137B Formulas Solutions (1)

7/28/2019 137B Formulas Solutions (1)

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Physics 137B

Served by Roger Griffith

Nutritional Facts:

Serving size: 1 Semester (16 weeks)

Servings per container: many problems and solutions

Chapter1: Time-Independent Pertubation Theory

1.1 Non-Degenerate Pertubation Theory

Pertubation theory is a systematic procedure for obtaining approximate solutions to the perturbed prob-

lem, by building on the known exact solutions to the unperturbed case.

When a pertubation is applied to a system the new Hamiltonian is given as

H = H0 +H

where H0 is the unperturbed Hamiltonian and H is the pertubed Hamiltonian. The new wave functionis given by

|n = |0n + |1n + 2|2n + ...and the eigenvalues are given as

En = E0 + E1 + 2E2 + ...

and we find that the first-oder perubation to the energy is given as

E

1

n = 0

n|H|0

n (1)This is the fundemental result of first-order pertubation theory; as a practical matter, it may well be the

most important equation in quantum mechanics. It says that the first-order correction to the energy is the

expectation value of the pertubation, in the unperturbed state.

Now, the unperturbed wave functions constitute a complete set, so 1n (like any other function) can beexpressed as a linear combination of them

1n = m=n

c(n)m

0m (2)

and the coefficients are given by

c(n)m =

0m|H|0nE0l E0n

(3)

the wave functions are given by

1n = m=n

0m|H|0nE0m E0n

0m (4)

The perturbation theory often yields suprisingly accurate energies but the wave functions are notori-

ously poor.

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Second-Order Energies

Proceeding as before, we take the inner product of the second order equation with 0n

0n|H02n+ 0n|H1n = E0n0n|2n +E1n0n|1n +E2n0n|0n (5)

and the second-order pertubation to the energies is given by

E2n = m=n

|0m|H|0n|2E0n E0m

(6)

1.2 Degenerate Perubation Theory

Two-Fold Degeneracy

Suppose that we have a degenerate system, i.e

E0k = E0n = E

0

We must diaganolize the secular equationE0 +Waa Wab

Wba E0 +Wbb

where

Wi j 0i |H|0j, (i, j= a,b) (7)and setting the determinate equal to 0 we find a quadratic equation of the form

E1 =1

2 Waa +Wbb (Waa Wbb)2 + 4|Wab|2 (8)This is the fundemental result of degenerate pertubation theory; the two roots correspond to the two

perturbed energies.

Problems

Problem # 1

The unperturbed Hamiltonian of a 1-D simple harmonic oscillator is H0 =p2x

2m+ kx

2

2 .The energy levels

are E = (n + 1/2), where = (k/m)1/2 and n = 0,1,2,3, .. . A perubation H = ax4is nowturned on.

a) Calculate E10 and E11 , the first order corrections to the n = 0 and n = 1 energy states.

We know that the ground state and the first excited state wave functions for the simple harmonic

oscillator are given by

10 =m

1/4e

m2 x

2

and 01 =m

1/42m

1/2xe

m2 x

2

the first order pertubation is defined as

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E1n = n|H|n where H = ax4

therefore

E10 =

0|ax4|0

= 2am

1/2

Z

0

x4em

x2 dx

=24

32a

m

1/2

m

5/2=

3

4a

m

2

and now for the first order correction to the first excited state is

E11 = 1|ax4|1 = 2am

1/22m

Z0

x6em

x2 dx

=15

4m

m

1/2

m

7/2

=15

4

m

2

(b). Calculate E20 , the second order correction to the ground state energy.

First, we know that the second order correction to the ground state energy is given by

E2n = m=n

|n|ax4|m|2E0n E0m

and we know

x =

2m(a + a) x4 =

2

4m22 (a + a)4

so we nee to find

E2n = a2

16

m

41

m=0

|0|(a + a)4|m|2m

now we need to foil the ladder operators, which I will not present here, but we find

0|(a + a)4|1 = 00|(a + a)4|2 = [

3

3

2

1 +

2

2

2 +

2

1

1

1]

0|(a + a)4|3

= 0

0|(a + a)4|4 = 26

so we find that the second order correction to the ground state energy is

E20 = a2

16

m

41

(6

2)2

2+

(2

6)2

4

= 21

8a2

m

4

Problem # 2

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Two identical bosons are placed in an infinite square well. They interact weakly with one another, via

the potential

V(x1,x2) = aV0(x1 x2)where V0 is a constant with the dimensions of energy and a is the width of the well.

(a). First, ignoring the interaction between the particles, find the ground state and first excited state-

both the wave functions and the associated energies.

The wave functions for a particle in a infinite square well with width a is given by

n =

2

asinnx

a

we also know that we can write the combined wave function as a product of two wave functions, i.e

01(x1,x2) =1(x1)1(x2) =2

asinx1

a

sinx2

a

and the energies are given by

E01 = E1 +E2 = 2E1 = 22

ma2

The wave functions and associated energies for the first excited state are given by

02 =

2

a

sinx1

a

sin

2x2

a

+ sin

2x1

a

sinx2

a

and the energies are

E02 = E1 +E2 = (n21 + n

22)

22

2ma2=

5

2

22

ma2

(b). Use first-order pertubation theory to calculate the effect of the particle-particle interaction on theground state and first excited state energies.

We know that

E01 = 01|H|01= 4

aV0

Za0

Za0

sin2x1

a

sin2

x2a

(x1 x2)dx1dx2

= 4a

V0

Za0

sin4x

a

dx

if we let

u = xa

dx = adu

so we find

E01 = 4

V0

Z0

sin4(u)du

= 4

V0

1

32(12u)

0

E01 = 3

2V0

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and for the first excited state

E02 = 02|H |02

= 2a

V0

Za0

Za0

sinx1

a

sin

2x2

a

+ sin

2x1

a

sinx2

a

2(x1 x2)dx1dx2

= 8

aV0Z

a

0 sin

2xa

sin

22xa

dx

if we let

u =x

adx =

a

du

so we find

E012 = 8

V0

Z0

sin2(u) sin2(2u)du

= 8

V0

4 u

16

0

E

0

1 = 2V0Problem # 3

A particle of mass m moves in a square well potential

V(x) =

0 |x|< a2 |x|> a2

A delta function of height V0( ) and width x( 0) is introduced at x = 0 (thus the area under thepeak is A = V0x)

(a). Show that this pertubation has no effect on the odd-parity states.

We know that the total (even and odd) wave functions for the infinite square well are given by

n =

2a

sin

nxa

odd wavefunction

2a

cos

nxa

even wave function

and we know that the first-order pertubation to the odd wave functions is

E1odd = n|H |m

=

2

aAZ

a/2

a/2 sin2nx

a

(x)dx

=2

aA sin2(0) = 0

thus this pertubation has no effect on the odd-parity states.

(b). Find general formulas for the energy shift and the coefficients anm for the even-parity states.

We know that

E0n E0m = (n2 + m2)22

2ma2

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and the coefficients can be found using

anm =m|H |n

E0n E0mand

m|A(x)|n = 2a

AZ

a/2

0cos

mx

a

cos

nx

a

(x)dx

=2

aA cos2(0)

=2

aA

so the coefficient is

anm =4ma

22A

(n2 m2)and the energy shift is

E1n =2

aA

Za/2a/2

cos2

nx

a

(x)dx =

2

aA

Problem # 4

An electron of charge e moves in a one-dimensional harmonic oscillator potential; the unperturbedHamiltonian is

H =p2

2m+

1

2m2x2

A weak, uniform electric field E is applied in the positive x direction. Thus, the potential energy due

to the electric field is eEx.

(a). Write down the pertubation Hamiltonian H in terms ofa,a

and other quantities.We know that

H =p2

2m+

1

2m2x2 H = eEx x =

2m(a + a)

and thus

H =

2meE(a + a)

(b). Find the first-order correction to the ground state energyusing time-independent pertubation the-

ory.

The ground state wave function for the harmonic oscillator is

0 =m

1/4e

m2 x

2

thus the first order correction is

E10 = 0|H|0 =

2meE0|(a + a)|0 = 0

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(c). Find the second-order correction to the ground state energy using time-independent pertubation

theory

E2n = m=n

|m|H|n|2E0n E0m

E0n E0m = (n m)

letting n = 0 we get

E20 =

(eE)2

2m2 m=n|

m|(a + a)|0|2

m

the only allowed transition is where m = 1 and so we find

|1|(a + a)|0|2 = [

1]2 = 1

and so we find

E20 = (eE)2

2m2

Problem # 5

Consider a particle confined to a 2-D infinite square well defined by 0

x

L and 0

y

L. Write

down the eigenfunctions np (x,y) (where n and p are positive integers), and the eigenvalues in terms ofthe ground state energym E1.

The infinite square potential is

V(x,y) =

0 if 0 x L, 0 y L otherwise

the wave function is n =

2

Lsinnx

L

the eigenfunctins and the eigenvalues in terms of the ground state energy are

0np =2

L

sinnxLsinpy

L and E0np =

2

2

2mL2

(n2 + p2) =E1

2

(n2 + p2)

For the case where n = p , the eigenfunctions are usually two-fold degenerate. Calculate the change inthe enrgies under the pertubation H = E1 sin(x/L), where 1.

If we define our wavefunctions as

a = 0np =

2

Lsin

nx

L

sin

py

L

b =

0pn =

2

Lsinpx

L

sinny

L

ifWi j

0i |H

|0j

then

Waa = 0np |H|0np Wbb = 0pn|H|0pn Wab = 0np |H |0pnfor Waa we find

Waa = np|H |np = 4L2

E1

ZL0

sinx

L

sin2

nxL

dx

ZL0

sin2py

L

dy

using mathematica yields

Waa = E18

n2

4n2 1

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for Wbb we find

Wbb = pn|H |pn = 4L2

E1

ZL0

sinx

L

sin2

pxL

dx

ZL0

sin2ny

L

dy


Wbb

= E18

p2

4p2 1for Wab we find

Wbb = np|H|pn = 4L2

E1

ZL0

sinx

L

sinnx

L

sinpx

L

dx

ZL0

sinny

L

sinpy

L

dy


Wab = 0

to find the energies we use

E1 =1

2[Waa +Wbb (Waa Wbb)]

E1+ = E18

n2

4n2 1

E1 = E18

p2

4p2 1

Problem # 6

A two-dimensional isotropic oscillator has the Hamiltonian

H =

2

2m

2

x2

+2

y2+ k

2

(1 + bxy)(x2 +y2)

(a). If b=0, write down the energies of the three lowest levels, stating the degeneracy in each case.

The energies of the 2-D harmonic oscillator are given by

E0 = non-degenerate

E1 = 2 double-degenerate

E2 = 3 triply-degenerate

(b). If b is a small positive number such that b

1, find the first-order pertubation corrections to the

energies of the ground state and first excited state. Think raising and lowering operators

We know that the perturbing Hamiltonian is

H =

k

2bxy(x2 +y2)

and because the oscillator is isotropic we know that

x =

2m(ax + ax) y =

2m(ay + ay)

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and the perturbing Hamiltonian written in terms of raising and lowering operators is

H =

kb2

8m22[(ax + ax)(a

y + ay)[(a

x + ax)

2 + (ay + ay)2]]

this is a very tedious task to foil this equation, so I will skip this step. If you do not believe this solution

do it yourself. The first-order correction to the ground state energy is

E100 = 00|H|00 = 0

the first excited state is doubly degenerate

1a =01 and 1b =10

and by inspection we know that

Waa = 01|H|01 = Wbb = 10|H|10 = 0

and finally

Wab =kb2

8m2201|axay(ay)2 + axayaxax + axayax ax + axay ayay|10

looking at each case seperately we find

01|axay(ay)2|10 =

1

2

2

1 = 2

01|axay axax|10 =

2

2

1

1 = 2

01|axay axax|10 =

1

1

1

1 = 1

01|axay ayay|10 = 1111 = 1thus

Wab =3kb2

4m22= Wba

and so

Waa = Wbb = 0 Wab = Wba =3kb2

4m22

and to find the energies we use

E1 =1

2

Waa +Wbb

(Waa Wbb)2 + 4|Wab|2therefore

E1+ = +Wab =3kb2

4m22E1 = Wab =

3kb2

4m22E100 = 0

Problem # 7

For the harmonic oscillator [V(x) = (1/2)kx2], the allowed energies are

En = (n + 1/2), (n = 0,1,2,...)

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where =

k/m is the classical frequency. Now suppose the spring constant increases slightly:k (1 + )k. (Perhaps we cool the spring, so it becomes less flexible.)

(a). Find the exact new energies (trivial in this case). Expand your formula as a power series in , upto second order.

(b). Now calculate the first-order correction pertubation energy. What is H here? Comapre yourresults with part (a). Hint : It is not necessary-in fact, it is not permitted - to calculate a single integral in

doing this problem.

Problem # 8

Supose we put a delta-function bump in the center of the infinite square well:

H = (x a/2)

where is a constant.

(a). Find the first-order correction to the allowed energies. Explain why the energies are not perturbed

for even n.

(b). Find the second-order correction to the energies (E2

n ). You can sum the series explicitly, obtaining2m(/n)2for odd n.Problem # 9

Consider a quantum system described by the Hamiltonian

H = H0 +H

where H0 is the two-dimensional harmonic oscillator Hamiltonian

H0 =1

2m(p2x + p

2y) +

1

2m2(x2 +y2)

In parts (a) and (b) of the problem, the perturbing Hamiltonian H is given by

H = K pxpy

with K a constant.

(a). Evaluate the ground state energy to second order in K.

(b). Evaluate the energy of the state(s) whose unperturbed energy is 2 to first order in K.

Chapter 2: The Variational Method

2.1 Applying the Variational Principle

Suppose you want to calculate the ground state energy,Egs, for a system described by the Hamiltonian H,

but you are unable to. The variational principle will get you an upper bound for Egs, which is sometimes

all you need. Heres how it works: Pick any normalized function whatsoever; I claim that

Egs |H| H (9)To use the variational method you must follow these instructions

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You must guess a trial wave function for the ground state

(r,)

If the wave function is not already normalized, you must normalize it

1 =n|n

You must calculate the expectation ofH

H = |H|

once you have calculated the expectation values you must minimize it, i.e take the derivative withrespect to and solve for min

dHd

= 0

Once you have found min you must use it to find (r)

Then we must use min to find the upper bound on the ground state energy

E0 Hmin

Problems

Problem # 1

A particle of mass m is contained in the one-dimensional potential well

V(x) = |x|

Use the (normalized) gaussian

(x) =

2

1/4ex

2

as a trial function to find the upper bound on the ground state energy.

we know that

H = T + V and also d2

dx2(ex

2

) = ex2

(42x2 2)We will also be using the integral defined as

Z0

x2nex2/a2 dx =

(2n)!

n!

a2

2n+1(10)

to solve this problem. To find the expectation of the potential we use

V =

2

1/22

Z0

xe2x2

dx = 2

2

1/2

4=

2

and for the expectation of the kinetic energy term we use

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T = 2

m

2

1/2Z0

e2x2

(42x2 2)dx = 2

m

2

1/2

2

12

3

2

=

2

2m

thus

H = 2

2m+

2

anddH

d=

2

2m

2(2)3/2= 0 (11)

solving for gives us

=

2m2

42

1/3pluggin this into equation 2 and doing a bit of algebra we find that the upper bound to the ground state

energy is

Hmin = 32

22

2m

1/3

Problem # 2

Obtain an upper bound on the ground state energy of the hydrogen atom, using the variational tech-

nique with the wave function for the ground state of the three-dimensional oscillator as the trial function:

=2a

3/4

ear2

Obtain a numerical estimate (in eV) for the energy.

The Hamiltonian in 3-D is given by

H = 2

2m2 +V

where

2 =1

r2

r

r2

r

+

1

r2 sin

sin

+

1

r2 sin2

2

2

we know that the potential energy for the hydrogen atom is defined as

V(r) = e2

40r

and also

H = T+ V and 1r2

r

r2

r

(ear

2

) = ear2

(4a2r2 6a)

to find the expectation of the potential we use

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V =

2a

3/2e2

40

Z20

d

Z0

sin()dZ

0re2ar

2

dr =

2a

3/2e2

0

Z0

re2ar2

dr

and using the integral in equation 1 we find

V =

2a

3/2 e2

0

1

4a= e2

0

a

23

1/2


T = 2

m

2a

3/22

Z0

r2e2ar2

(4a2r2 6a)dr

= 2

m(4a)

2a

1/2 Z0

r2e2ar2

(4a2r2 6a)dr

= 2

m(4a)

2a

1/2 4a2 12 12

2a

5 12a 12

2a

3

=32a

2m

therefore

H = 32a

2m e

2

0

a23

1/2and

dHda

=32

2m e

2

20

1

23

1/2

1

a1/2= 0 (12)

and solving for a we find

a =m2e4

182043

plugging this into equation 2 we get

Hmin = 3e4m

363220 e

4m

63220=

e4m

3220

3

36 1

6

= 1

12

e4m

3220

therefore the upper bound to the ground state energy of the hydrogen atom is

H

min =

8

3 m

22 e

2

40

2

= 83

E1 =

11.54 eV

Problem # 3

A particle of mass m is bound by the three-dimensional potential V(r) = V0er/a where 2/mV0a2 =3/4. Use the variational method with the trial function er to estimate an upper bound on the lowestenergy eigenvalue.

Note: You should find a quartic equation in (a). You should show (by plugging in) that the real rootsare a = 0.5 and 4.369 102.

we know

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V(r) = V0er/a (r) = er 2

V0a2m=

3

4 V0 = 4

3

2

ma2

and we will also be using the integral defined as

Z0

xnex/adx = n!an+1 (13)

to solve this problem. To find the expectation of the potential we use

H = T + V and 1r2

r

r2

r

(er) = er(2 2

r)

first we need to normalize the test function

A2Z2

0d

Z0

sin()dZ

0r2e2rdr = 1

A

2

8 1

23

= 1

A =

3

1/2

thus

(r) =

3

1/2er

to find the expectation of the potential we use

V = V0 3

4

Z

0r2er(2+

1a

)dr = V08

a

2a + 1

3


T = 2

2m

3

4

Z0

e2r(2r2 2r)dr = 23

m2

1

4 1

2

=

22

2m

therefore, we are also plugging in for v0 we get

H = 22

2m8V0

a

2a + 13

=

22

2m 32

2

3ma2 a

2a + 13

taking the derivative we find

dHd

=

2

m 32

2

ma2

a

2a + 1

2(2a + 1)a 22a2(2a + 1)2

= 0

we find a quartic equation of the form

(2a + 1)4 = 32a

and this equation satisfies both roots that are given, i.e a = 0.5 and a = 4.369 102

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Using a = 0.5 we find the upper bound to the ground state energy as

Hmin = 22

m

1

2 32a

3(2a + 1)3

letting = 0.5a and a = 0.5 we get

Hmin = 224ma2

if we use the other root a = 4.369 102 we find

Hmin = 2.65 104 2

ma2

which is not an upper bound on the energy. So we use a = 0.5 to find our upperbound for the energy.Problem # 4

Consider a particle with mass m moving in the one-dimensional potential

V(x) = x4

(a). Consider a single-paramter ansatz for the wave function consisting of ground state wave function

for a simple harmonic oscillator

with frequency , where is the variational parameter. Find the value of that minimizes H andobtain an upper bound on the energy of the ground state energy.

(b). Write down a one=paramter ansatz that you could use, with the variational principle, to obtain an

upper bound on the energy of the first-excited state. Explain the reasoning behind your choice of ansatz,

but do not go further with the calculation than writing down an ansatz.

Problem # 5Suppose youre given a quantum system whose Hamiltonian H0 admits two eigenstates, a (with

energy Ea), and b (with energy Eb). They are orthogonal, normalized, and non-degenerate (assume Ea isthe smaller of the two energies). Now turn on a pertubation H , with the following matrix elements:

a|H|a = b|H|b = 0 a|H|b = b|H|a = h

where h is some specified constant.

(a). Find the exact eigenvalues of the perturbed Hamiltonian.

(b). Estimate the energies of the perturbed system using second-order pertubation theory.

(c). Estimate the ground state energy of the perturbed system using the variational principle, with atrial of the form

= (cos )a + (sin )b

where is the adjustable paramter. Note : Writing the linear combiunation in this way is just a neatway to guarantee that is normalized.

(d). Compare your answers. Why is the variational principle so accurate, in this case?

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1.1 Electron Spin and Angular Momentum

[Lx,Ly] = iLz [Ly,Lz] = iLx [Lz,Lx] = iLy

it follows (as before) that the eigenvectors of L2 and Lz satisfy

L2|l m = 2l(l + 1)|l m Lz|l m = m|l m

and L|l m =

l(l + 1) m(m1)|l (m1)where L Lx iLz and

l = 0,1,2,3,... m = l,l + 1,..., l 1, lThe algebraic theory of spin is a carbon copy of the theory of orbital angular momentum, beggining

with the fundamental commutation relations:

[Sx,Sy] = iSz [Sy,Sz] = iSx [Sz,Sx] = iSy

it follows (as before) that the eigenvectors of S2 and Sz satisfy

S2

|s m = 2

s(s + 1)|s m Sz|s m = m|s mand

S|s m =

s(s + 1) m(m1)|s (m1)where S Sx iSz and

s = 0,1

2,1,

3

2,... m = s,s + 1,..., s 1,s

1.2 L-S Coupling

In light atoms (generally Z < 30), electron spin Si interact among themselves so they combine to form a

total angular momentum S. The same happens with orbital angular momentum li forming a single orbitalangular momentum L. The interaction between the quantum numbers L and S is called Russell-Saunders

coupling. Then L and S add together to form a total angular momentum J

J= |l + s| where l = 0 then J=1

2for fermions and mj = ml + ms

where l and s are

l = l1 + l2 + ... and s = s1 + s2 + ...

the eigenfunctions and eigenvalues are given as

J2 = jmj|J2|j,mj = j(j+ 1)2jjmjmjJz = jmj|Jz|j,mj = mjjjmjmjJ = jmj|J|j,mj =

j(j+ 1) mj(mj1)jjmjmj1

a couple of other useful relations is

Jy = 12i

(J+ J) Jx = 12

(J+ +J)

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1.3 Spin Orbit Energy

The pertubation Hamiltonian is given as

Hrel = p4

8m20c2

the spin orbit energy is given by

ESL =|En|

2

n(2l + 1)

1

l+1 if j= l + 1/2

1l

if j= l 1/2 where e2

40c 1

137is the fine structure constant

the Relativistic correction to the energy is given by

Erel = 54

2|En|

n

therefore the total energy is given as

ESL + Erel = 2

|En|n

3

4n 1

j+ 1/2

1.4 Hyperfine Interaction in H

The eigenfunctions ang eigenvalues are given as

F =I+ J whereI is the nuclear spin interaction

and

I2

= i(i + 1)2

Iz = mi |mi| iand the magnetic moment is

i =giN

I z = giNmiand we also know

F2 = f(f+ 1)2Fz = mf |mf| f

Problems

Problem # 1

The electron confiuguration of oxygen is 1s22s22p4 and the spin of its nucleus is 5/2

(a). If no electrons are excited out of their subshells, what are the possible spectral terms.

(b). What is the ground state of oxygen?

(c). For each possible overall atomic state (including both the electron states given by (a) and the

nucleus), how many energy levels are there due to the hyperfine splitting?

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(a). First we know that the 1s22s22p4 state is equivalent to 4 p electrons and we can treat this problem

as 2 p holes.

l1 = l2 = 1 s1 = s2 =1

2so

l = l1 + l2 =

2 l1 + l2

1

0 l1 l2s = s1 + s2 =

1 s1 + s2

0 s1 s2j = s + l

The first thing we need to know is, how many possible states are there? Since we know that these are

equivalent electrons we can put the first electron in any of the 6 states allowed in the P shell and the

second electron can only go into any of the 5 other states. thus the number of possible states is

6 5 = 30 possible statesthis is not total correct due to the Pauli Exclusion Principle and the fact that these are indistinguishable

particles, and we find that the total number of allowed states are given by

30

2!= 15 allowed states

we know

s1 = s2 =1

2l1 = l2 = 1

s = s1 s2 = 1 symmetric0 anti-symmetric

l = l1 l2 =

2 symmetric

1 anti-symmetric

0 symmetric

[Note: the maximum value of s, l have symmetric wave functions]

we want an overall anti-symmetric wave function, so this means that when the spin is symmetric we

can only have an angular momentum that is anti-symmetric etc. so

s = 1 l = 1 s = 0 l = 2,0

The over-all table is given as this, but we will only look at the allowed states

s l j 2s+1Lj mj = (2j + 1)0 0 0 1S0 1

1 0 1 3S1 3

1 1 2,1,0 3P2,1,0 9

0 1 1 1P1 3

1 2 3,2,1 3D3,2,1 15

0 2 2 1D2 5

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and the number of allowed states is given by

1S0,3P2,1,0.

1D2 total number of states is 15

(b). What is the ground state of oxygen?

To find the ground state, we must use Hunds rules.

1. Maximize s

2. Maximize l

3. if shell is > 1/2 full then j = l + s ,if shell is < 1/2 full then j = l s. We know that our shell ismore than half filled and we must use the former equation for j.

These are simple to apply now that we know what the allowed states are 1S0,3P2,1,0.

1D2

lmax = 1 smax = 1

a representation of this is given by

mlms 1 0 -1

+ 12

1

2

and so

jmax = l + s = 2so the ground state of oxygen is

3P2

(c). For each possible overall atomic state (including both the electron states given by (a) and the

nucleus), how many energy levels are there due to the hyperfine splitting?

First we know that

Eh f JI JI= 12

[F2 I2 J2] where F =I+S

and

JI = 12

[F2I2J2] = 2

2[f(f+ 1) i(i + 1) j(j+ 1)]

[Note: # ofEh f levels is the # of f values].

Now

i =5

2j=

2

1

0

f =

92, 7

2, 5

2, 3

272 ,

52 ,

32

52

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so when

j = 2 the hyperfine interaction breaks this into 5 levels

j = 1 the hyperfine interaction breaks this into 3 levels

j = 0 the hyperfine interaction does not split the levels

so in summary

1S0 has 1 hyperfine level3P0 has 1 hyperfine level3P1 has 3 hyperfine level3P2 has 5 hyperfine level1D2 has 5 hyperfine level

Problem # 2

A hydrogen atom in the n = 3 state(a). Construct a table displaying in columns all possible sets of values for the quantum numbers

l, ml, ms, j and mj(b). If we include the effects of the proton spin, what are the possible values of ( f,mf)?

(a). Our values are given as

l = n 1 =

2

1

0

for l = 2 ml =

2

1

0

12

for l = 1 ml =

1

0

1for l = 0 ml = 0 ms =

1

2s =

1

2

Since we are adding s = 12

to l, we have jvalues ofl 1/2 for l = 0 and j= 1/2 for l = 0. In all cases,the z-component is conserved: ms + ml = mj. In cases where mj = (l + s), only j= l + s is possible.

The table is given asl ml ms j mj2 2 1/2 5/2 5/2

2 2 -1/2 5/2,3/2 3/2

2 1 1/2 5/2,3/2 3/2

2 1 -1/2 5/2,3/2 1/2

2 0 1/2 5/2,3/2 1/2

2 0 -1/2 5/2,3/2 -1/2

2 -1 1/2 5/2,3/2 -1/2

2 -1 -1/2 5/2,3/2 -3/2

2 -2 1/2 5/2,3/2 -3/2

2 -2 -1/2 5/2 -5/2

1 1 1/2 3/2 3/2

1 1 -1/2 3/2,1/2 1/2

1 0 1/2 3/2,1/2 1/2

1 0 -1/2 3/2,1/2 -1/2

1 -1 1/2 3/2,1/2 -1/2

1 -1 -1/2 3/2 -3/2

0 0 1/2 1/2 1/2

0 0 -1/2 1/2 -1/2

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(b). If we include the effects of the proton spin, what are the possible values of ( f,mf)?The total angular momentum, F is the sum of the electrons total angular momentum Jand the protons

spin. Since sp = 1/2and jranges from 1/2 to 5/2, the possible values for f are 5/21/2=3,2; 3/21/2=2,1:and 1/21/2=1,0. The possible values of mf for a given f are f, f 1,...,f, so the possible values of(f,mf) are

f mf3 3,2,1,0,-1,-2,-3

2 2,1,0,-1,-2

1 1,0,-1

0 0

Problem # 3

Find all spectral terms , and idenfify the ground state configuration for the following unfilled subshells.

Check that the total number of states agrees with the expected degeneracy.

(a). nd1nd1 [with n = n](b). p1d2

(c). p4

(a). Here, the two electrons are inequivalent (different n), so there is no exclusion principle to worry

about. The possible spin states are S= 0,1, since we are adding two spin 1/2 particles, and the possibleorbital angular momentum states are L = 4,3,2,1,0. Any combination of these is possible, so we have.

s l j 2s+1Lj mj = (2j + 1)

0 0 0 1S0 1

0 1 1 1P1 3

0 2 2,1,0 1D2 5

0 3 3 1F3 7

0 4 4,2,1 1G4 9

1 0 1 3S1 3

1 1 2,1,0 3P2,1,0 9

1 2 3,2,1 3D3,2,1 15

1 3 4,3,2 1F4,3,2 21

1 4 5,4,3 3G5,4,3 27

If we sum all of the degenaries, we find a total of 100 states. Since there are 10 available states for

each electron, the expected number of states is 102 = 100, which agrees. Using Hunds rules, we maximizeS 1, L 4, and minimize J |L S| = 3to get a term of3G3.

(b). This case is similar in that we are again dealing with inequivalent electrons. Again the possible

spin states are S= 0,1, but the possible orbital angular momentum states are L = 3,2,1 since one of theelectrons now has l = 1. Constructing a similar table we have

s l j 2s+1Lj mj = (2j + 1)

0 1 1 1P1 3

0 2 2,1,0 1D2 5

0 3 3 1F3 7

1 1 2,1,0 3P2,1,0 9

1 2 3,2,1 3D3,2,1 15

1 3 4,3,2 1F4,3,2 21

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The total number of states is now 60, which agrees with our expectation of 6[p]x10[d]=60. For the

ground state, we maximize S 1, which then requires L 3, and since the subshell is less than half-full,minimize J 2, giving 3F2.

(c). We now have four electrons, but we can treat this as two holes. The possible spin configurations are

, as usual, a symmetric S= 1 triplet and an anti-symmetric S= 0 singlet. The possible angular momentumstates are L = 2,1,0. Which alternate symmetry: 2 and 0 are symmetric, 1 is anti-symmetric. Since

we are dealing with fermions, we must have over-all anti-symmetric wavefunctions, so we combine spinsymmetric states with anti-symmetric orbital angular momentum states and anti-symmetric spin states

with symmetric orbital angular momentum states.


equivalent electrons we can put the first electron in any of the 6 states allowed in the P shell and the


6 5 = 30 possible statesthis is not total correct due to the Pauli Exclusion Principle and the fact that these are indistinguishable

particles, and we find that the total number of allowed states are given by30


we know

s1 = s2 =1

2l1 = l2 = 1

s = s1 s2 =

1 symmetric

0 anti-symmetricl = l1 l2 =

2 symmetric

1 anti-symmetric

0 symmetric




s = 1 l = 1 s = 0 l = 2,0

The over-all table is given as this, but we will only look at the allowed states

s

l

j

2s+1Lj

mj

= (2j+ 1)

0 0 0 1S0 1

1 1 2,1,0 3P2,1,0 9

0 2 2 1D2 5

and the number of allowed states is given by

1S0,3P2,1,0.

1D2 total number of states is 15

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This gives us 15 states, which agrees with our expectations: six places to put the first electron, five

to put the second, and division by two to avoid double counting gives us 15 states. For the ground state,

we maximize S 1, which then requires L 1, and since the subshell is more than half-full, maximizeJ 2, giving 3P2.

Problem # 4

A beam of hydrogen atoms, emitted from an oven 400 K, passes through a Stern-Gerlach magnet oflength 1 m. The magnetic field gradient is 10 tesla/m.

Calculate the seperation of the two beams as they emerge from the magnet.

The energy of the electron is defined as

E =1

2mv2 = 2kBT

we also know that the potential and the force are defined as that the atom is experiencing are

U =

B and F =

U =B

therefore the force in the z direction is given by

Fz = zB

zwhere z = gs eSz

2m

we know that gs 2 for our case. We find that the force in the z direction is

Fz = e

2m

B

z

to find the z we have to use the classical kinematic equations.

Fz = meaz az =Fz

mz =e

2m2e

B

z

we also know that

z

2= z0 +V0,zt+

1

2azt

2 z0 = 0 V0,z = 0

therefore

z = azt2

we know that

az =e

2m2e

B

zand t =

me

4kbT

thus

z =e

me8kbT

B

z= 4.21 103m

Problem # 5

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For the hydrogen atom, the quantum number j can take only the values 1/2, 3/2, 5/2, ..., depending on

the value ofl.

Obtain the matrices J2,Jz,J+,J,Jy for l = 0,1

ifl = 0 then ml = 0 and mj = ml + ms = 12 . The number of matrix elements is described by 2j+ 1,

where j= l + s and s = 12 , therefore j=12 and the number of matrix elements for l = 0 is 2.

for J2

we find

j,mj|J2|j,mj = j(j+ 1)2jjmjmjtherefore we know that mj = (1/2,1/2) and mj = (1/2,1/2) , so we find

J2 =3

4

2

1 0

0 1

for Jz we find

j,

m

j|J

z|j,

mj =

mj

jj

mjmj

where mj = (1/2,1/2) and mj = (1/2,1/2), so we find

Jz =1

2

1 00 1

to find J+,J we need to use

j,mj|J|j,mj =j(j+ 1) mj(mj1)jjmjmj+1

to find J+ we use mj = 12 because we are using the raising operator. Using

j,mj|J|j,mj =

j(j+ 1) mj(mj+ 1)jjmjmj+1

where mj = (

1/2,1/2) and mj = (

1/2,1/2), we find

J+ =

0 1

0 0

and J is just the transpose ofJ+ which yields

J =

0 0

1 0

and finally Jy is found by using

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Jy =

1

2i

(J+ J) =

2i

0 1

0 0

0 0

1 0

therefore

Jy =

2 0 i

i 0

and also

Jx =

1

2

(J+ +J) =

2

0 1

0 0

+

0 0

1 0

therefore

Jx =

2

0 1

1 0

now ifl = 1 then ml = 1,0,1 and mj = ml + ms = 32 ,

12 ,

12 ,

32 . The number of matrix elements is

described by 2j+ 1, where j= l s and s = 12 , therefore j=32 ,

12

and the number of matrix elements for

l = 1 is 6., 2 for the case where j = 12 and 4 for the case where j=32 . We have already done the case

where j= 12 so we can just add these matrix to the 4x4 matrix.

for J2 we find

j,mj|J2|j,mj = j(j+ 1)2jjmjmjtherefore we know that mj= (1/2,1/2,3/2,1/2,1/2,3/2) and mj= (1/2,1/2,3/2,1/2,1/2,3

, so we find

J2 =1

4

2

3 0 0 0 0 0

0 3 0 0 0 0

0 0 15 0 0 0

0 0 0 15 0 0

0 0 0 0 15 0

0 0 0 0 0 15

for Jz we find

j,mj|Jz|j,mj = mjjjmjmjwhere

Jz =1

2

1 0 0 0 0 00 1 0 0 0 0

0 0 3 0 0 00 0 0 1 0 00 0 0 0 1 0

0 0 0 0 0 3

to find J+,J we need to use

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j,mj|J|j,mj =

j(j+ 1) mj(mj+ 1)jjmjmj+1where j= 1/2,3/2 and mj= (1/2,1/2,3/2,1/2,1/2,3/2) and mj = (1/2,1/2,3/2,1/2,1/2,

For J+ we find

J+ =

0 1 0 0 0 0

0 0 0 0 0 0

0 0 0

3 0 0

0 0 0 0 2 0

0 0 0 0 0

3

0 0 0 0 0 0

and J is just the transpose ofJ+ which yields

J =

0 0 0 0 0 0

1 0 0 0 0 0

0 0 0 0 0 00 0 3 0 0 00 0 0 2 0 0

0 0 0 0

3 0

and finally Jy is found by using

Jy =1

2i(J+ J) =

2i

0 1 0 0 0 0

0 0 0 0 0 0

0 0 0

3 0 0

0 0 0 0 2 0

0 0 0 0 0 30 0 0 0 0 0

0 0 0 0 0 0

1 0 0 0 0 0

0 0 0 0 0 0

0 0

3 0 0 0

0 0 0 2 0 0

0 0 0 0

3 0

therefore

Jy =

2

0 i 0 0 0 0i 0 0 0 0 0

0 0 0 i3 0 00 0

3 0 2i 0

0 0 0 2i 0 i30 0 0 0 i3 0

and also

Jx =

1

2

(J+ +J) =

2

0 1 0 0 0 0

0 0 0 0 0 0

0 0 0

3 0 0

0 0 0 0 2 0

0 0 0 0 0

3

0 0 0 0 0 0

+

0 0 0 0 0 0

1 0 0 0 0 0

0 0 0 0 0 0

0 0

3 0 0 0

0 0 0 2 0 0

0 0 0 0

3 0

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therefore

Jx =

2

0 1 0 0 0 0

1 0 0 0 0 0

0 0 0

3 0 0

0 0

3 0 2 0

0 0 0 2 0 30 0 0 0

3 0

Problem # 6

Write down all the possible eigenkets |j,mj corresponding to the case l = 3 , s = 1/2

ifl = 3 and s = 1/2 then j= ls = 7/2,5/2 and ml = 3,2,1,0,1,2,3 and ms = 1/2. Thereforeif j = 7/2 we find mj = ml + ms = 7/2,5/2,3/2,1/2,1/2,3/2,5/2,7/2 . If j= 1/2 then mj =ml + ms = 5/2,3/2,1/2,1/2,3/2,5/2

therefore for j= 7/2 we find all the eigenkets to be

|j,mj =72 ,72 , 72 ,52 , 72 ,32 , 72 ,12 , 72 , 12 , 72 , 32 , 72 , 52 , 72 , 52and for j= 5/2 we find all the eigenkets to be

|j,mj =52 ,52

,

52 ,32,

52 ,12,

52 , 12,

52 , 32,

52 , 52

Suppose now that the spin-orbit coupling is turned off so that ml and ms become good quantum num-

bers. For each of the eigenkets |j,mj just obtained, write down the corresponding eigenkets |ml,ms

72 ,72 = 3,1272 ,52

=

3,12,

2,12

72 ,32

=

2,12,

1,12

72 ,12

=

1,12,

0,12

5

2,

5

2

=

3,1

2

,

2,

1

2

52 ,32

=2,12

,1,12

52 ,12

=

1,12,

0,12

Problem # 7

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The relativistic correction to the ground state energy of the hydrogen atom can be found using a Hamil-

tonian of the form H = p4/8m30c2 , where p is the electron momentum and m0 is the electron rest mass.Calculate the energy correction using first-order pertubation theory. [Hint: |p4| = p2|p2 , sincep2 is Hermitian.

E10 = 100|H|100and p2 and the ground state wave function is

p2 = 22 and 100 = 1a3

er/a a =40

2

me2 e

2

40c(14)

therefore

E10 = 100|H|100 =

2

8m30c22100|2100

and

2100 = 1r2

r

r2

r

er/a = er/a

1

a2 2

ar

I left out the constant that comes along with 100 because it is not needed in the last calculation. So

we find

E10 =

4

8m30c2

1

a2 2

ar

2100|100 if we let =

4

8m30c2

thus

E10 = a3Z2

0d

Z0

sin dZ

0r2e2r/a

1

a2 2

ar

2dr and

Z20

dZ

0sin d = 4

= a3

4

Z0

r2e2r/a

1

a4+

4

r2a2 4

ra3

dr

= 4a3

Z0

r2e2r/a

1

a4+

4

r2a2 4

ra3

dr

= 4a7

Z0

r2e2r/adr +16

a5

Z0

e2r/adr 16a6

Z0

re2r/adr

Using

Z

0= xnex/adx = n! an+1

we find the first-order purtubation to be

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E10 = 4

a7

2!

a3

8

+

16

a5

a2

16

a6

a2

4

= 5a4

E10 = 584

m30c2a4

= 54

m

22

e2

40

2e2

40c

2= 5

4|E1|

2

Identical particles

There are two types of particles, bosons and fermions. Bosons have symmetric wave functions under

particle exchange and fermions have anti-symmetric wave functions under particle exchange

|S =1

2

|(1)|(2) + |(1)|(2)

|A = 12

|(1)|(2) |(1)|(2)

|T = |(1)|(2) or |T = |(1)|(2)The Slater determinent is given by

1N!

a(1) a(2) a(3)b(1) b(2) b(3)

c(1)

c(2)

c(3)

where the total # of possible terms is given by N!.

The Pauli Exclusion Principle says that you can never have two identical fermions occupy the same

state.

Problem # 8

Consider a 1-D simple harmonic. Use pertubation theory to calculate the shift in the ground state

energy introduce if one uses the relativistic expression (correct to first order) for the kinetic energy. [Use

the same trick as for the hydrogen atom!]

We know that the ground state wavefunction for the 1-D simple harmonic oscillator is given by

0 =m

1/4e

m2 x

2

we also know that the first order relativistic correction in terms of the kinetic energy is

Hr =

p4

8m3c2and p2 = 22

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thus the first order correction to the ground state wavefunction is given by

E10 = 0|H|0 =

4

8m3c220|20

where

20 =m

1/4 2

x2 (e

m

2 x2

) =m

1/4m

e

m

2 x2 m

x2 1

so we find

E10 =

4

8m3c2

m

2

m

x2 1

20|0

= 4

8m3c2

m

2m

1/22

Z0

m

2x4 2 m

x2

em

x2 dx +

Z

em

x2 dx

=

2m22

8m3c2 m

1/2

3

4

m22

2

m5/2

m

m3/2

+

m1/2

=

22

8mc2

m

1/234

m

1/2

m

1/2+

m

1/2

E10 = 3

32

22

mc2

Problem # 9

Calculate the wavelength, in centimeters, of the photon emitted under a hyperfine transition in theground state (n = 1) ofdeuterium . Deuterium is heavy hydrogen, with an extra neutron in the nucleus.The proton and neutron bindtogether to form a deuteron, with spin 1 and magnetic moment d =

gde2md

Sd;

the deuteron g-factor gd = 1.71. [Hint: This is intended to be a very short problem-just scale down fromhydrogen with appropriate g-factors and masses.]

we know

d =gde

2mdSd and gd = 1.71

and the wavelength of the emmited photon under a hyperfine splitting is given as

d =hc

Edand d =

EhEd

h Eh =0gpe

2

3mpmea3h = 21 cm

to find the hyperfine splitting for the deuteron we use

E1h f =0gde

2

3mdmea3SdSe

but we know

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SdSe =1

2(S2 S2e S2d) S2n = 2sn(sn + 1)

we have two special cases

S2

=342 when s =

12

154

2 when s = 12 S2

e =

3

42

S2

d = 22

and so we find

E1h f =0gde

2

2

3mdmea3

1 (singlet)+ 12 (triplet)

therefore

Ed =0gde

2

2

2mdmea3

therefore the wavelength of the emmited photon is

d =2gpmd

3gdmph =

2(2)(5.91)

3(1.71)h = 4.36(21 cm) = 91.53 cm

Problem # 10

Find all the spectral terms of the states that arise from the Russell-Saunders coupling of 3 inequivalent

p electrons.We know that for three inequivalent electrons in a p state we have

l1 = 1

l2 = 1

l3 = 1

s1 = s2 = s3 =

1

2

we also know that the number of possible states is given by

(2l1 + 1)(2s1 + 1) (2l1 + 1)(2s1 + 1) (2l1 + 1)(2s1 + 1) = 216 states

and also

s = s1 + s2 =

0

1+ s3 =

12

232

1 l = l1 + l2 =

2

1

0

+ l3 =

3

2 21 30

The n notation is simply the multiplicity of the given value, also

S,P,D,F,... = (l = 0,1,2,3,4,...) respectivily

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and the table is given by

s l j 2s+1Lj m

j

3/2 3 9/2,7/2,5/2,3/2 4F9/2,7/2,5/2,3/2 2813/2 2 7/2,5/2,3/2,1/2 4D7/2,5/2,3/2,1/2 2023/2 1 5/2,3/2,1/2

4

P5/2,3/2,1/2 1233/2 0 3/2 4S3/2 411/2 3 7/2,5/2 2F7/2,5/2 1421/2 2 5/2,3/2 2D5/2,3/2 1041/2 1 3/2,1/2 2P3/2,1/2 661/2 0 1/2 2S1/2 22

and when we add all the possible number of states we find a total of 216 possible states.

Problem # 11Find all the spectral terms allowed by the Pauli Exclusion Principle that arises from the

Russell-Saunders coupling of 8 equivalentd electrons. [Note: This is not as bad as it may seem! The

allowed states for holes (i.e. missing electrons) are precisely those for electrons. Thus, the spectral

terms are just those for 2 equivalent d electrons]


equivalent electrons we can put the first electron in any of the 10 states allowed in the D shell and the


9

10 = 90 possible states

this is not total correct due to the Pauli Exclusion Principle and the fact that these are indistinguishable

particles, and we find that the total number of allowed states are given by

90


we know

s1 = s2 =1

2l1 = l2 = 2

s = s1 s2 =

1 symmetric

0 anti-symmetricl = l1 l2 =

4 symmetric

3 anti-symmetric

2 symmetric

1 anti-symmetric

0 symmetric




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s = 1 l = 3,1 s = 0 l = 4,2,0


s l j 2s+1Lj mj1 3 4,3,2 3F4,3,2 21

1 1 2,1,0 3P2,1,0 90 4 4 1G4 9

0 2 2 1D2 5

0 0 0 1S0 1

and we find a total of 45 possible states

Problem # 12

The rare earth elements Nd, Gd, Ho may each be triply ionized to form Nd+++, Gd+++, and Ho+++.

The configurations for these ions are

Nd+++ 4f35s2p6

Gd+++ 4f75s2p6

Ho+++ 4f105s2p6

thus, the 5sp subshells are full, and all the action arises from the 4f subshells.

Hunds rules are given as

1. Full shells and subshells do not contribute to total S, the total spin angular momentum and L, thetotal orbital angular momentum quantum numbers.

2. The term with maximum multiplicity (maximum S) has the lowest energy level.

3. For a given multiplicity, the term with the largest value ofL has the lowest energy.

4. For atoms with less then half-filled shells, the level with the lowest value ofJ lies lowest in energy.

Otherwise, if the outermost shell is more than half-filled, the term with the highest value of J is the

one with the lowest energy.

(i). Use Hunds rules to obtain the spectral terms for each ion in the ground state.

for Nd+++ we find

mlms 3 2 1 0 -1 -2 -3

+ 12 1

2

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therefore

maximum S =3

2

maximum L = 6

and since this shell is less than half-filled we find the lowest value ofJ

minimumJ = |L S| = 92

therefore the spectral term is given by

4I9/2

for Gd+++we find

mlms 3 2 1 0 -1 -2 -3

+ 12

12

therefore

maximum S =7

2

maximum L = 0

and since this shell is half-filled we find J to be

J = S =7

2


8S7/2

for Ho+++we find

ml

ms 3 2 1 0 -1 -2 -3

+ 12 1

2

therefore

maximum S = 2maximum L = 6

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and since this shell is more than half-filled we find the highest value ofJ

maximum J = |L + S| = 8


5I8

(ii). Calculate the g-values and effective magnetic moments for these ions

The g-values and effective magnetic moments are given by

g =

1 +

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

and i = gB

j(j+ 1) B =

e

2m

for Nd+++ we find

s =3

2l = 6 j=

9

2

so we find

g =

1 +

92 (

112 ) +

32 (

52 ) 6(7)

2 92 (112 )

=

8

11i = 6

11

e

m

11

for Gd+++ we find

s =7

2l = 0 j=

7

2

so we find

g =

1 +

72

( 72

) + 72

( 72

)

2 72 (72 )

= 2 i = 3

2

e

m

7

for Ho+++ we find

s = 2 l = 6 j= 8

so we find

g =

1 +

72 + 6 42144

=

5

4i = 15

4

e

m

2

Problem # 13

In an atom with LS coupling, the relative seperations between adjacent energy levels of a particular

multiplet are 4:3:2:1. Assign the quantum numbers s, l, j. Repeat for 7:5:3.

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to find the quantum numbers we must use the Lande equation

E = 2k(j+ 1)

so for the first problem (Figure a) we must use

E2 = 2E1

2k(j+ 2) = 4k(j+ 1)

jmin = 0

therefore

jmax = 4 jmin = 0

thus

jmax = l + s = 4

jmin = |l s| = 0

and we find

j = 4,3,2,1,0 l = 2 s = 2

so for the second problem (Figure b) we must use

E2 =5

3E1

2k(j+ 2) =10

3k(j+ 1)

jmin =1

2

therefore

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jmax =7

2jmin =

1

2

thus

jmax =

l+

s=

7

2

jmin = |l s| = 12

and we find

j =7

2,

5

2,

3

2,

1

2l = 2 s =

3

2

Problem # 14

Consider a single d electron (l = 2). Calculate the allowed values of l, j,mjand g, presenting yourresults in a table. Draw a diagram showing the energy levels for a single d electron and a single p electron

(of lower energy) in a weak magnetic field. Draw arrows to indicate all the allowed electronic dipole

transitions. Label the various levels in terms of mjand the transitions in terms ofmj. [Relavent selectionrules: l = 1,j= 0,1,mj = 0,1]

We know that

j= |l s| thus l = 2 s =1

2j=

5

2,

3

2


l j mj g

2 52 52 65

2 52

32

65

2 52

12

65

2 52

- 12

65

2 52

- 22

65

2 52

- 52

65

2 32

32

45

2 32

12

45

2 3

2

- 1

2

4

52 3

2- 3

245

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Figure 1: Possible transitions from D to P shells

Problem # 15

Consider a single p electron. Obtain all the possible kets |j,mj, in terms of the kets |ml,ms with theappropriate Clebsch-Gordan coefficients.

since we are considering an electron in a p shell we know that

l = 1 j=3

2,

1

2

and the mj values are given as

j =3

2mj =

3

2,

1

2,1

2,3

2

j =1

2mj =

1

2,1

2

and the eigenkets are given as

|j,mj = |ml,ms

32 ,32 = 1,1232 ,12

=

1,120,12

12 ,12

=

1,120,12

we will need to use

|j,mj = m j=ml +ms

|ml,msml,ms|j,mj

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where ml,ms|j,mj is the Clebsh-Gordon coefficient, to find our C-G coefficients. We will also needto use raising and lowering operators.

J|j,mj =

j(j+ 1) mj(mj1)|j,mj1

L|l,ml = l(l + 1) ml(ml 1)|l,ml 1

S|s,ms = s(s + 1) ms(ms 1)|s,ms 1and we also know that

J = L + S J|j,mj = L|l,ml + S|s,mswe know that the C-G coefficients for32 ,32

=

1,12

is 1, now for the rest of the eigenkets we find

J32 , 32

= L

1, 12

+ S1, 12

using the raising and lowering relationships we find

3

32 , 12

=

2

0, 12

+ 1

1,12

therefore

32 , 12

=

2

3

0, 12

+1

3

1,12

and next we use

J+

32 ,32

= L+

1,12

+ S+

1,12

we find

3 3

2

,

1

2 = 2 0,

1

2+ 1 1,

1

2

therefore

32 ,12

=

2

3

0,12

+1

3

1, 12

For the other eigenkets we must take into account orthonormality and we can write the remaining

eigenkets as a linear combination of two eigenstates. i.e.

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12 , 12

=

1,12

+

0, 12

1

2,1

2

=

1, 1

2

+

0,1

2

we can make1

2, 1

2

orthogonal to

32, 1

2

by saying

1

2,

1

2

32 , 12

= 0

thus we find

1,1

2

+

0,

1

2

2

3

0,

1

2

+

13

1,1

2

= 0

231,

1

2 |0,1

2 + 1

30,

1

2 |1,1

2 +

31,

1

2 |1,1

2 + 2

30,1

2 |0,1

2 = 0we can see that the first two 0 due to orthonormality and so

3

=

2

3and we know ||2 + ||2 = 1

therefore

=

2

3 =

13

and so we find

12 , 12

=

2

3

1,12

+1

3

0, 12

to find the last C-G coefficients 1

2,1

2

32 ,12

= 0

thus we find

1, 12

+ 0,1223

0,12

+ 13

1, 12

= 0

31, 1

2|1, 1

2 +

2

30,1

2|0,1

2 +

2

31, 1

2|0,1

2 +

30,1

2|, 1

2 = 0

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we can see that the last two 0 due to orthonormality and so

3

=

2

3and we know ||2 + ||2 = 1

therefore

= 23 = 13and so we find

12 ,12

=

2

3

1, 12

+1

3

0,12

Problem # 16

Two particles are confined in a cubical box of sides length a. Determine the lowest energy level (wave

function,energy eigenvalues, degeneracy) for each spin state in the cases where the two particles are

(a). a proton and a neutron(b). two alpha particles [ an alpha particle is a helium nucleus: two protons and two neutrons]

(c). two neutrons

Chapter 5: WKB Approximation

The WKB method is a technique for obtaining approximate solutions to the time-independent Schrodinger

equation in one dimensional (the same basic idea can be applied to many other differential equations, and

to the radial part of the Schrodinger equation in three dimensions). It is particularly useful in calculating

bound state energies and tunneling rates through potential barriers.The phase change is given by

=1

Zx2x1

2m[EV(x)]dx =

n +

1

2

and the tunneling rate can be found using

ln Tn = 2

Zx2x1

2m[V(x) E]dx

ProblemsProblem # 1

A particle of mass m and electric charge q is confined within a one-dimensional well, with

V(x) =

V0 |x|< a/20 |x|> a/2

assume that V0 22ma2 , i.e. the well is deep.

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(a). Using WKB approximation, find the energy quantization condition for abound state E< 0.(b). Estimate the number of bound states of the system.

A weak external electric field 0 (in the x-direction) is applied to the system. Choose the zero of theelectrostatic potential so that ((x = 0) = 0. Assume that V0 q0a.

(c). Now the bound states are no longer stable. Explain why.

(d). Find the barrier penetration factor for the ground state. To leading order, you can assume that the

particle sits at x = 0 with an energy V0(a). Using WKB approximation, find the energy quantization condition for abound state E< 0.

Here we have sharp walls on both sides, so the energy quantization condition is

n=

Z2m(EV(x)dx =

Za/2a/2

2m(E+V0)dx =

2m(E+V0)a, n = 1,2,3,...

and solving for the energy we find

E =

V0

+1

2n

a

2

(b). Estimate the number of bound states of the system.

To estimate the number of bound states, we note that the condition is E< 0, so the highest energybound state corresponds to

V0 + 12m

nbound

a

2= 0

solving for nbound, we find

nbound =a

2mV0(c). Now the bound states are no longer stable. Explain why.

Turning on the electric field adds a term q0x to the potential, which gives it the form (exaggerated,assumes q > 0). The bound states are no longer bounded from below a trapped particle may reduce itsenergy by tunnelling out to x .

(d). Find the barrier penetration factor for the ground state. To leading order, you can assume that the

particle sits at x = 0 with an energy V0Since V0 22ma2 , we approximate (as specified) the ground state energy as

E V0 +1

2m

a2 V0

The tunneling factor is given by e2, where

=1

Zx1x0

2m(V(x)E)dx

here, the tunneling region begins at x0 = a/2 and ends at the classical turning points given by

V(x1) = E q0x1 = V0 x1 = V0q0

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we now evaluate

=1

Zx1x0

2m(q0x +V0)dx = 1

2

3

2m

q0 (0 (qx0 +V0)3/2 2

2m

3q0(V0)

3/2

where we use V0 q0a in the approximation. Our tunneling probability is then

T exp

42m3q0

(V0)3/2

Problem # 2

A particle is bound in a spherically symmetric logarithmic potential

V(r) = V0 ln(r/a)

where V0 and a are constants. If the particle has no angular momentum (l = 0) we can treat this a s onedimensional problem in the radial coordinate. Use WKB approximation to find the radial energy levels

of the bound states with no angular momentum. Because r varies from 0 , r = 0 acts as an infinitepotential barrier in this 1-D problem, so the phase change is given by = (n + 34 ) instead of(n +

12 ).

Hint: Z0

xexdx =

2

We know that

=1

Zr00

2m(EV(r))dr =

n +

3

4

n = 0,1,2,...

=1

Z

r0

02m(EV0 ln(r/a))dr

The turning point r is given by

E = kE+V(r) kE = 0 E = V0 ln(r/a) r0 = eE/V0

so we find

1

Zr00

2m(EV0 ln(r/a))dr = 1

Zr00

2mV0(ln(r0/a) ln(r/a))dr

=1

2mV0Zr0

0 ln(r0/r)drIf we let

x = ln(r0/r)

r = r0ex

dr = r0exdx

so we find1

2mV0

Z0

xexdx =

2

2mV0

aeE/V0 =

n +

3

4

pi

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thus

eE/V0 =2

(n + 3/4)

a

2mV0

and finally

En

= V0

lna2

mV0n + 3

4 = V

0lnn + 3

4+V

0ln

a2

mV0

Problem # 3

A pareticle of mass m moves in the potential V(x) = |x|, where is a constant. Use the WKBapproximation to find the energy levels, En.

the first thing we know is that

|x| = E thus x = E

x1 = E

x2 =

E

and we know that the energies can be found using

=

1

Zx2

x1 2m(EV(x))1/2

dx =

n +

1

2

thus we know1

ZE/E/

2m(E |x|)1/2dx =

n +

1

2

so we find2

ZE/0

2m(Ex)1/2dx

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if we let

u = Exdu = dx

du

= dx

thus

2

Z0E

2m(u)1/2du =

4

2m

3u3/2

0E

=4

2m

3E3/2

and so we find the allowed energies to be

4

2m

3E

3/2n =

n +

1

2

En =

3

4

2m

n +

1

2

2/3

Problem # 4Use the WKB approximation to find the allowed energies En of an infinite square well with a shelf,

of height V0 , extending half-way across:

V(x) =

V0, if 0 < x < a/2

0, if a/2 < x < a

otherwise

Express your answer in terms ofV0 and E0n (n)2/2ma2 (the nth allowed energy for the unper-

turbed infinite square well, with no shelf). Assume that E01 >V0, but do not assume that En V0.

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the allowed energies can be found using

=1

Zx2x1

2m(EV(x))1/2dx =

n +

1

2

since we know that n is proportional to n+1/2 we can just write this as

= 1

Zx2x1

2m(EV(x))1/2dx = nto find what the integral is we can split it up into to pieces i.e

2m

Za/20

(EV0)1/2dx +Za

a/2E1/2

dx =

2m

a

2

(EV0)1/2 +E1/2

= n

thus 2m

a

2

(EV0)1/2 +E1/2

= n

and so we find

(EV0)

1/2

+E1/22

= 4n222

2ma2 = 4E0n

foiling the left side we find

2E+ 2

E(EV0)V0 = 4E0nif we rearrange this equation and square both sides we find

(2

E(EV0)2 = (4E+V0 2E)2and this simplifies to

4E2 4EV0 = 16(E0n)2 + 4E0nV0 8E0nE+ 4E0nV0 +V20 4EV0 8EE0n + 4E2

thus we find16EE0n = 16(E

0n )

2 + 8E0nV0 +V2

0

and so we find

En = E0n +

V0

2+

V2016E0n

Problem # 5

In the process known as cold emmision, electrons are drawn from the surface of a room-temperature

metal by an external electric field . The various energies and potentials are indicated in the figure. Thesurface of the metal is the plane x = 0, and the potential for x > 0 (vacuum) is:

V(x) = +EF exwhere EF is the Fermi level of the electron gas in the metal and is the work function. Use WKB

approximation to show that the transmission coefficient is given by

T = exp

4

3

2m

3/2

e

Cold emmision is the principle underlying the field ion microscope.

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since we know that the transmission coefficient can be found using

T = e2 R

x2x1

2m(V(x)

E)1/2dx

we can write this as

ln T = 2

Zx2x1

2m(V(x) E)1/2dx

and since we know what the potential is, we get

2

Z/e0

2m( +EFex EF)1/2dx = 2

Z/e0

2m( ex)1/2dx

if we let

u = exdu = edx

due

= dx

substituting this into the above equation we get

2

2m

e

Z0

(u)1/2du =

4

2m

3eu3/2

0

= 4

2m

3e3/2

thus we find the transmission cofficient to be

ln T = 4

2m

3e3/2

and solving for T we find

T = exp

4

3

2m

3/2

e

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Problem # 6

Use the WKB approximation in the formZr2r1

p(r)dr = (n 1/2)

to estimate the bound state energies for hydrogen. Dont forget the centrifugal term in the effective

potential. The following integral may help:Zb

a

1

x

(x a)(b x)dx =

2(

b a)2

Note that you recover the Bohr energy levels when n 1 and n 1/2. answer

Enl 13.6 eV

[n (1/2) +

l(l + 1)]2

Chapter 6: Time-Dependent Pertubation Theory

6.1 General Formalismthis is an exact solution but totally useless

iak= an(t)k|H |neiknt where kn =1

[E0k E0n ]

the problem is that ak depends on the time dependence of all an(t). We can find ak by using

ak(t) = 1

Zt0

H

k jeik jt

dt

where H are the appropriate matrix elements and k is the final state of the system and j is the initialstate.

6.2 Harmonic Pertubations

The transition probability/unit time to any one of the final states |k is given by

W =1

t

k

|ak(t)|2

if we assume many states we can write this as an integral k

R(Ek)dEk

W =1

t

Zak(t)(Ek)dEk where dEk = d()

Fermis Golden Rule #1 is given by

W |k|2|ii|1|j|2

Fermis Golden Rule #2 is given by

W =2

|k||j|2(Ek)

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6.3 Interaction of Atomic Electron With E-M Radiation

The Hamiltonian for this interaction is given by

H

k j = eE0k|r|jcos tand the transition rates are given by

Wk j =e2

302 |k|r|j|2

(k j)

6.4 Selection Rules for Electric Dipole Transitions

The selection rules are as follows

= 1 for x,y, andz axis

m = 0,1 0 for the z axis and 1 for the x and y axis

S = 0

The spin cannot flip in an electric dipole interaction.

The polarization of an E-M wave

m = 0 only if k|z|j = 0i.e. E field is along z-axis, and ,

m = 1 only if k|x|j = k|y|j = 0also, the component ofE-field are out of phase by /2.

The Weak Field Limit: S L CouplingThe quantum numbers n,, j,mj can be expressed

|n j mj

in terms of |n m ms

will give us a new set of selection rules

= 1 for x,y, andz

j = 0,1 (cannot go from 0 0)mj = 0,1 (not from 0 0 ifj= 0)

S = 0

6.5 Spontaneous Emmision: Einstein A and B coefficients

The transmission and absorption rates are given by

RT = [A +B()]N2 transmission rate

RA = B()N1 absorption rate

where N1 and N2 are the number of atoms in states 1 and 2. Now we need to find the A and B

coefficients.

A = B()

e/kbT 1

=e23

3c30|1|r|2|2

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6.6 Higher Order Transitions

We have the following pertubation for transitions for an E-M field given by

H = eE0k|reikr|jcos t

and if

k r

1 eikr = 1

we have an electric diploe transition. If the selection rules does not allow this process we say it is

forbidden. But perhaps the process is allowed in higher order transition

k|reikr|j = k|r[1 + ik r + 12!

(ik r)2 + ..]|jwhere the first order term is simply the electron dipole transition and the second order term is the

electric quadrupole transition which is ~r2. In principle you have a very large number of higher order

terms.

6.7 Life Times and Line Widths

Natural Broadening

The transition rate is defined as

RT =1

where is the lifetime

109s electric dipole transitions 106s electric quadrupole transition

The uncertanty principle say

E

E

this means that the emitted photon is not sharp but is spread out

has a width1

Collision

The path length is given by

l =1

4nwhere is the cross-sectional area of the gas and n is the number density of the gas. The time between

collisions is thus given by

c =l

vv is the average velocity

and the rate is given by

RT =1

cThe equipartition theorem states

|v| =

3

2kbT

where kb is the Boltzman constant and T is the temperature.

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Problems

Problem # 1

An infinite one-dimensional square well [V=0, for |x| a2 , V= elsewhere] contains a particle of massm in the ground state. At time t= 0 a pertubation H =A(x) is turned on: (x) is the Dirac delta function.

Calculate the initial rate of transition into the first (n=2) and (n=3) excited states using time dependent

pertubation theory.

[NOTE: Find coefficients a2(t) and a3(t) using the integral expressions we derived in class and appro-priate matrix elements. Look carefulle at the parity of the two matrix elements before evaluating them!

Initial means t 1, where is the relavent energy differece.

We are given

H = A(x) E =

and the wave functions for the one-dimensional infinite square well are given by

odd =2

a sinnx

a

even =2

a cosnx

a

The integral expression derived in class is given by

ak(t) = i

Zt0

H

k jeik jt

dt k j =

1

E0k E0n

where kis the final state of the system and jis the initial state.

We need to find the matrix elements H k j . (we also know that n = 1 is an even wave function and n = 2is an odd wavefunction)

For the n : 1 2 we find

H

21 = 2|A(x)|1 =2A

a

Za/2

a/2cos

x

a

sin

2x

a

(x)dx =

2A

acos(0) sin(0) = 0

thus we know that

a2(t) = i

Zt0

H

21ei21t

dt = 0

For the n : 1 3 we find

H

31 = 3|A(x)|1 =2A

a

Za/2a/2

cos

x

a cos

3x

a

(x)dx =

2A

acos(0) cos(0) =

2A

a

we know what 31 is

31 =1

E03 E01

=

42

ma2

we will need to use the following identity for the next part

2i sin(31t/2) = ei31t/2 ei31t/2

we find the coefficient to be

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a3(t) = i

Zt0

H

31ei31t

dt = 2Ai

a

Zt0

ei31tdt = 2A

a

ei31t1

31t

t

= 2Aa

ei31t/2

ei31t/2 ei31t/2

31t

t = 2A

aeit/2

i sin(31t/2)

31t/2

t

so we get

a3(t) = 2Aa

ei31t/2

i sin(31t/2)

31t/2

t

The rate of transition is given by

|a3(t)|2

t=

2A

at

2sin(31t/2)

31t/2

2since 31t 1 sin(31t/2)

31t/2 1

we get

|a3(t)|2

t=

4A2

2a2t

Problem # 2

A hydrogen atom in its ground state is placed between the parallel plates of a capacitor. The z-axis

of the atom in perpendicular to the capacitor plates. At time t = 0, a uniform electric field =0et/ isapplied to the atom. The perturbing Hamiltonian is thus

H = er 0et/ = ez0et/ and since z is given by z = r cos()

we find the Hamiltonian to beH

= er cos()0et/

and the three hydrogen wave functions needed for this problem are given as

100 =1a30

er/a0 200 =1

2a0

1

2a0

1 r

2a0

er/2a0 210 =

12

1

4a5/20

cos()er/2a0

(a) Show that the electron has zero probability of being excited into the 2s state nlm = 200

what we are looking for is |a200(t)|2 so we must use the formula

a200(t) = i

Zt0

H

k jeik jt

dt

where

H

200,100 = 200| er cos()0et/|100 = e0et/200|r cos()|100

= e0et/

8a30

Z20

d

Z0

cos() sin()dZ

0r3

1 r2a0

e3r/2a0 dr

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but since we know that Z0

cos() sin()d = 0

causes

H

200,100 = 0

and thus

a200(t) = i

Zt0

H

k jeik jt

dt = 0

and if you square this number, you still get 0 thus the probability of this transition happening is nill.

(b) Show that after time t, the probability that the atom is in the 2p state nlm = 210 is

|a210(t)|2 =

215

310e220a

20

2(2 + 1/2)

we must first find the matrix element H 210,100 which is given by

H

210,100 = 210|er cos()0et/|100 = e0et/210|r cos()|100

= e0et/

8a40

Z20

d

Z0

cos2() sin()dZ

0r4e3r/2a0 dr

and so we find the integrals to be

Z20

d

Z0

cos2() sin()d letting u = cos() du = sin()d

thus we find Z2

0d

Z

0cos2() sin()d = 2

Z11

u2du =43

and for the other integral we find

Z0

r4e3r/2a0 dr = 4!

2a0

3

5putting this all together we get

H

210,100 = e0e

t/

8a40

4

34!

2a0

3 5

= 28a0

35

2e0e

t/

now that we have this solution we can find the coefficient. Plugging this into the equation derived in

class we find

a210(t) = i

Zt0

H

k jeik jt

dt

=ie0

a02

28

35

Zt0

et(i1/)dt

Zt0

et(i1/)dt =

1

i 1/

et(i1/) 1

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but since we know that

for t et(i1/) 0and so we find the coefficient to be given as

a210(t) = ie0

a02

28

35 1

i

1/

and the probability is given by

|a210(t)|2 = a210(t)a210(t)

=215

310e220a

20

2(2 + 1/2)

Problem # 3

Consider a particle of charge q and mass m, which is in simple harmonic motion along the x-axis so

that its Hamiltonian is given by

H0 =p2

2m+

1

2kx2

A homogenius electric field (t) is switched on at t = 0, so that the system is perturbed by the interac-tion

H = qx(t)if(t) has the form

(t) = 0et/

where 0 and are constants, and if the oscillator is in the grounds state for t 0, find the probabilitythat it will be found in an excited state as t , to first order in pertubation theory.

Problem # 4

A hydrogen atom initially (i.e. at t ) in its ground state is exposed to a pulse of ultraviolet lightthat can be thought of as a time varying electric field which points precisely in the z-direction at all times

and whose magnitude is given by

|E(t)| = E0 cos(t)et2/T2

with E0,, and T all constants. The frequency satisfies

=8

9(13.6 eV)

The constant T is large compared to all other timescales in the problem. At t , the atom is, ingeneral, in some superposition of energy eigenstates. Working to first order in time-dependent pertubation

theory, what are all the states that could arise in this superposition? (Specify state(s) by their quantum

numbers n, l, and m.)

6.8 Adiabatic Approximation

We can begin by writing the new Hamiltonian as

H(t) = H0(t) + H(t)

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the Adiabatic approximation is used ifH(t) varies very slowly

ak = n

ank| nei(nk) n = 1

Zt0

En()d

and ifk= n we find

k| n = k|H|n

En Ekthe Bohr angular frequency is given by

kn =EkEn

and

ak = n=k

an

k jk|H|nei

Rt0 k j()d

6.9 Sudden Approximation

if a system is in specific eigenstate at t= 0 and H0 H1 instantaneously we use sudden approximationmethod. Suppose that the eigenfuntion

| |then

t < 0 we find |(r, t) = n

an|neiEnt/

t > 0 we find ||(r, t) =

b|

Problems

Problem # 1

Consider a single electron in an atom in a high magnetic filed so that we can ignore the spin-orbit

interaction (Paschen-Bach effect). We investigate the magnetic dipole transition using the perturbing

HamiltonianB Lcos t, whereB = iBx + jBy + kBz is a constant andL is the angular momentum operator.

By considering appropriate matrix elements, show that the selection rules for magnetic dipole transi-tions are = 0, m = 0,1. Which of these apply when L is (i) parallel, and (ii) perpendicular to thez-axis? [Hint: Express Lx and Ly in terms of L+ and L.]

We know that

H =B Lcos t = (BxLx +ByLy +BzLz) cos t

thus we know that

k|H |j = cos t[Bxk|Lx|j +Byk|Ly|j +Bzk|Lz|j]

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forL parallel to the z-axis we find

BxLx = ByLy = 0

and

k|H|j = Bzk|Lz|jcos t = Bzll|Lz|mlml = Bz cos tmllm l mlthus the selection rules whenL is parallel to the z-axis

l = 0 m = 0

ifL is perpendicular to the z-axis we find

BzLz = 0

and

k|H|j = cos t[Bxk|Lx|j +Byk|Ly|j]we know that Lx and Ly can be expressed by lowering and raising operators

Ly =1

2i (L+ L) Lx =1

2 (L+ +L)

since we know that the relationship for finding expectation values using lowering and raising operatots

is given by

lm|L|lm =

l(l + 1) ml(ml 1)llml ml1thus

lm|L+|lm =

l(l + 1) ml(ml + 1)llml ml +1and

lm|L

|lm

=l(l + 1) ml(ml 1)llml ml1

thus the selection rules whenL is perpendicular to the z-axis are

l = 0 m = 1

Problem # 2

Consider a one-dimensional simple harmonic oscillator, in which the particle has a charge q, acted

upon a homogeneus time-dependent electric field E(t)=E0 exp[(t/)2] where E0 and are constants.Thus, the perturbing Hamiltonian is -qE(t)x. Assuming that dE(t)/dt is small, and that at t= the oscilla-tor is in the ground state, use adiabatic approximation to obtain the probability that it will be found in an

excited state as t . [Hint: as usual with problems on the harmonic oscillator, write the displacement interms of a, a. Then look carefully at the matrix elements to see which are zero. You should come up with

a slightly messy integral, which you can solve by completing the square on the exponent.]

we know that

H =p2

2m+

1

2m2x2 qE(t)x

if we let

k= m2

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we find

H =p2

2m+

1

2kx2 qE(t)x

we can complete the square and we find

H =p2

2m

+1

2

kx qE(t)

k

2

qE(t)

k

2

letting x0 = qE(t)k

we get

H =p2

2m+

1

2k

(x x0)2 x20

now we need to find

H

t= k x0(x x0) k x0x = k x0x =

2mm2 x0(a

+ a)

and soH

t =

2m

2qt

2 E0e(t/)

2

(a + a)

and

k|H|0 =

2m

2qt

2E0e

(t/)2k|(a + a)|0and we also know

10 =E1 E0

=

(1 + 1/2) (1/2)

=

this tells us that the only allowed transition is from n = 0 n = 1. So the probabilty transition is foundby using

P10 = |a1(t)|2 = 1 Z

t

0

H

t eitdt

2

so we can find

a1(t) =1

2m

2q

2E0

Z

te(t/)2

ei10tdt

if we let

K=1

2m

2q

2E0

we find

a10(t) = K

Z

te((t/)2i10t)dt

integrals of this form can be solved by completing the square, i.e

Z

e(ax2+bx)dx let y a[x + (b/2a)] then (ax2 + bx) = y2 (b2/4a)

if we apply this to our problem we know that

a =1

2b = i10

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we get

a10(t) = KZ

(y i2/2)ey2(2/4)dy

which gives us

a10(t) =

e(

2/4)K

Z

yey2

dy + e(2/4)K

Z

i2

2

ey2

dy

but we know that Z

yey2

dy = 0

so the integral simplyfies to

a10(t) = e(2/4)K3

Z

i

2ey

2

dy = e(2/4)K3i

2

thus the probability is given by

|a10(t)|2

= e(10

2/2)

K2

6

2

4

=q2E20

2

2me(10

2/2)

Problem # 3

The Hydrogen nucleus of mass 3 (tritium) is radioactive, and decays into a helium nucleus of mass 3

by -decay. (The helium atom is thus singly ionized.) Suppose the hydrogen atom is initially in its groundstate. Use sudden approximation to calculate the probabilities that the helium will be in the 2s and each of

the 2p states.

we have a reaction where tritium decays into helium-3

31H 32H by decay n p+ + e + e

the states are given by

state =

1s n = 1 l = 0 ml = 0

2s n = 1 l = 0 ml = 0

2p n = 2 l = 1 ml = 1,0

For helium, spherical harmonics are unchanged ( me) but the radial wave functions are changedbecause they depend on the potential, which depend on z. (z hydrogen=1, z helium=2)

100 =1

a

3/20

er/a0 200 =1

4

2

z

a0

3/22 zr

a0

ezr/2a0

210 =1

4

2

z

a0

3/2zr

a0ezr/2a0 cos 211 =

1

8

z

a0

3/2zr

a0ezr/2a0 sin ei

the probability that 1s 2s is given by

P = |200|100|2

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so we first need to find

200|100 =Z

0200100r2drd =

1

a30

2

Z0

2 2r

a0

e2r/a0 r2dr

this can be solve using

Z

0xne

x/a = n!an+1

so we find

200|100 = 4a30

2

2 a0

2

3 6a0

a02

4= 1

2

and the probability is given by

|200|100|2 = 14

and the probability that 1s

2p is not allowed, because

|210|100| =Z

0210100r2dr

Z20

d

Z0

sin cos d

Z0

sin cos d = 0

therefore

|210|100|2 = 0and we also find

|211|100| =Z

0211100r2dr

Z2

0d

Z

0sin2 d

Z0

sin2 d = 0

therefore

|211|100|2 = 0Problem # 4

An electron is held in a spin up state along the z

axis by a magnetic filed, B = B0 z. An experimenter

would like to reverse the electron spin adiabatically.

(a). Explain why, if he had the choice he would slowly rotate the magnetic field until it pointed in the

-z direction, as opposed to slowly reversing the field.

(b). Since it is expensive to rotate a magnet, and cheap to just reverse the current, the experimenter

decides it is almost as good just to reverse the field, provided he places the apparatus in a weak magnet

with its field B1 pointing some direction in the xy plane, what is his reasoning?

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(c). Now suppose the weak permanent magnetic field described in (b) is along the xdirection witha magnitude B1 B0.The experimenter slowly reverses the magnetic field along the zdirection in thefollowing manner:

Bz(t) = B0 tT

for T t T.Write down the Hamiltonian of the system during the reversion process.

(d). How large does T have to be for the adiabatic condition to be valid?

Problem # 5

A particle with charge q sits in the 1D harmonic oscillator ground state. At t = 0 a constant electricfield is turned on in an amount of time much less than 1, where is the angular frequency of theoscillator. Calculate the probability that it will be found in the ground state of the new potential.

Chapter 7: Scattering and Born Approximation

7.1 Scattering

Imagine a particle incident on some scattering center. it comes in with energy Ethe scattering cross-section

is given by

=S

Iwhere S=

prob

secand I= prob flux = v|i|

2

and the differential cross section is given as

D() =d

d=

S()

I

where S() is the probability/second that particle scatters into a unit solid angle at angle . We areassuming that scattering center has all the mass. We also know that the total cross section is given as

Z

D()d

7.1 Born Approximation

If we take a cube of side L, the density of the states is given by

(kf) =m

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137B Formulas Solutions (1)