math for quantum chemistry

8/3/2019 Math for Quantum Chemistry

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University of Connecticut

DigitalCommons@UConn

Chemistry Education Materials Department of Chemistry

10-9-2007

Introductory Mathematics for QuantumChemistry

Carl W. DavidUniversity of Connecticut, [email protected]

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Recommended CitationDavid, Carl W., "Introductory Mathematics for Quantum Chemistry" (2007). Chemistry Education Materials. Paper 50.http://digitalcommons.uconn.edu/chem_educ/50
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Introductory Mathematics for Quantum Chemistry

C. W. DavidDepartment of Chemistry

University of ConnecticutStorrs, Connecticut 06269-3060

(Dated: October 9, 2007)

I. SYNOPSIS

Since much of the mathematics needed for quantumchemistry is not covered in the first two years of calculus,a short introduction to those methods is presented here.Elegance has been eschewed in the Einstein spirit.

II. ORTHOGONAL FUNCTIONS

A. Dirac Notation

The notation employed in beginning discussions of

quantum chemistry becomes cumbersome when the com-plexity of problems increase and the need to see struc-

ture in equations dominates. Dirac introduced a notationwhich eliminates the ever-redundant from discourse.He defined a wave function as a ket.

The ket is defined as |index > , where index issome human chosen index or indicator whose purpose isto label the state (or the function) under discussion. Fora particle in a box, a potential ket is

n(x) = |n > Nn sinnx

L

=

2

Lsin

nxL

(2.1)

where the domain is 0 x L, Nn is a normalizationconstant, and n is a quantum number, i.e., an index

chosen by us.For the hydrogen atom, an appropriate ket might be

n=3,=1,m=1(,,) = |3, 1,1 >= N3,1,1(4 )e sin e (2.2)

which is a 3p electrons state () is a dimensionless radius.A harmonic oscillator might have an appropriate ket

of the form

3(x) = |3 >= N3 83x3 12x e2x2/2 (2.3)i.e., the ket notation allows us to indicate inside thefunny delimiters the essential characteristics of the in-volved wavefunction without extraneous elements whichmight distract attention from the overall viewpoint beingpropounded!

B. The Dot Product

Returning to the hydrogen atom, the following vector

< 3, 1,

1

|= N3,1,1(4

)e sin e+ (2.4)

which is the complex conjugate of the original wave func-tion [1] , is known as a bra vector, and the notation

< 3, 1,1|3, 1,1 >= 1 (2.5)is a shorthand for

all space

3,1,13,1,1d = 1 (2.6)

(i.e., the normalization equation, required so that theprobability of finding the electron somewhere is 1, cer-tain(!)), which itself is shorthand for the complicatedthree-dimensional integral, usually display is spherical

polar coordinates.One sees that constructing the braket [2] consists of

two separate concepts, making the bra from the ketby using the complex conjugate, juxtaposing the two inthe proper order, and integrating over the appropriatedomain. The analogy of this process is the dot productof elementary vector calculus, and it is here where we getour mental images of the processes described.

In three space, x, y, and z, we have unit vectors i, j, k,so that an arbitrary vector as:

R1 = Xi + Yj + Zk

could be re-written as

|R1 >= X|i > +Y|j > +Z|k >In this space, one doesnt integrate over all space, oneadds up over all components, thus

< R1|R1 >= R1 R1 = X2 + Y2 + Z2

since |i > is orthogonal (perpendicular) to |j > and|k >, etc..

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In the same sense,

|arb >=

n

cn

2

Lsin

nxL

(2.7)

where |arb > is an arbitrary state. This is our normalentry point for Fourier Series.

Before we go there, let us re-introduce orthogonality,i.e.,

< n|m >= 0 =

domain

nmd ; m = n (2.8)

where the domain is problem specific, as is the volumeelement d. This orthogonality is the analog of the ideathat

i j = 0as an example in normal vector calculus.

III. FOURIER SERIES

A. Particle in a Box vis-a-vis Fourier Series

The normal introduction to orthogonal functions is viaFourier Series, but in the context of quantum chemistry,we can re-phrase that into the Particle in a Box solutionto the Schrodinger Equation, i.e.,

h2

2m

2

x2+ 0 = E (3.1)

in the box (domain) 0 x L. The solutions are knownto be

n(x) = 2L sin nxL = |n > (3.2)where n is an integer greater than zero, and L remainsthe size of the box.

Now, a Fourier Series is an expansion of a function(periodic) over the same (repeated) domain as above, inthe form

|function >=

i

ci|i >=

i

ci

2

Lsin

ix

L(3.3)

(where we changed index for no particular reason fromn to i, other than to remind you that the index is adummy variable). The question is, what is the optimal,best, nicest, etc., etc., etc., value for each of the ci in thisexpansion?

B. A Minimum Error Approximation

To answer this question, we look for a measure of errorbetween the function and the expansion. To do this we

define a truncated expansion:

Sm =

i=mi=0

ci

2

Lsin

ix

L(3.4)

where, contrary to the particle in a box, we have em-ployed the i=0 particle in a box basis function. Now, weform an error at the point x:

|function >

Sm(x) = err(x) (3.5)

and notice that if we added up this error for every valueof x in the domain 0 x L we would be approachingthe concept we desire, a measure of the error, the dif-ference between the function and its approximation, thetruncated series.

But, you say, sometimes the error (err(x)) is positive,and sometimes its negative, depending on circumstances.We want a measure which counts either effect properly,and the answer is to define a new error

(|function > Sm(x))2 = ERR(x) [err(x)]2 (3.6)where clearly, the l.h.s. is positive definite, and thereforecan not give rise to fortuitous cancelation.

S (x)m

x

|function>

0 L

ERR(x)

ERR(x)

ERR (x)2

domain

FIG. 1: Error involved in approximating a function.

So, adding up ERR(x) at each point x in the domain,we haveL0

(|function > Sm(x))2 dx =L0

ERR(x)dx ERRORm(3.7)


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ERRORm is now a postive definite number which mea-sures the error committed in truncating the approximate

series at m. We re-write this explicitly to show that sum-mation:

L0

|function >

m

i=0ci

2

Lsin

ix

L

2dx =

L0

ERR(x)dx ERRORm (3.8)

and form the partial

ERRORmcj

(3.9)

holding all other ci constant, where i = 0...j...m, whichis

ERRORmj

= 2L

0

|function >

mi=0

ci

2

Lsin

ix

L

2

Lsin

j

Ldx

(3.10)

which we set equal to zero. This means that we are searching for that value of cj which makes the ERROR anextremum (of course, we want a minimum). This leads to the equation

L0

|function >

sin

jx

L

dx =

L0

m

i=0

ci

2

Lsin

ix

L

sin

j x

L

dx (3.11)

and, since one can exchange summation and integration, one has

L0

|function >

sinjx

L

dx =

mi=0

L0

ci

2

Lsin

ix

L

sin

j x

Ldx (3.12)

But the right hand side of this equation simplifies because the sines are orthogonal to each other over this domain,so the r.h.s. becomes L

0

|function >

sinjx

L

dx = cj

L0

2

Lsin

j x

L

sin

j x

Ldx (3.13)

since the only survivor on the r.h.s. is the i = j term.

cj =

L0|function > sin jxL dxL

0

2L sin

jxL

sin jxL dx

(3.14)

Multiplying top and bottom by

2/L one has

cj = < function|j >< j|j > (3.15)

which is fairly cute in its compactness. Of course, if |j >is normalized (as it is in our example) then

cj =< function|j > (3.16)which is even more compact!

C. Completeness

What if we wanted to approximate a function of theform

sin7x

L(3.17)

and i=7 were omitted from the summations (Equation3.3), i.e., the summation ran past this particular har-monic. It would look like:

+ c6 sin 6xL

+ c8 sin8x

L+

Clearly, since each sine is orthogonal to every othersine, the seventh sine

sin 7xL

would have no expansion

in this series, i.e., each ci would be zero! We can not
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leave out any term in the series, i.e., it must be complete,before one can assert that one can expand any functionin terms of these sines (and cosines, if need be). Saidanother way,

sin7x

L= c0 + c1 sin 1x

L+ c2 sin

2x

L+

since the one term on the r.h.s which is needed, was, by

assumption, absent!

IV. LADDER OPERATORS

A. Laddering in the Space of the Particle in a Box

We start by remembering the trigonometric definitionof the sine of the sum of two angles, i.e.,

sin( + ) = sin cos + cos sin (4.1)

We seek an operator, M+ which takes a state |n > andladders it up to the next state, i.e.,

|n+1 >. Symbolically,

we define the operator through the statement

M+|n > |n + 1 > (4.2)

and we will require the inverse, i.e.,

M|n > |n 1 > (4.3)where the down ladder operator is lowering us from n ton 1.

One sees immediately that there is a lower bound tothe ladder, i.e., that one can not ladder down from thelowest state, implying

M

|lowest >= 0 (4.4)

B. Form of the UpLadder Operator

What must the form ofM+ be, in order that it functionproperly? Since

sin

(n + 1)x

L

= cos

xL

sin

nx

L+cos

nxL

sin

xL

One has, defining the up-ladder operator M+

M+

sinnx

L

sin

(n + 1)x

L

= cos

xL

sin

nx

L+s

which looks like

M+|n > cosx

L

|n > +

L

n

sin

xL

|n >x

= |n+1 >

The Ln takes out the constant generated by the partialderivative.

What this is saying is that there is an operator, M+,which has a special form involving a multipier functionand a partial derivative, whose total effect on an eigen-function is to ladder it from n to n + 1.

M+ = cosx

L

+ L

n

sin

xL

x

written as an operator.

C. The DownLadder Operator

We need the down operator also, i.e.,

sin

(n 1)x

L

= cos

xL

sin

nx

Lsin

xL

cos

nx

L

making use of the even and odd properties of cosines andsines, respectively.

M

|n >

cos xL |n >

L

n sinx

L

|n >

x |n

1 >

M = cosx

L

L

n

sin

xL

x

D. Ladder Operators in the Operator

Representation

Stated another way, in the traditional position-momentum language, one has

M+ = cosx

L

L

n

sin

x

L

pxh

(4.5)

M = cosx

L

+L

n sinx

L

px

h

(4.6)


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E. Ladder Operators and the Hamiltonian

We need to show that this ladder operator works prop-erly with the Hamiltonian, i.e.,

p2x2m

Hop (4.7)

and to do this we assume

Hop|n >= |n >and ask what happens when we operate from the left withM+ (for example). We obtain

M+Hop|n >= M+|n > (4.8)

where we want to exchange the order on the l.h.s. sothat we can ascertain what the eigenvalue of the oper-ator M+|n > i,, i.e., we need the commutator of theHamiltonian with the ladder operator,

[Hop, M+] HopM+ M+Hop

Remembering that

[px, x] = h

one forms

[Hop, M+] =

p2x2m

cos

x

L+

L

n

sin

x

L

x

cosx

L+

L

n

sin

x

L

x

p2x2m

(4.9)

F. Some Precursor Commutators

First, one needs [px, cosxL ],

[px, cosx

L] h

xcos

xL

cos

xL

h

x

(4.10)

which works out to be

[px, cosx

L] = h

Lsin

xL

= px cos

x

L cos x

Lpx

(4.11)

or, in the most useful form for future work:

px cosx

L

= cos

xL

px + h

Lsin

xL

(4.12)

One needs the same commutator involving the appro-priate sine also:

[px, sinx

L

] = h

Lcos

xL

= px sin

xL

sin

xL

px

(4.13)

px sinx

L

= sin

xL

px h

Lcos

xL

(4.14)

Next, one needs [p2x, cos

xL

] (and its sine equivalent).

[p2x, cosx

L

] = p2x cos

xL

cos

xL

p2x (4.15)

= pxpx cos xL cos xL p2xwhich we re-write, factoring, as

= px

cos

xL

px + h

Lsin

xL

cos

xL

p2x

which we re-write as

= px cosx

L

px + h

L

px sin

xL

cos

xL

p2x

=

h L

sinx

L

+ cosx

L

px

px + h Lpx sin

xL cos x

L

p2x

i.e.,

= h

Lsin

xL

px + cos

xL

p2x + h

L

sin

xL

px h

Lcos

xL

cos

xL

p2x

= h

Lsin

xL

px + h

L

sin

xL

px h

Lcos

xL


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[p2x, cosx

L

] = 2h

Lsin

xL

px + h

2

L

2cos

xL

p2x cosx

L

= 2h

Lsin

xL

px + h

2

L

2cos

xL

+ cos

xL

p2x

Now one needs [p2x

, sin xL ][p2x, sin

xL

] = p2x sin

xL

sin

xL

p2x (4.16)

= px

px sin

xL

sin

xL

p2x

= px

sin

xL

px h

Lcos

xL

sin

xL

p2x

which we re-write as

=

px sinx

L

px h

L

px cos

xL

sin

xL

p2x

= sinx

L px h

L cos x

L px h

L px cos x

L sinx

L p2x= h

L

cos

xL

px

h

Lpx cos

xL

i.e.,

= h L

cos

xL

px

h

L

cos

x

Lpx + h

Lsin

xL

[p2x, sinx

L

] = 2h

Lcos

xL

px + h

2

L

2sin

xL

p2x sinx

L

= 2h

Lcos

xL

px + h

2 L

2sin

xL

+ sin

xL

p2x

We had, in Equation 4.8 the following,

M+Hop|n >= M+|n >which we now wish to reverse. We need the commutatorof the Hamiltonian with the UpLadder Operator [3].

[Hop, M+] =

p2x2m

cos

x

L L

n sin

x

L pxh

M+Hop

which becomes

[Hop, M+] =

p2x2m

cosx

L

p

2x

2m

L

n

sin

xL

pxhM+Hop (4.17)

expanding,

[Hop, M+] =

1

2mp2x cos

xL

1

2m

L

np2x sin

xL

pxhM+Hop

and using the previously obtained commutators,
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[Hop, M+] =

1

2m

2h

Lsin

xL

px + h

2

L

2cos

xL

+ cos

xL

p2x

12m

L

n

2h

Lcos

xL

px h2

L

2sin

xL

+ sin

xL

p2x

pxhM+Hop (4.18)

which expands, at first, to

[Hop, M+] = 1

2m

2h

Lsin

xL

px + h

2

L

2cos

xL

1

2m

L

n

2h

Lcos

xL

px h2

L

2sin

xL

pxh

(4.19)

[Hop, M+] =

1

2m

2h

Lsin

xL

px + h

2

L

2cos

xL

+1

2m

1

n

2cos

xL

px h

L

sin

xL

px (4.20)

2m[Hop, M+

] = 2h

L sinx

Lpx + h2 L2 cos xL 2n cos xL p2x 1n h L sinxL px (4.21)

2m[Hop, M+] =

2h +

1

n

h

Lsin

xL

px + h

2

L

2cos

xL

+

2

ncos

xL

p2x (4.22)

2m[Hop, M+] = h

2h +

1

n

h

L

sin

xL

pxh

+ h2

L

2cos

xL

+

2

ncos

xL

p2x (4.23)

and substituting for

sin

xL

pxh

one has

2m[Hop, M+] = h2(1 + 2n)L

2

M+ cosx

L + h2

L2

cos xL +2

ncos xL p2x (4.24)

[Hop, M+] = h2

(1 + 2n)

2m

L

2 M+ cos

xL

+ h2

1

2m

L

2cos

xL

+

2

ncos

xL

p2x2m

(4.25)

[Hop, M+] = h2

(1 + 2n)

2m

L

2 M+

h2 (1 + 2n)2m

L

2 cos

xL

+ h2

1

2m

L

2cos

xL

+

2

ncos

xL

p2x2m

(4.26)

[Hop, M+] = h2

(1 + 2n)

2m

L2

M+

+

h2

2m

L2

(2n)cos

x

L +

2

ncos

x

L Hop (4.27)

1. Alternative Formulation

2

x2

cos

xL

+

L

nsin

xL

x

f =


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8

x

cos

xL

x

L

sinx

L

+

L

nsin

xL

2x2

+1

ncos

xL

x

f =

L

sinx

L

fx

+ cosx

L

2fx2

L2

cos xL f L sinxL fx+

1

ncos

xL

2fx2

+L

nsin

xL

3fx3

nL

sinx

L

fx

+1

ncos

xL

2fx2

(4.28)

So the commutator would be:

2

x2

cos

xL

+

L

nsin

xL

x

f

cos

xL

+

L

nsin

xL

x

2f

x2=

2L

sinx

L

fx

L2

cos x

L f+

2

ncos

xL

2fx2

nL

sinx

L

fx

(4.29)

or

=

2

L

nL

sin

xL

x

L

2cos

xL

+

2

ncos

xL

2x2

(4.30)

=

2

L

nL

n

L

M+ cos

xL

L

2cos

xL

+

2

ncos

xL

2x2

(4.31)

2mh2

[H, M+] =

L

2(2n 1)

M+ cos

xL

L

2cos

xL

+

2

ncos

xL

2x2

(4.32)

2mh2

[H, M+] =

L

2(2n + 1) M+ +

L

2(2n)cos

xL

+

2

ncos

xL

2x2

(4.33)

[H, M+] = h2

2m

L

2(2n + 1) M+ +

L

2(2n)cos

xL

+

2

ncos

xL

2x2

(4.34)

[H, M+] = h2

2mL2 (2n + 1) M+ + 2nL2 cos xL 2n cos xL h

2

2m

2

x2 (4.35)

H|n >= n|n >and, operating from the left with M+ one has

M+H|n >= nM+|n >= n|n + 1 >


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9

and since [H, M+] = HM+ M+H, one hasHM+|n > [H, M+]|n >= n|n + 1 >

H|n + 1 > [H, M+]|n >= n|n + 1 >

H|n + 1 > +

h2

2m

L2

(2n + 1) M+ + 2nL2

cos xL |n > +2

ncos xL

h2

2m

2

x2 |n >= n

|n + 1 >

H|n + 1 > + h2

2m

L

2(2n + 1) M+ + 2n

L

2cos

xL

|n > 2

ncos

xL

n|n >= n|n + 1 >

H|n + 1 > + h2

2m

L

2(2n + 1)M+ +

2n

L

2 4m

nh2n

cos

xL

|n >= n|n + 1 >

H|n + 1 > + h2

2m

L

2(2n + 1)M+

|n > +

2n

L

2 4m

nh2n

cos

xL

|n >= n|n + 1 >

H|n + 1 > + h2

2m

L

2(2n + 1)

|n + 1 > +

2n

L

2 4m

nh2n

cos

xL

|n >= n|n + 1 >

H|n + 1 > +

2n

L

2 4m

nh2n

cos

xL

|n >= n|n + 1 > + h

2

2m

L

2(2n + 1)

|n + 1 >

From this we conclude that, if the elements in squarebrackets cancelled perfectly, M+|n + 1 > would be aneigenfunction, i.e., if

+

2n

L

2 4m

nh2n

= 0 (4.36)

i.e.,

n =n2h22

2mL2

then

H|n + 1 >= n +h2

2m

L2

(2n + 1) |n + 1 >Sines and cosines form a complete orthonormal set, suit-able for Fourier Series expansion, where ortho meansthat each member is orthogonal to every other member,and normal means that one can normalize them (if onewishes). There are other such sets, and each of themshows up in standard Quantum Chemistry, so each needsto be addressed separately.

V. HERMITE POLYNOMIALS

The relevant Schrodinger Equation is

h2

2

2

z2 +

k

2z2 = E (5.1)

where k is the force constant (dynes/cm) and is thereduced mass (grams). Cross multiplying, one has

2

z2 k

h2z2 = 2

h2E (5.2)

which would be simplified if the constants could be sup-pressed. To do this we change variable, from z to some-thing else, say x, where z = x. Then

z=

x

z

x=

1

xso

1

2

2

x2 k

h22x2 = 2

h2E (5.3)

and

2

x2 k

h24x2 = 2 2

h2E (5.4)


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10

which demands that we treat

1 =k

h24

=

1

kh2

1/4=

h2

k

1/4

With this choice, the differential equation becomes

2

x2 x2 = (5.5)

where

=22E

h2=

2

h2

k E

h2=

2E

k

h

A. A Guessed Solution

The easiest solution to this differential equation is

ex2

2

which leads to = 1 and

E =h

2

k

B. Laddering up on the Guessed Solution

Given

0 = |0 >= ex2

2

with = 1, it is possible to generate the next solution byusing

N+ = x

+ x (5.6)

as an operator, which ladders up from the ground (n=0)state to the next one (n=1) To see this we apply N+ to0 obtaining

N+0 = N+|0 >=

x+ x

e

x2

2 = (x) 0 + x0 = 2xex2/2 = 1 = |1 > (5.7)

Doing this operation again, one has

N+1 = N+

|1 >=

x+ x 2xex

2

2 = (

2+4x2)ex2/2

(5.8)etc., etc., etc..

C. Generating Hermites differential equation

Assuming

= ex2/2H(x)

where H(x) is going to become a Hermite polynomial.One then has

d

dx= xex2/2H(x) + ex2/2 dH(x)

dx

and

d2

dx2

= ex2/2H(x) + x2ex2/2H(x) 2xex2/2 dH(x)dx

+ ex2/2 d2

H(x)dx2

From Equation 5.5 one has,

2

x2 x2 = ex2/2H(x) 2xex2/2 dH(x)

dx+ ex

2/2 d2H(x)

dx2= ex2/2H(x) (5.9)

or

H(x) 2x dH(x)dx

+d2H(x)

dx2= H(x) (5.10)

which we re-write in normal lexicographical order

d2H(x)

dx2 2x dH(x)

dx (1 )H(x) = 0 (5.11)
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D. An alternative solution scheme

Starting with

dy

dx+ 2xy = 0 (5.12)

one has

dyy = 2xdx

so, integrating each side separately, one has

ny = x2 + nC

or, inverting the logarithm,

y = Cex2

We now differentiate Equation 5.12, obtaining

d2y

dx2+ 2

d(xy)

dx=

d2y

dx2+ 2x

dy

dx+ 2y

dx

dx=

d2y

dx2+ 2x

dy

dx+ 2y = 0 ; n = 0 (5.13)

Doing this again, i.e., differentiating this (second) equation (Equation 5.13), one has

d d2y

dx2

dx+

d2x d(y)dxdx

+ 2dy

dx=

d2

dydx

dx2

+ 2xd

dydx

dx

+ 4

dy

dx

= 0 ; n = 1

which is the same equation, (but with a 4 multiplier ofthe last term) applied to the first derivative of y. Takethe derivative again:

d

d2 dydxdx2 + 2x

d dydxdx + 4

dydx

dx

= 0

i.e.,

d2

d2ydx2

dx2

+ 2xd

d2ydx2

dx

+ 6

d2y

dx2

= 0

d2f(x)

dx2+ 2x

df(x)

dx+ 6f(x) = 0 ; n = 2

f(x) has the form g(x)ex2

where g(x) is a polynomialin x.

d2g(x)ex2

dx2+ 2x

dg(x)ex2

dx+ 2(n + 1)g(x)ex

2

= 0

i.e.,

g(x) 4xg(x) 2g(x) + 4x2g(x) + 2xg(x) 4x2g(x) + 2(n + 1)g(x) ex2 = 0

or

g(x) 2xg(x) + 2ng(x) = 0

and we had (see Equation 5.11)

H(x) 2xH(x) (1 )H(x) = 0

which leads to

2n = 1 +

i.e.,

= 1 + 2n =2E

/k

h

i.e.,

E = h(n + 1/2)

k

E. Frobenius Method

The most straight forward technique for handling theHermite differential equation is the method of Frobenius.We assume a power series Ansatz (ignoring the indicialequation argument here), i.e.,

=i=0

aixi
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and substitute this into Equation 5.11, obtaining

2

x2=i=2

i(i 1)aixi2

2x x

= 2i=1

iaixi

( 1) = ( 1)i aixi = 0

i.e.,

2

x2= 2(1)a2 + (3)(2)a3x + (4)(3)a4x

2 +

2x x

= 2a1x1 2a2x2 2a3x3 ( 1) = ( 1)a0 + ( 1)a1x + ( 1)a2x2 = 0which leads to

(3)(2)a3 + ( 1)a1 2a1 = 0 (odd)

(4)(3)a4 2a2 + ( 1)a2 = 0 (even)

(5)(4)a5 2a3 + ( 1)a3 = 0 (odd)etc., which shows a clear distinction between the evenand the odd powers of x. We can solve these equationssequentially.

We obtain

a3 =2 + 1

(3)(2)a1

a4 =2 + 1

(4)(3) a2 =2 + 1 (4)(3) 1 (2)(1)

i.e.,

a4 =

(3 )(1 )(4)(3)(2)(1)

etc..

This set of even (or odd) coefficients leads to a serieswhich itself converges unto a function which grows topositive infinity as x varies, leading one to require thatthe series be terminated, becoming a polynomial.

We leave the rest to you and your textbook.

VI. LEGENDRE POLYNOMIALS

A. From Laplaces Equation

There are so many different ways to introduce Leg-endre Polynomials that one searchs for a path into thissubject most suitable for chemists.

Consider Laplaces Equation:

2 = 2

x2+

2

y2+

2

z2= 0 (6.1)

which is a partial differential equation for (x,y,z). Forour purposes, the solutions can be obtained by guessing,

starting with = 1, and proceeding to = x, = y and = z. It takes only a little more imagination to obtain = xy and its associates = yz and = xz. It takesjust a little more imagination to guess = x2 y2, butonce done, one immediately guesses its two companions = x2 z2 and = y2 z2.

Reviewing, there was one simple (order 0) solution,three not so simple, but not particularly difficult solu-tions (of order 1) and six slightly more complicated solu-tions of order 2. Note that the first third-order solutionone might guess would be = xyz!

If one looks at these solutions, knowing that they arequantum mechanically important, the 1, x, y, and z so-

lutions, accompanied by the xy, xz, yz , and x2

y2

so-lutions suggest something having to do with wave func-tions, where the first function (1) is associated in someway with s-orbitals, the next three (x, y, and z) are as-sociated with p-orbitals, and the next (of order 2) areassociated with d-orbitals. If all this is true, the per-ceptive student will wonder where is the dz2 orbital, thefifth one, and how come there are six functions of ordertwo listed, when, if memory serves correctly, there are 5d-orbitals! Ah.

y2 z2 + x2 y

2can be rewritten as

y2 z2 + x2 y2 + z2 z2

which is

x2 + y2 + z2 3z2

or as

r2 3z2

which combines the last two (linearly dependent) solu-tions into one, the one of choice, which has been employedfor more than 50 years as the dz2 orbital.

In Spherical Polar Coordinates, these solutions become
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1 1x r sin cos y r sin sin z r cos

xy r sin cos r sin sin = r2 sin2 cos sin

xz r sin cos r cos = r2

sin cos cos yz r sin sin r cos = r2 sin cos sin

x2 y2 r2 sin2 cos2 r2 sin2 sin2 = r2 sin2 cos2 sin2 r2 3z2 r2 3r2 cos2 = r2 1 3cos2

... (6.2)

Now, Laplaces equation in spherical polar coordinatesis

2 =

1

r2

r2 r

r+

1

r2 sin2 sin sin

+

2

2 = 0(6.3)

Next, we notice that our earlier solutions were all of theform

= rY,m(r,,) (6.4)where specifies the order, 0, 1, 2, etc., and the func-tion of angles is dependent on this order, and on anotherquantum number as yet unspecified. Substituting thisform into the spherical polar form of Laplaces equationresults in a new equation for the angular parts alone,which we know from our previous results. We substituteEquation 6.4 into Equation 6.3 to see what happens, ob-

taining

2rY,m =1

r2r2

rY,mr

r

+1

r2 sin2

sin sin rY,m

+

2rY,m2

= 0

which, by the nature of partial differentiation, becomes

2rY,m = Y,m1

r2r2 r

r

r

+r1

r2 sin2

sin

sin Y,m

+

2Y,m2

= 0 (6.5)

and the first term in Equation 6.5 is

1

r2r2 r

r

r=

1

r2( + 1)r (6.6)

so, substituting into Equation 6.5 we obtain

2

r

Y,m = ( + 1)r2

Y,m + r2 1

sin2 sin sin

Y,m

+2

Y,m

2 = 0 (6.7)

which, of course, upon cancelling common terms, gives

(+1)Y,m+1

sin2

sin

sin Y,m

+

2Y,m2

= 0

(6.8)

Laplaces Equation in Spherical Polar Coordinates on theunit sphere, i.e. Legendres Equation!
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What then are these angular solutions we found?

= 0; 1

= 1; sin cos

= 1; sin sin

= 1;cos

= 2; sin2 cos sin

= 2; sin cos cos = 2; sin cos sin

= 2;

sin2

cos2 sin2 = 2;

1 3cos2

... (6.9)

and if you check back concerning their origin, you willsee where the Hydrogenic Orbitals naming pattern comesfrom for s, p and d orbitals.

These functions have been written in real form, andthey have complex (not complicated) forms which al-low a different simplification:

1s = 0; 1

2px

= 1; sin

e + e

22py = 1; sin

e e

2

2pz = 1;cos

3dxy = 2; sin2

e + e

2

e e

2

3dxz = 2; sin cos e + e

2

3dyz = 2; sin cos

e e2

3dx2y2 = 2;sin2

e + e

2 2

e e

2 2

3dz2 = 2;

1 3cos2

... (6.10)

Once they have been written in imaginary form, we cancreate intelligent linear combinations of them which illu-minate their underlying structure. Consider 2px + 2pywhich becomes

sin e

(employing DeMoivres theorem) aside from irrelevantconstants. Next consider 2px 2py which becomes

sin e

We have a trio of functions

2pm=1 = sin e

2pm=0 = cos

2pm=+1 = sin e+ (6.11)

(6.12)

which correspond to the m values expected of p-orbitals.

You will see that the same trick can be applied todxz and dyz .

Finally, the same trick, almost, can be applied to dxyand dx2y2 with the m = 0 value reserved for dz2 , all byitself.


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B. From the Hydrogen Atom

Perhaps the next best place to introduce Spherical Har-monics and Legendre Polynomials is the Hydrogen Atom,since its eigenfunctions have angular components whichare known to be Legendre Polynomials.

The Schrodinger Equation for the H-atom is

h2

2me2|n,,m > Ze2

r |n,,m >= En|n,,m >(6.13)

We know that this equation is variable separable, in

the sense that

|n,,m >= Rn,(r)Y,m(, )

Here, Rn, is the radial wave function, and Y,m is theangular wave function (a function of two variables), Since

2 = 1r2r

2

rr + 1r2 sin2

sin sin + 2

2

2|n,,m >= 2Rn,(r)Y,m(, ) =Yr2

r2 Rrr

+R

r2 sin2

sin

sin Y

+2Y2

which means that the Schrodinger Equation has the form

h2

2me Y

r2

r2 Rr

r

+R

r2

sin2

sin sin Y

+2Y

2

Ze2

r

R

Y= EnR

YDividing through by Rn,Y,m (abreviated as RY) we have

h2

2me

1

Rr2r2 Rr

r+

1

r2Ysin2

sin

sin Y

+2Y2

Ze

2

r= En

Multiplying through by r2 shows that the expected variable separation has resulted in a proper segregation of theangles from the radius.

1

R

r2 Rrr

+1

Ysin2

sin

sin Y

+2Y2

+

2mZe2r

h2= 2m

h2Enr

2

We can see this by noting that the underbracketed partof this last equation is a pure function of angles, with noradius explicitly evident.

It is now standard to obtain (recover?) the equation(6.8)

1

Ysin2

sin

sin Y

+2Y2

= ( + 1) (6.14)

where the minus sign (which is essentially arbitrary) isdemanded by convention.

Cross multipilying by

Y, one has

1

sin2

sin

sin Y

+2Y2

= ( + 1)Y (6.15)

which is written in traditional eigenfunction/eigenvalueform.

C. Another Form for the Legendre Differential

Equation

A change of variables can throw this equation (Equa-tion 6.15) into a special form, which is sometimes illumi-

nating. We write

= cos

and

= sin =

1 cos2 =

1 2

so

=

which is, substituting into Equation 6.15

=

1 2
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1

(1 2)

1 2

1 2

1 2

1 2 Y

+2Y2

= ( + 1)Y (6.16)

where the (+1) form is a different version of a constant,looking ahead to future results!

(1 2) Y

+2Y2

= ( + 1)Y (6.17)

where

Y,m(, ) = emS,m()For m = 0 we have

(1 2) S,0

= ( + 1)S,0 (6.18)

i.e., Legendres equation.

D. Schmidt Orthogonalization

We assume polynomials in n and seek orthonormalcombinations which can be generated starting with a con-stant (n=0) term. Then the normalization integral

< 0|0 >=1+1

20d = 1

[4] implies that

|0 >=

1

2

We now seek a function |1 > orthogonal to |0 > whichitself is normalizeable. We have

0 =< 1|0 >=1+1

1

1

2d

which has solution 1 = N1x. Normalizing, we have

< 1|1 >=1+1

21d = 1 =

1+1

N21 2d

which yields

1 = N212

3

i.e.,

|1 >=

3

2x

To proceed, we need to generate a function |2 > whichis not only normalizeable but orthogonal to

|0 > and

|1 >.

< 2|0 >=1+1

2

1

2d = 0

and

< 2|1 >=1+1

2

2

3d = 0

where |2 > is a polynomial of order 2 in , i.e.,2 = |2 >= a2 + b

Substituting, we have

< 2|0 >= 1+1

(a2 + b)1

2d = 0

and

< 2|1 >=1+1

(a2 + b)

3

2d = 0

We obtain two equations in two unknowns, a and b,1

2

2a

3+ 2b

= 0

and the other integral vanishes automatically. i.e.,

a

3

+ b = 0

i.e., a = 3b so|2 >= 3b2 + b

in unnormalized form. Normalizing gives

< 2|2 >=1+1

(3b2 + b)2d = 1

i.e.,

< 2|2 >= b21+1

(32 + 1)2d = 1

which gives

b = 11

+1(32 + 1)2d

where, of course, we recognize the sign ambiguity in thisresult.

One can continue forever with this Schmidt Orthog-onalization, but the idea is clear, and there are betterways, so why continue?
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E. Frobenius Solution to Legendres Equation

We start a Frobenius solution without worrying aboutthe technical details of the indicial equation, and justassert that the proposed solution Ansatz will be

S,0() = y() =

n=0cn

n = c0+c1+c22+ (6.19)

which we substitute into the differential equation:

(1 2) y()

= ( + 1)y() (6.20)

which leads to

y()

2 y()

+ ( + 1)y() = 0 (6.21)

2y()

2 22y()

2 2y()

+(+ 1)y() = 0 (6.22)

so that when we feed the Ansatz into this differentialequation we obtain

(2)(1)c2 + (3)(2)c3 + (4)(3)c42 + (5)(4)c5

3 2 (2)(1)c2 + (3)(2)c3 + (4)(3)c42 + (5)(4)c53

2 c1 + (2)c2 + (3)c32 + (4)c43 + +( + 1) c0 + c1 + c22 + c33 + = 0 (6.23)

(2)(1)c2 + (3)(2)c3 + (4)(3)c42 + (5)(4)c5

3 (2)(1)c22 (3)(2)c33 (4)(3)c44 (5)(4)c55

2c1 2(2)c22 2(3)c33 2(4)c44 +( + 1)c0 + ( + 1)c1 + ( + 1)c2

2 + ( + 1)c33 + = 0 (6.24)

which, in standard Frobenius form, we separately equateto zero (power by power) to achieve the appropriate re-cursion relationships. Note that there is an even and anodd set, based on starting the c0 or c1, which correspond

to the two arbitrary constants associated with a secondorder differential equation. We obtain

(2)(1)c2 + ( + 1)c0 = 0

+(3)(2)c3 2c1 + ( + 1)c1 = 0+(4)(3)c4

2 (2)(1)c22 2(2)c22 + ( + 1)c22 = 0(5)(4)c5

3 (3)(2)c33 2(3)c33 + ( + 1)c33 = 0(4)(3)c44 2(4)c45 + etc = 0

(2)(1)c2 + ( + 1)c0 = 0

+(3)(2)c3 2c1 + ( + 1)c1 = 0+(4)(3)c4 (2)(1)c2 2(2)c2 + ( + 1)c2 = 0

(5)(4)c5

(3)(2)c3

2(3)c3 + ( + 1)c3 = 0

(4)(3)c44 2(4)c45 etc + = 0 (6.25)

(2)(1)c2 = ( + 1)c0+(3)(2)c3 = (2 ( + 1))c1

+(4)(3)c4 = ((2)(1) + 2(2) ( + 1))c2(5)(4)c5((3)(2) + 2(3) ( + 1))c3

(4)(3)c44 2(4)c45 etc + = 0 (6.26)

c2 = ( + 1)2!

c0

c3 = (2 + ( + 1))(3)(2)

c1

c4 = ((2)(1) 2(2) + ( + 1))(4)(3)

c2

c5 = ((3)(2) 2(3) + ( + 1))(5)(4)

c3

etc. (6.27)

which we re-write in terms of c0 and c1 only, i.e.,

c2 = ( + 1)2!

c0

c3 = (2 + ( + 1))(3)(2)

c1 +

c4 = ((2)(1)

2(2) + ( + 1))

(4)(3) (( + 1)

2! c0c5 =

((3)(2) 2(3) + ( + 1))

(5)(4)

(2 + ( + 1))

(3)(2)

c1

etc.(6.28)

so S,0() = f1c0 + f2c1 where f1 and f2 are power seriesbased on the above set of coefficients. For an even series,declare c1 = 0 and choose an value which truncates the


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power series into a polynomial. Do the opposite for anodd solution.

F. Rodriques Formula

Rodriques formula is

S,0 P() =1

2!

d(2

1)

d

where Legendres Equation is

(1 2) P()d2

2 dPd

+ ( + 1)P() = 0

To show this, we start by defining

g (2 1)

and find that

dgd

= 2(2 1)1

and

d2gd2

= 2(2 1)1 + 42( 1)(2 1)2

We now form (construct)

(1 2) d2g

d2= 2(2 1) 42( 1)(2 1)1

2( 1) dgd

= 42( 1)(2 1)1

+2g = 2(2 1) (6.29)The r.h.s. of this equation set adds up to zero, and one obtains on the left:

A() = (1 2) d2g

d2+ 2( 1) dg

d+ 2g = 0 (6.30)

Defining the l.h.s of this equation as A(), we form

dA()

d=

(1 2) d

3

d3 2 d

2

d2+ 2( 1) d

2

d2+ 2( 1) d

d+ 2

d

d

g

= (1 2) d2d2

(2 4) dd

+ (4 2) dgd

d2A()

d2=

(1 2) d

2

d2 (2 6) d

d+ (6 6)

d2gd2

d3A()

d3=

(1 2) d

2

d2 (2 8) d

d+ (8 12)

d3gd3

Continuing, one notices that the changing coefficients areregular in their appearance, so that the following table,which summarizes the pattern of coefficients,

= 1 2 4 = 2 2 ( + 1) 4 2 = (3 + ) 2 =6 if = 1 = 2 2

6 = 2

2

( + 1) 6

6 = (3 + )

2

= 6 if = 2

= 3 2 8 = 2 2 ( + 1) 8 12 = (3 + ) 2 = 12 if = 3...

......

...

= 2 2 ( + 1) = 2 8 12 = (3 + ) 2 = ( + 1)

leads to generalization by which one finally obtains

A(x)

=

(1 2)

2

2 2

+ ( + 1)

g()

d

(6.31)

so, if

dgd

KP()


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with K constant, then

P() =1

K

dgd

=1

K

d(2 1)d

Here,

2!

is chosen for Ks value to make the normalization auto-matic.

G. Generating Function for Legendre Polynomials

The technically correct generating function for Legen-dre polynomials is via the equation

11 2xu + u2 =

0

Pn(x)un (6.32)

Here, we expand the denominator using the binomial the-orem,

1

(1 + y)m = 1my+m(m + 1)

2! y2

m(m + 1)(m + 2)

3! y3

+

where m = 12

and the series converges when y < 1. Notice

that it is an alternating series. Identifying y = u2 2xuwe have

1

(1 2xu + u2)(1/2) = 1(1/2)(u22xu)+ (1/2)((1/2) + 1)

2!(u22xu)2 (1/2)((1/2) + 1)((1/2) + 2)

3!(u22xu)3+

which we now re-arrange in powers of u (in the mode required by Equation 6.32), obtaining

1

(1 2xu + u2)1/2 = 1u2

2+ xu +

(3/4)

2!(u4 4xu3 + 4x2u2) (1/2)(3/2)(5/2)

3!(u2 2xu)3 +

1. Alternative Generating Function Method

Another method of introducing Legendre Polynomialsis through the generating function. Since this methodis very important in Quantum Mechanical computationsconcerning poly-electronic atoms and molecules, it isworth our attention. When one considers the Hamilto-nian of the Helium Atoms electrons, one has

Ze2

r1 Ze

2

r2+

e2

r12(6.33)

where r12 is the distance between electron 1 and elec-tron 2, i.e., it is the electron-electron repulsion term. Weexamine this term in this discussion. We can write thiselectron-electron repulsion term as

1

r12=

1

r21 + r22 2r1r2 cos (6.34)

where r1 and r2 are the distances from the nucleus toelectrons 1 and 2 respectively. is the angle betweenthe vectors from the nucleus to electron 1 and electron2. It is required that we do this is two domains, one inwhich r1 > r2 and one in which r2 > r1. This is done for

convergence reasons (vide infra). For the first case, wedefine

=r2r1

so that < 1, and Equation 6.34 becomes

1

r12=

1

r1

11 + 2 2cos (6.35)

which we now expand in a power series in cos (whichwill converge while zeta < 1). We have

1

r1

11 + 2 2cos =

1

r1

1 + 11!

d

1

1+22 cos

d cos

cos=0

cos +1

2!

d2

11+22 cos

d cos 2

cos =0

cos2 +

(6.36)
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It is customary to change notation from cos to , so

1

r12=

1

r1

11 + 2 2 (6.37)

which we now expand in a power series in (which willconverge while < 1). We have

1

r1

11 + 2 2 =

1

r11 + 11!

d1

1+22d

=0

+1

2!

d21

1+22d2

=0

2 + + (6.38)

which we leave in this form, ready to evaluate and equateto earlier versions of this expansion.

H. The Expansion of a Finite Dipole in Legendre

Polynomials

There is yet another way to see Legendre Polynomials

in action, through the expansion of the potential energyof point dipoles. To start, we assume that we have adipole at the origin, with its positive charge (q) at (0,0,-a/2) and its negative charge (q) at (0,0,+a/2), so that

the bond length is a, and therefore the dipole mo-ment is qa.

At some point P(x,y,z), located (also) at r,,, wehave that the potential energy due to these two pointcharges is

U(x , y , z, a) =q

x2 + y2 + (z a/2)2

+q

x2 + y2 + (z + a/2)2which is just Coulombs law.

If we expand this potential energy as a function of a,the bond distance, we have

U(x , y , z, a) = U(x,y,z, 0) +1

1!

dU

da

a=0

a +1

2!

d2U

da2

a=0

a2 +1

3!

d3U

da3

a=0

a3 +

All we need do, now, is evaluate these derivatives. We have, for the first

1

1!

dU

da a=0 = q1

22(z a/2)(1/2)

(x2

+ y2

+ (z a/2)2

))3/2 +

1

22(z + a/2)(1/2)

(x2

+ y2

+ (z + a/2)2

))3/2

which is, in the limit a 0,

q z

r3

so, we have, so far,

U(x , y , z, a) = 0 qa z

r3

+ = qa

cos

r2

+

i.e., to the dipolar form.

For the second derivative, we take the derivative of thefirst derivative:

1

2!

d dUdada

a=0

= q 12 2!

d

(za/2)

(x2+y2+(za/2)2))3/2

+

(z+a/2)(x2+y2+(z+a/2)2))3/2

da

which equals zero. The next term gives

qcos

3 5cos2

8r4a3

and so it goes.


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21

I. Alternative Formulations for Angular

Momentum Operators

To proceed, it is of value to inspect the angular mo-mentum operator in terms of angles rather than Carte-sian coordinates.

Remember that

x = r sin cos

y = r sin sin

z = r cos

We start with a feast of partial derivatives:r

x

y,z

= sin cos (6.39)

r

y

x,z

= sin sin (6.40)

rz

x,y

= cos (6.41)

x

y,z

=cos cos

r(6.42)

y

x,z

=cos sin

r(6.43)

z x,y = sin

r(6.44)

and finally

x

y,z

= sin r sin

(6.45)

y

x,z

=cos

r sin (6.46)

z

x,y

= 0 (6.47)

which we employ on the defined x-component of the an-gular momentum, thus

Lx ypz zpy = h sin sin z

h cos

y

where

z=

r

z

x,y

r+

z

x,y

+

z

x,y

= (cos )

r+

sin

r

+ (0)

and

y=

r

y

x,z

r+

y

x,z

+

y

x,z

= (sin sin )

r+

cos sin

r

+

cos

r sin

and parenthetically,

x=

r

x

y,z

r+

x

y,z

+

x

y,z

= (sin cos )

r+

cos cos

r

+

sin

r sin

so,

Lx ypz zpy = hr sin sin

(cos )

r+

sin

r

hr cos

(sin sin )

r+

cos sin

r

+

cos

r sin

(6.48)

At constant r, the partial with respect to r looses mean- ing, and one has

Lxh = r sin sin

+

sin

r

cos

cos sin

r

+

cos

r sin

(6.49)


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22

which leads to (combining the partial derivative termsand cancelling the r terms)

Lx = h

sin

+

cos cos

sin

(6.50)

and similar arguments lead to

Ly = hcos

cos sin

sin

(6.51)Lz = h

(6.52)

with

L2 = h2

1

sin2

sin

+

2

2

(6.53)

We see that the operator associated with LegendresEquation (multiplied by a constant) has emerged, mean-ing that Legendre Polynomials and angular momentum

are intimately asssociated together.

J. Ladder Operator for Angular Momentum

Defining

L+ = Lx + Ly

and

L = Lx Ly

and knowing

LxLyLyLz = (ypzzpy)(zpxxpz)(zpxxpz)(ypzzpy)

which is

= ypzzpx ypzxpz zpyzpx + zpyxpz zpxypz+zpxzpy + xpzypz xpzzpy = ypzzpx

which is, reordering:

= ypzzpx zpxypz ypzxpz + xpzypzzpyzpx + zpxzpy + zpyxpz xpzzpy

which is, cancelling:

= ypzzpx zpxypz + zpyxpz xpzzpy= ypx(pzz zpz) + xpy(zpz pzz)

= ypx(h) + xpy(h)= h(xpy ypx) = hLz = [Lx, Ly]

that

[Lx, Ly] = hLz (6.54)

[Ly, Lz] = hLx (6.55)

[Lz, Lx] = hLy (6.56)

and knowing

[Lx, L+] = Lx(Lx + Ly) (Lx + Ly)Lx

which is, expanding:

= LxLx + LxLy LxLx LyLx= LxLx LxLx + (LxLy LyLx)

= +(hLz)

= hLz

we derive that

[Lx, L+] = hLz (6.57)

[Ly, L+] = hLz (6.58)

[Lz, L+] = hL+ (6.59)

and its companion

[Lz, L] = hL+ (6.60)

This last equation tell us that L+ ladders us up, and L

ladders us down on Lz components. If

Lz| >= something| >


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23

then

LLz| >= somethingL| >or

[LzL + h]| >= somethingL| >

LzL

| >= (something h)L

| >which means that where ever we were, we are now lowerby h.

L

=

h

h

h

h

K

K h

K2

K3

K4

K5

FIG. 2: Laddering Down on the z-component of angular mo-mentum

Next, we need an expression for L+L, not the com-mutator. It is:

L+L = (Lx + Ly)(Lx Ly)

= L2x + L2y + (LyLx LxLy)

or

= L2x + L2y + L

2z + (LyLx LxLy) L2z

or

= L2x + L2y + L + z

2 + (hLz) L2zi.e.

L+L = L2 + hLz L2zand repeating this for the other order, one has

LL+ = L2 hLz L2zIf

Lz| >= Kh| >

with no intentional restriction on the value of K, and

L2| >= Kh2L+| > (6.61)

LL+| > +hLz| > +L2z| >= Kh2| >

If this particular ket is the highest one, |hi >, then lad-dering up on it must result in destruction, so we have

+hLz|hi > +L2z|hi >= Kh2|hi >

+(hi)h2|hi > +(hi)2h2|hi >= Kh2|hi >

which means that

+(hi) + (hi)2 = K

while, working down from the low ket one has

L+L|lo > hLz|lo > +L2z|lo >= Kh2|lo >

which is

(lo)h2Lz|lo > +(lo)2h2|lo >= Kh2|lo >

i.e.,

(lo) + (lo)2 = K

which means

(lo) + (lo)2 = K = +(hi) + (hi)2 = (lo)((lo) 1) = (hi)((hi) + 1)

which can only be true if (lo) = - (hi). Say (hi) = 7,then K = 7*8 and -7(-7-1) which is the same! Note that

if (hi) = 7.1 then (lo) can not be achieved by steppingdown integral multiples of h.


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VII. LAGUERRE POLYNOMIALS

The radial equation for the H-atom is

h2

2

d2

dr2+

2

r

d

dr ( + 1)

r2

R(r) Ze

2

rR(r) = ER(r)

which we need to bring to dimensionless form before pro-ceeding (text book form). Cross multiplying, and defin-ing = E we have

d2

dr2+

2

r

d

dr ( + 1)

r2

R(r)+

2Ze2

h2rR(r)2

h2R(r) = 0

and where we are going to only solve for states with > 0,i.e., negative energy states.

Defining a dimensionless distance, = r we have

d

dr=

d

dr

d

d=

d

d

so that the equation becomes

2 d2

d2+ 2

d

d ( + 1)

2

R()+ 2Ze2

h2R()2

h2R() = 0

which is, upon dividing through by 2,d2

d2+

2

d

d ( + 1)

2

R()+

2Ze2

h2R() 2

h22R() = 0

Now, we choose as

2 =2

h2

sod2

d2+

2

d

d ( + 1)

2

R() +

2Ze2

R()R() = 0

(7.1)To continue, we re-start our discussion with Laguerres

differential equation:

xd2y

dx2+ (1 x) dy

dx+ y = 0 (7.2)

To show that this equation is related to Equation 7.1 wedifferentiate Equation 7.2

dxd2ydx2 + (1 x) dydx + y = 0dx (7.3)

which gives

y + xy y + (1 x)y + y = 0

which is

xy + (2 x)y + ( 1)y = 0

or x

d2

dx2+ (2 x) d

dx+ ( 1)

dy

dx= 0 (7.4)

Doing it again (differentiating), we obtain

d (xy + (2 x)y + ( 1)y = 0)dx

which leads to

y + xy y + (2 x)y + ( 1)y = 0

which finally becomesx

d2

dx2+ (3 x) d

dx+ ( 2)

d2y

dx2= 0 (7.5)

Generalizing, we havex

d2

dx2+ (k + 1 x) d

dx+ ( k)

dky

dxk= 0 (7.6)

VIII. PART 2

Consider Equation 7.1d2

d2+

2

d

dR() ( + 1

2

R()+

2Ze2

h2R()R() = 0

(8.1)if we re-write it as

d2

d2+ 2

d

d ( + 1

R() +

2Ze2

h2R()R() = 0

(8.2)

(for comparison with the following):

xy + 2y +

n k 1

2 x

4 k

2 14x

y = 0 (8.3)

Notice the similarity if x, i.e., powers of x, x1 etc.,

2Ze2

h2 n k 1

2(8.4)

x

4(8.5)

k2 14x

( + 1)

(8.6)

We force the asymptotic form of the solution y(x) tobe exponentially decreasing, i.e.,

y = ex/2x(k1)/2v(x) (8.7)
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and ask what equation v(x) solves. We do this in twosteps, first assuming

y = ex/2w(x)

and then assuming that w(x) is

w = x(k1)/2v(x)

So, assuming the first part of Equation 8.7, we have

y(x) =k 1

2x(k3)/2w(x) + x(k1)/2w(x)

and

y(x) =k 1

2

k 32

x(k5)/2w(x) +k 1

2x(k3)/2w(x) +

k 12

x(k3)/2w(x) + x(k1)/2w(x) = 0

which we now substitute into Equation 8.3 to obtain

xy =k 1

2

k 32

x(k3)/2w(x) + (k

1)x(k1)/2w(x) + x(k+1)/2w(x)

2y = 2k 1

2x(k3)/2w(x) + 2x(k1)/2w(x)

ny = nx(k1)/2w(x)

k 12

y = k 12

x(k1)/2w

x4

y = x(k+1)/2

4w

k2 14x

y = k2 1

4x(k3)/2w 0 (8.8)

or

xw + (k + 1)w +

n k 1

2 x

4

w = 0 (8.9)

IX.

Now we let

w = ex/2v(x)

(as noted before) to obtain

w = 12

ex/2v + ex/2v

w =1

4ex/2v ex/2v + ex/2v

so, substituting into Equation 8.8 we have

xw = ex/2x

4v xv + xv

(k + 1)w = ex/2

k + 1

2v + (k + 1)v

n k 12

x4

w = ex/2

n k 1

2 x

4

v (9.1)

= 0 (9.2)

so, v solves Equation 7.6 if = n. Expanding the r.h.s.of Equation 9.1 we have

x

4v

x

4+xv+(k +1

x)v+n

k 12

k + 1

2 v = 0i.e.,

xv + (k + 1 x)v + (n k)v = 0which is Equation 7.6, i.e.,

v =dky

dxk
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and

y = ex/2x(k1)/2dky

dxk

or

w =1

4ex/2v ex/2v + ex/2v

so, substituting into Equation 8.8 we have

xw = ex/2x

4v xv + xv

(k + 1)w = ex/2

k + 1

2v + (k + 1)v

n k 12

x4

w = ex/2

n k 1

2 x

4

v (9.3)

= 0 (9.4)

so, v solves Equation 7.6 if = n. Expanding the r.h.s.of Equation 9.3 we have

x4

v x4

+xv+(k +1x)v+n k 12

k + 12

v = 0i.e.,

xv + (k + 1 x)v + (n k)v = 0

which is Equation 7.6, i.e.,

v =dky

dxk

and

y = ex/2x(k1)/2 dk

y

dxk

or

R() = e/2(k1)/2)Lkn()

where y and R() are solutions to Laguerres Equationof degree n. Wow.

X. PART 3

Now, all we need do is solve Laguerres differential

equation.

xy + (1 x)y + y = 0

where is a constant (to be discovered). We let

y =?

=0

ax

and proceed as normal

xy = 2a2x + (3)(2)a3x2 + (4)(3)a4x

3 + +y = (2)a2x + (3)a3x

2 + (4)a4x3 +

xy = a2x (2)a2x2 (3)a3x3 +y = a0 + a1x + a2x

2 + = 0 (10.1)which yields

a1 = a0a2 =

1 4

a1

a3 =2

9a2

a4 =3

16a3 (10.2)

or, in general,

aj+1 =j

(j + 1)2aj

which means

a1 = 1 a0

a2 = (1 )(4)(1)

a0

a3 = (2 )(1 )(9)(4)(1)

a0

a4 = (3 )(2 )(1 )(16)(9)(4)(1)

a0

(10.3)which finally is

aj =

j1k=0(k )

jk=1(k2)

a0

and

aj+1 = (j ) j1k=0(k )

(j + 1)2jk=1(k2)

a0 =j

(j + 1)2aj

which implies that

aj+1aj

=j

(j + 1)2 1

j

as j . This is the behaviour of y = ex, which wouldoverpower the previous Ansatz, so we must have trunca-tion through an appropriate choice of (i.e., = n).

XI.

If were an integer, then as j increased, and passedinto we would have a zero numerator in the expression

aj+1 =(j )(j + 1)2

aj
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and all higher as would be zero (i.e., not a power seriesbut a polynomial instead)! But

2 =2

h2= 2E

h2

so, from Equation 8.6 we have

k2 14

= ( + 1)

k2 1 = 42 + 4

k = 2 + 1

so

k2 12

=2 + 1 1

2=

and therefore Equation 8.3 tells us thatn k 1

2

= n = 2Ze

2

h2

implies

=2Ze2

h2(n

)2

2 = 2Eh2

=Z2e4

h4(n )2

i.e.,

E = Z2e4

2h2(n )2

[1] the only thing that happens in making the complex con-jugate is that each is changed to (and each ).[2] the bracket, hah, hah.

[3] notice the rewriting into momentum language![4] The limits correspond to = 0 to =
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math for quantum chemistry

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