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Spring 2001 PHYS 385 Quantum Physics Course Summary

This course summary organizes the basic ideas contained in the lecture notes, the textbook, and the coursehandouts. It is not intended as a substitute for these materials, and it skips many important details. Any discussionof core concepts is necessarily abridged. The emphasis is on putting most of the important equations and conceptsin one place, to serve as a launching point for further study.

1 Bohr-Sommerfeld Quantization

1.1 Quanta

Planck introduced the constant that now bears his name to eliminate the ultraviolet divergence in the black-bodyradiation spectrum. It defines a fundamental scale of action, that allows us to relate kinematic and wave propertiesof quantum mechanical objects.

h ' 6.626× 10−27erg s ~ ' 1.055× 10−27erg s (1a)p = h/λ p = ~k (1b)

E = hν E = ~ω (1c)

Quantum mechanical waves obey the superposition principle, and therefore may exhibit quantum mechanicalinterference.

Experimentally, the emission and absorption spectra of atoms were known to exhibit discrete lines, which satis-fied the Rydberg-Ritz combination principle, that ν = Ai −Aj , where the Ai and Aj were empirically determinedterms. This discrete spectrum could not be explained by classical models, and we now understand these terms asthe energy levels of the atom. The arithmetic relationships among certain groups of lines were identified as earlyas 1885, when Balmer showed that certain lines were predicted by the formula hν = Ry(1/22 − 1/n2) (Balmer,unaware of h, focused on the wavelength). Similar series are associated with Lyman, Paschen, Brackett, and Pfund.

Other experimental indications of quantization:

Franck-Hertz Energy lost by electrons in inelastic collisions with atomsStern-Gerlach Quantization of magnetic moment, spatial quantizationPhotoelectric effect UV photon in, electron out: photons as kinematic particlesCompton effect Inelastic X-ray scattering: photon kinematics (Compton wavelength λC ≡ h/mc ' 0.024A)

1.2 The Bohr Atom

Bohr proposed that the electrons in atoms could only exist in certain well-defined, stable orbits, which satisfied theBohr-Sommerfeld quantization condition,

∮

p · dq = nh, n ∈ N, (2)

where p is the momentum and q is the position coordinate of an electron in three-dimensional space; the integralis performed over some closed orbit in phase space p,q. Considering the electron as a wave with wavelengthλ = h/p, this Bohr-Sommerfeld quantization condition ensures that the wave is described by a function that issingle-valued.

Although the Bohr atom is a crude approximation of the full solution to the Schrodinger equation in a central

1


1/r potential, it does produce several correct results:

L = n~ Angular momentum (3a)

a0 ≡~2

me2 ' 0.53A Bohr radius (3b)

En = −Z2

n2

(

e2

2a0

)

Energy (3c)

Ry =e2

2a0' 13.6 eV Rydberg (3d)

vn

c=

Zn

α Relativistic effects (3e)

α ≡ e2

~c '1

137Fine structure constant (3f)

µ = nµB Magnetic dipole moment (3g)

µB =e~

2mc' 9.3× 10−21 erg/G Bohr magneton (3h)

Bohr required that his formulation of quantum mechanics would satisfy the tested principles of classical me-chanics, in the limit of large quantum numbers, where the scale of action set by Planck’s constant is small. This isknown as the correspondence principle.

2


2 Wave Mechanics

2.1 The Wave Function

Wave mechanics refers to Schrodinger’s formalism of quantum mechanics, in which a quantum mechanical state isrepresented by a complex wave function, which is frequently denoted by ψ(r, t). Born interpreted the wave functionas a complex probability amplitude. Given an ensemble of particles, all described by the same wave function, theprobability density for finding one particle at position r at time t is given by the norm squared of the wave function:

P(r, t) ≡ |ψ(r, t)|2∫

dr |ψ(r, t)|2. (4)

The integral in the denominator is taken to be over all space. The wave function may be multiplied by an arbitrarycomplex constant, and this probability density will remain invariant; therefore, we are generally free to choose aproper normalization of the wave function, such that

∫

drP(r, t) =∫

dr |ψ(r, t)|2 = 1. (5)

This normalization condition will set the magnitude of the wave function at every point in space, but it still leavesus free to choose the complex phase of the wave function. In general, we choose this phase to simplify calculationsin any particular problem.

The wave function obeys the superposition principle, and therefore exhibits quantum interference: if one possiblestate of a particle is given by ψ1 = |ψ1|eiφ1 , and another possible state is given by ψ2 = |ψ2|eiφ2 , so that the totalwave function is given by ψ = ψ1 + ψ2,

|ψ|2 = |ψ1 + ψ2|2 = |ψ1|2 + |ψ2|2 + 2|ψ1| |ψ2| cos(φ1 − φ2). (6)

2.2 The Schrodinger Equation

The wave function evolves in time according to a partial differential equation, the time-dependent Schrodingerequation,

i~∂ψ(r, t)∂t

= − ~2

2m∇2ψ(r, t) + V (r)ψ(r, t) = Hψ(r, t), (7)

where H is the differential operator that corresponds to the Hamiltonian of the system (note that the Hamiltonian ofcharged particles in a magnetic field has a slightly different form). To obtain the quantum mechanical Hamiltonianoperator from the classical Hamiltonian, which is a function of position and momentum, we make the substitution

p → ~i∇. (8)

An eigenfunction of the Hamiltonian is a function ψ(r, t) that satisfies the time-independent Schrodinger equa-tion,

Hψ(r, t) = − ~2

2m∇2ψ(r, t) + V (r)ψ(r, t) = Eψ(r, t). (9)

States which are described by eigenfunctions of the Hamiltonian are called stationary states, and evolve in timeaccording to

ψ(r, t) = ψ(r) exp(−iEt/~). (10)

The probability density P(r, t) of a quantum mechanical system must be a conserved quantity. We may definea probability current density,

S(r, t) =~

2miψ∗(r, t)∇ψ(r, t)− ψ(r, t)∇ψ∗(r, t). (11)

A wave function that evolves according to the time-dependent Schrodinger equation will then necessarily satisfythe continuity equation

∂P(r, t)∂t

+∇ · S(r, t) = 0. (12)

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2.3 The Free Particle and Eigenfunctions of the Momentum Operator

The stationary states of a free particle in one dimension may be written as plane waves:

ϕp(x, t) = ϕp(x)e−iEt/~, with (13a)

ϕp(x) =1

√

(2π~)eipx/~ and (13b)

E =p2

2m. (13c)

The factor of√

(2π~) gives the proper “normalization” of the plane-wave basis:∫ ∞

−∞dxϕ∗p(x)ϕp′(x) =

12π~

∫ ∞

−∞dx ei(p′−p)x/~ = δ(p− p′). (14)

These plane waves are simultaneous eigenfunctions of the Hamiltonian, H = p2/2m, and the momentum operator,p = (h/i)∂/∂x. This is possible because [H, p] = 0. The energy eigenvalues of the plane wave states are doublydegenerate: Ep = E−p. By labeling a state according to its momentum quantum number, we have a unique basisin which to expand any one-dimensional wave function. Thus, we have expressed the wave function as a functionof x, t, and labelled it by its (continuous) momentum quantum number p. We did not use this notation at the timethat we developed Fourier analysis, but now you have seen other basis sets labelled by other quantum numbers,and the meaning here is the same.

It is important to remember that any individual plane-wave basis function has a constant amplitude over allspace, and so is not actually normalizable. Thus, it is not a legitimate candidate for a physically admissible quantummechanical wave function. To construct a normalizable wave function, we must truncate the wave function in space,which necessarily implies that the wave function is a superposition of plane waves with different p values. We writesuch a superposition as ψ(p), calling it the momentum-space representation of the wave function. It is the Fouriertransform of the real-space, or coordinate, representation ψ(x):

ψ(x) =1√2π~

∫ ∞

−∞dp eipx/~ψ(p) ψ(p) =

1√2π~

∫ ∞

−∞dx e−ipx/~ψ(x), (15a)

ψ(x) =∫ ∞

−∞dpψ(p)ϕp(x) ψ(p) =

∫ ∞

−∞dxϕ∗p(x)ψ(x). (15b)

Written in this way, we may understand ψ(p) as the coefficients of ψ(x) in the basis ϕp(x).The momentum and coordinate representations of the wave function obey Parseval’s theorem:

∫ ∞

−∞dx |ψ(x)|2 =

∫ ∞

−∞dp |ψ(p)|2. (16)

The Fourier transform relations between x and p representations of the wave function lead directly to the HeisenbergUncertainty Relation:

(∆x)rms(∆p)rms ≥~2. (17)

Additional Fourier transform relations exist between the time and frequency (or energy) domain, and betweenreal-space and momentum-space in d > 1 dimensions:

f(t) =1√2π

∫ ∞

−∞dω e−iωtF (ω) F (ω) =

1√2π

∫ ∞

−∞dt eiωtf(t), (18a)

f(t) =1√2π~

∫ ∞

−∞dω e−iEt/~F (E) F (E) =

1√2π~

∫ ∞

−∞dt eiEt/~f(t), (18b)

ψ(x) =1

√

(2π~)d

∫

allspace

dp eip·x/~ψ(p) ψ(p) =1

√

(2π~)d

∫

allspace

dx e−ip·x/~ψ(x). (18c)

4


In d > 1, the Heisenberg Uncertainty Relation applies only to xi and pi, where i indicates a particular coordinateaxis. There is no problem with performing a measurement of the position along the x-axis and the momentumalong the y-axis, for example.

A wavepacket evolves in time according to the time-dependent Schodinger equation. To characterize the motionof the wavepacket, two velocities prove useful,

vϕ =ωk

vg =∂ω∂k

, (19)

where vϕ is called the phase velocity and vg is called the group velocity. The group velocity of the wavepacketcorresponds to the classical motion of the particle.

We found that a 1D gaussian wavepacket is a wavepacket of minimum uncertainty, and will remain a gaussianas it evolves, though with a time-dependent width. If the state at t = 0 has a real-space representation

ψ(x, t = 0) = [πa2]−1/4 exp[−x2/2a2] exp[ip0x/~], (20)

then its momentum-space representation is

ψ(p) = [π~2/a2]−1/4 exp[−a2(p− p0)2/2~2]. (21)

Note that the rms width of the momentum distribution is given by

(∆p)2rms =∫ ∞

−∞dp p2|ψ(p)|2 =

~2

2a2 , (22)

while the rms width of the probability distribution in real space is (∆x)rms = a/√

2.The momentum-space representation is appropriate for calculating the temporal evolution of the wavepacket,

because ψ(p) tells us the amplitude of any particular plane wave ϕp with energy E = p2/2m. When we allow thesystem to evolve in time, we multiply each plane wave by its associated exponential factor e−iEt/~:

ψ(p, t) = [π~2/a2]−1/4 exp[−a2(p− p0)2/2~2] exp[−iEt/~]. (23)

Then, if we want to know the real-space width of the wavepacket, we back-transform to get ψ(x, t), which we maywrite in the form

ψ(x, t) = π[a(t)]2−1/4 exp−(x− vgt)2/2[a(t)]2 exp[ip0(x− vϕt)/~], (24)

where [a(t)]2 = a2 + i~t/m. Thus, the envelope of the wavepacket moves at the group velocity vg = p0/m, whilethe underlying wavefronts move at the phase velocity vϕ = p0/2m. The probability distribution |ψ(x, t)|2 will thenspread in time, according to

(∆x)rms =a√2

√

1 +(

~tma2

)2

. (25)

For t ma/~, the width remains approximately constant, growing only quadratically with time. However, fort ma/~, the width grows linearly with time, at a rate ~/

√2ma = ~/2m(∆x)rms ' (∆p)rms/m = (∆v)rms.

2.4 The Particle in a One-Dimensional Box (Infinite Well)

For a particle in a box, Schrodinger equation in the box is that of a free particle, and we have a general solution ofthe form ψ(x) = A+eipx/~ + A−e−ipx/~ = A+eikx + A−e−ikx. The boundary conditions at the walls of the box areψ(0) = ψ(L) = 0, which are met by normalized solutions of the form

ψn =

√

2L

sinnπxL

, n > 0, (26)

with energy eigenvalues

En =~2

2m

(nπL

)2=~2k2

n

2m, kn =

nπL

. (27)

5


The general solution of Schodinger’s equation in the infinite well is of the form:

ψ(x) =∞∑

n=1

cnψn(x). (28)

If we are given some wave function ψ(x), then we can find the coefficient cn of ψn(x) by taking the inner productof ψ(x) with ψn(x):

cn =∫ ∞

−∞dxψ∗n(x)ψ(x). (29)

Once we have found the spectral decomposition of the wave function ψ(x), we can find the time dependence of thestate by multiplying each basis state by the appropriate exponential factor e−iEt/~:

ψ(x, t) =∞∑

n=1

cn e−iEnt/~ ψn(x). (30)

The probability density for finding the particle at position x at time t will be

P(x, t) = |ψ(x, t)|2 =∞∑

m,n=1

c∗mcne−i(En−Em)t/~ψ∗m(x)ψn(x), (31)

which is oscillatory whenever the state ψ(x, t) involves more than one eigenstate of the Hamiltonian.

2.5 General Methodology

Our approach to the particle in a box illustrates a general approach to all problems in wave mechanics that wehave encountered in the course. Let us summarize the general procedure, for an arbitrary system with a discretespectrum (for a continuous spectrum, we would replace the the summations by integrals):

1. Write the classical Hamiltonian H(p,x).

2. Make an operator replacement for the classical variables.

3. Find the eigenvalues En and eigenfunctions ϕn(x) of the time-independent Schrodinger equation

Hϕn(x) = Enϕn(x).

4. Normalize the eigenfunctions.

5. Orthogonalize degenerate eigenfunctions (this was not necessary for the particle in a box and 1D harmonicoscillator, which have nondegenerate spectra, but was necessary for higher dimensional systems with angularmomentum).

6. Given an arbitrary state ψ(x) at t = 0, find the spectral decomposition of ψ in terms of the energy eigenstates:

ψ(x) =∑

cnϕn(x),

cn = (ϕn(x), ψ(x)).

where (f, g) is the inner product of the two functions f and g.

7. Each energy eigenfunction evolves according to e−iEnt/~, where En is the eigenvalue of the Hamiltonianassociated with the eigenfunction ϕn(x):

ψ(x, t) =∑

cn e−iEnt/~ ϕn(x).

An exception to this general methodology is spin, which has no classical analogue, and is naturally represented interms of states in an abstract vector space, not as functions of spatial coordinates. Even here, however, the analysisfrom point (3) onward may be employed with a simple substitution of Dirac kets (|+〉, |−〉 for S = 1/2) instead ofeigenfunctions.

6


3 The Mathematical Structure of Quantum Mechanics

3.1 Quantum States as Elements of a Vector Space

In wave mechanics, we found that the TISE could be understood as an eigenvalue equation, with the stationarystates corresponding to the eigenfunctions of the Hamiltonian. Any physically admissible wave function may be de-composed into a superposition of these eigenfunctions. We also found that in this spectral decomposition of the wavefunction the coefficient cn of a particular eigenfunction ϕn(x) was given by the overlap integral cn =

∫

dx ϕ∗n(x)ψ(x).Mathematically, we can think of the eigenfunctions of the Hamiltonian as vectors in an abstract vector space, definedby an inner product (ψa(x), ψb(x)) =

∫

dxψ∗a(x)ψb(x). There are potentially infinite dimensions in our quantummechanical vector space, one for each distinct eigenfunction of the Hamiltonian, but its properties are very similarto those of the more familiar three-dimensional vector space, where the inner product is just the dot product.

To formalize the relationship between quantum mechanical states and abstract vectors, we introduced Diracnotation, which is a commonly used representation of quantum mechanical states. In Dirac notation, a quantumstate is written as a “ket,” |ψ〉, which we wrote as the wave function ψ(x) in wave mechanics. In Dirac notation,the TISE is H|ϕn〉 = En|ϕn〉. Associated with each ket is a linear operator that we call a “bra,” which acts onkets to give a number. If the kets |ψa〉, |psib〉 are associated with wave functions ψa(x), ψb(x), then we have

〈ψa|ψb〉 = (ψa(x), ψb(x)) =∫ ∞

−∞dxψ∗a(x)ψb(x). (32)

We can also write a general state |ψ〉 in terms of some basis of states |ϕn〉, usually the eigenstates of the Hamiltonian:

|ψ〉 =∑

allstates

cn|ϕn〉. (33)

Here, as before, we write the spectral decomposition for a discrete basis; for a continuous basis we replace the sumsby integrals over the appropriate variable(s).

Given a particular orthonormal basis |ϕn〉 for the state space, we may write a general state in terms of thecoefficients cn in its spectral decomposition, just as we write a three-dimensional vector in terms of its coefficients(vx, vy, vz) in the expression v = vxx + vy y + vz z. For kets, we write the coefficients as a column vector, and forbras, we write them as a row vector:

〈ψ| =(

c∗1 c∗2 c∗3 . . .)

=(

〈ψ|ϕ1〉〈ψ|ϕ2〉〈ψ|ϕ3〉 . . .)

, |ψ〉 =

c1

c2

c3...

=

〈ϕ1|ψ〉〈ϕ2|ψ〉〈ϕ3|ψ〉

...

. (34)

By writing them in this way, the inner product is just the matrix product of the row vector with the column vector:

〈ψa|ψb〉 =(

c∗a1 c∗a2 c∗a3 . . .)

cb1

cb2

cb3...

=∑

n

c∗ancbn. (35)

We can describe the relationship between the bra and the ket by saying that one is the adjoint of the other,where the adjoint operator is written as a superscript dagger:

〈ψ| = (|ψ〉)† (36)

In the matrix representation of the states (row and column vectors are 1xN and Nx1 matrices, respectively), theadjoint of a matrix is the complex conjugate of its transpose:

M † = (MT)∗ M †ij = M∗

ji. (37)

7


3.2 Observables and Hermitian Operators

A linear operator provides a linear correspondence for every state in a quantum mechanical system to some otherstate: given a linear operator O,

|ψ′〉 = O|ψ〉, and (38a)

O(a1|ψ1〉+ a2|ψ2〉) = a1O|ψ1〉+ a2O|ψ2〉. (38b)

These operators are associative under multiplication and addition, and commutative under addition, but they arenot commutative under multiplication: O1O2 6= O2O1.

The projection operator is a particularly simple example of an operator, that projects an arbitrary state ontosome subspace of the state space. For a one-dimensional subspace of all states proportional to |ψS〉, the projectionoperator PψS may be written as

PψS = |ψS〉〈ψS |. (39)

This has a matrix representation given by multiplying the column vector representation of |ψS〉 by the row vectorrepresentation of 〈ψS |. If |ψS〉 is one of the basis states, |ϕn〉, then the matrix representation of PψS has only onenonzero element at position (n, n).

An orthonormal basis of a quantum mechanical state space satisfies the closure relation, which we write belowfor different types of bases:

1 =∑

n

|ϕn〉〈ϕn| Discrete (40a)

1 =∫

dα |ϕ(α)〉〈ϕ(α)| Continuous (40b)

1 =∑

n

|ϕn〉〈ϕn|+∫

dα|ϕ(α)〉〈ϕ(α)| Mixed (40c)

Mixed bases include the states of the finite potential well and of the hydrogen atom (where the continuum corre-sponds to ionized states).

In quantum mechanics, each observable property of a physical system must be represented mathematically bya Hermitian operator, which is self-adjoint; that is, given an inner product (ψa, ψb) between two states in our statespace, a Hermitian operator H satisfies (ψa, Hψb) = (Hψa, ψb). Alternatively, we have

H = H†, Hij = H∗ji. (41)

The adjoint of an operator product is(AB)† = B†A†. (42)

The form of the operator representation may vary, depending on the representation of the state space. Thus, ifwe choose to represent a particle in one dimension as a wave function in coordinate space, ψ(x), the Hermitianoperator associated with momentum is p = (~/i)d/dx; if we represent it as a wave function in momentum space,ψ(p), the Hermitian operator is just the numerical value of the momentum at which the function is evaluated:p = p. From here on, we will drop the additional “hat” notation, O, for operators, unless the operator character isnot evident from the context. Given an arbitrary linear operator O, not necessarily Hermitian, we can form twodifferent Hermitian operators:

H =12(O + O†), H =

12i

(O −O†). (43)

In a matrix representation of the state space, Hermitian operators are square matrices. The individual matrixelements for a discrete basis may be written in Dirac notation as Aij = 〈ϕi|A|ϕj〉, indicating that the matrixelement may be calculated by finding the inner product of |ϕi〉 with A|ϕj〉. In wave mechanics, these matrixelements are simply integrals, in which the differential operator A is sandwiched between two basis wavefunctions:

Aij =∫

dxϕ∗i (x)Aϕj(x). (44)

8


The matrix elements are arranged in matrix format in the usual way,

A =

A11 A12 . . . A1j . . .A21 A22 . . . A2j . . ....

......

Ai1 Ai2 . . . Aij . . ....

......

(45)

In a continuous basis (plane waves, for example) the matrix elements A(p, p′) are functions of continuous variablesp and p′, and the matrix representation of operators becomes more awkward. The idea, nonetheless, is the same,and the term “matrix element” continues to be used.

Linear operators may be characterized by their eigenvalues, and their corresponding eigenvectors. For a generallinear operator A, an eigenvector |ϕλ〉 of A, with eigenvalue λ, satisfies the eigenvalue equation

A|ϕλ〉 = λ|ϕλ〉. (46)

Given a matrix representation of A, the eigenvalues λ of A will be given by the following characteristic equation:

det[A− λ1] =

∣

∣

∣

∣

∣

∣

∣

A11 − λ A12 . . .A21 A22 − λ . . ....

...

∣

∣

∣

∣

∣

∣

∣

= 0. (47)

A defining characteristic of Hermitian operators is that they have real eigenvalues. This is the reason that observ-ables must be represented by them. Once we know the eigenvalues of A by solving its characteristic equation, wemay find an eigenvector |ϕn〉 associated with a particular eigenvalue λn by solving the system of equations givenby Eq. 46:

A11 A12 . . .A21 A22 . . ....

...

c1c2...

= λn

c1

c2...

. (48)

If two eigenvectors of a Hermitian operator have different eigenvalues, then these eigenvectors are orthogonal.We say that two states are distinct if they are not parallel; that is, they are not related by a simple constant factor.

For example, when we choose a baisis of N states to span an N -dimensional state space, each of the N states mustbe distinct. An eigenvalue that is associated with one and only one distinct eigenvector is called nondegenerate, andwhen we solve the system of equations given by Eq. 48, we will get N−1 equations in N unknowns that completelyconstrains the relationship among the coefficients cn. When several distinct eigenvectors satisfy the eigenvalueequation, Eq. 46 for the same eigenvalue, the eigenvalue is called degenerate, and its degeneracy g is given by thenumber of distinct eigenvectors associated with it. Eq. 48 then gives N − g equations in N unknowns, and we mustchoose an appropriate basis for the degenerate subspace. In general, we must use the Gramm-Schmidt procedureto find an orthogonal basis for the degenerate subspace:

1. Pick arbitrary coefficients c1, c2, . . . .

2. Normalize this state. This will be the first basis vector, |ϕ(1)n 〉, corresponding to eigenvalue λn.

3. Pick new, different coefficients c′1, c′2, . . . . Call the state defined by these coefficients |ψ〉.

4. From |ψ〉, create a new state |ψ′〉 that is orthogonal to |ϕ(1)n 〉, by projecting out any component parallel to it:

|ψ′〉 = (1− Pψ(1)n

)|ψ〉.

5. Normalize |ψ′〉 to get the next basis vector, |ϕ(2)n 〉:

|ϕ(2)n 〉 =

|ψ′〉√

〈ψ′|ψ′〉.

9


6. Repeat steps (3)-(5) until you have found an orthonormal basis for the degenerate subspace.

Frequently, however, we may find an orthogonal basis for some degenerate subspace of A by identifying someother Hermitian operator B that commutes with A. The plane wave basis for a free particle provides a particularlysimple example: |ψp〉 and |ψ−p〉 have the same eigenvalue in H = p2/2m, but have different eigenvalues under p,so they form an orthogonal basis for the two-fold degenerate subspace corresponding to eigenvalues p2/2m of theHamiltonian. Similarly, we chose the 2` + 1 degenerate eigenstates of L2 to simultaneously satsify an eigenvalueequation with Lz, each with a different eigenvalue m, so they were necessarily orthogonal.

3.3 Postulates of Quantum Mechanics

1. A physical system at time t0 is defined by a state |ψ(t0)〉

2. Every measurable physical quantity A is associated with a Hermitian operator A acting on |ψ(t0)〉.

3. The only possible result of a measurement of A is one of the eigenvalues of A.

4. When A is measured on a normalized state |ψ〉, the probability P(an) (or probability density P(a) ) ofobtaining the eigenvalue an (a) of A is

P(an) =gn∑

i=1

|〈ψ(i)n |ψ〉|2 (Discrete, degeneracy gn)

P(a) = |〈ψ(a)|ψ〉|2 (Continuous)

5. If the measurement of A on |ψ〉 gives the result an, the state immediately after the measurement is thenormalized projection of |ψ〉 onto the subspace associated with an,

Pϕan|ψ〉

√

〈ψ|Pϕan|ψ〉

.

6. The time evolution of |ψ(t)〉 is given by the time dependent Schrodinger equation,

i~ ddt|ψ(t)〉 = H|ψ(t)〉,

where H is the Hamiltonian of the system.

3.4 Expectation Values and Commutation Relations

The expectation value of an observable A for a state |ψ〉 is the weighted average of all possible outcomes of themeasurement of A on a physical system described by |ψ〉. It is written as 〈A〉 or A, and is given by

〈A〉 = 〈ψ|A|ψ〉 =∑

n

|cn|2an, (49)

where the cn are the coefficients that define |ψ〉 in terms of the eigenstates of A, and the an are the associatedeigenvalues of A. In wave mechanics, it is sometimes more straightforward to use the form

〈A〉 =∫

dxψ∗(x)Aψ(x). (50)

It is important to recognize that the expectation value of an operator may not be an eigenvalue of the operator, sothat an individual measurement will never produce the expectation value as a result. The expectation value of theoperator is what would be obtained on average in a measurement on a large number of identically prepared states.

When two operators do not commute, it is not possible to perform their associated measurements with infiniteprecision. This is related to the projection postulate (5). If [A,B] 6= 0, then the eigenstates of B can not be made

10


parallel to the eigenstates of A for all circumstances, and the order in which we perform our projection operationswill matter. We know from the Fourier analysis of wave mechanics that x and p can not be measured simultaneouslywith infinite precision, and in fact their associated operators do not commute:

[x, p] = i~. (51)

We may generalize the Heisenberg Uncertainty Principle to other observables by defining for an operator A anrms uncertainty (∆A)rms, defined as the expectation value of the measurement error:

(∆A)rms = 〈A2〉 − 〈A〉2. (52)

Then the generalized uncertainty principle is

(∆A)rms(∆B)rms ≥12|〈[A,B]〉|. (53)

The expectation value of an operator A may be seen to evolve in time according to

ddt〈A〉 =

i~ 〈[H,A]〉+ 〈∂A

∂t〉, (54)

where the expectation values are calculated at fixed time. This leads directly to Ehrenfest’s theorem,

ddt〈p〉 = −〈dV

dt〉, (55)

which is the quantum mechanical equivalent of F = ma. We may also see from Eq. 54 that any observable thatcommutes with the Hamiltonian is a conserved quantity. This has important consequences in the application ofsymmetry principles to quantum mechanical problems. For example, the Hamiltonian of the particle-in-a-boxpotential (centered at the origin) is symmetric about x = 0, so the physical system must be invariant under theparity operator P : x → −x. Thus, [H,P ] = 0, and a quantum state with definite parity will remain with thatparity for all time. Additionally, the particle-in-a-box states may be labelled by quantum numbers ±1, dependingon their parity; this is not necessary to identify the states here, because the spectrum of the Hamiltonian isnondegenerate. However, for many situations, degenerate quantum states may be identified by their eigenvaluesunder some symmetry-related operator (i.e., the two-dimensional particle-in-a-box).

Commuting observables also play an important role in defining the natural basis of eigenstates for any physicalproblem. Since all eigenstates of a Hermitian operator with different eigenvalues are orthogonal, it is natural totry to express the basis states for a physical problem in terms of relevant observables. If observable A commuteswith the Hamiltonian, then an eigenstate of A with eigenvalue a will remain in the same eigenstate, with the sameeigenvalue, as time evolves. The eigenvalues of A are said to be good quantum numbers of the system, and wemay label our states by these eigenvalues without worrying that the labels will lose their meaning as the systemevolves. When we have a set of commuting observables, and the results of their measurement completely determinesthe quantum states of the system, we have a complete set of commuting observables. For the hydrogen atom, forexample, the observables H, L2, and Lz form a complete set of commuting observables. For plane waves in onedimension, H and p form a complete set of commuting observables.

11


4 One-Dimensional Problems

4.1 General Features

If we consider a particle of mass m in an arbitrary one-dimensional (1D) potential, there are a few things that wecan say about it without actually solving the TISE. As shown in Fig. 1,

for a potential that is bounded above on the left by V =

x

V(x)

Ea

Eb

Ec

Ed

Vmax

Figure 1: Energy scales of a general 1D potential

0, on the right by V = Vmax, and bounded from below byV = Vmin, there are four characteristic energy scales, eachwith different types of states:

1. E > Vmax (Ea): the eigenvalues of the TISE form a con-tinuum, which is two-fold degenerate, with one rightwardtravelling state and one leftward travelling state.

2. 0 < E < Vmax (Eb): the eigenvalues of the TISE form anondegenerate continuum.

3. Vmin < E < 0 (Ec): the eigenvalues are discrete, nonde-generate bound states.

4. E < Vmin (Ed): there is no solution to the TISE.

4.2 Piecewise Constant Potentials

For a constant potential V over some domain D, the TISE is

− ~2

2md2ψ(x)

dx2 = (E − V )ψ(x), x ∈ D, (56)

which has three different categories of solution:

1. E > V , travelling wave solutions:

ψ(x) = A(+)eikx + A(−)e−ikx, k =

√

2m~2 (E − V ). (57)

2. E < V , exponentially decaying solutions:

ψ(x) = B(+)eρx + B(−)e−ρx, ρ =

√

2m~2 (V − E). (58)

3. E = V , linear solutions:

ψ(x) = Cx + D. (59)

At the boundaries between domains with different constant potentials, the wave function must satisfy boundaryconditions. At the discontinuities in the potential, the wave function must be both continuous and differentiable.By matching up the solutions for each domain using these boundary conditions, the complete solution to the TISEfor the piecewise constant potential may be obtained. Where the potential is symmetric about some point, thenthe wave functions of the particle in the potential may have well-defined parity, and this may be used to providefurther conditions on the wave functions in different domains.

The basic potentials that we considered were:

12


1. The finite potential well. This has both bound state solutions, which we found by solving a transcendentalequation with a graphical method. The continuum solutions behave similarly to the scattering solutions ofthe barrier potential, discussed below.

2. The potential step. This has a continuum of solutions, and we found the coefficients of transmission andreflection at the step. If we consider the left hand side of the step Region 1, and the right hand side Region 2,then the transmission and reflection coefficients for a particle moving from left to right are

T =k2

k1

∣

∣

∣

∣

∣

A(+)2

A(+)1

∣

∣

∣

∣

∣

2

and R =

∣

∣

∣

∣

∣

A(−)1

A(+)1

∣

∣

∣

∣

∣

2

. (60)

The factor of k2/k1 in the transmission coefficient arises because the coefficient is defined in terms of a ratioof the probability current density S on each side of the step. The current density is proportional to the speed,which is different on either side of the step, so the transmission coefficient includes the ratio of the speeds inaddition to the square of the ratio of the amplitudes.

3. The barrier potential. We discussed two types of solutions: scattering for E > V0, and tunnelling for E < V0,where V0 is the potential height of the barrier. The scattering solutions are valid for both positive and negativevalues of V0. We found that for the scattering solutions, the transmission coefficient of the barrier exhibitedresonances, like the resonances of a Fabry-Perot cavity in optics:

T =4E(E − V0)

4E(E − V0) + V 20 sin2[

√

2m(E − V0)L/~]. (61)

For the tunneling solutions are given by this equation, after some manipulation to account for the fact thatthe argument of the sine function becomes imaginary:

T =4E(V0 − E)

4E(V0 − E) + V 20 sinh2[

√

2m(V0 − E)L/~]∼ 16E(V0 − E)

V 20

e−2ρL, (62)

where ρ =√

2m(V0 −E)/~2 as usual and L is the barrier width.

4.3 The Harmonic Oscillator

The Hamiltonian of the Harmonic oscillator is

H =p2

2m+

12kx2 =

p2

2m+

12mω2x2. (63)

We found it useful to use dimensionless operators,

X = βx, P =1β~p, H =

1~ωH, (64)

where β =√

mω/~. Then

H =12(X2 + P2), (65)

with commutation relations[X,P] = i. (66)

To solve for the eigenvalues and eigenvectors of the harmonic oscillator, we defined the annihilation and creationoperators,

a =1√2(X + iP), a† =

1√2(X− iP). (67)

13


In terms of the original operators x and p, we write

a =√

mω2~ x + i

1√2m~ω

p, a† =√

mω2~ x− i

1√2m~ω

p. (68)

The creation and annihilation operators obey the commutation relation

[a, a†] = 1. (69)

If we further define the number operator,N = a†a, (70)

Then the harmonic oscillator hamiltonian may be written simply as

H =(

N +12

)

~ω. (71)

The eigenvectors of the Hamiltonian operator are manifestly eigenvectors of the number operator, and if we let nbe the eigenvalue of N associated with an eigenstate |ϕn〉, the TISE is just

H|ϕn〉 =(

n +12

)

|ϕn〉. (72)

By examining the properties of N , a, a†, and their various commutation relations, we were able to prove thatn ∈ N. Thus, the energy spectrum of the harmonic oscillator Hamiltonian is

En =(

n +12

)

~ω n ∈ N. (73)

We found that in the basis of Hamiltonian eigenstates, the matrix elements of a and a† are

amn =√

nδm,n−1, (74a)

a†mn =√

n + 1δm,n+1. (74b)

Thus, given an arbitrary state |ϕn〉, we can use the creation and annihilation operators to generate all othereigenstates of the Hamiltonian. Writing Eqs. 68 in the differential operator form of wave mechanics allows us tocalculate the wave functions ϕn(x), first by solving a|ϕ0〉 = 0, then using the creation operator to generate thehigher energy states, via

|ϕn〉 =1√n!

(a†)n|ϕ0〉. (75)

The first few eigenfunctions are

ϕ0(x) =(mω

π~

)1/4exp

(

−12

mω~ x2

)

, (76a)

ϕ1(x) =[

4π

(mω~

)3]1/4

x exp(

−12

mω~ x2

)

(76b)

ϕ2(x) =( mω

4π~

)1/4 (

2mω~ x2 − 1

)

exp(

−12

mω~ x2

)

(76c)

By inverting Eqs. 68, we can also find the matrix elements of x and p in the same basis:

xmn =

√

~2mω

[√n + 1 δm,n+1 +

√n δm,n−1

]

, (77a)

pmn = i

√

m~ω2

[√n + 1 δm,n+1 −

√n δm,n−1

]

, (77b)

14


In addition to the operator method for finding the eigenfunctions of the harmonic oscillator, we solved thedifferential equation given by the TISE in wave mechanics. With the substitutions ξ =

√

mω/~x and ε = E/~ω,this differential equation is

d2ψ(ξ)dξ2 + (2ε− ξ2)ψ(ξ) = 0. (78)

We solved this equation in the limit ξ → 0 first, to find solutions of the gaussian form ψ(ξ) ∝ e−ξ2/2. We thenwrote the general solution as a product of this gaussian with a polynomial function, ψ(ξ) = h(ξ)e−ξ2/2, with

h(ξ) =∞∑

m=0

amξm, (79)

and where h(ξ) dominates the behavior of the eigenfunction for ξ → 0. The differential equation then yields arecursion relation that relates the am for different values of m:

(m + 1)(m + 2)am+2 = (2m− 2ε + 1)am. (80)

From the recursion relation we were able to find the eigenfunctions of even and odd parity, for a discrete set ofεm = m + 1/2 that terminate the polynomial sum in Eq. 79.

15


5 Angular Momentum

5.1 General Properties

We define angular momentum in quantum mechanics just as we do in classical mechanics,

L = r× p. (81)

This definition, together with the definitionL± = Lx ± iLy, (82)

imply the following useful commutation relations:

[Ly, Lz] = i~Lx [Lz, L±] = ±~L±[Lz, Lx] = i~Ly [L+, L−] = 2~Lz (83)

[Lx, Ly] = i~Lz [L2, Lz] = 0

[L2, L±] = 0

Due to the nonzero commutators among the various components of L, it is impossible to measure all threecomponents of the angular momentum simultaneously. In fact, if we measure the component of L along one axis, ingeneral the projection of the state along that axis will disrupt any subsequent measurements of L along a differentaxis. This is known as spatial quantization. By convention, we choose Lz to be the quantization axis of angularmomentum, and express states of angular momentum in terms of the simultaneous eigenvectors of Lz and L2, whichcommutes with Lz. The eigenvalue equations are

L2|ϕ`m〉 = `(` + 1)~2|ϕ`m〉 (84a)

Lz|ϕ`m〉 = m~|ϕ`m〉 (84b)

Through operator methods similar to those that we used to solve the harmonic oscillator, we found the followingconditions on `, m:

2` ∈ N, m1 −m2 ∈ Z, (85a)` ≥m ≥ −`, 2` + 1 allowed values. (85b)

We found that the L± operators act much as the operators a and a† act in the state space of the harmonic oscillator,by transforming the states |ϕ`m〉 into the states |ϕ`m±1〉. The matrix elements are

〈ϕ`m|Lz|ϕ`′m′〉 = m~ δ``′δmm′ (86a)

〈ϕ`m|L±|ϕ`′m′〉 = ~√

`(` + 1)−m(m∓ 1) δ``′δmm′±1 (86b)

5.2 Wave Functions

It is most convenient to express the eigenfunctions of angular momentum in a spherical coordinate system,shown in Fig. 2. The relationship to cartesian coordinates is given by

x = r sin θ cos φ (87a)

y = r sin θ sin φ (87b)

z = r cos θ (87c)

The differential operator form of Lz in spherical coordinates is simply

Lz =hi

∂∂φ

. (88)

16


Consequently, the eigenfunctions of Lz are simply

Φm(φ) = eimφ/√

2π. (89)

The operator L2 expressed in spherical coordinates is

L2 = −~2[

1sin θ

∂∂θ

(

sin θ∂∂θ

)

+1

sin2 θ∂2

∂φ2

]

. (90)

This is obtained directly from the Laplacian in spherical coordinates, which is

∇2 =1r2

∂∂r

(

r2 ∂∂r

)

+1

r2 sin θ∂∂θ

(

sin θ∂∂θ

)

+1

r2 sin2 θ∂2

∂φ2 . (91)

The relationship between Eqs. 90 and 91 is given by the classical relation

p2 = p2r + L2/r2. (92)

For any spherically symmetric potential, it is useful to ex-

x

y

z

θ

φ

r

Figure 2: Spherical coordinates

press the eigenfunctions of the total Hamiltonian in terms ofa separation of variables:

ψ(r, θ, φ) = R(r)Θ(θ)Φ(φ), (93)

where each of the three product functions satisfy separate, or-dinary differential equations, which may or may not be cou-pled. In the solution to Lzψ(r, θ, φ) = m~ψ(r, θ, φ), for ex-ample, the solution for Φm is independent of the functions Rand Θ. The solutions of the eigenvalue equation L2ψ(r, θ, φ) =`(` + 1)~2ψ(r, θ, φ) do not constrain R, and we may write theeigenvalue equation in terms of functions, called the sphericalharmonics, that depend only on θ and φ, and are simultane-ously eigenfunctions of Lz with eigenvalue m~: Y`m(θ, φ) =Θ(θ)Φ(φ). These functions describe orbital angular momen-tum only, and therefore are restricted to integer values of `and m. Half-integral values of angular momentum describeonly internal angular momentum, or spin, which we representas abstract vectors, not as wave functions of spatial coordinates.

To generate the spherical harmonics, we used a method similar to the one we used in finding the eigenfunctionsfor the harmonic oscillator: we note that L+Y``(θ, φ) = 0, and express L+ as a differential operator in sphericalcoordinates to obtain a simple differential equation, that we solve. With the solution Y`` in hand, we can use thedifferential operator L− to generate the remaining m values. The differential operators are

L± ≡ Lx ± iLy = ~e±iφ(

± ∂∂θ

+ i cot θ∂∂φ

)

, (94)

and the first few normalized spherical harmonics obtained from this procedure are

Y00 =1√4π

Y22 =

√

1532π

sin2 θe2iφ

Y11 = −√

38π

sin θeiφ Y21 = −√

158π

sin θ cos θeiφ (95)

Y10 =

√

34π

cos θ Y20 =

√

516π

(3 cos2 θ − 1),

17


andY`−m = (−1)m Y`m. (96)

We may express any function of angular coordinates in terms of the spherical harmonics, according to the usualexpansion:

f(θ, φ) =∞∑

`=0

∑

m=−`

a`mY`m(θ, φ), (97a)

a`m =∫

dΩY ∗`m(θ, φ)f(θ, φ) (97b)

=∫ 2π

0dφ

∫ π

0dθ sin θ Y ∗

`m(θ, φ)f(θ, φ) (97c)

Since there are now two quantum numbers, ` and m, the probability of obtaining `(` + 1)~2 in a measurementof L2 is a sum,

P (`) =∑

m=−`

|a`m|2, (98)

as is the probability of obtaining m~ in a measurement of Lz,

P (m) =∞∑

`=0

|a`m|2. (99)

We also have expressions for expectation values that involve double summations:

〈L2〉 =∞∑

`=0

∑

m=−`

`(` + 1)~2 |a`m|2 (100a)

〈Lz〉 =∞∑

`=0

∑

m=−`

m~ |a`m|2 (100b)

5.3 Radial Equation

To completely solve the TISE in spherical coordinates, we must find the functions R(r). The Hamiltonian is

H =p2

r

2m+

L2

2mr2 + V (r) (101a)

= − ~2

2m1r2

∂∂r

(

r2 ∂∂r

)

+L2

2mr2 + V (r), (101b)

and if we write ψ(r, θ, φ) = R(r)Y`m(θ, φ), then the TISE may be transformed, using the eigenvalue equation forL2 Y`m, into an ordinary differential equation for R(r),

[

− ~2

2m1r2

ddr

(

r2 ddr

)

+~2`(` + 1)

2mr2 + V (r)]

Rk`(r) = Ek`Rk`(r), (102)

where we have introduced the subscripted Ek` and Rk`(r) to identify the eigenvalue equation that is being solved(`) as well as the particular eigenvalue associated with that solution (k).

By introducing a new function,uk`(r) = rRk`(r), (103)

we can convert Eq. 102 into another, simpler equation,(

− ~2

2md2

dr2 +~2`(` + 1)

2mr2 + V (r))

uk`(r) = Ek`uk`(r), (104)

18


which looks just like the 1D TISE except that we have an effective potential

Veff (r) = V (r) +`(` + 1)~2

2mr2 , (105)

and we have an additional boundary condition,

uk`(r → 0) → 0. (106)

5.4 The Magnetic Moment of a Charged Particle with Angular Momentum

The Hamiltonian of a charged particle in a magnetic field B is written in terms of the vector potential A associatedwith the magnetic field,

B = ∇×A; (107)

the Hamiltonian isH =

12m

∣

∣

∣p−ecA

∣

∣

∣

2+ V (r), (108)

where e is the electric charge of the particle, negative for an electron, and the units are CGS. We are free to use agauge transformation,

A → A′ = A +∇χ(r), (109)

where χ(r) is an arbitrary function of the spatial coordinates. Thus we are free to write the Hamiltonian in theCoulomb gauge, for which

∇ ·A = 0. (110)

With this choice of gauge, the Hamiltonian takes the form

H = − ~2

2m∇2 − e~

imcA · ∇+

e2

2mc2 A2 + V (r). (111)

For a constant magnetic field, the vector potential in the Coulomb gauge is A = (r×B)/2, and for a field B = Bz z,the Hamiltonian becomes

H = − ~2

2m∇2 − e

2mcLzBz +

e2

8mc2 B2z (x2 + y2) + V (r). (112)

The second and third terms on the right hand side of Eq. 112 are responsible for paramagnetism and diamagnetism,respectively. Since Lz is in units of ~, the paramagnetic term provides the basis for quantizing the magnetic momentof a charged particle with angular momentum. The quantum of magnetic moment is the Bohr magneton, and iswritten as

µB =e~

2mc. (113)

For electrons, this is a negative number, and the magnetic moment points antiparallel to the magnetic field.The Hamiltonian of a particle with spin in a magnetic field is analagous to the paramagnetic term in Eq. 112:

Hspin = −gµB

~ S ·B, (114)

where the gyromagnetic ratio or g-factor g = 2.0023 ' 2 for electrons. For an orbital magnetic moment the g-factoris g = 1.

19


6 The Hydrogen Atom

6.1 Effective One-Body Problem

The hydrogen atom is composed of a proton and an electron, interacting via a central 1/r potential. The Hamil-tonian is

H =p21

2m1+

p22

2m2+ V (r1 − r2), (115)

and the overall wave function of the two particles may be written as a product of single-particle wave functions:

ψ(r1, r2) = ψ1(r1)ψ2(r2). (116)

To find the solutions of the TISE, it is useful to make a coordinate transformation into center-of-mass-coordinates,

M = m1 + m2 µ =m1m2

m1 + m2(117a)

R =m1r1 + m2r2

m1 + m2r = r1 − r2 (117b)

P = p1 + p2 p =m2p1 −m1p2

m1 + m2(117c)

The transformed Hamilton is then

H =P 2

2M+

p2

2µ+ V (r), (118)

which has solutions of the form ψ(R, r) = ϕP(R)ψµ(r) where ϕP(R) = eiP·R/√

2π~ are just the eigenfunctionsof a free particle (which is the whole hydrogen atom). This reduces the two-particle Hamiltonian to an effectivesingle-particle Hamiltonian, the eigenfunctions of which are the internal states of the hydrogen atom.

6.2 Solutions to the Radial Equation for a 1/r Potential

Our method for solving the radial equation, Eq. 102, for the 1/r Coulomb potential was similar to the methoddescribed in Sec. 4.3 for solving the differential equation given by the harmonic oscillator Hamiltonian. The radialequation for uk`(r) is

(

− ~2

2md2

dr2 +~2`(` + 1)

2mr2 − Ze2

r

)

uk`(r) = Ek`uk`(r), (119)

which we can place in dimensionless form with the following substitutions:

a0 =~2

µe2 , EI =e2

2a0, ρ =

ra0

, λ2n` = −En`

EI, (120)

where we have changed the radial quantum number from k to n in anticipation of the integer values of n. Withthese substitutions, the radial equation becomes

(

d2

dρ2 −`(` + 1)

ρ2 +2Zρ− λ2

n`

)

uk`(ρ) = 0, (121)

which in the limit ρ → 0 becomes(

d2

dρ2 − λ2n`

)

un`(ρ) = 0. (122)

The solution to Eq. 122 is justun`(ρ → 0) ∝ e−λn`ρ, (123)

so to find the solution for all ρ we writeun`(ρ) = T (ρ)e−λn`ρ, (124)

20


where

T (ρ) = ρs∞∑

q=0

aqρq. (125)

We obtain s = ` + 1 from the boundary condition of Eq. 106. When we plug Eq. 124 into Eq. 121, we get therecursion relation

q(q + 2` + 1)aq = 2[λn`(` + q)− Z]aq−1. (126)

This recursion relation will generate a series T (ρ) that will not be normalizable if it is not terminated. Thus, theonly good solutions to the radial equation are those for which aq = 0 for some q, for this to occur we must haveλn`(` + q)− Z = 0. Returning to the dimensionful quantities in Eq. 120, we have

En` = −Z2

n2

e2

2a0, n ∈ N, n ≥ ` + 1. (127)

Thus we have solved for the energy spectrum of the hydrogen atom, which has levels that are n2-fold degeneratefor each value of n, due to the multiplicity of angular momentum states for each radial quantum number.

A few of the properly normalized solutions to the radial equation are listed below, together with its name inspectroscopic notation.

1s R10(r) =(

Za0

)3/2

2 exp(−Zr/a0) (128a)

2s R20(r) =(

Z2a0

)3/2 (

2− Zra0

)

exp(−Zr/2a0) (128b)

2p R21(r) =(

Z2a0

)3/2 1√3

Zra0

exp(−Zr/2a0) (128c)

3s R30(r) =(

Z3a0

)3/2

2

[

1− 23

Zra0

+227

(

Zra0

)2]

exp(−Zr/3a0) (128d)

3p R31(r) =(

Z3a0

)3/2 4√

23

Zra0

(

1− 16

Zra0

)

exp(−Zr/3a0) (128e)

3d R32(r) =(

Z3a0

)3/2 227

√

25

(

Zra0

)2

exp(−Zr/3a0) (128f)

(128g)

6.3 Hydrogenic Wave Functions, Matrix Elements and Selection Rules

The solution of the radial equation may be combined with the spherical harmonics to give the hydrogenic wavefunctions,

ψn`m(r, θ, φ) = Rn`(r)Y`m(θ, φ). (129)

Just as with other systems, we may describe observable operators for the hydrogen atom in a matrix representation,with matrix elements given by integrals over three-dimensional space:

〈ψn`m|A|ψn′`′m′〉 =∫

d3rψ∗n`m(r, θ, φ)Aψn′`′m′(r, θ, φ) (130a)

=∫ ∞

−∞dr r2

∫ π

0dθ sin θ

∫ 2π

0dφψ∗n`m(r, θ, φ)Aψn′`′m′(r, θ, φ) (130b)

The spherical symmetry of the 1/r potential implies that the eigenfunctions of the Hamiltonian will have well-defined parity, where in three dimensions the parity operator changes the sign of all three cartesian coordinateaxes: Pψ(r) = ψ(−r). We have

P |ψn`m〉 = (−1)`|ψn`m〉 (131)

21


If an operator has well-defined parity, also, then we may immediately identify nonzero matrix elements as thosewhich do not change sign when the parity operator acts on both the operator and the states. In particular, theinteraction of electromagnetic radiation with atoms is dominated by the interaction of the electric field with theelectric dipole of the atom. For the atom to possess an electric dipole, it must be in a superposition of states whichhave opposite parity, since the electric dipole operator is odd under parity. This is an example of a selection ruleon atomic transitions. Upon further analysis of the properties of the electric dipole operator, we may obtain thefollowing electric dipole selection rules:

∆` = ±1 (132)

∆m = 0, ±1. (133)

These selection rules, together with the principle of the conservation of angular momentum, imply that electro-magnetic carries angular momentum, and that it is quantized in units of Lz = ±~, where we choose z to be alongthe direction of propogation.

6.4 The Pauli Exclusion Principle and the Periodic Table

We found that systems of identical particles exist in quantum states that possess well-defined symmetry underparticle exchange. For a system of two particles, we write |ψ(1, 2)〉 to indicate a particular labelling for the particles.The behavior of “Particle 1” in the state |ψ(1, 2)〉 may be entirely different from the behavior of “Particle 1” in thestate |ψ(2, 1)〉. The symmetry of the system under particle exchange, however, provides important constraints onthe allowed eigenstates of the Hamiltonian, and empirically we know that the nature of these constraints dependon the spin of the particles. Half-integral spin systems (s = 1/2, 3/2, 5/2,. . . ) are called fermions, and possessstates that are antisymmetric under particle exchange. Integral spin systems are called bosons and are symmetricunder particle exchange. For two particles, the symmetric and antisymmetric states are

For bosons: |ψ(s)(1, 2)〉 =1√2[|ψ(1, 2)〉+ |ψ(2, 1)〉] (134)

For fermions: |ψ(a)(1, 2)〉 =1√2[|ψ(1, 2)〉 − |ψ(2, 1)〉] (135)

The antisymmetry of fermions under particle exchange is responsible for the Pauli Exclusion Principle, thatstates that no two identical fermions may occupy the same quantum state. If we have a many-particle Hamiltonianthat possesses no interactions between the particles, then we saw that we can write the many-particle state as aproduct of single-particle states:

|ψ(1, 2, . . . , N)〉 = ψα1(1) ψα2(2) . . . ψαN (N), (136a)|ψ(2, 1, . . . , N)〉 = ψα1(2) ψα2(1) . . . ψαN (N). (136b)

Here, the αn are the quantum numbers that identify the particular single-particle state, including the spin projectionquantum number. The fully antisymmetric state may be written as a determinant of a matrix of single-particlewave functions, called a Slater determinant:

ψ(a)(1, 2, . . . , N) =

∣

∣

∣

∣

∣

∣

∣

∣

∣

ψα1(1) ψα1(2) . . . ψα1(N)ψα2(1) ψα2(2) . . . ψα2(N)

......

...ψαN (1) ψαN (2) . . . ψαN (N)

∣

∣

∣

∣

∣

∣

∣

∣

∣

(137)

The Pauli Exclusion Principle is expressed here by the fact that if any two single-particle states in this determinantare the same, then two rows of the matrix will be identical, and the determinant will be zero.

One of the many consequences of the Pauli exclusion principle is the structure of the periodic table. As thenuclear charge increases, so that we need to add more electrons for charge neutrality, each new electron must go intoa different quantum state. Since the spin projection quantum number of an electron may take two different values,each hydrogenic wave function may hold two electrons. For low Z atoms, the electronic states fill up “in order,”

22


that is, 1s, 2s, 2p, 3s, 3p. But beyond this, the interaction of the electrons with each other become important, andin the transition metal series and beyond, the structure is more complicated (or richer, depending on the eye of thebeholder).

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7 Spin and Two-Level Systems

While it is possible to define spherical harmonic functions Y 12 ±

12, we found that these functions were no longer

connected by the differential operator representations of L±, as they were for integral values of `, m. This forcedus to use the abstract vector representation of spin, where the basis |+〉, |−〉 is defined in terms of the eigenstatesof the Sz operator:

“Spin-up” Sz|+〉 =~2|+〉 |+〉 =

(

10

)

(138a)

“Spin-down” Sz|−〉 = −~2|+〉 |−〉 =

(

01

)

(138b)

In this basis, we may write S = ~2σ, where the components of σ are the three Pauli matrices:

σx =(

0 11 0

)

σy =(

0 −ii 0

)

σz =(

1 00 −1

)

(139)

The Pauli matrices exhibit commutation relations that are similar to the operators of angular momentum,

[σy, σz] = 2iσx, (140a)

[σz, σx] = 2iσy, (140b)

[σx, σy] = 2iσz. (140c)

Spatial quantization results from these commutation relations. That is, if we know that the system is in aneigenstate of Sz, say |+〉, then it must be in a superposition of eigenstates of Sx, or of any other operator thatprojects the spin along an axis different from z. Consequently, if we apply a magnetic field along an axis that isdifferent from the original quantization axis, the spin will precess around the magnetic field, and oscillat between |+〉and |−〉 with a frequency ω = ∆E/~ = 2µBB/~. Our treatment of this precession followed the general methodologylaid out in Sec. 2.5.

Although in any particular case it is fairly straightforward to solve for the eigenvalues and eigenvectors of a2× 2 Hermitian matrix, we discussed a general method of solution. We can write the Hamiltonian matrix as a sumof two matrices, one proportional to 1 and the other proportional to a traceless matrix that we denote by K,

H =12(H11 + H22)1 +

12(H11 −H22)K, (141)

where

K =

(

1 2H12H11−H22

2H21H11−H22

−1

)

. (142)

The all states are eigenstates of 1, so the eigenstates of the Hamiltonian are determined by the eigenstates of K:

H|ψ±〉 = E±|ψ±〉, (143a)

K|ψ±〉 = κ±|ψ±〉. (143b)

By defining angles 0 ≤ θ < π and 0 ≤ φ < 2π such that

tan θ =2|H21|

H11 −H22, H21 = |H21|eiφ, (144)

we have

K =(

1 tan θe−iφ

tan θeiφ −1

)

, (145)

which has the characteristic equation

|K − κ1| = κ2 − 1− tan2 θ = 0, (146)

24


with solutions

κ± = ± 1cos θ

= ±√

(H11 −H22)2 + 4|H12|2H11 −H22

. (147)

Thus the energy eigenvalues are

E± =12(H11 + H22)±

12

√

(H11 −H22)2 + 4|H12|2, (148)

with associated eigenvectors

|ψ+〉 =(

cos θ2 e−iφ/2

sin θ2 eiφ/2

)

|ψ−〉 =(

− sin θ2 e−iφ/2

cos θ2 eiφ/2

)

(149)

Note that the angles θ, φ may be interpreted as the azimuthal and meridional angles, respectively, of a fictitiousspin with respect to the quantization axis. For θ = 0, |ψ+〉 = |+〉, while for θ = π, |ψ+〉 = |−〉.

The Pauli matrices, together with 1, form a basis for the space of all 2 × 2 Hermitian matrices. Thus, anytwo-state system may be described by the formalism that we have developed for spin. We discussed how thisformalism could be applied to a variety of physical systems, including the inversion oscillations of the ammoniamolecule, the bonding and antibonding states of the H+ ion, and the resonating valence bond of benzene. In eachof these examples, we identified two states of the system, |+〉, |−〉, which would be degenerate if the Hamiltonianexhibited no coupling between them. As we turned on some coupling, however, represented by off-diagonal matrixelements in the Hamiltonian, we found that the eigenstates of the Hamiltonian became superpositions of the originaltwo states, |ψ+〉, |ψ−〉 and the energy of one state rose while the other fell in energy; for H11 = H22 = E andH12 = −V , we found energy levels E± = E ± V . We saw that if the system is placed in a superposition of stateswith different energies, then the system will precess, just as the spin system does in a magnetic field, at a frequencyω = ∆E/~.

We also discussed the phenomenon of level anticrossing in the application of an electric field to the ammoniamolecule. Here, the effect of an electric field may be represented as a diagonal matrix in basis |+〉, |−〉, butit has off-diagonal matrix elements in the basis |ψ+〉, |ψ−〉. Thus, when the field is zero, the eigenstates ofthe Hamiltonian are |ψ+〉, |ψ−〉, but as the field is turned on, the eigenstates of the Hamiltonian transformcontinuously into |+〉, |−〉 at high fields, E V/d, where E is the electric field, V is the intrinsic couplingbetween the two states |+〉 and |−〉, and d is the electric dipole strength of the ammonia molecule.

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