



ml (148)




ml (149)



radic









and (150)








Qk middot x j =

983163


(153)





19








Qkψ = σkψ (159)



σk = plusmn1 (160)




α3

β 3γ 3

=

αβ γ + π

(162)




α2

β 2γ 2

=


(164)

20


α1

β 1γ 1

=

(α3)2(β 3)2(γ 3)2

=


(165)



mminusl


mminusl


ml (166)




σ1 = (minus1) j

σ2 = (minus1) j+l

σ3 = (minus1)l (168)

34 Rotations of L j




from (104) we findd




d

dφY 2 = Y 1 (172)


d2

dφ2Y 1 = minusY 1 (173)

21










Q3L1 Qdagger

3 = minusL1

Q3L2 Qdagger

3 = minusL2

Q3L3 Qdagger

3 = L3

(177)




H = H

hc = AL2

a + BL2b + C L2

c (178)


A =

4πcI a B =

4πcI b C =

4πcI c (179)




H = BL21 + C L2

2 + AL23

= B + C

2 L middot L +

2AminusB minus C

2L23 +

C minus B

4 (L2

+ + L2minus) (181)


22

41 Quantum numbers





Eσ1σ3 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm (183)


P = Q1 (184)


p = σ1 = plusmn1 (185)



|k⟩ =

991770

2 j + 1

8π2 D jlowast



L3|k⟩ = k|k⟩ (187)

Lplusmn|k⟩ =radic

( j plusmn k)( j ∓ k + 1)|k ∓ 1⟩ (188)

L middot L|k⟩ = j( j + 1)|k⟩ (189)


23


|ψn⟩ =sum

k

|k⟩vkn (190)



k



H ik = ⟨i|H |k⟩ = 2 j + 1

8π2

int

D jmiHD jlowast



⟨k|H |k⟩ = 1

2

983131


(193)


4

radic



4

radic





|1⟩ =

100

harr991770

3

8π2D1lowast

m1

|0⟩ =

010

harr991770

3

8π2D1lowast

m0

|minus 1⟩ =

001

harr991770

3

8π2D1lowast

mminus1 (196)


24

L middot L =

2 0 00 2 00 0 2

L23 =

1 0 00 0 00 0 1

L2+ =

0 0 20 0 00 0 0

L2minus =

0 0 00 0 02 0 0

(197)


H = 1

2


(198)


|ψx⟩ = 1radic

2

101

|ψy⟩ = 1radic

2

10

minus1

|ψz⟩ =

010

(199)







Q1 =


minus1 0 0

Q2 =

0 0 10 minus1 01 0 0

Q3 =


(202)





25

44 State labels








minusminus jm











L2aψ

kakc jm = k2



a = 0 1 4 9 j2 Except





26


ka =


(208)





L2cψ

kakc jm = k2






mplusmnka If we


kc =


(210)


kakc =


1 j minus 11 j0 j

(211)

27




ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka


852009

(213)



kakc jm (214)


σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)


σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)


σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)


ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)


References




28



I =sum

k

I kxkxk (29)


Iminus1 =sum

k

xkxk





H =sum

micro

1


sum

micro

1


= 1

2

sum

micro


2

sum

micro


= 1

2

sum

micro


2ΩiI ijΩ j (32)


H = 1


2



21 Euler angles



6














0 0 1

(40)









j = xprimekBkj (43)



j or B =



(44)




j (45)



jxprimekBminus1




jk = Bkj (46)

7




kC kj (47)



j or C =


0 0 1

(48)





















C = x jxkC jk (54)



orX = ABC (56)


X =



(57)



8

22 Spin matrices




U =

1 0 00 1 00 0 1

(60)


S 1 = 1

i

0 0 00 0 10 minus1 0

S 2 = 1

i

0 0 minus10 0 01 0 0

S 3 = 1

i

0 1 0minus1 0 0

0 0 0

(61)


(S i) jk = ϵijk

i (62)


S 21 =

0 0 00 1 00 0 1

S 22 =

1 0 00 0 00 0 1

S 23 =

1 0 00 1 00 0 0

(63)


S iS i = 2U = s(s + 1)U (64)








9






2 = xprime jxprime



k(S 3) jk (69)







2(Mmiddot)2 +

1







3 (75)






Ω = Ω jx j (78)


Ω1

Ω2

Ω3

=


α

β γ

(79)

10


α

β γ

= 1

sin β


cosα sinα 0

Ω1

Ω2

Ω3

(80)







Ω = Ω jx j (83)


Ω1

Ω2

Ω3

=


cos β 0 1

α

β γ

(84)


α

β γ

= 1

sin β



Ω1

Ω2

Ω3

(85)



ψ = ψ(α β γ t) (86)


i part

parttψ = Hψ =

2


11







T δtψ =

983080

1minus δt part

partt

983081

ψ = (1 + iδtH )ψ (91)

T δrψ =

983080








δ φ middot L = 1

i

983080

δα part

partα + δβ

part

partβ + δγ

part

partγ

983081

(95)


L1 = 1

i sin β

983080



part

partβ + cosα

part

partγ

983081

(96)

L2 = 1

i sin β

983080



part

partβ + sin α

part

partγ

983081

(97)

L3 = 1

i

part

partα (98)

12



i sinβ

983080

minus cos β part


part

partβ +

part

partγ

983081

(99)


L1 = 1

i sin β

983080

minus cos γ part


part


part

partγ

983081

(100)

L2 = 1

i sin β

983080

sin γ part


part


part

partγ

983081

(101)

L3 = 1

i

part

partγ (102)



i sin β

983080

minus part


part

partβ + cosβ

part

partγ

983081

(103)








[L j Lk] = 0 (106)




13




























= 0 (116)

14


L middot L = L jL j

= X jrLrX jsLs



= LrLr (117)

3 Spherical Basis


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(118)


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(119)



2


0radic

2 0

(120)





2

minus1 i 0

0 0radic

21 i 0

(122)



or





or



15

From (120) (40) (42) and (48) we find

N daggerAN =


N daggerBN =


2

1minuscosβ2


21minuscosβ

2sinβradic

2

1+cosβ2

N daggerCN =


(125)



D1(α β γ ) =





2 eiγ


2 eiα 1+cosβ

2 ei(α+γ )

(127)





lowastν (128)






16




radic



D jml(α β γ ) = ⟨ jm|R| jl⟩ (131)







int 2π

0

dα

int π

0

sin βdβ

int 2π

0

dγD jlowastmlD

jprime

mprimelprime = 8π2



ml(α β γ )


ml



ml in (94) we find


ml (135)


ml (136)



ml (137)

17



minusmminusl (138)



ml (139)




ml (140)



radic






| jl = R| jl⟩ or ⟨ jm| = jm|R (142)





J 3| jl = l| jl

J plusmn| jl =radic



D jml = ⟨ jm|R| jl⟩ = jm|R| jl (145)





ml (147)

18



ml (148)




ml (149)



radic









and (150)








Qk middot x j =

983163


(153)





19








Qkψ = σkψ (159)



σk = plusmn1 (160)




α3

β 3γ 3

=

αβ γ + π

(162)




α2

β 2γ 2

=


(164)

20


α1

β 1γ 1

=

(α3)2(β 3)2(γ 3)2

=


(165)



mminusl


mminusl


ml (166)




σ1 = (minus1) j

σ2 = (minus1) j+l

σ3 = (minus1)l (168)

34 Rotations of L j




from (104) we findd




d

dφY 2 = Y 1 (172)


d2

dφ2Y 1 = minusY 1 (173)

21










Q3L1 Qdagger

3 = minusL1

Q3L2 Qdagger

3 = minusL2

Q3L3 Qdagger

3 = L3

(177)




H = H

hc = AL2

a + BL2b + C L2

c (178)


A =

4πcI a B =

4πcI b C =

4πcI c (179)




H = BL21 + C L2

2 + AL23

= B + C

2 L middot L +

2AminusB minus C

2L23 +

C minus B

4 (L2

+ + L2minus) (181)


22

41 Quantum numbers





Eσ1σ3 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm (183)


P = Q1 (184)


p = σ1 = plusmn1 (185)



|k⟩ =

991770

2 j + 1

8π2 D jlowast



L3|k⟩ = k|k⟩ (187)

Lplusmn|k⟩ =radic

( j plusmn k)( j ∓ k + 1)|k ∓ 1⟩ (188)

L middot L|k⟩ = j( j + 1)|k⟩ (189)


23


|ψn⟩ =sum

k

|k⟩vkn (190)



k



H ik = ⟨i|H |k⟩ = 2 j + 1

8π2

int

D jmiHD jlowast



⟨k|H |k⟩ = 1

2

983131


(193)


4

radic



4

radic





|1⟩ =

100

harr991770

3

8π2D1lowast

m1

|0⟩ =

010

harr991770

3

8π2D1lowast

m0

|minus 1⟩ =

001

harr991770

3

8π2D1lowast

mminus1 (196)


24

L middot L =

2 0 00 2 00 0 2

L23 =

1 0 00 0 00 0 1

L2+ =

0 0 20 0 00 0 0

L2minus =

0 0 00 0 02 0 0

(197)


H = 1

2


(198)


|ψx⟩ = 1radic

2

101

|ψy⟩ = 1radic

2

10

minus1

|ψz⟩ =

010

(199)







Q1 =


minus1 0 0

Q2 =

0 0 10 minus1 01 0 0

Q3 =


(202)





25

44 State labels








minusminus jm











L2aψ

kakc jm = k2



a = 0 1 4 9 j2 Except





26


ka =


(208)





L2cψ

kakc jm = k2






mplusmnka If we


kc =


(210)


kakc =


1 j minus 11 j0 j

(211)

27




ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka


852009

(213)



kakc jm (214)


σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)


σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)


σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)


ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)


References




28














0 0 1

(40)









j = xprimekBkj (43)



j or B =



(44)




j (45)



jxprimekBminus1




jk = Bkj (46)

7




kC kj (47)



j or C =


0 0 1

(48)





















C = x jxkC jk (54)



orX = ABC (56)


X =



(57)



8

22 Spin matrices




U =

1 0 00 1 00 0 1

(60)


S 1 = 1

i

0 0 00 0 10 minus1 0

S 2 = 1

i

0 0 minus10 0 01 0 0

S 3 = 1

i

0 1 0minus1 0 0

0 0 0

(61)


(S i) jk = ϵijk

i (62)


S 21 =

0 0 00 1 00 0 1

S 22 =

1 0 00 0 00 0 1

S 23 =

1 0 00 1 00 0 0

(63)


S iS i = 2U = s(s + 1)U (64)








9






2 = xprime jxprime



k(S 3) jk (69)







2(Mmiddot)2 +

1







3 (75)






Ω = Ω jx j (78)


Ω1

Ω2

Ω3

=


α

β γ

(79)

10


α

β γ

= 1

sin β


cosα sinα 0

Ω1

Ω2

Ω3

(80)







Ω = Ω jx j (83)


Ω1

Ω2

Ω3

=


cos β 0 1

α

β γ

(84)


α

β γ

= 1

sin β



Ω1

Ω2

Ω3

(85)



ψ = ψ(α β γ t) (86)


i part

parttψ = Hψ =

2


11







T δtψ =

983080

1minus δt part

partt

983081

ψ = (1 + iδtH )ψ (91)

T δrψ =

983080








δ φ middot L = 1

i

983080

δα part

partα + δβ

part

partβ + δγ

part

partγ

983081

(95)


L1 = 1

i sin β

983080



part

partβ + cosα

part

partγ

983081

(96)

L2 = 1

i sin β

983080



part

partβ + sin α

part

partγ

983081

(97)

L3 = 1

i

part

partα (98)

12



i sinβ

983080

minus cos β part


part

partβ +

part

partγ

983081

(99)


L1 = 1

i sin β

983080

minus cos γ part


part


part

partγ

983081

(100)

L2 = 1

i sin β

983080

sin γ part


part


part

partγ

983081

(101)

L3 = 1

i

part

partγ (102)



i sin β

983080

minus part


part

partβ + cosβ

part

partγ

983081

(103)








[L j Lk] = 0 (106)




13




























= 0 (116)

14


L middot L = L jL j

= X jrLrX jsLs



= LrLr (117)

3 Spherical Basis


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(118)


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(119)



2


0radic

2 0

(120)





2

minus1 i 0

0 0radic

21 i 0

(122)



or





or



15

From (120) (40) (42) and (48) we find

N daggerAN =


N daggerBN =


2

1minuscosβ2


21minuscosβ

2sinβradic

2

1+cosβ2

N daggerCN =


(125)



D1(α β γ ) =





2 eiγ


2 eiα 1+cosβ

2 ei(α+γ )

(127)





lowastν (128)






16




radic



D jml(α β γ ) = ⟨ jm|R| jl⟩ (131)







int 2π

0

dα

int π

0

sin βdβ

int 2π

0

dγD jlowastmlD

jprime

mprimelprime = 8π2



ml(α β γ )


ml



ml in (94) we find


ml (135)


ml (136)



ml (137)

17



minusmminusl (138)



ml (139)




ml (140)



radic






| jl = R| jl⟩ or ⟨ jm| = jm|R (142)





J 3| jl = l| jl

J plusmn| jl =radic



D jml = ⟨ jm|R| jl⟩ = jm|R| jl (145)





ml (147)

18



ml (148)




ml (149)



radic









and (150)








Qk middot x j =

983163


(153)





19








Qkψ = σkψ (159)



σk = plusmn1 (160)




α3

β 3γ 3

=

αβ γ + π

(162)




α2

β 2γ 2

=


(164)

20


α1

β 1γ 1

=

(α3)2(β 3)2(γ 3)2

=


(165)



mminusl


mminusl


ml (166)




σ1 = (minus1) j

σ2 = (minus1) j+l

σ3 = (minus1)l (168)

34 Rotations of L j




from (104) we findd




d

dφY 2 = Y 1 (172)


d2

dφ2Y 1 = minusY 1 (173)

21










Q3L1 Qdagger

3 = minusL1

Q3L2 Qdagger

3 = minusL2

Q3L3 Qdagger

3 = L3

(177)




H = H

hc = AL2

a + BL2b + C L2

c (178)


A =

4πcI a B =

4πcI b C =

4πcI c (179)




H = BL21 + C L2

2 + AL23

= B + C

2 L middot L +

2AminusB minus C

2L23 +

C minus B

4 (L2

+ + L2minus) (181)


22

41 Quantum numbers





Eσ1σ3 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm (183)


P = Q1 (184)


p = σ1 = plusmn1 (185)



|k⟩ =

991770

2 j + 1

8π2 D jlowast



L3|k⟩ = k|k⟩ (187)

Lplusmn|k⟩ =radic

( j plusmn k)( j ∓ k + 1)|k ∓ 1⟩ (188)

L middot L|k⟩ = j( j + 1)|k⟩ (189)


23


|ψn⟩ =sum

k

|k⟩vkn (190)



k



H ik = ⟨i|H |k⟩ = 2 j + 1

8π2

int

D jmiHD jlowast



⟨k|H |k⟩ = 1

2

983131


(193)


4

radic



4

radic





|1⟩ =

100

harr991770

3

8π2D1lowast

m1

|0⟩ =

010

harr991770

3

8π2D1lowast

m0

|minus 1⟩ =

001

harr991770

3

8π2D1lowast

mminus1 (196)


24

L middot L =

2 0 00 2 00 0 2

L23 =

1 0 00 0 00 0 1

L2+ =

0 0 20 0 00 0 0

L2minus =

0 0 00 0 02 0 0

(197)


H = 1

2


(198)


|ψx⟩ = 1radic

2

101

|ψy⟩ = 1radic

2

10

minus1

|ψz⟩ =

010

(199)







Q1 =


minus1 0 0

Q2 =

0 0 10 minus1 01 0 0

Q3 =


(202)





25

44 State labels








minusminus jm











L2aψ

kakc jm = k2



a = 0 1 4 9 j2 Except





26


ka =


(208)





L2cψ

kakc jm = k2






mplusmnka If we


kc =


(210)


kakc =


1 j minus 11 j0 j

(211)

27




ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka


852009

(213)



kakc jm (214)


σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)


σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)


σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)


ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)


References




28




kC kj (47)



j or C =


0 0 1

(48)





















C = x jxkC jk (54)



orX = ABC (56)


X =



(57)



8

22 Spin matrices




U =

1 0 00 1 00 0 1

(60)


S 1 = 1

i

0 0 00 0 10 minus1 0

S 2 = 1

i

0 0 minus10 0 01 0 0

S 3 = 1

i

0 1 0minus1 0 0

0 0 0

(61)


(S i) jk = ϵijk

i (62)


S 21 =

0 0 00 1 00 0 1

S 22 =

1 0 00 0 00 0 1

S 23 =

1 0 00 1 00 0 0

(63)


S iS i = 2U = s(s + 1)U (64)








9






2 = xprime jxprime



k(S 3) jk (69)







2(Mmiddot)2 +

1







3 (75)






Ω = Ω jx j (78)


Ω1

Ω2

Ω3

=


α

β γ

(79)

10


α

β γ

= 1

sin β


cosα sinα 0

Ω1

Ω2

Ω3

(80)







Ω = Ω jx j (83)


Ω1

Ω2

Ω3

=


cos β 0 1

α

β γ

(84)


α

β γ

= 1

sin β



Ω1

Ω2

Ω3

(85)



ψ = ψ(α β γ t) (86)


i part

parttψ = Hψ =

2


11







T δtψ =

983080

1minus δt part

partt

983081

ψ = (1 + iδtH )ψ (91)

T δrψ =

983080








δ φ middot L = 1

i

983080

δα part

partα + δβ

part

partβ + δγ

part

partγ

983081

(95)


L1 = 1

i sin β

983080



part

partβ + cosα

part

partγ

983081

(96)

L2 = 1

i sin β

983080



part

partβ + sin α

part

partγ

983081

(97)

L3 = 1

i

part

partα (98)

12



i sinβ

983080

minus cos β part


part

partβ +

part

partγ

983081

(99)


L1 = 1

i sin β

983080

minus cos γ part


part


part

partγ

983081

(100)

L2 = 1

i sin β

983080

sin γ part


part


part

partγ

983081

(101)

L3 = 1

i

part

partγ (102)



i sin β

983080

minus part


part

partβ + cosβ

part

partγ

983081

(103)








[L j Lk] = 0 (106)




13




























= 0 (116)

14


L middot L = L jL j

= X jrLrX jsLs



= LrLr (117)

3 Spherical Basis


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(118)


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(119)



2


0radic

2 0

(120)





2

minus1 i 0

0 0radic

21 i 0

(122)



or





or



15

From (120) (40) (42) and (48) we find

N daggerAN =


N daggerBN =


2

1minuscosβ2


21minuscosβ

2sinβradic

2

1+cosβ2

N daggerCN =


(125)



D1(α β γ ) =





2 eiγ


2 eiα 1+cosβ

2 ei(α+γ )

(127)





lowastν (128)






16




radic



D jml(α β γ ) = ⟨ jm|R| jl⟩ (131)







int 2π

0

dα

int π

0

sin βdβ

int 2π

0

dγD jlowastmlD

jprime

mprimelprime = 8π2



ml(α β γ )


ml



ml in (94) we find


ml (135)


ml (136)



ml (137)

17



minusmminusl (138)



ml (139)




ml (140)



radic






| jl = R| jl⟩ or ⟨ jm| = jm|R (142)





J 3| jl = l| jl

J plusmn| jl =radic



D jml = ⟨ jm|R| jl⟩ = jm|R| jl (145)





ml (147)

18



ml (148)




ml (149)



radic









and (150)








Qk middot x j =

983163


(153)





19








Qkψ = σkψ (159)



σk = plusmn1 (160)




α3

β 3γ 3

=

αβ γ + π

(162)




α2

β 2γ 2

=


(164)

20


α1

β 1γ 1

=

(α3)2(β 3)2(γ 3)2

=


(165)



mminusl


mminusl


ml (166)




σ1 = (minus1) j

σ2 = (minus1) j+l

σ3 = (minus1)l (168)

34 Rotations of L j




from (104) we findd




d

dφY 2 = Y 1 (172)


d2

dφ2Y 1 = minusY 1 (173)

21










Q3L1 Qdagger

3 = minusL1

Q3L2 Qdagger

3 = minusL2

Q3L3 Qdagger

3 = L3

(177)




H = H

hc = AL2

a + BL2b + C L2

c (178)


A =

4πcI a B =

4πcI b C =

4πcI c (179)




H = BL21 + C L2

2 + AL23

= B + C

2 L middot L +

2AminusB minus C

2L23 +

C minus B

4 (L2

+ + L2minus) (181)


22

41 Quantum numbers





Eσ1σ3 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm (183)


P = Q1 (184)


p = σ1 = plusmn1 (185)



|k⟩ =

991770

2 j + 1

8π2 D jlowast



L3|k⟩ = k|k⟩ (187)

Lplusmn|k⟩ =radic

( j plusmn k)( j ∓ k + 1)|k ∓ 1⟩ (188)

L middot L|k⟩ = j( j + 1)|k⟩ (189)


23


|ψn⟩ =sum

k

|k⟩vkn (190)



k



H ik = ⟨i|H |k⟩ = 2 j + 1

8π2

int

D jmiHD jlowast



⟨k|H |k⟩ = 1

2

983131


(193)


4

radic



4

radic





|1⟩ =

100

harr991770

3

8π2D1lowast

m1

|0⟩ =

010

harr991770

3

8π2D1lowast

m0

|minus 1⟩ =

001

harr991770

3

8π2D1lowast

mminus1 (196)


24

L middot L =

2 0 00 2 00 0 2

L23 =

1 0 00 0 00 0 1

L2+ =

0 0 20 0 00 0 0

L2minus =

0 0 00 0 02 0 0

(197)


H = 1

2


(198)


|ψx⟩ = 1radic

2

101

|ψy⟩ = 1radic

2

10

minus1

|ψz⟩ =

010

(199)







Q1 =


minus1 0 0

Q2 =

0 0 10 minus1 01 0 0

Q3 =


(202)





25

44 State labels








minusminus jm











L2aψ

kakc jm = k2



a = 0 1 4 9 j2 Except





26


ka =


(208)





L2cψ

kakc jm = k2






mplusmnka If we


kc =


(210)


kakc =


1 j minus 11 j0 j

(211)

27




ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka


852009

(213)



kakc jm (214)


σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)


σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)


σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)


ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)


References




28

22 Spin matrices




U =

1 0 00 1 00 0 1

(60)


S 1 = 1

i

0 0 00 0 10 minus1 0

S 2 = 1

i

0 0 minus10 0 01 0 0

S 3 = 1

i

0 1 0minus1 0 0

0 0 0

(61)


(S i) jk = ϵijk

i (62)


S 21 =

0 0 00 1 00 0 1

S 22 =

1 0 00 0 00 0 1

S 23 =

1 0 00 1 00 0 0

(63)


S iS i = 2U = s(s + 1)U (64)








9






2 = xprime jxprime



k(S 3) jk (69)







2(Mmiddot)2 +

1







3 (75)






Ω = Ω jx j (78)


Ω1

Ω2

Ω3

=


α

β γ

(79)

10


α

β γ

= 1

sin β


cosα sinα 0

Ω1

Ω2

Ω3

(80)







Ω = Ω jx j (83)


Ω1

Ω2

Ω3

=


cos β 0 1

α

β γ

(84)


α

β γ

= 1

sin β



Ω1

Ω2

Ω3

(85)



ψ = ψ(α β γ t) (86)


i part

parttψ = Hψ =

2


11







T δtψ =

983080

1minus δt part

partt

983081

ψ = (1 + iδtH )ψ (91)

T δrψ =

983080








δ φ middot L = 1

i

983080

δα part

partα + δβ

part

partβ + δγ

part

partγ

983081

(95)


L1 = 1

i sin β

983080



part

partβ + cosα

part

partγ

983081

(96)

L2 = 1

i sin β

983080



part

partβ + sin α

part

partγ

983081

(97)

L3 = 1

i

part

partα (98)

12



i sinβ

983080

minus cos β part


part

partβ +

part

partγ

983081

(99)


L1 = 1

i sin β

983080

minus cos γ part


part


part

partγ

983081

(100)

L2 = 1

i sin β

983080

sin γ part


part


part

partγ

983081

(101)

L3 = 1

i

part

partγ (102)



i sin β

983080

minus part


part

partβ + cosβ

part

partγ

983081

(103)








[L j Lk] = 0 (106)




13




























= 0 (116)

14


L middot L = L jL j

= X jrLrX jsLs



= LrLr (117)

3 Spherical Basis


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(118)


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(119)



2


0radic

2 0

(120)





2

minus1 i 0

0 0radic

21 i 0

(122)



or





or



15

From (120) (40) (42) and (48) we find

N daggerAN =


N daggerBN =


2

1minuscosβ2


21minuscosβ

2sinβradic

2

1+cosβ2

N daggerCN =


(125)



D1(α β γ ) =





2 eiγ


2 eiα 1+cosβ

2 ei(α+γ )

(127)





lowastν (128)






16




radic



D jml(α β γ ) = ⟨ jm|R| jl⟩ (131)







int 2π

0

dα

int π

0

sin βdβ

int 2π

0

dγD jlowastmlD

jprime

mprimelprime = 8π2



ml(α β γ )


ml



ml in (94) we find


ml (135)


ml (136)



ml (137)

17



minusmminusl (138)



ml (139)




ml (140)



radic






| jl = R| jl⟩ or ⟨ jm| = jm|R (142)





J 3| jl = l| jl

J plusmn| jl =radic



D jml = ⟨ jm|R| jl⟩ = jm|R| jl (145)





ml (147)

18



ml (148)




ml (149)



radic









and (150)








Qk middot x j =

983163


(153)





19








Qkψ = σkψ (159)



σk = plusmn1 (160)




α3

β 3γ 3

=

αβ γ + π

(162)




α2

β 2γ 2

=


(164)

20


α1

β 1γ 1

=

(α3)2(β 3)2(γ 3)2

=


(165)



mminusl


mminusl


ml (166)




σ1 = (minus1) j

σ2 = (minus1) j+l

σ3 = (minus1)l (168)

34 Rotations of L j




from (104) we findd




d

dφY 2 = Y 1 (172)


d2

dφ2Y 1 = minusY 1 (173)

21










Q3L1 Qdagger

3 = minusL1

Q3L2 Qdagger

3 = minusL2

Q3L3 Qdagger

3 = L3

(177)




H = H

hc = AL2

a + BL2b + C L2

c (178)


A =

4πcI a B =

4πcI b C =

4πcI c (179)




H = BL21 + C L2

2 + AL23

= B + C

2 L middot L +

2AminusB minus C

2L23 +

C minus B

4 (L2

+ + L2minus) (181)


22

41 Quantum numbers





Eσ1σ3 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm (183)


P = Q1 (184)


p = σ1 = plusmn1 (185)



|k⟩ =

991770

2 j + 1

8π2 D jlowast



L3|k⟩ = k|k⟩ (187)

Lplusmn|k⟩ =radic

( j plusmn k)( j ∓ k + 1)|k ∓ 1⟩ (188)

L middot L|k⟩ = j( j + 1)|k⟩ (189)


23


|ψn⟩ =sum

k

|k⟩vkn (190)



k



H ik = ⟨i|H |k⟩ = 2 j + 1

8π2

int

D jmiHD jlowast



⟨k|H |k⟩ = 1

2

983131


(193)


4

radic



4

radic





|1⟩ =

100

harr991770

3

8π2D1lowast

m1

|0⟩ =

010

harr991770

3

8π2D1lowast

m0

|minus 1⟩ =

001

harr991770

3

8π2D1lowast

mminus1 (196)


24

L middot L =

2 0 00 2 00 0 2

L23 =

1 0 00 0 00 0 1

L2+ =

0 0 20 0 00 0 0

L2minus =

0 0 00 0 02 0 0

(197)


H = 1

2


(198)


|ψx⟩ = 1radic

2

101

|ψy⟩ = 1radic

2

10

minus1

|ψz⟩ =

010

(199)







Q1 =


minus1 0 0

Q2 =

0 0 10 minus1 01 0 0

Q3 =


(202)





25

44 State labels








minusminus jm











L2aψ

kakc jm = k2



a = 0 1 4 9 j2 Except





26


ka =


(208)





L2cψ

kakc jm = k2






mplusmnka If we


kc =


(210)


kakc =


1 j minus 11 j0 j

(211)

27




ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka


852009

(213)



kakc jm (214)


σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)


σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)


σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)


ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)


References




28






2 = xprime jxprime



k(S 3) jk (69)







2(Mmiddot)2 +

1







3 (75)






Ω = Ω jx j (78)


Ω1

Ω2

Ω3

=


α

β γ

(79)

10


α

β γ

= 1

sin β


cosα sinα 0

Ω1

Ω2

Ω3

(80)







Ω = Ω jx j (83)


Ω1

Ω2

Ω3

=


cos β 0 1

α

β γ

(84)


α

β γ

= 1

sin β



Ω1

Ω2

Ω3

(85)



ψ = ψ(α β γ t) (86)


i part

parttψ = Hψ =

2


11







T δtψ =

983080

1minus δt part

partt

983081

ψ = (1 + iδtH )ψ (91)

T δrψ =

983080








δ φ middot L = 1

i

983080

δα part

partα + δβ

part

partβ + δγ

part

partγ

983081

(95)


L1 = 1

i sin β

983080



part

partβ + cosα

part

partγ

983081

(96)

L2 = 1

i sin β

983080



part

partβ + sin α

part

partγ

983081

(97)

L3 = 1

i

part

partα (98)

12



i sinβ

983080

minus cos β part


part

partβ +

part

partγ

983081

(99)


L1 = 1

i sin β

983080

minus cos γ part


part


part

partγ

983081

(100)

L2 = 1

i sin β

983080

sin γ part


part


part

partγ

983081

(101)

L3 = 1

i

part

partγ (102)



i sin β

983080

minus part


part

partβ + cosβ

part

partγ

983081

(103)








[L j Lk] = 0 (106)




13




























= 0 (116)

14


L middot L = L jL j

= X jrLrX jsLs



= LrLr (117)

3 Spherical Basis


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(118)


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(119)



2


0radic

2 0

(120)





2

minus1 i 0

0 0radic

21 i 0

(122)



or





or



15

From (120) (40) (42) and (48) we find

N daggerAN =


N daggerBN =


2

1minuscosβ2


21minuscosβ

2sinβradic

2

1+cosβ2

N daggerCN =


(125)



D1(α β γ ) =





2 eiγ


2 eiα 1+cosβ

2 ei(α+γ )

(127)





lowastν (128)






16




radic



D jml(α β γ ) = ⟨ jm|R| jl⟩ (131)







int 2π

0

dα

int π

0

sin βdβ

int 2π

0

dγD jlowastmlD

jprime

mprimelprime = 8π2



ml(α β γ )


ml



ml in (94) we find


ml (135)


ml (136)



ml (137)

17



minusmminusl (138)



ml (139)




ml (140)



radic






| jl = R| jl⟩ or ⟨ jm| = jm|R (142)





J 3| jl = l| jl

J plusmn| jl =radic



D jml = ⟨ jm|R| jl⟩ = jm|R| jl (145)





ml (147)

18



ml (148)




ml (149)



radic









and (150)








Qk middot x j =

983163


(153)





19








Qkψ = σkψ (159)



σk = plusmn1 (160)




α3

β 3γ 3

=

αβ γ + π

(162)




α2

β 2γ 2

=


(164)

20


α1

β 1γ 1

=

(α3)2(β 3)2(γ 3)2

=


(165)



mminusl


mminusl


ml (166)




σ1 = (minus1) j

σ2 = (minus1) j+l

σ3 = (minus1)l (168)

34 Rotations of L j




from (104) we findd




d

dφY 2 = Y 1 (172)


d2

dφ2Y 1 = minusY 1 (173)

21










Q3L1 Qdagger

3 = minusL1

Q3L2 Qdagger

3 = minusL2

Q3L3 Qdagger

3 = L3

(177)




H = H

hc = AL2

a + BL2b + C L2

c (178)


A =

4πcI a B =

4πcI b C =

4πcI c (179)




H = BL21 + C L2

2 + AL23

= B + C

2 L middot L +

2AminusB minus C

2L23 +

C minus B

4 (L2

+ + L2minus) (181)


22

41 Quantum numbers





Eσ1σ3 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm (183)


P = Q1 (184)


p = σ1 = plusmn1 (185)



|k⟩ =

991770

2 j + 1

8π2 D jlowast



L3|k⟩ = k|k⟩ (187)

Lplusmn|k⟩ =radic

( j plusmn k)( j ∓ k + 1)|k ∓ 1⟩ (188)

L middot L|k⟩ = j( j + 1)|k⟩ (189)


23


|ψn⟩ =sum

k

|k⟩vkn (190)



k



H ik = ⟨i|H |k⟩ = 2 j + 1

8π2

int

D jmiHD jlowast



⟨k|H |k⟩ = 1

2

983131


(193)


4

radic



4

radic





|1⟩ =

100

harr991770

3

8π2D1lowast

m1

|0⟩ =

010

harr991770

3

8π2D1lowast

m0

|minus 1⟩ =

001

harr991770

3

8π2D1lowast

mminus1 (196)


24

L middot L =

2 0 00 2 00 0 2

L23 =

1 0 00 0 00 0 1

L2+ =

0 0 20 0 00 0 0

L2minus =

0 0 00 0 02 0 0

(197)


H = 1

2


(198)


|ψx⟩ = 1radic

2

101

|ψy⟩ = 1radic

2

10

minus1

|ψz⟩ =

010

(199)







Q1 =


minus1 0 0

Q2 =

0 0 10 minus1 01 0 0

Q3 =


(202)





25

44 State labels








minusminus jm











L2aψ

kakc jm = k2



a = 0 1 4 9 j2 Except





26


ka =


(208)





L2cψ

kakc jm = k2






mplusmnka If we


kc =


(210)


kakc =


1 j minus 11 j0 j

(211)

27




ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka


852009

(213)



kakc jm (214)


σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)


σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)


σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)


ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)


References




28


α

β γ

= 1

sin β


cosα sinα 0

Ω1

Ω2

Ω3

(80)







Ω = Ω jx j (83)


Ω1

Ω2

Ω3

=


cos β 0 1

α

β γ

(84)


α

β γ

= 1

sin β



Ω1

Ω2

Ω3

(85)



ψ = ψ(α β γ t) (86)


i part

parttψ = Hψ =

2


11







T δtψ =

983080

1minus δt part

partt

983081

ψ = (1 + iδtH )ψ (91)

T δrψ =

983080








δ φ middot L = 1

i

983080

δα part

partα + δβ

part

partβ + δγ

part

partγ

983081

(95)


L1 = 1

i sin β

983080



part

partβ + cosα

part

partγ

983081

(96)

L2 = 1

i sin β

983080



part

partβ + sin α

part

partγ

983081

(97)

L3 = 1

i

part

partα (98)

12



i sinβ

983080

minus cos β part


part

partβ +

part

partγ

983081

(99)


L1 = 1

i sin β

983080

minus cos γ part


part


part

partγ

983081

(100)

L2 = 1

i sin β

983080

sin γ part


part


part

partγ

983081

(101)

L3 = 1

i

part

partγ (102)



i sin β

983080

minus part


part

partβ + cosβ

part

partγ

983081

(103)








[L j Lk] = 0 (106)




13




























= 0 (116)

14


L middot L = L jL j

= X jrLrX jsLs



= LrLr (117)

3 Spherical Basis


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(118)


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(119)



2


0radic

2 0

(120)





2

minus1 i 0

0 0radic

21 i 0

(122)



or





or



15

From (120) (40) (42) and (48) we find

N daggerAN =


N daggerBN =


2

1minuscosβ2


21minuscosβ

2sinβradic

2

1+cosβ2

N daggerCN =


(125)



D1(α β γ ) =





2 eiγ


2 eiα 1+cosβ

2 ei(α+γ )

(127)





lowastν (128)






16




radic



D jml(α β γ ) = ⟨ jm|R| jl⟩ (131)







int 2π

0

dα

int π

0

sin βdβ

int 2π

0

dγD jlowastmlD

jprime

mprimelprime = 8π2



ml(α β γ )


ml



ml in (94) we find


ml (135)


ml (136)



ml (137)

17



minusmminusl (138)



ml (139)




ml (140)



radic






| jl = R| jl⟩ or ⟨ jm| = jm|R (142)





J 3| jl = l| jl

J plusmn| jl =radic



D jml = ⟨ jm|R| jl⟩ = jm|R| jl (145)





ml (147)

18



ml (148)




ml (149)



radic









and (150)








Qk middot x j =

983163


(153)





19








Qkψ = σkψ (159)



σk = plusmn1 (160)




α3

β 3γ 3

=

αβ γ + π

(162)




α2

β 2γ 2

=


(164)

20


α1

β 1γ 1

=

(α3)2(β 3)2(γ 3)2

=


(165)



mminusl


mminusl


ml (166)




σ1 = (minus1) j

σ2 = (minus1) j+l

σ3 = (minus1)l (168)

34 Rotations of L j




from (104) we findd




d

dφY 2 = Y 1 (172)


d2

dφ2Y 1 = minusY 1 (173)

21










Q3L1 Qdagger

3 = minusL1

Q3L2 Qdagger

3 = minusL2

Q3L3 Qdagger

3 = L3

(177)




H = H

hc = AL2

a + BL2b + C L2

c (178)


A =

4πcI a B =

4πcI b C =

4πcI c (179)




H = BL21 + C L2

2 + AL23

= B + C

2 L middot L +

2AminusB minus C

2L23 +

C minus B

4 (L2

+ + L2minus) (181)


22

41 Quantum numbers





Eσ1σ3 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm (183)


P = Q1 (184)


p = σ1 = plusmn1 (185)



|k⟩ =

991770

2 j + 1

8π2 D jlowast



L3|k⟩ = k|k⟩ (187)

Lplusmn|k⟩ =radic

( j plusmn k)( j ∓ k + 1)|k ∓ 1⟩ (188)

L middot L|k⟩ = j( j + 1)|k⟩ (189)


23


|ψn⟩ =sum

k

|k⟩vkn (190)



k



H ik = ⟨i|H |k⟩ = 2 j + 1

8π2

int

D jmiHD jlowast



⟨k|H |k⟩ = 1

2

983131


(193)


4

radic



4

radic





|1⟩ =

100

harr991770

3

8π2D1lowast

m1

|0⟩ =

010

harr991770

3

8π2D1lowast

m0

|minus 1⟩ =

001

harr991770

3

8π2D1lowast

mminus1 (196)


24

L middot L =

2 0 00 2 00 0 2

L23 =

1 0 00 0 00 0 1

L2+ =

0 0 20 0 00 0 0

L2minus =

0 0 00 0 02 0 0

(197)


H = 1

2


(198)


|ψx⟩ = 1radic

2

101

|ψy⟩ = 1radic

2

10

minus1

|ψz⟩ =

010

(199)







Q1 =


minus1 0 0

Q2 =

0 0 10 minus1 01 0 0

Q3 =


(202)





25

44 State labels








minusminus jm











L2aψ

kakc jm = k2



a = 0 1 4 9 j2 Except





26


ka =


(208)





L2cψ

kakc jm = k2






mplusmnka If we


kc =


(210)


kakc =


1 j minus 11 j0 j

(211)

27




ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka


852009

(213)



kakc jm (214)


σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)


σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)


σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)


ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)


References




28







T δtψ =

983080

1minus δt part

partt

983081

ψ = (1 + iδtH )ψ (91)

T δrψ =

983080








δ φ middot L = 1

i

983080

δα part

partα + δβ

part

partβ + δγ

part

partγ

983081

(95)


L1 = 1

i sin β

983080



part

partβ + cosα

part

partγ

983081

(96)

L2 = 1

i sin β

983080



part

partβ + sin α

part

partγ

983081

(97)

L3 = 1

i

part

partα (98)

12



i sinβ

983080

minus cos β part


part

partβ +

part

partγ

983081

(99)


L1 = 1

i sin β

983080

minus cos γ part


part


part

partγ

983081

(100)

L2 = 1

i sin β

983080

sin γ part


part


part

partγ

983081

(101)

L3 = 1

i

part

partγ (102)



i sin β

983080

minus part


part

partβ + cosβ

part

partγ

983081

(103)








[L j Lk] = 0 (106)




13




























= 0 (116)

14


L middot L = L jL j

= X jrLrX jsLs



= LrLr (117)

3 Spherical Basis


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(118)


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(119)



2


0radic

2 0

(120)





2

minus1 i 0

0 0radic

21 i 0

(122)



or





or



15

From (120) (40) (42) and (48) we find

N daggerAN =


N daggerBN =


2

1minuscosβ2


21minuscosβ

2sinβradic

2

1+cosβ2

N daggerCN =


(125)



D1(α β γ ) =





2 eiγ


2 eiα 1+cosβ

2 ei(α+γ )

(127)





lowastν (128)






16




radic



D jml(α β γ ) = ⟨ jm|R| jl⟩ (131)







int 2π

0

dα

int π

0

sin βdβ

int 2π

0

dγD jlowastmlD

jprime

mprimelprime = 8π2



ml(α β γ )


ml



ml in (94) we find


ml (135)


ml (136)



ml (137)

17



minusmminusl (138)



ml (139)




ml (140)



radic






| jl = R| jl⟩ or ⟨ jm| = jm|R (142)





J 3| jl = l| jl

J plusmn| jl =radic



D jml = ⟨ jm|R| jl⟩ = jm|R| jl (145)





ml (147)

18



ml (148)




ml (149)



radic









and (150)








Qk middot x j =

983163


(153)





19








Qkψ = σkψ (159)



σk = plusmn1 (160)




α3

β 3γ 3

=

αβ γ + π

(162)




α2

β 2γ 2

=


(164)

20


α1

β 1γ 1

=

(α3)2(β 3)2(γ 3)2

=


(165)



mminusl


mminusl


ml (166)




σ1 = (minus1) j

σ2 = (minus1) j+l

σ3 = (minus1)l (168)

34 Rotations of L j




from (104) we findd




d

dφY 2 = Y 1 (172)


d2

dφ2Y 1 = minusY 1 (173)

21










Q3L1 Qdagger

3 = minusL1

Q3L2 Qdagger

3 = minusL2

Q3L3 Qdagger

3 = L3

(177)




H = H

hc = AL2

a + BL2b + C L2

c (178)


A =

4πcI a B =

4πcI b C =

4πcI c (179)




H = BL21 + C L2

2 + AL23

= B + C

2 L middot L +

2AminusB minus C

2L23 +

C minus B

4 (L2

+ + L2minus) (181)


22

41 Quantum numbers





Eσ1σ3 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm (183)


P = Q1 (184)


p = σ1 = plusmn1 (185)



|k⟩ =

991770

2 j + 1

8π2 D jlowast



L3|k⟩ = k|k⟩ (187)

Lplusmn|k⟩ =radic

( j plusmn k)( j ∓ k + 1)|k ∓ 1⟩ (188)

L middot L|k⟩ = j( j + 1)|k⟩ (189)


23


|ψn⟩ =sum

k

|k⟩vkn (190)



k



H ik = ⟨i|H |k⟩ = 2 j + 1

8π2

int

D jmiHD jlowast



⟨k|H |k⟩ = 1

2

983131


(193)


4

radic



4

radic





|1⟩ =

100

harr991770

3

8π2D1lowast

m1

|0⟩ =

010

harr991770

3

8π2D1lowast

m0

|minus 1⟩ =

001

harr991770

3

8π2D1lowast

mminus1 (196)


24

L middot L =

2 0 00 2 00 0 2

L23 =

1 0 00 0 00 0 1

L2+ =

0 0 20 0 00 0 0

L2minus =

0 0 00 0 02 0 0

(197)


H = 1

2


(198)


|ψx⟩ = 1radic

2

101

|ψy⟩ = 1radic

2

10

minus1

|ψz⟩ =

010

(199)







Q1 =


minus1 0 0

Q2 =

0 0 10 minus1 01 0 0

Q3 =


(202)





25

44 State labels








minusminus jm











L2aψ

kakc jm = k2



a = 0 1 4 9 j2 Except





26


ka =


(208)





L2cψ

kakc jm = k2






mplusmnka If we


kc =


(210)


kakc =


1 j minus 11 j0 j

(211)

27




ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka


852009

(213)



kakc jm (214)


σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)


σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)


σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)


ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)


References




28



i sinβ

983080

minus cos β part


part

partβ +

part

partγ

983081

(99)


L1 = 1

i sin β

983080

minus cos γ part


part


part

partγ

983081

(100)

L2 = 1

i sin β

983080

sin γ part


part


part

partγ

983081

(101)

L3 = 1

i

part

partγ (102)



i sin β

983080

minus part


part

partβ + cosβ

part

partγ

983081

(103)








[L j Lk] = 0 (106)




13




























= 0 (116)

14


L middot L = L jL j

= X jrLrX jsLs



= LrLr (117)

3 Spherical Basis


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(118)


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(119)



2


0radic

2 0

(120)





2

minus1 i 0

0 0radic

21 i 0

(122)



or





or



15

From (120) (40) (42) and (48) we find

N daggerAN =


N daggerBN =


2

1minuscosβ2


21minuscosβ

2sinβradic

2

1+cosβ2

N daggerCN =


(125)



D1(α β γ ) =





2 eiγ


2 eiα 1+cosβ

2 ei(α+γ )

(127)





lowastν (128)






16




radic



D jml(α β γ ) = ⟨ jm|R| jl⟩ (131)







int 2π

0

dα

int π

0

sin βdβ

int 2π

0

dγD jlowastmlD

jprime

mprimelprime = 8π2



ml(α β γ )


ml



ml in (94) we find


ml (135)


ml (136)



ml (137)

17



minusmminusl (138)



ml (139)




ml (140)



radic






| jl = R| jl⟩ or ⟨ jm| = jm|R (142)





J 3| jl = l| jl

J plusmn| jl =radic



D jml = ⟨ jm|R| jl⟩ = jm|R| jl (145)





ml (147)

18



ml (148)




ml (149)



radic









and (150)








Qk middot x j =

983163


(153)





19








Qkψ = σkψ (159)



σk = plusmn1 (160)




α3

β 3γ 3

=

αβ γ + π

(162)




α2

β 2γ 2

=


(164)

20


α1

β 1γ 1

=

(α3)2(β 3)2(γ 3)2

=


(165)



mminusl


mminusl


ml (166)




σ1 = (minus1) j

σ2 = (minus1) j+l

σ3 = (minus1)l (168)

34 Rotations of L j




from (104) we findd




d

dφY 2 = Y 1 (172)


d2

dφ2Y 1 = minusY 1 (173)

21










Q3L1 Qdagger

3 = minusL1

Q3L2 Qdagger

3 = minusL2

Q3L3 Qdagger

3 = L3

(177)




H = H

hc = AL2

a + BL2b + C L2

c (178)


A =

4πcI a B =

4πcI b C =

4πcI c (179)




H = BL21 + C L2

2 + AL23

= B + C

2 L middot L +

2AminusB minus C

2L23 +

C minus B

4 (L2

+ + L2minus) (181)


22

41 Quantum numbers





Eσ1σ3 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm (183)


P = Q1 (184)


p = σ1 = plusmn1 (185)



|k⟩ =

991770

2 j + 1

8π2 D jlowast



L3|k⟩ = k|k⟩ (187)

Lplusmn|k⟩ =radic

( j plusmn k)( j ∓ k + 1)|k ∓ 1⟩ (188)

L middot L|k⟩ = j( j + 1)|k⟩ (189)


23


|ψn⟩ =sum

k

|k⟩vkn (190)



k



H ik = ⟨i|H |k⟩ = 2 j + 1

8π2

int

D jmiHD jlowast



⟨k|H |k⟩ = 1

2

983131


(193)


4

radic



4

radic





|1⟩ =

100

harr991770

3

8π2D1lowast

m1

|0⟩ =

010

harr991770

3

8π2D1lowast

m0

|minus 1⟩ =

001

harr991770

3

8π2D1lowast

mminus1 (196)


24

L middot L =

2 0 00 2 00 0 2

L23 =

1 0 00 0 00 0 1

L2+ =

0 0 20 0 00 0 0

L2minus =

0 0 00 0 02 0 0

(197)


H = 1

2


(198)


|ψx⟩ = 1radic

2

101

|ψy⟩ = 1radic

2

10

minus1

|ψz⟩ =

010

(199)







Q1 =


minus1 0 0

Q2 =

0 0 10 minus1 01 0 0

Q3 =


(202)





25

44 State labels








minusminus jm











L2aψ

kakc jm = k2



a = 0 1 4 9 j2 Except





26


ka =


(208)





L2cψ

kakc jm = k2






mplusmnka If we


kc =


(210)


kakc =


1 j minus 11 j0 j

(211)

27




ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka


852009

(213)



kakc jm (214)


σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)


σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)


σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)


ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)


References




28




























= 0 (116)

14


L middot L = L jL j

= X jrLrX jsLs



= LrLr (117)

3 Spherical Basis


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(118)


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(119)



2


0radic

2 0

(120)





2

minus1 i 0

0 0radic

21 i 0

(122)



or





or



15

From (120) (40) (42) and (48) we find

N daggerAN =


N daggerBN =


2

1minuscosβ2


21minuscosβ

2sinβradic

2

1+cosβ2

N daggerCN =


(125)



D1(α β γ ) =





2 eiγ


2 eiα 1+cosβ

2 ei(α+γ )

(127)





lowastν (128)






16




radic



D jml(α β γ ) = ⟨ jm|R| jl⟩ (131)







int 2π

0

dα

int π

0

sin βdβ

int 2π

0

dγD jlowastmlD

jprime

mprimelprime = 8π2



ml(α β γ )


ml



ml in (94) we find


ml (135)


ml (136)



ml (137)

17



minusmminusl (138)



ml (139)




ml (140)



radic






| jl = R| jl⟩ or ⟨ jm| = jm|R (142)





J 3| jl = l| jl

J plusmn| jl =radic



D jml = ⟨ jm|R| jl⟩ = jm|R| jl (145)





ml (147)

18



ml (148)




ml (149)



radic









and (150)








Qk middot x j =

983163


(153)





19








Qkψ = σkψ (159)



σk = plusmn1 (160)




α3

β 3γ 3

=

αβ γ + π

(162)




α2

β 2γ 2

=


(164)

20


α1

β 1γ 1

=

(α3)2(β 3)2(γ 3)2

=


(165)



mminusl


mminusl


ml (166)




σ1 = (minus1) j

σ2 = (minus1) j+l

σ3 = (minus1)l (168)

34 Rotations of L j




from (104) we findd




d

dφY 2 = Y 1 (172)


d2

dφ2Y 1 = minusY 1 (173)

21










Q3L1 Qdagger

3 = minusL1

Q3L2 Qdagger

3 = minusL2

Q3L3 Qdagger

3 = L3

(177)




H = H

hc = AL2

a + BL2b + C L2

c (178)


A =

4πcI a B =

4πcI b C =

4πcI c (179)




H = BL21 + C L2

2 + AL23

= B + C

2 L middot L +

2AminusB minus C

2L23 +

C minus B

4 (L2

+ + L2minus) (181)


22

41 Quantum numbers





Eσ1σ3 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm (183)


P = Q1 (184)


p = σ1 = plusmn1 (185)



|k⟩ =

991770

2 j + 1

8π2 D jlowast



L3|k⟩ = k|k⟩ (187)

Lplusmn|k⟩ =radic

( j plusmn k)( j ∓ k + 1)|k ∓ 1⟩ (188)

L middot L|k⟩ = j( j + 1)|k⟩ (189)


23


|ψn⟩ =sum

k

|k⟩vkn (190)



k



H ik = ⟨i|H |k⟩ = 2 j + 1

8π2

int

D jmiHD jlowast



⟨k|H |k⟩ = 1

2

983131


(193)


4

radic



4

radic





|1⟩ =

100

harr991770

3

8π2D1lowast

m1

|0⟩ =

010

harr991770

3

8π2D1lowast

m0

|minus 1⟩ =

001

harr991770

3

8π2D1lowast

mminus1 (196)


24

L middot L =

2 0 00 2 00 0 2

L23 =

1 0 00 0 00 0 1

L2+ =

0 0 20 0 00 0 0

L2minus =

0 0 00 0 02 0 0

(197)


H = 1

2


(198)


|ψx⟩ = 1radic

2

101

|ψy⟩ = 1radic

2

10

minus1

|ψz⟩ =

010

(199)







Q1 =


minus1 0 0

Q2 =

0 0 10 minus1 01 0 0

Q3 =


(202)





25

44 State labels








minusminus jm











L2aψ

kakc jm = k2



a = 0 1 4 9 j2 Except





26


ka =


(208)





L2cψ

kakc jm = k2






mplusmnka If we


kc =


(210)


kakc =


1 j minus 11 j0 j

(211)

27




ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka


852009

(213)



kakc jm (214)


σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)


σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)


σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)


ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)


References




28


L middot L = L jL j

= X jrLrX jsLs



= LrLr (117)

3 Spherical Basis


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(118)


ξ1 = x1 + ix2

minusradic


x1 minus ix2radic 2

(119)



2


0radic

2 0

(120)





2

minus1 i 0

0 0radic

21 i 0

(122)



or





or



15

From (120) (40) (42) and (48) we find

N daggerAN =


N daggerBN =


2

1minuscosβ2


21minuscosβ

2sinβradic

2

1+cosβ2

N daggerCN =


(125)



D1(α β γ ) =





2 eiγ


2 eiα 1+cosβ

2 ei(α+γ )

(127)





lowastν (128)






16




radic



D jml(α β γ ) = ⟨ jm|R| jl⟩ (131)







int 2π

0

dα

int π

0

sin βdβ

int 2π

0

dγD jlowastmlD

jprime

mprimelprime = 8π2



ml(α β γ )


ml



ml in (94) we find


ml (135)


ml (136)



ml (137)

17



minusmminusl (138)



ml (139)




ml (140)



radic






| jl = R| jl⟩ or ⟨ jm| = jm|R (142)





J 3| jl = l| jl

J plusmn| jl =radic



D jml = ⟨ jm|R| jl⟩ = jm|R| jl (145)





ml (147)

18



ml (148)




ml (149)



radic









and (150)








Qk middot x j =

983163


(153)





19








Qkψ = σkψ (159)



σk = plusmn1 (160)




α3

β 3γ 3

=

αβ γ + π

(162)




α2

β 2γ 2

=


(164)

20


α1

β 1γ 1

=

(α3)2(β 3)2(γ 3)2

=


(165)



mminusl


mminusl


ml (166)




σ1 = (minus1) j

σ2 = (minus1) j+l

σ3 = (minus1)l (168)

34 Rotations of L j




from (104) we findd




d

dφY 2 = Y 1 (172)


d2

dφ2Y 1 = minusY 1 (173)

21










Q3L1 Qdagger

3 = minusL1

Q3L2 Qdagger

3 = minusL2

Q3L3 Qdagger

3 = L3

(177)




H = H

hc = AL2

a + BL2b + C L2

c (178)


A =

4πcI a B =

4πcI b C =

4πcI c (179)




H = BL21 + C L2

2 + AL23

= B + C

2 L middot L +

2AminusB minus C

2L23 +

C minus B

4 (L2

+ + L2minus) (181)


22

41 Quantum numbers





Eσ1σ3 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm (183)


P = Q1 (184)


p = σ1 = plusmn1 (185)



|k⟩ =

991770

2 j + 1

8π2 D jlowast



L3|k⟩ = k|k⟩ (187)

Lplusmn|k⟩ =radic

( j plusmn k)( j ∓ k + 1)|k ∓ 1⟩ (188)

L middot L|k⟩ = j( j + 1)|k⟩ (189)


23


|ψn⟩ =sum

k

|k⟩vkn (190)



k



H ik = ⟨i|H |k⟩ = 2 j + 1

8π2

int

D jmiHD jlowast



⟨k|H |k⟩ = 1

2

983131


(193)


4

radic



4

radic





|1⟩ =

100

harr991770

3

8π2D1lowast

m1

|0⟩ =

010

harr991770

3

8π2D1lowast

m0

|minus 1⟩ =

001

harr991770

3

8π2D1lowast

mminus1 (196)


24

L middot L =

2 0 00 2 00 0 2

L23 =

1 0 00 0 00 0 1

L2+ =

0 0 20 0 00 0 0

L2minus =

0 0 00 0 02 0 0

(197)


H = 1

2


(198)


|ψx⟩ = 1radic

2

101

|ψy⟩ = 1radic

2

10

minus1

|ψz⟩ =

010

(199)







Q1 =


minus1 0 0

Q2 =

0 0 10 minus1 01 0 0

Q3 =


(202)





25

44 State labels








minusminus jm











L2aψ

kakc jm = k2



a = 0 1 4 9 j2 Except





26


ka =


(208)





L2cψ

kakc jm = k2






mplusmnka If we


kc =


(210)


kakc =


1 j minus 11 j0 j

(211)

27




ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka


852009

(213)



kakc jm (214)


σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)


σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)


σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)


ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)


References




28

From (120) (40) (42) and (48) we find

N daggerAN =


N daggerBN =


2

1minuscosβ2


21minuscosβ

2sinβradic

2

1+cosβ2

N daggerCN =


(125)



D1(α β γ ) =





2 eiγ


2 eiα 1+cosβ

2 ei(α+γ )

(127)





lowastν (128)






16




radic



D jml(α β γ ) = ⟨ jm|R| jl⟩ (131)







int 2π

0

dα

int π

0

sin βdβ

int 2π

0

dγD jlowastmlD

jprime

mprimelprime = 8π2



ml(α β γ )


ml



ml in (94) we find


ml (135)


ml (136)



ml (137)

17



minusmminusl (138)



ml (139)




ml (140)



radic






| jl = R| jl⟩ or ⟨ jm| = jm|R (142)





J 3| jl = l| jl

J plusmn| jl =radic



D jml = ⟨ jm|R| jl⟩ = jm|R| jl (145)





ml (147)

18



ml (148)




ml (149)



radic









and (150)








Qk middot x j =

983163


(153)





19








Qkψ = σkψ (159)



σk = plusmn1 (160)




α3

β 3γ 3

=

αβ γ + π

(162)




α2

β 2γ 2

=


(164)

20


α1

β 1γ 1

=

(α3)2(β 3)2(γ 3)2

=


(165)



mminusl


mminusl


ml (166)




σ1 = (minus1) j

σ2 = (minus1) j+l

σ3 = (minus1)l (168)

34 Rotations of L j




from (104) we findd




d

dφY 2 = Y 1 (172)


d2

dφ2Y 1 = minusY 1 (173)

21










Q3L1 Qdagger

3 = minusL1

Q3L2 Qdagger

3 = minusL2

Q3L3 Qdagger

3 = L3

(177)




H = H

hc = AL2

a + BL2b + C L2

c (178)


A =

4πcI a B =

4πcI b C =

4πcI c (179)




H = BL21 + C L2

2 + AL23

= B + C

2 L middot L +

2AminusB minus C

2L23 +

C minus B

4 (L2

+ + L2minus) (181)


22

41 Quantum numbers





Eσ1σ3 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm (183)


P = Q1 (184)


p = σ1 = plusmn1 (185)



|k⟩ =

991770

2 j + 1

8π2 D jlowast



L3|k⟩ = k|k⟩ (187)

Lplusmn|k⟩ =radic

( j plusmn k)( j ∓ k + 1)|k ∓ 1⟩ (188)

L middot L|k⟩ = j( j + 1)|k⟩ (189)


23


|ψn⟩ =sum

k

|k⟩vkn (190)



k



H ik = ⟨i|H |k⟩ = 2 j + 1

8π2

int

D jmiHD jlowast



⟨k|H |k⟩ = 1

2

983131


(193)


4

radic



4

radic





|1⟩ =

100

harr991770

3

8π2D1lowast

m1

|0⟩ =

010

harr991770

3

8π2D1lowast

m0

|minus 1⟩ =

001

harr991770

3

8π2D1lowast

mminus1 (196)


24

L middot L =

2 0 00 2 00 0 2

L23 =

1 0 00 0 00 0 1

L2+ =

0 0 20 0 00 0 0

L2minus =

0 0 00 0 02 0 0

(197)


H = 1

2


(198)


|ψx⟩ = 1radic

2

101

|ψy⟩ = 1radic

2

10

minus1

|ψz⟩ =

010

(199)







Q1 =


minus1 0 0

Q2 =

0 0 10 minus1 01 0 0

Q3 =


(202)





25

44 State labels








minusminus jm











L2aψ

kakc jm = k2



a = 0 1 4 9 j2 Except





26


ka =


(208)





L2cψ

kakc jm = k2






mplusmnka If we


kc =


(210)


kakc =


1 j minus 11 j0 j

(211)

27




ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka


852009

(213)



kakc jm (214)


σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)


σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)


σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)


ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)


References




28




radic



D jml(α β γ ) = ⟨ jm|R| jl⟩ (131)







int 2π

0

dα

int π

0

sin βdβ

int 2π

0

dγD jlowastmlD

jprime

mprimelprime = 8π2



ml(α β γ )


ml



ml in (94) we find


ml (135)


ml (136)



ml (137)

17



minusmminusl (138)



ml (139)




ml (140)



radic






| jl = R| jl⟩ or ⟨ jm| = jm|R (142)





J 3| jl = l| jl

J plusmn| jl =radic



D jml = ⟨ jm|R| jl⟩ = jm|R| jl (145)





ml (147)

18



ml (148)




ml (149)



radic









and (150)








Qk middot x j =

983163


(153)





19








Qkψ = σkψ (159)



σk = plusmn1 (160)




α3

β 3γ 3

=

αβ γ + π

(162)




α2

β 2γ 2

=


(164)

20


α1

β 1γ 1

=

(α3)2(β 3)2(γ 3)2

=


(165)



mminusl


mminusl


ml (166)




σ1 = (minus1) j

σ2 = (minus1) j+l

σ3 = (minus1)l (168)

34 Rotations of L j




from (104) we findd




d

dφY 2 = Y 1 (172)


d2

dφ2Y 1 = minusY 1 (173)

21










Q3L1 Qdagger

3 = minusL1

Q3L2 Qdagger

3 = minusL2

Q3L3 Qdagger

3 = L3

(177)




H = H

hc = AL2

a + BL2b + C L2

c (178)


A =

4πcI a B =

4πcI b C =

4πcI c (179)




H = BL21 + C L2

2 + AL23

= B + C

2 L middot L +

2AminusB minus C

2L23 +

C minus B

4 (L2

+ + L2minus) (181)


22

41 Quantum numbers





Eσ1σ3 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm (183)


P = Q1 (184)


p = σ1 = plusmn1 (185)



|k⟩ =

991770

2 j + 1

8π2 D jlowast



L3|k⟩ = k|k⟩ (187)

Lplusmn|k⟩ =radic

( j plusmn k)( j ∓ k + 1)|k ∓ 1⟩ (188)

L middot L|k⟩ = j( j + 1)|k⟩ (189)


23


|ψn⟩ =sum

k

|k⟩vkn (190)



k



H ik = ⟨i|H |k⟩ = 2 j + 1

8π2

int

D jmiHD jlowast



⟨k|H |k⟩ = 1

2

983131


(193)


4

radic



4

radic





|1⟩ =

100

harr991770

3

8π2D1lowast

m1

|0⟩ =

010

harr991770

3

8π2D1lowast

m0

|minus 1⟩ =

001

harr991770

3

8π2D1lowast

mminus1 (196)


24

L middot L =

2 0 00 2 00 0 2

L23 =

1 0 00 0 00 0 1

L2+ =

0 0 20 0 00 0 0

L2minus =

0 0 00 0 02 0 0

(197)


H = 1

2


(198)


|ψx⟩ = 1radic

2

101

|ψy⟩ = 1radic

2

10

minus1

|ψz⟩ =

010

(199)







Q1 =


minus1 0 0

Q2 =

0 0 10 minus1 01 0 0

Q3 =


(202)





25

44 State labels








minusminus jm











L2aψ

kakc jm = k2



a = 0 1 4 9 j2 Except





26


ka =


(208)





L2cψ

kakc jm = k2






mplusmnka If we


kc =


(210)


kakc =


1 j minus 11 j0 j

(211)

27




ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka


852009

(213)



kakc jm (214)


σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)


σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)


σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)


ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)


References




28



minusmminusl (138)



ml (139)




ml (140)



radic






| jl = R| jl⟩ or ⟨ jm| = jm|R (142)





J 3| jl = l| jl

J plusmn| jl =radic



D jml = ⟨ jm|R| jl⟩ = jm|R| jl (145)





ml (147)

18



ml (148)




ml (149)



radic









and (150)








Qk middot x j =

983163


(153)





19








Qkψ = σkψ (159)



σk = plusmn1 (160)




α3

β 3γ 3

=

αβ γ + π

(162)




α2

β 2γ 2

=


(164)

20


α1

β 1γ 1

=

(α3)2(β 3)2(γ 3)2

=


(165)



mminusl


mminusl


ml (166)




σ1 = (minus1) j

σ2 = (minus1) j+l

σ3 = (minus1)l (168)

34 Rotations of L j




from (104) we findd




d

dφY 2 = Y 1 (172)


d2

dφ2Y 1 = minusY 1 (173)

21










Q3L1 Qdagger

3 = minusL1

Q3L2 Qdagger

3 = minusL2

Q3L3 Qdagger

3 = L3

(177)




H = H

hc = AL2

a + BL2b + C L2

c (178)


A =

4πcI a B =

4πcI b C =

4πcI c (179)




H = BL21 + C L2

2 + AL23

= B + C

2 L middot L +

2AminusB minus C

2L23 +

C minus B

4 (L2

+ + L2minus) (181)


22

41 Quantum numbers





Eσ1σ3 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm (183)


P = Q1 (184)


p = σ1 = plusmn1 (185)



|k⟩ =

991770

2 j + 1

8π2 D jlowast



L3|k⟩ = k|k⟩ (187)

Lplusmn|k⟩ =radic

( j plusmn k)( j ∓ k + 1)|k ∓ 1⟩ (188)

L middot L|k⟩ = j( j + 1)|k⟩ (189)


23


|ψn⟩ =sum

k

|k⟩vkn (190)



k



H ik = ⟨i|H |k⟩ = 2 j + 1

8π2

int

D jmiHD jlowast



⟨k|H |k⟩ = 1

2

983131


(193)


4

radic



4

radic





|1⟩ =

100

harr991770

3

8π2D1lowast

m1

|0⟩ =

010

harr991770

3

8π2D1lowast

m0

|minus 1⟩ =

001

harr991770

3

8π2D1lowast

mminus1 (196)


24

L middot L =

2 0 00 2 00 0 2

L23 =

1 0 00 0 00 0 1

L2+ =

0 0 20 0 00 0 0

L2minus =

0 0 00 0 02 0 0

(197)


H = 1

2


(198)


|ψx⟩ = 1radic

2

101

|ψy⟩ = 1radic

2

10

minus1

|ψz⟩ =

010

(199)







Q1 =


minus1 0 0

Q2 =

0 0 10 minus1 01 0 0

Q3 =


(202)





25

44 State labels








minusminus jm











L2aψ

kakc jm = k2



a = 0 1 4 9 j2 Except





26


ka =


(208)





L2cψ

kakc jm = k2






mplusmnka If we


kc =


(210)


kakc =


1 j minus 11 j0 j

(211)

27




ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka


852009

(213)



kakc jm (214)


σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)


σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)


σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)


ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)


References




28



ml (148)




ml (149)



radic









and (150)








Qk middot x j =

983163


(153)





19








Qkψ = σkψ (159)



σk = plusmn1 (160)




α3

β 3γ 3

=

αβ γ + π

(162)




α2

β 2γ 2

=


(164)

20


α1

β 1γ 1

=

(α3)2(β 3)2(γ 3)2

=


(165)



mminusl


mminusl


ml (166)




σ1 = (minus1) j

σ2 = (minus1) j+l

σ3 = (minus1)l (168)

34 Rotations of L j




from (104) we findd




d

dφY 2 = Y 1 (172)


d2

dφ2Y 1 = minusY 1 (173)

21










Q3L1 Qdagger

3 = minusL1

Q3L2 Qdagger

3 = minusL2

Q3L3 Qdagger

3 = L3

(177)




H = H

hc = AL2

a + BL2b + C L2

c (178)


A =

4πcI a B =

4πcI b C =

4πcI c (179)




H = BL21 + C L2

2 + AL23

= B + C

2 L middot L +

2AminusB minus C

2L23 +

C minus B

4 (L2

+ + L2minus) (181)


22

41 Quantum numbers





Eσ1σ3 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm (183)


P = Q1 (184)


p = σ1 = plusmn1 (185)



|k⟩ =

991770

2 j + 1

8π2 D jlowast



L3|k⟩ = k|k⟩ (187)

Lplusmn|k⟩ =radic

( j plusmn k)( j ∓ k + 1)|k ∓ 1⟩ (188)

L middot L|k⟩ = j( j + 1)|k⟩ (189)


23


|ψn⟩ =sum

k

|k⟩vkn (190)



k



H ik = ⟨i|H |k⟩ = 2 j + 1

8π2

int

D jmiHD jlowast



⟨k|H |k⟩ = 1

2

983131


(193)


4

radic



4

radic





|1⟩ =

100

harr991770

3

8π2D1lowast

m1

|0⟩ =

010

harr991770

3

8π2D1lowast

m0

|minus 1⟩ =

001

harr991770

3

8π2D1lowast

mminus1 (196)


24

L middot L =

2 0 00 2 00 0 2

L23 =

1 0 00 0 00 0 1

L2+ =

0 0 20 0 00 0 0

L2minus =

0 0 00 0 02 0 0

(197)


H = 1

2


(198)


|ψx⟩ = 1radic

2

101

|ψy⟩ = 1radic

2

10

minus1

|ψz⟩ =

010

(199)







Q1 =


minus1 0 0

Q2 =

0 0 10 minus1 01 0 0

Q3 =


(202)





25

44 State labels








minusminus jm











L2aψ

kakc jm = k2



a = 0 1 4 9 j2 Except





26


ka =


(208)





L2cψ

kakc jm = k2






mplusmnka If we


kc =


(210)


kakc =


1 j minus 11 j0 j

(211)

27




ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka


852009

(213)



kakc jm (214)


σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)


σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)


σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)


ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)


References




28








Qkψ = σkψ (159)



σk = plusmn1 (160)




α3

β 3γ 3

=

αβ γ + π

(162)




α2

β 2γ 2

=


(164)

20


α1

β 1γ 1

=

(α3)2(β 3)2(γ 3)2

=


(165)



mminusl


mminusl


ml (166)




σ1 = (minus1) j

σ2 = (minus1) j+l

σ3 = (minus1)l (168)

34 Rotations of L j




from (104) we findd




d

dφY 2 = Y 1 (172)


d2

dφ2Y 1 = minusY 1 (173)

21










Q3L1 Qdagger

3 = minusL1

Q3L2 Qdagger

3 = minusL2

Q3L3 Qdagger

3 = L3

(177)




H = H

hc = AL2

a + BL2b + C L2

c (178)


A =

4πcI a B =

4πcI b C =

4πcI c (179)




H = BL21 + C L2

2 + AL23

= B + C

2 L middot L +

2AminusB minus C

2L23 +

C minus B

4 (L2

+ + L2minus) (181)


22

41 Quantum numbers





Eσ1σ3 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm (183)


P = Q1 (184)


p = σ1 = plusmn1 (185)



|k⟩ =

991770

2 j + 1

8π2 D jlowast



L3|k⟩ = k|k⟩ (187)

Lplusmn|k⟩ =radic

( j plusmn k)( j ∓ k + 1)|k ∓ 1⟩ (188)

L middot L|k⟩ = j( j + 1)|k⟩ (189)


23


|ψn⟩ =sum

k

|k⟩vkn (190)



k



H ik = ⟨i|H |k⟩ = 2 j + 1

8π2

int

D jmiHD jlowast



⟨k|H |k⟩ = 1

2

983131


(193)


4

radic



4

radic





|1⟩ =

100

harr991770

3

8π2D1lowast

m1

|0⟩ =

010

harr991770

3

8π2D1lowast

m0

|minus 1⟩ =

001

harr991770

3

8π2D1lowast

mminus1 (196)


24

L middot L =

2 0 00 2 00 0 2

L23 =

1 0 00 0 00 0 1

L2+ =

0 0 20 0 00 0 0

L2minus =

0 0 00 0 02 0 0

(197)


H = 1

2


(198)


|ψx⟩ = 1radic

2

101

|ψy⟩ = 1radic

2

10

minus1

|ψz⟩ =

010

(199)







Q1 =


minus1 0 0

Q2 =

0 0 10 minus1 01 0 0

Q3 =


(202)





25

44 State labels








minusminus jm











L2aψ

kakc jm = k2



a = 0 1 4 9 j2 Except





26


ka =


(208)





L2cψ

kakc jm = k2






mplusmnka If we


kc =


(210)


kakc =


1 j minus 11 j0 j

(211)

27




ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka


852009

(213)



kakc jm (214)


σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)


σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)


σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)


ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)


References




28


α1

β 1γ 1

=

(α3)2(β 3)2(γ 3)2

=


(165)



mminusl


mminusl


ml (166)




σ1 = (minus1) j

σ2 = (minus1) j+l

σ3 = (minus1)l (168)

34 Rotations of L j




from (104) we findd




d

dφY 2 = Y 1 (172)


d2

dφ2Y 1 = minusY 1 (173)

21










Q3L1 Qdagger

3 = minusL1

Q3L2 Qdagger

3 = minusL2

Q3L3 Qdagger

3 = L3

(177)




H = H

hc = AL2

a + BL2b + C L2

c (178)


A =

4πcI a B =

4πcI b C =

4πcI c (179)




H = BL21 + C L2

2 + AL23

= B + C

2 L middot L +

2AminusB minus C

2L23 +

C minus B

4 (L2

+ + L2minus) (181)


22

41 Quantum numbers





Eσ1σ3 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm (183)


P = Q1 (184)


p = σ1 = plusmn1 (185)



|k⟩ =

991770

2 j + 1

8π2 D jlowast



L3|k⟩ = k|k⟩ (187)

Lplusmn|k⟩ =radic

( j plusmn k)( j ∓ k + 1)|k ∓ 1⟩ (188)

L middot L|k⟩ = j( j + 1)|k⟩ (189)


23


|ψn⟩ =sum

k

|k⟩vkn (190)



k



H ik = ⟨i|H |k⟩ = 2 j + 1

8π2

int

D jmiHD jlowast



⟨k|H |k⟩ = 1

2

983131


(193)


4

radic



4

radic





|1⟩ =

100

harr991770

3

8π2D1lowast

m1

|0⟩ =

010

harr991770

3

8π2D1lowast

m0

|minus 1⟩ =

001

harr991770

3

8π2D1lowast

mminus1 (196)


24

L middot L =

2 0 00 2 00 0 2

L23 =

1 0 00 0 00 0 1

L2+ =

0 0 20 0 00 0 0

L2minus =

0 0 00 0 02 0 0

(197)


H = 1

2


(198)


|ψx⟩ = 1radic

2

101

|ψy⟩ = 1radic

2

10

minus1

|ψz⟩ =

010

(199)







Q1 =


minus1 0 0

Q2 =

0 0 10 minus1 01 0 0

Q3 =


(202)





25

44 State labels








minusminus jm











L2aψ

kakc jm = k2



a = 0 1 4 9 j2 Except





26


ka =


(208)





L2cψ

kakc jm = k2






mplusmnka If we


kc =


(210)


kakc =


1 j minus 11 j0 j

(211)

27




ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka


852009

(213)



kakc jm (214)


σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)


σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)


σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)


ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)


References




28










Q3L1 Qdagger

3 = minusL1

Q3L2 Qdagger

3 = minusL2

Q3L3 Qdagger

3 = L3

(177)




H = H

hc = AL2

a + BL2b + C L2

c (178)


A =

4πcI a B =

4πcI b C =

4πcI c (179)




H = BL21 + C L2

2 + AL23

= B + C

2 L middot L +

2AminusB minus C

2L23 +

C minus B

4 (L2

+ + L2minus) (181)


22

41 Quantum numbers





Eσ1σ3 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm (183)


P = Q1 (184)


p = σ1 = plusmn1 (185)



|k⟩ =

991770

2 j + 1

8π2 D jlowast



L3|k⟩ = k|k⟩ (187)

Lplusmn|k⟩ =radic

( j plusmn k)( j ∓ k + 1)|k ∓ 1⟩ (188)

L middot L|k⟩ = j( j + 1)|k⟩ (189)


23


|ψn⟩ =sum

k

|k⟩vkn (190)



k



H ik = ⟨i|H |k⟩ = 2 j + 1

8π2

int

D jmiHD jlowast



⟨k|H |k⟩ = 1

2

983131


(193)


4

radic



4

radic





|1⟩ =

100

harr991770

3

8π2D1lowast

m1

|0⟩ =

010

harr991770

3

8π2D1lowast

m0

|minus 1⟩ =

001

harr991770

3

8π2D1lowast

mminus1 (196)


24

L middot L =

2 0 00 2 00 0 2

L23 =

1 0 00 0 00 0 1

L2+ =

0 0 20 0 00 0 0

L2minus =

0 0 00 0 02 0 0

(197)


H = 1

2


(198)


|ψx⟩ = 1radic

2

101

|ψy⟩ = 1radic

2

10

minus1

|ψz⟩ =

010

(199)







Q1 =


minus1 0 0

Q2 =

0 0 10 minus1 01 0 0

Q3 =


(202)





25

44 State labels








minusminus jm











L2aψ

kakc jm = k2



a = 0 1 4 9 j2 Except





26


ka =


(208)





L2cψ

kakc jm = k2






mplusmnka If we


kc =


(210)


kakc =


1 j minus 11 j0 j

(211)

27




ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka


852009

(213)



kakc jm (214)


σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)


σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)


σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)


ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)


References




28

41 Quantum numbers





Eσ1σ3 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm


Eσ1σ2 jm (183)


P = Q1 (184)


p = σ1 = plusmn1 (185)



|k⟩ =

991770

2 j + 1

8π2 D jlowast



L3|k⟩ = k|k⟩ (187)

Lplusmn|k⟩ =radic

( j plusmn k)( j ∓ k + 1)|k ∓ 1⟩ (188)

L middot L|k⟩ = j( j + 1)|k⟩ (189)


23


|ψn⟩ =sum

k

|k⟩vkn (190)



k



H ik = ⟨i|H |k⟩ = 2 j + 1

8π2

int

D jmiHD jlowast



⟨k|H |k⟩ = 1

2

983131


(193)


4

radic



4

radic





|1⟩ =

100

harr991770

3

8π2D1lowast

m1

|0⟩ =

010

harr991770

3

8π2D1lowast

m0

|minus 1⟩ =

001

harr991770

3

8π2D1lowast

mminus1 (196)


24

L middot L =

2 0 00 2 00 0 2

L23 =

1 0 00 0 00 0 1

L2+ =

0 0 20 0 00 0 0

L2minus =

0 0 00 0 02 0 0

(197)


H = 1

2


(198)


|ψx⟩ = 1radic

2

101

|ψy⟩ = 1radic

2

10

minus1

|ψz⟩ =

010

(199)







Q1 =


minus1 0 0

Q2 =

0 0 10 minus1 01 0 0

Q3 =


(202)





25

44 State labels








minusminus jm











L2aψ

kakc jm = k2



a = 0 1 4 9 j2 Except





26


ka =


(208)





L2cψ

kakc jm = k2






mplusmnka If we


kc =


(210)


kakc =


1 j minus 11 j0 j

(211)

27




ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka


852009

(213)



kakc jm (214)


σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)


σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)


σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)


ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)


References




28


|ψn⟩ =sum

k

|k⟩vkn (190)



k



H ik = ⟨i|H |k⟩ = 2 j + 1

8π2

int

D jmiHD jlowast



⟨k|H |k⟩ = 1

2

983131


(193)


4

radic



4

radic





|1⟩ =

100

harr991770

3

8π2D1lowast

m1

|0⟩ =

010

harr991770

3

8π2D1lowast

m0

|minus 1⟩ =

001

harr991770

3

8π2D1lowast

mminus1 (196)


24

L middot L =

2 0 00 2 00 0 2

L23 =

1 0 00 0 00 0 1

L2+ =

0 0 20 0 00 0 0

L2minus =

0 0 00 0 02 0 0

(197)


H = 1

2


(198)


|ψx⟩ = 1radic

2

101

|ψy⟩ = 1radic

2

10

minus1

|ψz⟩ =

010

(199)







Q1 =


minus1 0 0

Q2 =

0 0 10 minus1 01 0 0

Q3 =


(202)





25

44 State labels








minusminus jm











L2aψ

kakc jm = k2



a = 0 1 4 9 j2 Except





26


ka =


(208)





L2cψ

kakc jm = k2






mplusmnka If we


kc =


(210)


kakc =


1 j minus 11 j0 j

(211)

27




ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka


852009

(213)



kakc jm (214)


σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)


σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)


σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)


ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)


References




28

L middot L =

2 0 00 2 00 0 2

L23 =

1 0 00 0 00 0 1

L2+ =

0 0 20 0 00 0 0

L2minus =

0 0 00 0 02 0 0

(197)


H = 1

2


(198)


|ψx⟩ = 1radic

2

101

|ψy⟩ = 1radic

2

10

minus1

|ψz⟩ =

010

(199)







Q1 =


minus1 0 0

Q2 =

0 0 10 minus1 01 0 0

Q3 =


(202)





25

44 State labels








minusminus jm











L2aψ

kakc jm = k2



a = 0 1 4 9 j2 Except





26


ka =


(208)





L2cψ

kakc jm = k2






mplusmnka If we


kc =


(210)


kakc =


1 j minus 11 j0 j

(211)

27




ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka


852009

(213)



kakc jm (214)


σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)


σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)


σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)


ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)


References




28

44 State labels








minusminus jm











L2aψ

kakc jm = k2



a = 0 1 4 9 j2 Except





26


ka =


(208)





L2cψ

kakc jm = k2






mplusmnka If we


kc =


(210)


kakc =


1 j minus 11 j0 j

(211)

27




ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka


852009

(213)



kakc jm (214)


σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)


σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)


σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)


ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)


References




28


ka =


(208)





L2cψ

kakc jm = k2






mplusmnka If we


kc =


(210)


kakc =


1 j minus 11 j0 j

(211)

27




ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka


852009

(213)



kakc jm (214)


σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)


σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)


σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)


ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)


References




28




ψkakc jm =

991770

2 j + 1

16π2

852008

D jlowastmka


852009

(213)



kakc jm (214)


σ1σ2σ3

=

σ1(minus1)kaσ1

(minus1)ka

(215)


σ1σ2σ3

=

(minus1)kc

σ2σ3

(216)


σ1σ2σ3

=

(minus1)kc

(minus1)ka+kc

(minus1)ka

(217)


ψkakc jm = ψ

0kc jm =

991770

[ j]

8π2D jlowast

m0 (218)


References




28

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