better equation sheet qm

7/27/2019 Better Equation Sheet Qm

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Chapter 1: The wave Function

Schrodinger Equation

i

t =

2

2m

2

x2 + V (1)

Probability

Pab =

ba

|(x, t)|2dx (2)

For Normalization

1 =

|(x, t)|2dx, where |(x, t)|2 =

(3)once is normalized it stays normalized forall time

d

dt

|(x, t)|2dx = 0 (4)

Operators for momentum and position in gen-eral form

x =

xdx (5)

p =

i

x

dx (6)

Q(x, p) =

Q

x,

i

x

dx (7)

Expectation value of kinetic energy

T = 2

2m

2

x2dx (8)

de Broglie formula

p = h

= 2

(9)

uncertainty principle

xp 2

(10)

probability current

J(x, t) i2m

{

x

x

(11)

Chapter 2: Stationary States

seperation of variables

(x, t) = (x)(t) = (x)eiEt/ (12)

d

dt= iE

(13)

(t) = eiEt/ (14)

(x, t) = (x)eiEt/ (15)

Time Independent Schrodinger equation

22m

d2dx2

+ V = E (16)

Hamiltonian

H(x, p) =p2

2m+ V(x) (17)

substitute p = (/i)(/x) and the corre-sponding Hamiltonian operator is

H = 2

2m

d2

dx2+ V(x) = E (18)

H = Hdx = E (19)You can always write the general solution of

the Schrodinger equation as a linear combinationof seperable solutions

(x, 0) =n=1

cnn(x) (20)

(x, t) =

n=1

cnn(x)eiEnt/ =

n=1

cnn(x, t)

(21)

The Infinite Square Well

V(x) =

0, if 0xa, otherwise (22)

plugging this into the TISE and we get

d2

dx2= k2 (23)

k

2mE

(24)

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(x) = A sin kx (25)

kn =n

a(26)

En =n222

2ma2(27)

after normalization, the general TISE solu-tions for the infinite square well is

n(x) =

2

asinn

ax

(28)

as a collection, the functions n(x) have someinteresting properties

1. They are alternately even and odd, withrespect to the center of the well.

2. As you go up in energy, each succesivestate has one more node (zero crossing)

3. They are mutually orthogonal, in thesense that

m(x)n(x)dx = 0 (29)

Kronecker deltam(x)

n(x)dx = mn (30)

mn = 0, if m=n1, if m=n (31)we say that s are orthonormal

4. They are complete, in the sense that anyother function, f(x), can be expressed as a linearcombination of them:

f(x) =n=1

cnn(x) =

2

a

n=1

cn sinn

ax

(32)

the cns can fe found using Fouriers trick

cn =

n(x)f(x)dx (33)

=

2

a

a0

sinn

ax

(x, 0)dx (34)

Pn = |cn|2 (35)n=1

|cn|2 = 1 (36)

H =n=1

|cn|2En (37)

The harmonic Oscillator

V(x) =1

2kx2 (38)

km

(39)

V(x) =1

2m2x2 (40)

rewriting equation 18 in a more suggestiveform

1

2m[p2 + (mx)2] = E (41)

we can use raising and lowering operators tosolve this problem

a+ 12m

(ip + mx) (42)

a 12m

(ip + mx) (43)

aa+ =1

2m[p2 + (mx)2 imw(xp px)]

(44)

commutator

[A, B] AB BA (45)

[x, p]f(x) = if(x) (46)

[x, p] = i (47)

aa+ =1

H +

1

2(48)

H =

aa+ 1

2

(49)

a+a =1

H 1

2(50)

[a

, a+] = 1 (51)

H =

a+a +

1

2

(52)

In terms ofa, then the Schrodinger equationof the harmonic oscillator takes the form

aa 1

2

= E (53)

H(a+) = (E+ )(a+) (54)

H(a) = (E )(a) (55)

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there must be a lowest state and this occurswhen

a0 = 0 (56)

this means

12m

d

dx+ mx

0 = 0 (57)

the genral solution for the ground state of theharmonic oscillator is

0(x) =m

e

m2

x2 (58)

E0 =

1

2

(59)

En =

n +

1

2

(60)

n = An(a+)n0(x) (61)

a+an = nn, aa+n = (n + 1)n (62)

a+n =

n + 1n+1, an =

nn1 (63)

n =1n!

(a+)n0 (64)

x =

2m(a+ + a) (65)

p = i

m

2(a+ a) (66)

The Free particle

V(x) = 0 everywhere

(x, t) = Aeik(xk2m

t + Beik(x+k2m

t) (67)

k(x, t) = Aei(kx k2

2mt (68)

k

2mE

, with

k>0 travelingtotherightk


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d2

dx2= 2mE

2 = k2 (82)

where

k 2mE

(83)

for bound states E < 0 the general solution iswhen x < 0

(x) = Aekx + Bekx (84)

the first term blows up as x so wemust choose A=0

(x) = Bekx (85)

for bound states E < 0 the general solution iswhen x > 0

(x) = F ekx + Gekx (86)

the second term blows up as x so wemust choose G=0

(x) = F ekx (87)

It remains only to stitch these two functionstogether, using the appropriate boundery condi-tions at x = 0, the standard boundary conditions

for

1. is always continuous 2. d/dx is con-tinuous except at points where the potential isinfinite

in this case the first boundary condition tellsus that F = B, so

(x) =Bekx, (x0)Bekx, (x0) (88)

the second boundary condition gives

ddx =

2m

2

(0) (89)

Evidently the delat-function well, regardlessof its strength , has exactly one bound state

(x) =

m

em|x|

2

; E = m2

22(90)

What about scattering states, with E > 0?Forx < 0 the Schrodinger equation read

d2

dx2= 2mE

2 = k2 (91)

where

k

2mE

(92)

is real and positive. The general solution isfor x < 0

(x) = Aeikx + Beikx (93)

and this time we cannot rule out either term,since neither blows up. and similarly for x > 0

(x) = F eikx + Geikx (94)

for boundary condition number 1 this implies

F + G = A + B (95)

the derivatives are

d/dx=ik(Feikx+Geikx), for(x>0), so d/dx|+=ik(FG)d/dx=ik(Aeikx+Beikx), for(x


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The reflection coefficient is

R |B|2

|A|2 =2

1 + 2(102)

and the transmission coefficient is

T |F|2

|A|2 =1

1 + 2(103)

and the sum should be one

R + T = 1 (104)

R and T are functions of and hence func-tions ofE

R =

1

1 + (22E/m2) (105)

T =1

1 + (m2/22E)(106)

The Finite Square Well

V(x) =V0, for axa0, for |x|>a (107)

This potential admits both bound states andscattering states. we will look at the boundstates first. in the region x < a the potentialis zero, so the Schrodinger equation reads

d2

dx2= 2mE

2 = k2 (108)

where

k 2mE

(109)

the general solution as before is

(x) = Aekx + Bekx (110)

the first term blows up as x so wemust choose A=0

(x) = Bekx, for x < a (111)in the region a < x < a,V(x) = V0, and

the Schrodinger equation reads

2

2m

d2

dx2 V0 = E, or d

2

dx2= l2

(112)

where

l 2m(E+ V0)

(113)

and the general solution is

(x) = Csin(lx) + D cos(lx), for a < x < a(114)

when x > a the genral solution is

(x) = F ekx + Gekx (115)

the first term blows up as x so wemust choose G=0

(x) = F ekx, for x > a (116)

The next step is to impose boundary condi-tions: and d/dx continuous at a and a.but we can save a little time by noting that thispotential is an even function, so we can assumewith no loss of generality that the solutions areeither even or odd. since (x) = (x).

for even solutions use D cos(lx) and for oddsolutions use Csin(lx). I will show the even so-lutions

The continuity of (x), at x = a, says

F eka = D cos(la) (117)

and the continuity of d/dx, says

kF eka =

lD sin(la) (118)

deviding equation 118 by 117 , we find that

k = l tan(la) (119)

This is a formula for the allowed energies,since k and l are both functions of E. To solvefor E, we first adopt a nicer notation: Let

z la, and z0 a

2mV0 (120)

(k2 + l2) = 2mV0

2, so ka =

z20 z2 (121)

and now equation 119 read

tan(z) =

(z0/z)2 1 (122)This is a trancendental equation and can olny

be solved numerically or graphically. Two limit-ing cases are of special interest.

1. Wide, deep well. if z0 is very large,the intersections occur just slightly below zn =n/2, with n odd it follows that

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En + V0 = n222

2m(2a)2(123)

2. Shallow, narrow well. As z0 decreases,

there are fewer and fewer bound states, until fi-nally (for z0 < /2,where the lowest odd statedisappears) only one remains. It is interestingto note that there is always one bound state, nomatter how weak the well becomes.

Now moving on to scattering states whereE > 0

the general solution as before is

(x) = Aeikx + Beikx for (x < a) (124)

where(as usual)

k

2mE

(125)

inside the well, where V(x) = V0 and thegeneral solution is

(x) = Csin(lx) + D cos(lx), for a < x < a(126)

where as before

l 2m(E+ V0)

(127)

To the right, assuming there is no incomingwave in this region, we have

(x) = F eikx (128)

Here A is the incident amplitude, B is the re-flected amplitude, and F is the transmitted am-plitude.

There are four boundary conditions: Conti-nuity of (x) at a says

Aeika + Beika = Csin(la) + D cos(la) (129)

continuity of d/dx at a gives

ik[Aeika Beika] = l[Ccos(la) + D sin(la)](130)

continuity of (x) at +a yields

Csin(la) + D cos(la) = F eika (131)

continuity of d/dx at +a gives

l[Ccos(la) D sin(la)] = iF eika (132)

We can use two of these to eliminate C andD, and solve the remaining two for B and F:

B = isin(2la)

2kl(l2 k2)F (133)

F =e2ikaA

cos(2la) i (k2+l22kl sin(2la)(134)

The transmission coefficient (T = |F|2/|A|2)expressed in terms of the original variables, is

given by

T1 = 1 +V20

4E(E+ V0)sin2

2a

2m(E+ V0)

(135)

Notice that T = 1 (the well becomes trans-parent) whenever the sine is zero, which is tosay, when

2a

2m(En + V0) = n (136)

where n is an integer. The energies for perfecttransmission, then are given by

En + V0 =n222

2m(2a)2(137)

which happens to be the allowed energies forthe infinite square well

Chapter 3: Formalism

The collection of all functions of x constituesa vector space. To represent a possible physicalstate, the wave function must be normalized:

||2dx = 1 (138)

The set of all square-integrable functions,on a specified interval

f(x) such that

ba

|f(x)|2dx < (139)

this constitutes a (much smaller) vector space.mathematicians call it L2(a, b); physicist call itHilbert Space. In quantum mechanics, then

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Wave functions live in Hilbert Space

We define the inner product of two func-tions, f(x) and g(x), as follows

f|g ba

f(x)g(x)dx (140)

If f and g are both square-integrable (thatis, if they are both in Hilbert space),their innerproduct is guaranteed to exist (the integral inequation 140 converges to a finite number), Thisfollows from the integral Schwarz inequality

ba

f(x)g(x)dx

b

a

|f(x)|2dxba

|g(x)|2dx(141)

Notice in particular that

g|f = f|g (142)

Moreover, the inner product of f(x) with it-self,

f|f =ba

|f(x)|2dx 0 (143)

is real and non-negative; its zero only iff(x) = 0

A function is said to be normalized if its

inner product with itself is 1; two functions areorthogonal if their inner product is 0; and a setof functions, {fn}, is orthonormal if they arenormalized and mutually orthogonal;

fm|fn = mn (144)Finally, a set of functions is complete if any

other function (in Hilbert space) can be ex-pressed as a linear combination of them

f(x) =

n=1cnfn(x) (145)

if the functions {fn(x)} are orthonormal, thecoefficients are given by Fouriers trick

cn = fn|f (146)Observables

The expectation value of an observable Q(x, p)can be expressed very neatly in inner-productnotation:

Q = Qdx = |Q (147)

now the outcome of a measurement has got tobe real, and so, a fortiori, is the average of manymeasurements:

Q = Q (148)But the complex conjugate of an inner prod-

uct reverses the order

|Q = Q| (149)and this must hold true for any wave function

. Thus operators representing observableshavethe verty special property that

f|Qf = Qf|f for all f(x) (150)

we call such operator hermitian

Observables are represented by hermitian operators

Determinate States

Q = q (151)

This is the eigenvalue equation for the op-erator Q; is an eigenfunction of Q, and q isthe corresponding eigenvalue; thus

Determinate states are eigenfunctions of Q

Measurement of q on such a state is certain

to yield the eigenvalue q. The collection of alleigenvalues of an operator is called its spec-trum. Sometimes two (or more) linearly in-dependent eigenfunctions share the same eigen-value; in that case the spectrum is said to bedegenerate.

Eigenfunctions of Hermitian Operators

Discrete Spectra

Mathematically, the normalized eigenfunc-tions of a hermitian operato have two importantproperties.

Theorem 1: Their eigenvalues are real.

Qf = qf (152)

(i.e.,f(x) is an eigenfunction of Q, with eigen-value q), and

f|Qf = Qf|f (153)then

qf|f = q f|f (154)and q must be real

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Theorem 2: Eigenfunctions belonging todistinct eigenvalues are orthogonal.

Qf = qf, and Qg = qg (155)

and Q is hermitian. Then f|Qg = Qf|gso

qf|g = q f|g (156)again the inner product exist because the

eigenfunctions are in Hilbert space by assump-tion. but q is real so if q = q it must be thatf|g = 0.

Continuous Spectrum

If the spectrum of a hermitian operator is con-tinuous, the eigenfunctions are not normalizable,

and the proofs of Theorems 1 and 2 fail, becausethe inner products may not exist.

Generalized Statistical Interpretation

If you measure an observable Q(x, p) on a par-ticle in the state (x, t), you are certain to getone of the eigenvalues of the hermitian operator.If the spectrum is dicrete, the probability of get-ting the particular eigenvalue qn associated withthe orthonormalized eigenfunction fn(x) is

|cn|2, where cn = fn| (157)If the spectrum is continuous, with real eigen-

values q(z) and associated Dirac orthonormal-ized eigenfunctions fz(x), the probability of get-ting result in the range dz is

|c(z)|2dz, where c(z) = fz| (158)

The eigenfunctions of a hermitian operatorare complete, so that the wave function can bewritten as a linear combination of them

(x, t) = n

cnfn(x) (159)

Because the eigenfunctions are orthonormal,the coefficients are given by Fouriers trick

cn = fn| =

fn(x)(x, t)dx (160)

Of course, the total probability (summed overall possible outcomes) has got to equal one

n|cn|2 = 1 (161)

Similarly, the expectation value Q should bethe sum over all possible outcomes of the eigen-value times the probablity of getting that eigen-value

Q =n

qn|cn|2 (162)

The momentum space wave function,(p, t) is essentially the Fourier transform of theposition space wave function (x, t), which byPlancherels theorem is its inverse Fourier trans-form.

(p, t) =12

eipx/(x, t)dx (163)

(x, t) =

1

2

eipx/

(p, t)dp (164)

(165)

According to the generalized statistical inter-pretation, the probability that a measurement ofthe momentum would yield a result in the rangedp is

|(p, t)|2dp (166)The Uncertainty Principle

The generalized uncertainty principle is

2A2B

1

2i[A, B]

2(167)

The energy-time uncertainty principle

d

dtQ = i

[H, Q] +

Q

dt

(168)

Chapter 4: Quantum Mechanics in ThreeDimensions

The generalization to three dimensions isstraightforward. Schrodingers equation says

i

t= H (169)

the Hamiltonian operator H is obtained from theclassical energy

1

2mv2 + V =

1

2m(p2x +p

2y +p

2z) + V (170)

by the standard prescription (applied now toy and z, as well as x)

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px i

x, py

i

y, pz

i

z(171)

or

p i (172)

thus

i

t=

2

2m2 + V (173)

where

2 2

x2+

2

y 2+

2

z 2(174)

is the Laplacian, in cartesian coordinates.The potential V and the wave function arenow functions of r = (x,y,z) and t.

the normalization condition reads||2d3r = 1 (175)

with the integral taken over all space. if thepotential is independent of time, there will be acomplete set of stationary states

n(r, t) = n(r)eiEnt/ (176)

where the spatial wave function n satisfiesthe time-independent Schrodinger

2

2m2 + V = E (177)

The general solution to the time-dependentSchrodinger equation is

(r, t) =

cnn(r)eiEnt/ (178)

Seperation Of variables Typically, the po-

tential is a function only of the distance fromthe origin. In that case it is natural to adoptspherical coordinates, (r,,). In sphericalcoordinates the Laplacian takes the form

2 = 1r2

r

r2

r

+

1

r2 sin

sin

+1

r2 sin2

2

2

(179)

We begin by looking for solutions that areseperable by products

(r,,) = R(r)Y(, ) (180)

,. The Angular Equation

sin

sin

Y

+

2Y

2= l(l + 1)sin2 Y

(181)

you might recognize this equation, it occursin the solution to Laplaces equation in classicalelectrodynamics. as always, we try seperation ofvariables

Y(, ) = ()() (182)

plugging this in and deviding by gives ustwo solutions

1

sin

d

d

sin

d

d

+ l(l + 1)sin2 = m2

(183)

1

d2

d2= m2 (184)

The equation is easy

d2

d2=

m2

() = eim (185)

Now when advances by 2, we return to thesame point in space, so it is natural to requirethat

( + 2) = () (186)

from this it follows that m must be an integer

m = 0, 1, 2,.... (187)The equation is not so simple, the solution

is

() = APml (cos ) (188)

where Pml is the associated Legendre func-tion, defined by

Pml (x) (1 x2)|m|/2

d

dx

|m|Pl(x) (189)

and Pl(x) is the lth Legendre polynomial,defined by the Rodrigues formula

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Pl(x) 12ll!

d

dx

l(x2 1)l (190)

Now, the volume element in spherical coordi-nates is

d3r = r2 sin drdd (191)

so the normalization condition becomes

||2r2drdd =

|R|2r2dr

|Y|2 sin dd = 1 (192)

It is convenient to normalize R and Y sepa-rately

0

|R|2r2dr = 1 (193)20

0

|Y|2 sin dd = 1 (194)

The normalized angular wave functions arecalled spherical harmonics

Yml (, ) =

(2l + 1)4

(l |m|)!(l + |m|)! e

imPml (cos )

(195)

where = (1)m for m 0 and = 1 form 0. As we shall prove later on, they areautomatically orthogonal.

The Radial Equation Notice that the an-gular part of the wave function, Y(, ), is thesame for all spherically symmetric potentials; theactual shape of the potential ,V(r), affects onlythe radial part of the wave function,R(r), whichis determined by

u(r) rR(r) (196)using this relationship we can now write the

radial equation as

2

2m

d2u

dr2+

V +

2

2m

l(l + 1)

r2

u = Eu (197)

the normalization condition becomes

0

|u|2dr = 1 (198)

this is as far as we can go until a specific po-tential V(r) is provided.

The infinite spherical well

V(r) =0, ifra, ifr>a (199)

Outside the well, the wave function is zero;inside the well, the radial equation says

d2u

dr2=

l(l + 1)

r2 k2

u (200)

where

k

2mE

(201)

as usual. Our problem is to solve this equa-

tion, subject to the boundary condition u(a) =0. The case l = 0 is easy

d2u

dr2= k2u u(r) = A sin(kr) + B cos(kr)

(202)

we must choose B = 0 because r 0 theradial wave function blows up. The boundarycondition then requires sin(ka) = 0, and henceka = n, for some integer n. The allowed ener-gies are evidently

En0 =n222

2ma2, (n = 1, 2, 3,...) (203)

the same as for the one-dimensional infinitesquare well. The general solution to equation200 (for an arbitrary integer l) is not as familiar

u(r) = Arjl(kr) + Brnl(kr) (204)

wherejl(x) is the spherical Bessel functionpf order l, and nl(x) is the spherical Neumannfunction of order l. They are defined as follows

jl(x) (x)l

1

x

d

dx

lsin x

x(205)

nl(x) (x)l

1

x

d

dx

lcos x

x(206)

Notice that the Bessel functions are finite atthe origin, but Neumann functions blow up atthe origin. Accordingly, we must have Bl = 0,and hence

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R(r) = Ajl(kr) (207)

There remains the boundary condition, R(a) =0. Evidently k must be chosen such that

jl(ka) = 0 (208)

the boundary condition requires that

k =1

anl (209)

where nl is the nth zero of the lth sphericalBessel function. The allowed energies then, aregiven by

Enl =

2

2ma2

2nl (210)

and the wave functions are

nlm(r,,) = Anl(nlr/a)Yml (, ) (211)

with the constant Anl to be detemined by nor-malization.

The Hydrogen Atom

From Coulombs law, the potential energy is

V(r) = e2

401r

(212)

and the radial equation says

2

2m

d2u

dr2+

e

2

40

1

r+

2

2m

l(l + 1)

r2

u = Eu

(213)

our problem is to solve this equation for u(r),and determine the allowed energies. Our firsttask is to tidy up the notation. Let

k 2mE

E = k222m

(214)

dividing equation 213 by E gives

1

k2d2u

dr2=

1 me

2

202k

1

(kr)+

l(l + 1)

(kr)2

u

(215)

This suggest that we introduce

kr and 0 me2

202k(216)

so that

d2u

d2= 1

0

+l(l + 1)

2 u (217)Next we examine the asymptotic form of the

solutions. As , the constant term in thebrackets dominate, so approximately

d2u

d2= u (218)

The general solution is

u() = Ae + Be (219)

but e blows up ( as ), so B = 0,evidently

u() Ae (220)for large . On the other hand, as 0 the

centrifugal term dominates, approximately then

d2u

d2=

l(l + 1)

2u (221)

the general solution is

u() = Cl+1 + Dl (222)

let D = 0, because l blows up as 0,thus

u() Cl+1 (223)The next step is to peel off the asymptotic

behaviour, introducing the new function v()

u() = l+1ev() (224)

Finally, we assume the solution, v(), can beexpressed as a power series in

v() = j=0

cjj (225)

cj+1 =

2(j + l + 1) 0

(j + 1)(j + 2l + 2)

cj (226)

This recursion formula determines the coef-ficients, and hence the function v(). Now letssee what the coefficients look like for large j (thiscorresponds to large , where the higher powerdominate . In this regime the recursion formulasays

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cj+1 = 2jj(j + 1)

cj =2

j + 1cj (227)

suppose for a moment that this were exact.Then

cj =2j

j!c0 (228)

so

v() = c0

j=0

2j

j!j (229)

and hence

u() = c0l+1

e

(230)which blows up at large . The series must

terminate. There must occur some maximal in-teger, jmax such that

cjmax+1 = 0 (231)

(and beyond which all coefficients vanish au-tomatically). equation 226 becomes

2(jmax + l + 1) 0 = 0 (232)Defining

n jmax + l + 1 (233)(the so-called principal quantum num-

ber), we have

0 = n (234)

But 0 determines E

E = k2

2

2m= me

4

822220(235)

so the allowed energies are

En =

m

22

e2

4

21

n2=

E1n2

n = 1, 2, ..

(236)

This is the famous Bohr Formula. combin-ing equation 216 and 234 we get

k =

me2

402

1

n=

1

an(237)

where

a 402

me2= 0.529x1010m (238)

is called the Bohr radius. it follows that

=r

an(239)

The spatial wave functions for hydrogen arelabeled by three quantum numbers (n,l, andm)

nlm(r,,) = Rnl(r)Yml (, ) (240)

and

Rnl(r) =1

r

l+1ev() (241)

and v() is a polynomial of degree jmax =n l 1 in , whose coefficients are determinedby the recursion formula

cj+1 =2(j + l + 1 n

(j + 1)(j + 2l + 2)cj (242)

The polynomial v() is a function well knownto mathematicians; apart from normalization, itcan be written as

v() = L2l+1nl1(2) (243)

where

Lpqp(x) (1)p

d

dx

pLq(x) (244)

is an associated Laguerre polynomial,and

Lq(x) ex

d

dx

q(exxq) (245)

is the qth Laguerre polynomial, the nor-

malized hydrogen wave functions are

nlm =

2

na

3(n l 1)!2n[(n + l)!]3

er/na

2r

na

l

L2l+1nl1(2r/na)

Yml (, ) (246)

The spectrum of Hydrogen

An electron may undergo a transition tosome other energy state, either by absorbing en-ergy or emmiting energy, the difference in energybetween the initial and final states are

13


14/25


15/25

Now,

LL = L2 L2z i(iLz) (267)or putting it the other way around

L2 = LL + L2z Lz (268)it follows that

L2ft = (LL+ + L2z + Lz)ft = 2l(l + 1)ft

(269)

and nence

= 2l(l + 1) (270)

This tells us the eigenvalue of L2 in terms ofthe maximum eigenvalue of Lz. Evidently theeigenvalues ofLz are m, where m goes from lto l in N integer steps. In particular, it followsthat l = l+N, and hence l = N/2, so l must bean integer or a half-integer. The eigenfunctionsare characterized by the numbers l and m

L2fml = 2l(l + 1)fml ; Lzf

ml = mf

ml (271)

where

l = 0, 1/2, 1, 3/2,...; m = l, l + 1,...,l 1, l(272)

Eigenfunctions

First of all we need to rewrite Lx, Ly, and Lzin spherical coordinates. Now, L=(/i)(r ) ,and the gradient, in spherical coordinates, is

= r r

+ 1

r

+

1

r sin

(273)

meanwhile, r=rr, so

L =

i

r(r r)

r+ (r )

+ (r ) 1

sin

(274)

But (r r) = 0, (r ) = , and (r ) = and hence

L =

i

1

sin

(275)

The unit vectors and can be resolved intotheir cartesian components

= (cos cos )i + (cos sin )j (sin )k(276)

= (sin )

i + (cos )

j (277)

Evidently

Lx =

i

sin

cos cot

(278)

Ly =

i

+cos

sin cot

(279)

and

Lz =

i

(280)

using the raising and lowering operators wefind

L2 = 2

1

sin

sin

+

1

sin2

2

2

(281)

conclusion: Spherical harmonics are eigen-functions of L2 and Lz. When we solved theSchrodinger equation by seperation of variables,we were inadvertently constructing simultaneouseigenfunctions of the three commuting operatorsH, L2, and Lz

H = E , L2 = 2l(l + 1), Lz = m(282)

incidentally, we can use equation 281 torewrite the Schrodinger equation

1

2mr2

2

r

r2

r

+ L2

+ V = E

(283)

SpinIn classical mechanics, a rigid object admits

two kinds of angular momentum: orbital (L =r p), associated with the motion of the centerof mass, and spin (S = I) , associated withmotion about the center of mass. The algebraictheory of spin is a carbon copy of the theory oforbital angular momentum, beggining with thefundamental commutation relations

[Sx, Sy] = iSz, [Sy, Sz] = iSx, [Sz, Sx] = iSy(284)

15


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it follows that the eigenvectors of S2 and Szsatisfy

S2

|sm

= 2s(s + 1)

|sm

; S

z|sm

= m

|sm

(285)and

S|sm =

s(s + 1) m(m 1)|s(m 1)(286)

where S SxiSy. But this time the eigen-functions are not spherical harmonics , and thereis no apriori reason to exclude the half-integervalues of s and m:

s = 0, 12

, 1, 32

,...; m = s, s + 1,...,s 1, s.(287)

Pi mesons have spin 1/2; photons have spin 1;deltas have spin 3/2; gravitons have spin 2; andso on.

Spin 1/2

By far the most important case is s = 1/2, forthis is the spin of the particles that make up ordi-nary matter( protons, neutrons, and electrons),as well as quark and leptons. The general stateof a spin-1/2 particle can be expressed as a two-element column matrix (or spinor);

=

a

b

= a+ + b (288)

with

+ =

1

0

(289)

representing spin up, and

=

0

1

(290)

for spin down. Meanwhile, the spin operatorsbecome 22 matrices, which we can work out bynoting their effect on + and .Equation 285says

S2+ =3

4

2+ and S2 =

3

4

2 (291)

we can write S2 in matrix for as

S2 =3

4

2

1 0

0 1 (292)

Similarly

Sz+ =

2+, Sz =

2 (293)

from which it follows that

Sz =

2

1 0

0 1

(294)

Meanwhile, equation 286 says

S+ = +, S+ = , S+ = S = 0(295)

so

S+ =

0 10 0

, S =

0 01 0

(296)

Now S = Sx iSy, so Sx = (1/2)(S+ + S)and Sy = (1/2i)(S+ S), and hence

Sx =

2

0 1

1 0

, Sy =

2

0 i

i 0

(297)

Since Sx, Sy, and Sz all carry a factor of/2,it is tidier to write S=(/2), where

x

0 1

1 0

, y

0 i

i 0

, z

1 0

0 1

(298)

These are the famous Pauli spin matrices.The eigenspinors of Sz are (of course)

+ =

1

0

,

eigenvalue +

2

(299)

=

0

1

,

eigenvalue

2

(300)

If you measure Sz on a particle in the generalstate (equation 288), you could get +/2, withprobability |a|2, or /2, with probability |b|2.Since they are the only possibilities

|a|2 + |b|2 = 1 (301)(i.e, the spinor must be normalized). But

what if instead, you chose to measure Sx? Whatare the possible results. According to the gener-alized statistical interpretation, we nee dto knowthe eigenvalues and eigenspinors of Sx, the char-acteristic equation is

16


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/2/2 = 0 2 = 22 2

Not suprisingly, the possible values for Sx arethe same as those for Sz . The normalized eigen-spinors of S

xare

(x)+ =

12

12

,

eigenvalue +

2

(302)

(x) = 1

21

2

,

eigenvalue

2

(303)

As the eigenvectors of a hermitian matrix,

they span the space; the generic spinor (equa-tion 288) can be expressed as a linear combina-tion of them

=

a + b

2

(x)+ +

a b

2

(x) (304)

If you measure Sx, the probability of getting+/2 is (1/2)|a + b|2, and the probability of get-ting /2 is (1/2)|a b|2.

Electron in a Magnetic Field

A spinning charged particle constitutes amagnetic dipole. Its magnetic dipole mo-ment,, is proportional to its spin angular mo-mentum, S:

= S (305)

the proportionality constant, , is called thegyromagnetic ratio. When a magnetic dipoleis placed in a magnetic field B, it experiences atourqe, x B which tends to line it up parallel tothe field. The energy associated with this tourqeis

H = B (306)so the Hamiltonian of a spinning charged par-

ticle, at rest in a magnetic field B is

H = B S (307)Larmor preccesion: Imagine a particle of

spin 1/2 at rest in a uniform magnetic field,which points in the z direction

B = B0k (308)

The Hamiltonian, in matrix form, is

H = B0Sz = B02

1 0

0

1 (309)

The eigenstates of H are the same as those ofSz

+, with energy E+ = (B0)/2, with energy E = +(B0)/2

since the Hamiltonian is time-independent,the general solution to the time-dependentSchrodinger equation is

i

t= H (310)

can be expressed in terms of the stationary

states

(t) = a+eiE+t/+beiEt/ =

aeiB0t/2

beiB0t/2

(311)

The constants a and b are determined by theinitial conditions

(0) =

a

b

(312)

With no essential loss of generality I will write

a = cos(/2) and b = sin(/2), Thus

(t) =

cos(/2)eiB0t/2

sin(/2)eiB0t/2

(313)

To get a feel for whats happening here, letscalculate the expectation value of S, as a func-tion of time

Sx = (t)Sx(t) = (cos(/2)eiB0t/2 sin(/2)eiB0t/

2

0 1

1 0cos(/2)eiB0t/2

sin(/2)eiB0t/2=

2sin cos(B0t) (314)

similarly,

Sy = (t)Sy(t) = 2

sin sin(B0t)

(315)

and

Sz = (t)Sy(t) = 2

cos (316)

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Evidently S is tilted at a constant angle to the z axis, and precesses about the field at theLarmor frequency

= B0 (317)Just as it would classically

Addition of Angular Momenta

Suppose now that we have twofor e spin-1/2particlesfor example, the electron and the pro-ton in the ground state of hydrogen. Each canhave spin up or spin down so there are four pos-sibilities in all

, , , (318)where the first row refers to the electron and

the second row to the proton. Question: Whatis the total angular momentum of the atom? Let

S S(1) + S(2) (319)Each of these four composite states is an

eigenstate of Szthe z-components simply add

Sz12 = (S(1)z +S

(2)z )12 = (S

(1)z chi1)2+1(S

(2)z 2)

= (m11)2+1(m22) = (m1+m2)12(320)

So m (the quantum number of the compositesystem) is just m1 + m2

: m = 1;: m = 0;: m = 0;: m = 1

At first galnce, this doesnt look right: m issupposed to advance in integer steps , from sto +s, so it appears that s = 1 but there is an

extra state with m = 0. One way to untanglethis problem is to apply the lowering operator,

S = S(1) + S

(2) to the state , using equation

296

S() = (S(1) ) + (S(2) ) (321)= () + () (322)= ( + ) (323)

Evidently the three states with s = 1 are (inthe notation

|s m

):

|11 =|10 = 1

2( + )

|1 1 =s = 1 (triplet)

This is called the triplet combination, for the

obvious reason. Meanwhile, the orthogonal statewith m = 0 carries s = 0;

|00 = 1

2()

s = 0 (singlet) (324)

(if you apply the raisong or lowering operatorto this state, youll get zero

I need to prove that the triplet states areeigenvectors of S2 with eigenvalue 22, and thesinglet is an eigenvector of S2 with eigenvalue 0.Now,

S2 = (S(1) + S(2)) (S(1) + S(2) (325)= (S(1))2 + (S(2))2 + 2S(1) S(2) (326)

Using equations 294 and 297, we have

S(1) + S(2)() = (S(1)x )(S(2)x )+ (S(1)y )(S(2)y ) + (S(1)z )(S(2)z )

= 2

2 + i

2 i

2 )

+

2)

2)

=

2

4(2 ) (327)

similarly

S(1) + S(2)() = 2

4(2 ) (328)

It follows that

S(1)S(2)|10 = 2

4

12

(2 +2 ) = 2

4|10

(329)

and

S(1)S(2)|00 = 2

4

12

(2 2 + ) = 32

4|00

(330)

Returning to equation 325-326 ( and usingequation 291), we conclude that

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S2|10 =

32

4+

32

4+ 2

2

4

|10 = 22|10

(331)

S2|00 =

32

4+

32

4 2 3

2

4

|00 = 0 (332)

What we have just done (combining spin 1/2with spin 1/2 to get spin 1 and spin 0) is the sim-plest example of a larger problem: if you com-bine spin s1 with spin s2, what total spins s canyou get? The answer is that you get every spinfrom (s+s2) down to (s1 s2) or (s2 s1), ifs2 > s1 in integer steps

s = (s1 +s2), (s1 +s21), (s1 +s22),..., |s1s2|(333)

The combined state |sm with total spin s andz-component m will be some linear combinationof the composite states |s1m1|s2m2

|sm =

m1+m2=m

Cs1s2sm1m2m|s1m1|s2m2 (334)

(because the z components add, the only com-posite states that contribute are those for which

m1 + m2 = m). The triplet combination andthe singlet are special cases of this general form,with s1 = s2 = 1/2 I used the notation = | 12 12 ,= |12 (12 ). The constants Cs1s2sm1m2m are calledClebsch-Gordan coefficients.

Chapter 5: Identical Particles

Two-Particle System

The state of a two-particle system is a func-tion of the coordinates of particle one (r1), thecoordinates of particle two (r2), and the time

(r1, r2, t) (335)

Its time evolution is deterimed (as always) bythe Schrodinger equation:

i

t= H (336)

Where H is the Hamiltonian for the wholesystem

H = 2

2m121

2

2m222 + V(r1, r2, t) (337)

The statistical interpretation carries over inthe obvious way

|(r1, r2, t)|2d3r1d3r2 (338)is the probability of finding particle 1 in the

volume d3r1 and particle 2 in the volume d3r2;

evidently must be normalized in such a waythat

|(r1, r2, t)|2d3r1d3r2 = 1 (339)

For time-independent potentials, we obtain acomplete set of solutions by seperation of vari-ables

(r1, r2, t) = (r1, r2)eiEt/ (340)

where the spatial wave function () satisfiesthe time-independent Schrodinger equation

2

2m121

2

2m222 + V = E (341)

and E is the total energy of the system

Bosons and Fermions

Suposse particle 1 is in the (one-particle) statea(r), and particle 2 is in the state b(r). (ignor-ing spin at the moment). In that case (r1, r2)is a simple product

(r1, r2) = a(r)b(r) (342)

Quantum mechanics neatly accomidates theexistance of particles that are indistinguishablein principle: We simply construct a wave func-tion that is non committal as to which particleis in which state. there are actually two ways todo it

(r1, r2) = A[a(r)b(r) b(r)a(r)] (343)Thus the theory admits to kinds of identical

particles: bosons, for which we use the plussign , and fermions, for which we use the minussign. Photons and mesons are bosons; protonsand electrons are fermions. It so happebns that

all particles with integerspin are bosons, and

all particles with half-integer spin are fermions

It follows, in paticular, that two identicalfermions cannot occupy the same state. For ifa = b, then

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(r1, r2) = A[a(r)a(r) a(r)a(r)] = 0(344)

This is the famous Pauli exclusion princi-ple. I assumed, for the sake of argument, thatone particle was in the state a and the other instate b, but there is a more general way to for-mulate the problem. Let us define the exchangeoperator, P, which interchanges the two parti-cles

P f(r1, r2) = f(r1, r2) (345)

Clearly P2 = 1, and it follows that the eigen-values ofP are 1. Now, if the two particles areidentical, the Hamiltonian must treat them the

same: m1 = m2 and V(r1, r2) = (r2, r1) . It fol-lows that P and H are compatible observables,

[P, H] = 0 (346)

and hence we can find a complete set of func-tions that are simultaneously eigenstates of both.That is to say, we can find solutions to theSchrodinger equation that are either symmetric(eigenvalue + 1) or antisymmetric (eigenvalue -1) under exchange

(r1, r2) =

(r1, r2) (347)

The symmetrization requirement statesthat, for identical particles, the wave function isnot merely alllowed, but required to satisfy equa-tion 347. This is the general statement, of whichequation 343 is a special case.

Exchange Forces

To give some sense of what the symmetriza-tion requirement does, we can suppose that oneparticle is in state a(x), and the other is instate b(x), and these two states are orthogonaland normalized. If the two particles are distin-

guishable, and number 1 is in state a, then thecombined wave function is

(x1, x2) = a(x1)b(x2) (348)

if they are identical bosons, the compositewave function is

+(x1, x2) =1

2[a(x1)b(x2) + b(x1)a(x2)]

(349)

and if they are identical fermions

(x1, x2) =1

2[a(x1)b(x2) b(x1)a(x2)]

(350)

Lets calculate the expectation value of thesquare of the separation distance between thetwo particles

(x1 x2)2 = x21 + x22 2x1x2 (351)

Case 1: Distinguishable particles. Forthe wave function in equation 348

x21 =

x21|a(x1)|2dx1

|b(x2)|2dx2 = x2a

(352)and

x22 =

|a(x1)|2dx1

x22|b(x2)|2dx2 = x2b(353)

and

x1x2 =

x1|a(x1)|2dx1

x2|b(x2)|2dx2 = xaxb(354)

in this case then

(x1 x2)2d = x2a + x2b 2xaxb (355)

Case 2: Identical particles. For the wavefunction in equations 349 and 350

x21 =1

2(

x21|a(x1)|2dx1

|b(x2)|2dx2

+

x21|b(x1)|2dx1

|a(x2)|2dx2

x21a(x1)b(x1)dx1 b(x2)a(x2)dx2

x21b(x1)a(x1)dx1

a(x2)

b(x2)dx2)

=1

2

x2a + x2b 0 0 = 12

x2a + x2b

(356)

similarly

x22 =1

2 x2b + x2a (357)

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(naturally, x22 = x21, since you cant tellthem apart.) But

x1x2 =1

2 (

x1|a(x1)|2dx1 x2|b(x2)|2dx2+

x1|b(x1)|2dx1

x2|a(x2)|2dx2

x1a(x1)b(x1)dx1

x2b(x2)

a(x2)dx2

x1b(x1)a(x1)dx1

x2a(x2)

b(x2)dx2)

=1

2(xaxb + xbxa xabxba xbaxab)

= xaxb |xab|2 (358)

where

psia(x1)|2dx1

x22|b(x2)|2dx2 = x2b (359)

and

x1x2 =

x1|a(x1)|2dx1

x2|b(x2)|2dx2 = xaxb(360)

in this case then

(x1 x2)2d = x2a + x2b 2xaxb (361)

Case 2: Identical particles. For the wavefunction in equations 349 and 350

x21 =1

2(

x21|a(x1)|2dx1

|b(x2)|2dx2

+

x21|b(x1)|2dx1

|a(x2)|2dx2

x21a(x1)b(x1)dx1 b(x2)a(x2)dx2

x21b(x1)a(x1)dx1

a(x2)

b(x2)dx2)

=1

2

x2a + x2b 0 0 = 12

x2a + x2b

(362)

similarly

x22 =1

2 x2b + x2a (363)

(naturally, x22 = x21, since you cant tellthem apart.) But

x1x2 =1

2 (

x1|a(x1)|2dx1 x2|b(x2)|2dx2+

x1|b(x1)|2dx1

x2|a(x2)|2dx2

x1a(x1)b(x1)dx1

x2b(x2)

a(x2)dx2

x1b(x1)a(x1)dx1

x2a(x2)

b(x2)dx2)

=1

2(xaxb + xbxa xabxba xbaxab)

= xaxb |xab|2 (364)

where

xab

xa(x)b(x)dx (365)

Evidently

(x1x2)2 = x2a+x2b2xaxb2|xab|2(366)

comparing equations 355 and 360, we see thatthe difference resides in the final term

(x)2 = (x)2d 2|xab|2 (367)

Atoms

A neutral atom, of atomic number Z, consistsof a heavy nucleus, with electric charge Ze, sor-rounded by Z electrons (mass m and charge e).The Hamiltonian for this system is

H =

Zj=1

2

2m2j

1

40

Ze2

rj

+1

2

1

40

Z

j=k

e2

|rj rk| (368)

The term in the curly brackets represents thekinetic plus potential of the jth electron, in theelectric field of the nucleus; the secon sum (whichruns over all values ofj and k exceptj = k) is thepotential associated with the multual repulsionof the electrons. (the factor of 1/2 in front cor-responds that the summation counts each pairtwice. The problem is to solve Schrodingersequation

21


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H = E (369)

for the wave function (r1, r2,...rZ). Becauseelectrons are identical fermions, however, not all

solutions are acceptable;only those for which thecomplete state (position and spin)

(r1, r2,...rZ)(s1, s2,...sZ) (370)

is antisymmetric with respect to interchangeof any two electrons. In particular, no two elec-trons can occupy the same state

Helium

After hydrogen, the simplest atom is helium(Z = 2). The Hamiltonian,

H =

2

2m21

1

40

2e2

r1

+

2

2m22

1

40

2e2

r2

+1

40

e2

|r1 r2| (371)

consists of two hydrogenicHamiltonians (withnuclear charge 2e), one for each electron 1 and 2,together with the final term describing the repul-sion of the two electrons. It is this last term that

causes all the the trouble. If we simply ignore it,the Schrodinger equation sperates, and the so-lutions can be written as products of hydrogenwave functions

(r1, r2) = nlm(r1)nlm(r2) (372)

only with half the Bohr radius and four timesthe Bohr energies. The total energy would be

E = 4(En + En) (373)

where En = 13.6/n2 eV. In particular, theground state would be

0(r1, r2) = 100(r1)100(r2) =8

a3e2(r1+r2)/a

(374)

and its energy would be

E0 = 8(13.6 eV) = 109 eV (375)

Solids

The Free Electron Gas

Suppose the object in question is a rectangu-lar solid, with dimensions lx, ly, lz, and imaginethat an electron inside experiences no force at

all, except at the impenetrable walls

V(x,y,z) =

0, if0 < x < lx, 0 < y < ly,and0 < z < lz

, otherwiseThe Schrodinger equation,

2

2m2 = E (376)

separates, in cartesian coordinates: (x,y,z) =X(x)Y(y)Z(z), with

2

2m

dX2

dx2= ExX (377)

2

2m

dY2

dy2= EyY (378)

2

2m

dZ2

dz2= EzZ (379)

and E = Ex + Ey + Ez. Letting

kx

2mEx

, ky

2mEy

, kz

2mEz

(380)

we obtain the general solutions

X(x) = Ax sin(kxx) + Bx cos(kxx) (381)

Y(y) = Ay sin(kyy) + Bx cos(kyy) (382)

Z(z) = Az sin(kzz) + Bx cos(kzz) (383)

(384)

The boundery conditions require X(0) =Y(0) = Z(0) = 0, so Bx = By = B + z = 0,and X(lx) = Y(ly) = Z(lz) = 0, so that

kxlx = nx, kyly = ny, kzlz = nz (385)

where each n is a positive integer

nx = 1, 2, 3, .., ny = 1, 2, 3, .., nz = 1, 2, 3, ..(386)

The normalized wave functions are

nxnynz =

8

lxlylzsin

nx

lxx

sin

ny

lyy

sin

nz

lzz

(387)

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xab

xa(x)b(x)dx (388)

Evidently

(x1x2)2 = x2a+x2b2xaxb2|xab|2(389)

comparing equations 355 and 360, we see thatthe difference resides in the final term

(x)2 = (x)2d 2|xab|2 (390)

Atoms

A neutral atom, of atomic number Z, consists

of a heavy nucleus, with electric charge Ze, sor-rounded by Z electrons (mass m and charge e).The Hamiltonian for this system is

H =

Zj=1

2

2m2j

1

40

Ze2

rj

+1

2

1

40

Zj=k

e2

|rj rk| (391)

The term in the curly brackets represents thekinetic plus potential of the jth electron, in theelectric field of the nucleus; the secon sum (whichruns over all values ofj and k exceptj = k) is thepotential associated with the multual repulsionof the electrons. (the factor of 1/2 in front cor-responds that the summation counts each pairtwice. The problem is to solve Schrodingersequation

H = E (392)

for the wave function (r1, r2,...rZ). Because

electrons are identical fermions, however, not allsolutions are acceptable;only those for which thecomplete state (position and spin)

(r1, r2,...rZ)(s1, s2,...sZ) (393)

is antisymmetric with respect to interchangeof any two electrons. In particular, no two elec-trons can occupy the same state

Helium

After hydrogen, the simplest atom is helium(Z = 2). The Hamiltonian,

H =

2

2m21

1

40

2e2

r1

+ 2

2m22 140

2e2

r2

+

1

40

e2

|r1 r2| (394)

consists of two hydrogenicHamiltonians (withnuclear charge 2e), one for each electron 1 and 2,together with the final term describing the repul-sion of the two electrons. It is this last term thatcauses all the the trouble. If we simply ignore it,the Schrodinger equation sperates, and the so-lutions can be written as products of hydrogen

wave functions

(r1, r2) = nlm(r1)nlm(r2) (395)

only with half the Bohr radius and four timesthe Bohr energies. The total energy would be

E = 4(En + En) (396)

where En = 13.6/n2 eV. In particular, theground state would be

0(r1, r2) = 100(r1)100(r2) =8

a3e2(r1+r2)/a

(397)

and its energy would be

E0 = 8(13.6 eV) = 109 eV (398)

Solids

The Free Electron Gas

Suppose the object in question is a rectangu-

lar solid, with dimensions lx, ly, lz, and imaginethat an electron inside experiences no force atall, except at the impenetrable walls

V(x,y,z) =

0, if 0 < x < lx, 0 < y < ly, and 0 < z < lz

, otherwiseThe Schrodinger equation,

2

2m2 = E (399)

separates, in cartesian coordinates: (x,y,z) =X(x)Y(y)Z(z), with

23


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2

2m

dX2

dx2= ExX (400)

2

2m

dY2

dy2 = EyY (401)

2

2m

dZ2

dz2= EzZ (402)

and E = Ex + Ey + Ez. Letting

kx

2mEx

, ky

2mEy

, kz

2mEz

(403)

we obtain the general solutions

X(x) = Ax sin(kxx) + Bx cos(kxx) (404)

Y(y) = Ay sin(kyy) + Bx cos(kyy) (405)

Z(z) = Az sin(kzz) + Bx cos(kzz) (406)

The boundery conditions require X(0) =Y(0) = Z(0) = 0, so Bx = By = B + z = 0,and X(lx) = Y(ly) = Z(lz) = 0, so that

kxlx = nx, kyly = ny, kzlz = nz (407)

where each n is a positive integer

nx = 1, 2, 3, .., ny = 1, 2, 3, .., nz = 1, 2, 3, ..(408)

The normalized wave functions are

nxnynz =

8

lxlylzsin

nx

lxx

sin

ny

lyy

sin

nz

lzz

(409)

and the allowed energies are

Enxnynz =

22

2m

n2xl2x

+n2xl2x

+n2xl2x

=

2k2

2m(410)

where k is the magnitude of the wave vector,k (kx, ky, kz).

If you imagine a three-dimensional space,with axes kx, ky, kz, and planes drawn in at kx =(/lx), (2/lx), (3/lx),..., at ky = (/ly), (2/ly), (3/ly),...,and at kz = (/lz), (2/lz), (3/lz),.... Eachblock in this grid, and hence each state, occu-pies a volume

3

lxlylz=

3

V(411)

of k-space, where V = lxlylz is the volume of

the object itself. Suppose the sample contains Natoms , and each contribute q electrons. Theywill fill up one octant of a sphere in k-space,whose radius ,kF is determined by the fact thateach pair of electrons require a volume 3/V

1

8

4

3k3F

=

N q

2

3

V

(412)

thus

kF = (32)1/3 (413)

where

N qV

(414)

is the free electron density ( the number offree electrons per unti volume)

The boundery seperating occupied and unoc-cupied states, in k-space, is called the Fermisurface. The corresponding energy is called theFermi energy, EF; for a free electron gas,

EF =

2

2m(32)2/3 (415)

The total energy of the electron gas can becalculated as follows; Each of these states carriesan energy 2k2/2m, so the energy of the shell is

dE =

2k2

2m

V

2k2dk (416)

and hence the total energy is

Etot =

2V

2m2

kF0

k4dk =

2k5FV

102m

=

2(32N q)5/3

102m V2/3

(417)

This quantum mechanical energy plays a rolerather analogous to the internal thermal energy(U) of an ordinary gas. In particular, it exerts apressure on the walls, for if the box expands byan amount dV, the total energy decreases

dEtot = 23

2(32N q)5/3

102mV5/3dV

= 23

EtotdV

V(418)

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and this shows up as work done on the out-side (dW = P dV) by the quantum pressure P.Evidently

P =2

3

EtotV

=2

3

2k5F102m

=(32)2/32

5m5/3

(419)

this is sometimes called degeneracy pres-sure strictly due to quantum effects.

better equation sheet qm

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