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Physics 137B
Served by Roger Griffith
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Chapter1: Time-Independent Pertubation Theory
1.1 Non-Degenerate Pertubation Theory
Pertubation theory is a systematic procedure for obtaining approximate solutions to the perturbed problem, 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 function
is given by
|ψn " = |ψ0n " + λ|ψ1n " + λ2 |ψ2n " + ...
and the eigenvalues are given as
En = E 0 + λE 1 + λ2 E 2 + ...
and we find that the first-oder perubation to the energy is given as
En1 = #ψ0n |H ! |ψ0n "
(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 ψ1n (like any other function) can be
expressed as a linear combination of them
ψ1n =
(n) 0
ψm
∑ cm
(2)
#ψ0m |H ! |ψ0n "
El0 − En0
(3)
#ψ0m |H ! |ψ0n " 0
ψm
Em0 − En0
m$=n
(4)
m$=n
and the coefficients are given by
(n)
cm =
the wave functions are given by
ψ1n =
∑
The perturbation theory often yields suprisingly accurate energies but the wave functions are notoriously poor.
1
Second-Order Energies
Proceeding as before, we take the inner product of the second order equation with ψ0n
#ψ0n |H 0 ψ2n " + #ψ0n |H ! ψ1n " = En0 #ψ0n |ψ2n " + En1 #ψ0n |ψ1n " + En2 #ψ0n |ψ0n "
(5)
and the second-order pertubation to the energies is given by
En2 =
|#ψ0m |H ! |ψ0n "|2
∑ En0 − Em0
m$=n
(6)
1.2 Degenerate Perubation Theory
Two-Fold Degeneracy
Suppose that we have a degenerate system, i.e
Ek0 = En0 = E 0
We must diaganolize the secular equation
"
! 0
E +Waa
Wab
Wba
E 0 +Wbb
where
Wi j ≡ #ψ0i |H ! |ψ0j ", (i, j = a, b)
(7)
and setting the determinate equal to 0 we find a quadratic equation of the form
!
"
#
1
1
2
2
Waa +Wbb ± (Waa −Wbb ) + 4|Wab |
E± =
2
(8)
This is the fundemental result of degenerate pertubation theory; the two roots correspond to the two
perturbed energies.
Problems
Problem # 1
2
p2
The unperturbed Hamiltonian of a 1-D simple harmonic oscillator is H0 = 2mx + kx2 .The energy levels
are E = !ω(n + 1/2), where ω = (k/m)1/2 and n = 0, 1, 2, 3, .. . A perubation H ! = ax4 is now
turned on.
a) Calculate E01 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
ψ10
=
$ mω %1/4
π!
2
− mω
2! x
e
and
ψ01
the first order pertubation is defined as
2
=
$ mω %1/4 & 2mω '1/2
π!
!
mω 2
xe− 2! x
En1 = #ψn |H ! |ψn "
where H ! = ax4
therefore
E01
4
= #0|ax |0" = 2a
$ mω %1/2 Z
∞
x4 e−
mω 2
! x
dx
π!
0
&
'
&
'
! 5/2
3
24 $ mω %1/2 √
! 2
π
=
a
= a
32
π!
mω
4
mω
and now for the first order correction to the first excited state is
E11
4
$ mω %1/2 & 2mω ' Z
∞
x6 e−
mω x2
!
dx
π!
!
0
&
'
&
'
15 $ mω % √ $ mω %1/2 ! 7/2
15 ! 2
=
=
π
4
!
π!
mω
4 mω
= #1|ax |1" = 2a
(b). Calculate E02 , 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
En2 =
and we know
x=
so we nee to find
En2
(
|#n|ax4 |m"|2
0
0
m$=n En − Em
∑
!
(a† + a)
2mω
a2
=−
16
&
!
mω
'4
x4 =
!2
(a† + a)4
4m2 ω2
1
|#0|(a† + a)4 |m"|2
∑
!ω m$=0
m
now we need to foil the ladder operators, which I will not present here, but we find
#0|(a† + a)4 |1" = 0
√ √ √ √
√ √ √
√ √ √ √
#0|(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" = 2 6
so we find that the second order correction to the ground state energy is
a2
E02 = −
16
&
!
mω
'4
) √
√ *
&
'
1 (6 2)2 (2 6)2
21 2 ! 4
= −
+
a
!ω
2
4
8!ω
mω
Problem # 2
3
Two identical bosons are placed in an infinite square well. They interact weakly with one another, via
the potential
V (x1 , x2 ) = −aV0 δ(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 stateboth the wave functions and the associated energies.
The wave functions for a particle in a infinite square well with width a is given by
(
2 $ nπx %
sin
ψn =
a
a
we also know that we can write the combined wave function as a product of two wave functions, i.e
2 $ πx1 % $ πx2 %
ψ01 (x1 , x2 ) = ψ1 (x1 )ψ1 (x2 ) = sin
sin
a
a
a
and the energies are given by
E10 = E1 + E2 = 2E1 =
π 2 !2
ma2
The wave functions and associated energies for the first excited state are given by
√ ! $
&
'
&
' $
"
2πx2
2πx1
2
πx1 %
πx2 %
0
sin
sin
+ sin
sin
ψ2 =
a
a
a
a
a
and the energies are
π 2 !2
5 π 2 !2
=
2ma2
2 ma2
(b). Use first-order pertubation theory to calculate the effect of the particle-particle interaction on the
ground state and first excited state energies.
E20 = E1 + E2 = (n21 + n22 )
We know that
if we let
E10 = #ψ01 |H ! |ψ01 "
Z aZ a
$ πx %
$ πx %
4
1
2
= − V0
sin2
δ(x1 − x2 )dx1 dx2
sin2
a
a
a
0 0
Z a
$ %
4
4 πx
= − V0
dx
sin
a
a
0
u=
πx
a
a
dx = du
π
so we find
E10
π
4
= − V0
sin4 (u)du
π
0
"π
!
1
4
(12u)
= − V0
π
32
0
E10 =
Z
3
− V0
2
4
and for the first excited state
E20 = #ψ02 |H ! |ψ02 "
&
'
&
' $
"
Z aZ a! $
2
πx1 %
2πx2
2πx1
πx2 % 2
= − V0
sin
sin
+ sin
sin
δ(x1 − x2 )dx1 dx2
a
a
a
a
a
0 0
&
'
Z a
$ %
8
2 2πx
2 πx
= − V0
sin
dx
sin
a
a
a
0
if we let
u=
πx
a
a
dx = du
π
so we find
0
E12
E10
π
8
sin2 (u) sin2 (2u)du
= − V0
π
0
8 + $ u %,π
= − V0 4
π
16 0
= −2V0
Z
Problem # 3
A particle of mass m moves in a square well potential
0
|x| <
V (x) =
∞
|x| >
a
2
a
2
A delta function of height V0 (→ ∞) and width Δx(→ 0) is introduced at x = 0 (thus the area under the
peak is A = V0 Δx)
(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
#
 2 sin 1 nπx 2
odd wavefunction
a
ψn = # a
1
2
 2 cos nπx
even wave function
a
a
and we know that the first-order pertubation to the odd wave functions is
1
Eodd
= #n|H ! |m"
Z a/2
$ nπx %
2
=
δ(x)dx
A
sin2
a
a
−a/2
2
A sin2 (0) = 0
=
a
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
En0 − Em0 = (n2 + m2 )
5
π 2 !2
2ma2
and the coefficients can be found using
#ψm |H ! |ψn "
En0 − Em0
anm =
and
$ nπx %
$ mπx %
a/2
2
cos
δ(x)dx
#ψm |Aδ(x)|ψn " =
A
cos
a
a
a
0
2
=
A cos2 (0)
a
2
A
=
a
Z
so the coefficient is
anm =
4ma
A
2
2
2
π ! (n − m2 )
and the energy shift is
2
En1 = A
a
Z a/2
−a/2
cos2
$ nπx %
a
δ(x)dx =
2
A
a
Problem # 4
An electron of charge −e moves in a one-dimensional harmonic oscillator potential; the unperturbed
Hamiltonian is
p2 1 2 2
H =
+ mω x
2m 2
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 of a, a†and other quantities.
We know that
(
!
p2 1
+ mω2 x2 H ! = eEx x =
(a† + a)
H =
2m 2
2mω
and thus
(
!
!
H =
eE(a† + a)
2mω
(b). Find the first-order correction to the ground state energyusing time-independent pertubation theory.
The ground state wave function for the harmonic oscillator is
ψ0 =
thus the first order correction is
E01 = #0|H ! |0" =
$ mω %1/4
π!
(
mω 2
e− 2! x
!
eE#0|(a† + a)|0" = 0
2mω
6
(c). Find the second-order correction to the ground state energy using time-independent pertubation
theory
|#m|H ! |n"|2
En2 = ∑
En0 − Em0 = (n − m)!ω
0 − E0
E
n
m
m$=n
letting n = 0 we get
E02 = −
|#m|(a† + a)|0"|2
(eE)2
2mω2 m$∑
m
=n
the only allowed transition is where m = 1 and so we find
√
|#1|(a† + a)|0"|2 = [ 1]2 = 1
and so we find
E02 = −
(eE)2
2mω2
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 ψnp (x, y) (where n and p are positive integers), and the eigenvalues in terms of
the ground state energym E1 .
The infinite square potential is
0
if 0 ≤ x ≤ L, 0 ≤ y ≤ L
V (x, y) =
∞ otherwise
the wave function is ψn =
(
$ nπx %
2
sin
L
L
the eigenfunctins and the eigenvalues in terms of the ground state energy are
ψ0np =
$ nπx % $ pπy %
π 2 !2 2
E1
2
0
sin
sin
and Enp
=
(n + p2 ) = (n2 + p2 )
2
L
L
L
2mL
2
For the case where n $= p , the eigenfunctions are usually two-fold degenerate. Calculate the change in
the enrgies under the pertubation H ! = λE1 sin(πx/L), where λ * 1.
If we define our wavefunctions as
$ nπx % $ pπy %
2
sin
sin
L
L
L
$ pπx % $ nπy %
2
sin
= ψ0pn = sin
L
L
L
ψa = ψ0np =
ψb
if Wi j ≡ #ψ0i |H ! |ψ0j " then
Waa = #ψ0np |H ! |ψ0np " Wbb = #ψ0pn |H ! |ψ0pn " Wab = #ψ0np |H ! |ψ0pn "
for Waa we find
4
Waa = #np|H |np" = 2 λE1
L
!
using mathematica yields
Z L
0
sin
$ πx %
L
8
Waa = λE1
π
7
&
2
sin
$ nπx %
n2
4n2 − 1
L
'
dx
Z L
0
sin2
$ pπy %
L
dy
for Wbb we find
4
Wbb = #pn|H |pn" = 2 λE1
L
!
Z L
0
sin
using mathematica yields
$ πx %
L
8
Wbb = λE1
π
for Wab we find
4
Wbb = #np|H |pn" = 2 λE1
L
!
using mathematica yields
Z L
0
sin
$ πx %
L
sin
&
2
sin
L
sin
sin2
dx
Z L
sin
L
p2
4p2 − 1
$ nπx %
dx
Z L
$ pπx %
0
$ nπy %
L
dy
'
$ pπx %
L
0
$ nπy %
L
sin
$ pπy %
L
dy
Wab = 0
to find the energies we use
1
1
E±
= [Waa +Wbb ± (Waa −Wbb )]
2
1
E+
1
E−
&
'
8
n2
= λE1
π 4n2 − 1
&
'
p2
8
= λE1
π 4p2 − 1
Problem # 6
A two-dimensional isotropic oscillator has the Hamiltonian
&
'
!2 ∂2
k
∂2
H =−
+ 2 + (1 + bxy)(x2 + y2 )
2
2m ∂x
∂y
2
(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 = !ω non-degenerate
E1 = 2!ω double-degenerate
E2 = 3!ω 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
k
2
H ! = bxy(x2 + y2 )
and because the oscillator is isotropic we know that
(
(
!
!
x=
(a†x + ax ) y =
(a† + ay )
2mω
2mω y
8
and the perturbing Hamiltonian written in terms of raising and lowering operators is
kb!2
[(a† + ax )(a†y + ay )[(a†x + ax )2 + (a†y + ay )2 ]]
8m2 ω2 x
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
H !=
1
E00
= #00|H ! |00" = 0
the first excited state is doubly degenerate
ψ1a = ψ01 and ψ1b = ψ10
and by inspection we know that
Waa = #01|H ! |01" = Wbb = #10|H ! |10" = 0
and finally
Wab =
kb!2
#01|axay (a†y )2 + ax a†y ax a†x + ax a†y a†x ax + ax a†y ay a†y |10"
2
2
8m ω
looking at each case seperately we find
√ √ √ √
1 2 2 1=2
√ √ √
√
#01|axa†y ax a†x |10" =
2 2 1 1=2
√ √ √ √
#01|axa†y a†x ax |10" =
1 1 1 1=1
√ √ √ √
†
†
1 1 1 1=1
#01|axay ay ay |10" =
#01|axay (a†y )2 |10" =
thus
Wab =
3kb!2
= Wba
4m2 ω2
and so
3kb!2
Waa = Wbb = 0 Wab = Wba =
4m2 ω2
and to find the energies we use
1
E±
!
"
#
1
=
Waa +Wbb ± (Waa −Wbb )2 + 4|Wab |2
2
therefore
1
E+
= +Wab =
3kb!2
4m2 ω2
1
E−
= −Wab = −
3kb!2
4m2 ω2
1
E00
=0
Problem # 7
For the harmonic oscillator [V (x) = (1/2)kx2 ], the allowed energies are
En = (n + 1/2)!ω,
9
(n = 0, 1, 2, ...)
3
where ω = k/m is the classical frequency. Now suppose the spring constant increases slightly:
k → (1 + ε)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 ε, up
to second order.
(b). Now calculate the first-order correction pertubation energy. What is H ! here? Comapre your
results 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 ! = αδ(x − a/2)
where α 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 (En2 ). You can sum the series explicitly, obtaining
−2m(α/π!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
1 2
(px + p2y ) + mω2 (x2 + y2 )
2m
2
In parts (a) and (b) of the problem, the perturbing Hamiltonian H ! is given by
H ! = K px py
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!ω 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 ψ whatsoever; I claim that
Egs ≤ #ψ|H|ψ" ≡ #H"
To use the variational method you must follow these instructions
10
(9)
• You must guess a trial wave function for the ground state
ψ(!r, λ)
• If the wave function is not already normalized, you must normalize it
1 = #ψn |ψn "
• You must calculate the expectation of #H"
#H" = #ψ|H|ψ"
• once you have calculated the expectation values you must minimize it, i.e take the derivative with
respect to λ and solve for λmin
d#H"
=0
dλ
• Once you have found λmin you must use it to find ψ(r)
• Then we must use λmin to find the upper bound on the ground state energy
E0 ≤ #H"min
Problems
Problem # 1
A particle of mass m is contained in the one-dimensional potential well
V (x) = β|x|
Use the (normalized) gaussian
ψ(x) =
&
2α
π
'1/4
2
e−αx
as a trial function to find the upper bound on the ground state energy.
we know that
#H" = #T " + #V "
and also
2
d 2 −αx2
(e
) = e−αx (4α2 x2 − 2α)
2
dx
We will also be using the integral defined as
√ (2n)! $ a %2n+1
π
n!
2
0
to solve this problem. To find the expectation of the potential we use
Z ∞
#V " =
&
2α
π
2 /a2
x2n e−x
'1/2
2β
Z ∞
0
dx =
−2αx2
xe
&
2α
dx = 2
π
and for the expectation of the kinetic energy term we use
11
'1/2
·
β
β
=√
4α
2απ
(10)
!2
#T " = −
m
=
!2 α
2m
&
2α
π
'1/2 Z
0
∞
−2αx2
e
!2
=−
m
(4α2 x2 − 2α)dx
&
2α
π
'1/2
√
)
π α2
&
1
√
2α
'3
α
−√
2α
*
thus
#H" =
β
!2
β
!2 α
d#H"
+√
=
−
=0
and
2m
dα
2m 2(2απ)3/2
2απ
(11)
solving for α gives us
α=
&
β 2 m2
!4 2π
'1/3
pluggin this into equation 2 and doing a bit of algebra we find that the upper bound to the ground state
energy is
3
#H"min =
2
&
'
β2!2
1/3
2mπ
Problem # 2
Obtain an upper bound on the ground state energy of the hydrogen atom, using the variational technique with the wave function for the ground state of the three-dimensional oscillator as the trial function:
ψ=
&
2a
π
'3/4
e−ar
2
Obtain a numerical estimate (in eV) for the energy.
The Hamiltonian in 3-D is given by
H=−
!2 2
∇ +V
2m
where
& 2 '
&
'
&
'
1
∂
1
∂
∂
1 ∂
2 ∂
r
+ 2
sin θ
+ 2 2
∇ = 2
r ∂r
∂r
r sin θ ∂θ
∂θ
r sin θ ∂φ2
2
we know that the potential energy for the hydrogen atom is defined as
V (r) = −
e2
4πε0 r
and also
#H" = #T " + #V "
&
'
2
2
1 ∂
2 ∂
r
(e−ar ) = e−ar (4a2 r2 − 6a)
and 2
r ∂r
∂r
to find the expectation of the potential we use
12
&
2a
#V " = −
π
'3/2
e2
4πε0
Z 2π
0
dφ
Z π
0
sin(θ)dθ
Z ∞
0
−2ar2
re
&
2a
dr = −
π
'3/2
e2
ε0
Z ∞
0
2
re−2ar dr
and using the integral in equation 1 we find
&
2a
#V " = −
π
'3/2
e2 1
e2 $ a %1/2
·
=−
ε0 4a
ε0 2π3
and for the expectation of the kinetic energy term we use
'
Z ∞
2
2a 3/2
2π
r2 e−2ar (4a2 r2 − 6a)dr
π
0
& '1/2 Z ∞
2
2
2a
!
r2 e−2ar (4a2 r2 − 6a)dr
= − (4a)
m
π
0
)
&
'5
&
'3 *
'1/2
&
2
√
!
1
1
2a
π 4a2 · 12 √
− 12a √
= − (4a)
m
π
2 2a
2 2a
!2
#T " = −
m
&
3!2 a
=
2m
therefore
3!2 a e2 $ a %1/2
d#H" 3!2
e2
#H" =
and
−
=
−
2m
ε0 2π3
da
2m 2ε0
and solving for a we find
a=
&
1
2π3
'1/2
·
1
a1/2
=0
(12)
m2 e4
18ε20!4 π3
plugging this into equation 2 we get
e4 m
e4 m
3e4 m
−
=
#H"min =
36π3 !2 ε20 6π3 !2 ε20 π3 !2 ε20
&
1
3
−
36 6
'
=−
1 e4 m
12 π3 !2 ε20
therefore the upper bound to the ground state energy of the hydrogen atom is
)
& 2 '2 *
8
8
m
e
#H"min = −
= E1 = −11.54 eV
2
3π 2! 4πε0
3π
Problem # 3
A particle of mass m is bound by the three-dimensional potential V (r) = −V0 e−r/a where !2 /mV0 a2 =
3/4. Use the variational method with the trial function e−αr to estimate an upper bound on the lowest
energy eigenvalue.
Note: You should find a quartic equation in (aα). You should show (by plugging in) that the real roots
are aα = 0.5 and 4.369 × 10−2.
we know
13
V (r) = −V0 e−r/a
3
4 !2
!2
=
⇒
V
=
0
V0 a2 m 4
3 ma2
ψ(r) = e−αr
and we will also be using the integral defined as
Z ∞
0
xn e−x/a dx = n!an+1
(13)
to solve this problem. To find the expectation of the potential we use
&
'
1 ∂
2α
2 ∂
#H" = #T " + #V " and 2
r
(e−αr ) = e−αr (α2 − )
r ∂r
∂r
r
first we need to normalize the test function
A2
Z 2π
0
dφ
Z π
sin(θ)dθ
0
Z ∞
r2 e−2αr dr = 1
0
& '3
1
2
=1
A 8π
2α
& 3 '1/2
α
A=
π
thus
ψ(r) =
to find the expectation of the potential we use
α3
#V " = −V0 · 4π
π
Z ∞
0
&
α3
π
'1/2
2 −r(2α+ 1a )
r e
e−αr
αa
dr = −V0 8
2αa + 1
and for the expectation of the kinetic energy term we use
! 2 α3
#T " = −
· 4π
2m π
Z ∞
0
−2αr
e
&
'3
"
!
! 2 α2
! 2 α3
1
1
=
(α r − 2αr)dr = −
·2
−
m
4α 2α
2m
2 2
therefore, we are also plugging in for v0 we get
'3
&
'3
&
!2 α2 32!2
αa
αa
! 2 α2
=
− 8V0
−
#H" =
2m
2αa + 1
2m
3ma2 2αa + 1
taking the derivative we find
d#H" !2 α 32!2
=
−
dα
m
ma2
we find a quartic equation of the form
&
αa
2αa + 1
'2 !
"
(2αa + 1)a − 2α2a2
=0
(2αa + 1)2
(2αa + 1)4 = 32αa
and this equation satisfies both roots that are given, i.e αa = 0.5 and αa = 4.369 × 10−2
14
Using αa = 0.5 we find the upper bound to the ground state energy as
!
"
! 2 α2 1
32αa
#H"min =
−
m 2 3(2αa + 1)3
letting α =
0.5
a
and αa = 0.5 we get
#H"min = −
!2
24ma2
if we use the other root αa = 4.369 × 10−2 we find
!2
ma2
which is not an upper bound on the energy. So we use αa = 0.5 to find our upperbound for the energy.
Problem # 4
Consider a particle with mass m moving in the one-dimensional potential
#H"min = 2.65 × 10−4
V (x) = λx4
(a). Consider a single-paramter ansatz for the wave function consisting of ground state wave function
for a simple harmonic oscillator
with frequency ω, where ωis the variational parameter. Find the value of ω that minimizes #H" and
obtain 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 # 5
Suppose you’re given a quantum system whose Hamiltonian H0 admits two eigenstates, ψa (with
energy Ea ), and ψb (with energy Eb ). They are orthogonal, normalized, and non-degenerate (assume Ea is
the smaller of the two energies). Now turn on a pertubation H ! , with the following matrix elements:
#ψa |H ! |ψa " = #ψb |H ! |ψb " = 0 #ψa |H ! |ψb " = #ψb |H ! |ψa " = 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 a
trial of the form
ψ = (cos φ)ψa + (sin φ)ψb
where φ is the adjustable paramter. Note : Writing the linear combiunation in this way is just a neat
way to guarantee that ψis normalized.
(d). Compare your answers. Why is the variational principle so accurate, in this case?
15
1.1 Electron Spin and Angular Momentum
[Lx , Ly ] = i!Lz
[Ly , Lz ] = i!Lx
it follows (as before) that the eigenvectors of
L2
[Lz , Lx ] = i!Ly
and Lz satisfy
L2 |l m" = !2 l(l + 1)|l m"
Lz |l m" = !m|l m"
and
where L± ≡ Lx ± iLz and
L± |l m" = !
3
l(l + 1) − m(m ± 1)|l (m ± 1)"
l = 0, 1, 2, 3, ...
m = −l, −l + 1, ..., l − 1, l
The algebraic theory of spin is a carbon copy of the theory of orbital angular momentum, beggining
with the fundamental commutation relations:
[Sx , Sy ] = i!Sz
[Sy, Sz ] = i!Sx
[Sz , Sx ] = i!Sy
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 m"
and
where S± ≡ Sx ± iSz and
3
S± |s m" = ! s(s + 1) − m(m ± 1)|s (m ± 1)"
1 3
s = 0, , 1, , ...
2 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 orbital
angular 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
for fermions
2
and m j = ml + ms
where l and s are
l = l1 + l2 + ... and s = s1 + s2 + ...
the eigenfunctions and eigenvalues are given as
#J 2 " = # j! m! j |J 2 | j, m j " = j( j + 1)!2 δ j! j δm! j m j
#Jz " = # j! m! j |Jz | j, m j " = m j !δ j! j δm! j m j
#
#J± " = # j! m! j |J± | j, m j " = j( j + 1) − m j (m j ± 1)!δ j! j δm! j m j ±1
a couple of other useful relations is
#Jy" =
1
(J+ − J− )
2i
16
1
#Jx " = (J+ + J− )
2
1.3 Spin Orbit Energy
The pertubation Hamiltonian is given as
Hrel = −
p4
8m20 c2
the spin orbit energy is given by
1
|En |α2
1
e2
l+1 if j = l + 1/2 where α ≡
#ΔE!S·!L " =
≈
is the fine structure constant
1
n(2l + 1) − l if j = l − 1/2
4πε0 !c 137
the Relativistic correction to the energy is given by
#ΔErel " =
5 α2 |En |
4 n
therefore the total energy is given as
α2 |En |
#ΔE!S·!L " + #ΔErel " =
n
&
3
1
−
4n j + 1/2
'
1.4 Hyperfine Interaction in H
The eigenfunctions ang eigenvalues are given as
F = !I + J! where !I is the nuclear spin interaction
and
#I 2 " = i(i + 1)!2
#Iz " = mi ! |mi | ≤ i
and the magnetic moment is
µi =
gi µN !
I
!
#µz " = gi µN mi
and we also know
#F 2 " = f ( f + 1)!2
#Fz" = m f ! |m f | ≤ f
Problems
Problem # 1
The electron confiuguration of oxygen is 1s2 2s2 2p4 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?
17
(a). First we know that the 1s2 2s2 2p4 state is equivalent to 4 p electrons and we can treat this problem
as 2 p holes.
l1 = l2 = 1
s1 = s2 =
1
2
so
l!


2
= l1 + l2 = 1


0
l1 + l2
l1 − l2
1
s! = s1 + s2 =
0
s1 + s2
s1 − s2
j ! = 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 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
30
= 15 allowed states
2!
we know
s1 = s2 =
1
2
l1 = l2 = 1


2
!
l = l1 ± l2 = 1


0
1 symmetric
s! = s1 ± s2 =
0 anti-symmetric
symmetric
anti-symmetric
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!
0
1
1
0
1
0
l!
0
0
1
1
2
2
j!
0
1
2,1,0
1
3,2,1
2
2s! +1 L
1S
3S
j!
0
1
3P
2,1,0
1P
1
3D
3,2,1
1D
2
18
m!j = ∑(2 j! + 1)
1
3
9
3
15
5
and the number of allowed states is given by
1
S0, 3 P2,1,0 . 1 D2 total number of states is 15
(b). What is the ground state of oxygen?
To find the ground state, we must use Hund’s 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 is
more 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 1 S0, 3 P2,1,0 . 1 D2
!
lmax
=1
s!max = 1
a representation of this is given by
ms
+ 21
− 21
and so
ml
0
↑
1
↑
-1
!
jmax
= l ! + s! = 2
so the ground state of oxygen is
3
P2
(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
1
Eh f ∝ #J! ·!I" J! ·!I = [F 2 − I 2 − J 2 ] where !F = !I + !S
2
and
[Note: # of Eh f
Now
1
!2
#J! ·!I" = [#F 2 " − #I 2" − #J 2 "] = [ f ( f + 1) − i(i + 1) − j( j + 1)]
2
2
levels is the # of f values].
i =
5
2


2
j= 1


0

9 7 5 3

2, 2, 2, 2
f = 27 , 52 , 32

5
2
19
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
1S
0
3P
0
3P
1
3P
2
1D
2
has 1 hyperfine level
has 1 hyperfine level
has 3 hyperfine level
has 5 hyperfine level
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 m j
(b). If we include the effects of the proton spin, what are the possible values of ( f , m f )?
(a). Our values are given as


2
l = n−1 = 1


0


2





1
for l = 2 ml = 0



−1



−2


1
for l = 1 ml = 0


−1
1
1
for l = 0 ml = 0 ms = ± s =
2
2
Since we are adding s = 21 to l, we have j values of l ± 1/2 for l $= 0 and j = 1/2 for l = 0. In all cases,
the z-component is conserved: ms + ml = m j . In cases where m j = ±(l + s), only j = l + s is possible.
The table is given as
l
2
2
2
2
2
2
2
2
2
2
1
1
1
1
1
1
0
0
ml
2
2
1
1
0
0
-1
-1
-2
-2
1
1
0
0
-1
-1
0
0
ms
1/2
-1/2
1/2
-1/2
1/2
-1/2
1/2
-1/2
1/2
-1/2
1/2
-1/2
1/2
-1/2
1/2
-1/2
1/2
-1/2
j
5/2
5/2,3/2
5/2,3/2
5/2,3/2
5/2,3/2
5/2,3/2
5/2,3/2
5/2,3/2
5/2,3/2
5/2
3/2
3/2,1/2
3/2,1/2
3/2,1/2
3/2,1/2
3/2
1/2
1/2
20
mj
5/2
3/2
3/2
1/2
1/2
-1/2
-1/2
-3/2
-3/2
-5/2
3/2
1/2
1/2
-1/2
-1/2
-3/2
1/2
-1/2
(b). If we include the effects of the proton spin, what are the possible values of ( f , m f )?
The total angular momentum, F is the sum of the electron’s total angular momentum J and the proton’s
spin. Since s p = 1/2and j ranges from 1/2 to 5/2, the possible values for f are 5/2±1/2=3,2; 3/2±1/2=2,1:
and 1/2±1/2=1,0. The possible values of m f for a given f are f , f − 1, ..., − f , so the possible values of
( f , m f ) are
f
3
2
1
0
mf
3,2,1,0,-1,-2,-3
2,1,0,-1,-2
1,0,-1
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). nd 1 n! d 1 [with n $= n! ]
(b). p1 d 2
(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 possible
orbital angular momentum states are L = 4, 3, 2, 1, 0. Any combination of these is possible, so we have.
s!
0
0
0
0
0
1
1
1
1
1
l!
0
1
2
3
4
0
1
2
3
4
j!
0
1
2,1,0
3
4,2,1
1
2,1,0
3,2,1
4,3,2
5,4,3
2s! +1 L
1S
j!
0
1P
1
1D
2
1F
3
1G
4
3S
1
3P
2,1,0
3D
3,2,1
1F
4,3,2
3G
5,4,3
m!j = ∑(2 j! + 1)
1
3
5
7
9
3
9
15
21
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 Hund’s rules, we maximize
S → 1, L → 4, and minimize J → |L − S| = 3to get a term of 3 G3 .
(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 the
electrons now has l = 1. Constructing a similar table we have
s!
0
0
0
1
1
1
l!
1
2
3
1
2
3
j!
1
2,1,0
3
2,1,0
3,2,1
4,3,2
2s! +1 L
1P
1
1D
2
1F
3
j!
3P
2,1,0
3D
3,2,1
1F
4,3,2
21
m!j = ∑(2 j! + 1)
3
5
7
9
15
21
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 3 F2 .
(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 momentum
states 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 spin
symmetric states with anti-symmetric orbital angular momentum states and anti-symmetric spin states
with symmetric orbital angular momentum states.
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 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
30
= 15 allowed states
2!
we know
s1 = s2 =
1
2
l1 = l2 = 1


2
!
l = l1 ± l2 = 1


0
1 symmetric
s! = s1 ± s2 =
0 anti-symmetric
symmetric
anti-symmetric
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!
0
1
0
l!
0
1
2
j!
0
2,1,0
2
2s! +1 L
1S
j!
0
3P
2,1,0
1D
2
m!j = ∑(2 j! + 1)
1
9
5
and the number of allowed states is given by
1
S0, 3 P2,1,0 . 1 D2 total number of states is 15
22
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, maximize
J → 2, giving 3 P2 .
Problem # 4
A beam of hydrogen atoms, emitted from an oven 400 K, passes through a Stern-Gerlach magnet of
length 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
1
E = mv2 = 2kB T
2
we also know that the potential and the force are defined as that the atom is experiencing are
U = −!µ!B and F = −!∇U = !∇!µ!B
therefore the force in the z direction is given by
eSz
∂B
where µz = −gs
∂z
2m
we know that gs ≈ 2 for our case. We find that the force in the z direction is
Fz = µz
e! ∂B
2m ∂z
to find the Δz we have to use the classical kinematic equations.
Fz = ±
Fz = me az ⇒ az =
Fz
e! ∂B
=
mz 2m2e ∂z
we also know that
1
Δz
= z0 +V0,zt + azt 2 z0 = 0 V0,z = 0
2
2
therefore
Δz = azt 2
we know that
e! ∂B
az =
and t =
2m2e ∂z
thus
Δz =
(
me
4kb T
e! ∂B
= 4.21 × 10−3m
me 8kb T ∂z
Problem # 5
23
For the hydrogen atom, the quantum number j can take only the values 1/2, 3/2, 5/2, ..., depending on
the value of l.
Obtain the matrices J 2 , Jz , J+, J− , Jy for l = 0, 1
if l = 0 then ml = 0 and m j = ml + ms = ± 21 . The number of matrix elements is described by 2 j + 1,
where j = l + s and s = 21 , therefore j = 12 and the number of matrix elements for l = 0 is 2.
for J 2 we find
!
# j! , m j |J 2 | j, m j " = j( j + 1)!2 δ j! j δm!j m j
therefore we know that m j = (−1/2, 1/2) and m!j = (−1/2, 1/2) , so we find
3
J = !2
4
2
&
1 0
0 1
'
for Jz we find
!
# j! , m j |Jz| j, m j " = m j !δ j! j δm!j m j
where m j = (−1/2, 1/2) and m!j = (−1/2, 1/2), so we find
1
Jz = !
2
&
−1 0
0 1
'
to find J+ , J− we need to use
!
# j! , m j |J± | j, m j " =
#
j( j + 1) − m j (m j ± 1)!δ j! j δm!j m j +1
#
j( j + 1) − m j (m j + 1)!δ j! j δm!j m j +1
to find J+ we use m j = − 12 because we are using the raising operator. Using
!
!
# j , m j |J± | j, m j " =
where m j = (−1/2, 1/2) and m!j = (−1/2, 1/2), we find
J+ = !
&
0 1
0 0
'
&
0 0
1 0
'
and J− is just the transpose of J+ which yields
J− = !
and finally Jy is found by using
24
therefore
& '
&!
" !
"'
1
!
0 1
0 0
Jy =
(J+ − J− ) =
−
0 0
1 0
2i
2i
!
Jy =
2
&
0 −i
i 0
'
and also
therefore
& '
&!
" !
"'
1
!
0 1
0 0
Jx =
(J+ + J− ) =
+
0 0
1 0
2
2
!
Jx =
2
&
0 1
1 0
'
now if l = 1 then ml = −1, 0, 1 and m j = ml + ms = − 32 , − 12 , 21 , 32 . The number of matrix elements is
described by 2 j + 1, where j = l ± s and s = 12 , therefore j = 23 , 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 = 21 so we can just add these matrix to the 4x4 matrix.
for J 2 we find
!
# j! , m j |J 2 | j, m j " = j( j + 1)!2 δ j! j δm!j m j
therefore we know that m j = (−1/2, 1/2, −3/2, −1/2, 1/2, 3/2) and m!j = (−1/2, 1/2, −3/2, −1/2, 1/2, 3/2)
, so we find

for Jz we find


1 2
2
J = ! 
4 


3
0
0
0
0
0
0 0 0 0 0
3 0 0 0 0
0 15 0 0 0
0 0 15 0 0
0 0 0 15 0
0 0 0 0 15








!
# j! , m j |Jz| j, m j " = m j !δ j! j δm!j m j
where

to find J+ , J− we need to use


1 
Jz = ! 
2 


−1
0
0
0
0
0
0 0
0 0 0
1 0
0 0 0
0 −3 0 0 0
0 0 −1 0 0
0 0
0 1 0
0 0
0 0 3
25








!
!
# j , m j |J± | j, m j " =
#
j( j + 1) − m j (m j + 1)!δ j! j δm!j m j +1
where j = 1/2, 3/2 and m j = (−1/2, 1/2, −3/2, −1/2, 1/2, 3/2) and m! j = (−1/2, 1/2, −3/2, −1/2, 1/2, 3/2).
For J+ we find




J+ = ! 



0
0
0
0
0
0
1
0
0
0
0
0
and J− is just the transpose of J+ which yields




J− = ! 



0
1
0
0
0
0

0 0 0 0
0 √0 0 0 

3 0 0 
0

0 0 2 √0 

0 0 0
3 
0 0 0 0
0 0 0 0
0 0 0 0
0 √0 0 0
3 0 0
0
0 0 2 √0
0 0 0
3
and finally Jy is found by using



& '

!
1

(J+ − J− ) = 
Jy =

2i
2i 


0
0
0
0
0
0
1
0
0
0
0
0
therefore

and also


!
Jy = 
2





& '

1
!

Jx =
(J+ + J− ) = 

2
2 


0
0
0
0
0
0








 
0 0 0 0

0 √0 0 0 
 

0
3 0 0 
−

0 0 2 √0  

0 0 0
3  
0 0 0 0
0
1
0
0
0
0
0 0 0 0
0 0 0 0
0 √0 0 0
0
3 0 0
0 0 2 √0
0 0 0
3
0 −i 0
0
0
0
i 0
0
0√
0
0
0 0 √0 −i 3 0
0
0 0
3
0
−2i
0√
0 0
0
2i
0 −i 3
√
0
0 0
0
0
i 3
0
0
0
0
0
0
1
0
0
0
0
0
 
0 0 0 0

0 √0 0 0 
 

0
3 0 0 
+

0 0 2 √0  

0 0 0
3  
0 0 0 0
26
0
1
0
0
0
0
0
0
0
0
0
0

0
0
0
0
0
0
















0 0 0 0
0 0 0 0
0 √0 0 0
0
3 0 0
0 0 2 √0
3
0 0 0







therefore



!
Jx = 
2


0
1
0
0
0
0

1 0
0
0
0
0 0 √0
0
0 

0 √0
3 0
0 

0
3 0
2 √0 

0 0
2 √0
3 
3 0
0 0
0
Problem # 6
Write down all the possible eigenkets | j, m j " corresponding to the case l = 3 , s = 1/2
if l = 3 and s = 1/2 then j = l ±s = 7/2, 5/2 and ml = −3, −2, −1, 0, 1, 2, 3 and ms = ±1/2. Therefore
if j = 7/2 we find m j = ml + ms = −7/2, −5/2, −3/2, −1/2, 1/2, 3/2, 5/2, 7/2 . If j = 1/2 then m j =
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
AB
C B
C B
C B
C B
C B
C B
C B
CD
B7 7 B7 5 B7 3 B7 1 B7 1 B7 3 B7 5 B7 5
| j, m j " = BB , −
,B ,−
,B ,−
,B ,−
,B ,
,B ,
,B ,
,B ,
2 2 B2 2 B2 2 B2 2 B2 2 B2 2 B2 2 B2 2
and for j = 5/2 we find all the eigenkets to be
C B
C B
C B
C B
C B
CD
AB
B5 5 B5 3 B5 1 B5 1 B5 3 B5 5
B
B
B
B
B
B
, ,−
, ,−
, ,
, ,
, ,
| j, m j " = B , −
2 2 B2 2 B2 2 B2 2 B2 2 B2 2
Suppose now that the spin-orbit coupling is turned off so that ml and ms become good quantum numbers. For each of the eigenkets | j, m j " just obtained, write down the corresponding eigenkets |ml , ms "
B
C B
C
B7 7
B
1
B ,±
B
B 2 2 = B±3, ± 2
B
C B
C B
C
B
B
B7 5
1
1
B
B
B ,±
B 2 2 = B±3, ∓ 2 , B±2, ± 2
B
C B
C B
C
B7 3
B
B
1
1
B ,±
B
B
B 2 2 = B±2, ∓ 2 , B±1, ± 2
B
C B
C B
C
B
B7 1
1 BB
1
B
B ,±
B 2 2 = B±1, ∓ 2 , B0, ± 2
B
C B
C B
C
B5 5
B
1 BB
1
B ,±
B
B 2 2 = B±3, ∓ 2 , B±2, ± 2
B
C B
C B
C
B
B5 3
1 BB
1
B
B ,±
B 2 2 = B±2, ∓ 2 , B±1, ± 2
B
C B
C B
C
B5 1
B
B
1
1
B ,±
B
B
B 2 2 = B±1, ∓ 2 , B0, ± 2
Problem # 7
27
The relativistic correction to the ground state energy of the hydrogen atom can be found using a Hamiltonian of the form H ! = −p4 /8m30 c2 , where p is the electron momentum and m0 is the electron rest mass.
Calculate the energy correction using first-order pertubation theory. [Hint: #ψ|p4 |ψ" = #p2 ψ|p2 ψ" , since
p2 is Hermitian.
E01 = #ψ100 |H ! |ψ100 "
and p2 and the ground state wave function is
p2 = −!2 ∇2 and ψ100 = √
1
πa3
e−r/a
a=
4πε0 !2
me2
α≡
e2
4πε0 !c
(14)
therefore
E01 = #ψ100 |H ! |ψ100 " = −
!2
#∇2 ψ100 |∇2 ψ100 "
8m30 c2
and
&
&
'
'
1 ∂
1
2
2 ∂
−r/a
−r/a
∇ ψ100 = 2
r
e
=e
−
r ∂r
∂r
a2 ar
2
I left out the constant that comes along with ψ100 because it is not needed in the last calculation. So
we find
E01
'
&
2 2
!4
1
ψ100 |ψ100 "
= − 3 2# 2 −
a
ar
8m0 c
if we let λ =
!4
8m30 c2
thus
E01
'
&
Z 2π Z π
Z 2π Z π
Z ∞
1
2 2
λ
2 −2r/a
dr
and
dφ
sin θdθ = 4π
dφ
sin θdθ
r e
−
=− 3
πa 0
a2 ar
0
0
0
0
&
'
Z ∞
1
4
4
λ
2 −2r/a
dr
r e
+
−
= − 3 4π
πa
a4 r2 a2 ra3
0
&
'
Z
4λ ∞ 2 −2r/a 1
4
4
=− 3
r e
+
−
dr
a 0
a4 r2 a2 ra3
Z
Z
Z
16λ ∞ −2r/a
4λ ∞ 2 −2r/a
16λ ∞ −2r/a
r e
dr + 5
=− 7
e
dr − 6
re
dr
a 0
a 0
a 0
Using
Z ∞
0
= xn e−x/a dx = n! an+1
we find the first-order purtubation to be
28
E01
E01
! & 3 '"
! "
a
16λ + a , 16λ a2
4λ
− 6
+ 5
= − 7 2!
a
8
2
4
a
a
5λ
= − 4
a
)
& 2 '2 * &
'2
m
e2
e
5
5 !4
5
= −
− 2
= − |E1 |α2
=
3
2
4
8 m0 c a
4
2! 4πε0
4πε0 !c
4
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
F
1 E
|ψ"S = √ |ψα (1)"|ψβ(2)" + |ψβ(1)"|ψα(2)"
2
F
1 E
|ψ"A = √ |ψα (1)"|ψβ(2)" − |ψβ(1)"|ψα(2)"
2
|ψT " = |ψα (1)"|ψβ(2)" or |ψT " = |ψβ (1)"|ψα(2)"
The Slater determinent is given by


ψa (1) ψa (2) ψa (3)
1
√  ψb (1) ψb (2) ψb (3) 
N! ψ (1) ψ (2) ψ (3)
c
c
c
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
ψ0 =
$ mω %1/4
mω 2
e− 2! x
π!
we also know that the first order relativistic correction in terms of the kinetic energy is
!
Hr = −
p4
8m3 c2
and p2 = −!2 ∇2
29
thus the first order correction to the ground state wavefunction is given by
E01 = #ψ0 |H ! |ψ0 " = −
!4
#∇2 ψ0 |∇2 ψ0 "
3
2
8m c
where
∇2 ψ0 =
so we find
$ mω %1/4 $ mω % mω 2 $ mω
$ mω %1/4 ∂2
%
2
− 2! x
2
− mω
2! x ) =
e
x
−
1
(e
π!
∂x2
π!
!
!
%2
!4 $ mω %2 $ mω 2
#
x
−
1
ψ0 |ψ0 "
8m3 c2 !
!
! Z ∞ !$
"
"
Z ∞
mω 2
!4 $ mω %2 $ mω %1/2
mω %2 4
mω 2 − mω x2
x
−
= − 3 2
2
x −2
x e ! dx +
e ! dx
8m c
!
π!
!
!
0
−∞
)
&
'
&
'
&
' *
! 5/2 mω ! 3/2
! 1/2
!2 m2 ω2 $ mω %1/2 √ 3 m2 ω2
−
+
π
= −
8m3 c2 π!
4 !2
mω
! mω
mω
*
) &
'
&
'
&
'
!2 ω2 $ mω %1/2 3 ! 1/2
! 1/2
! 1/2
= −
−
+
8mc2 !
4 mω
mω
mω
E01 = −
E01 =
−
3 !2 ω2
32 mc2
Problem # 9
Calculate the wavelength, in centimeters, of the photon emitted under a hyperfine transition in the
ground state (n = 1) of deuterium . Deuterium is “heavy” hydrogen, with an extra neutron in the nucleus.
gd e
The proton and neutron bindtogether to form a deuteron, with spin 1 and magnetic moment µd = 2m
Sd ;
d
the deuteron g-factor gd = 1.71. [Hint: This is intended to be a very short problem-just scale down from
hydrogen with appropriate g-factors and masses.]
we know
gd e
Sd and gd = 1.71
2md
and the wavelength of the emmited photon under a hyperfine splitting is given as
µd =
λd =
hc
ΔEd
and λd =
ΔEh
λh
ΔEd
ΔEh =
µ0 g p e2
3πm p me a3
to find the hyperfine splitting for the deuteron we use
Eh1 f =
µ0 gd e2
#Sd · Se "
3πmd me a3
but we know
30
λh = 21 cm
1
Sd · Se = (S2 − Se2 − Sd2 )
2
we have two special cases
-
S2 =
3 2
4!
15 2
4!
when s = − 12
when s = 21
Eh1 f
µ0 gd e2 !2
=
3πmd me a3
Sn2 = !2 sn (sn + 1)
3
S2e = !2
4
S2d = 2!2
and so we find
−1
+ 12
(singlet)
(triplet)
therefore
µ0 gd e2 !2
2πmd me a3
therefore the wavelength of the emmited photon is
ΔEd =
λd =
2g p md
2(2)(5.91)
λh =
λh = 4.36(21 cm) = 91.53 cm
3gd m p
3(1.71)
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
1
s! 1 = s!2 = s!3 =
l2 = 1

2

l3 = 1
we also know that the number of possible states is given by
and also
(2l1 + 1)(2s1 + 1) · (2l1 + 1)(2s1 + 1) · (2l1 + 1)(2s1 + 1) = 216 states
s! = s! 1 + s! 2 =
-
0
1
+ s! 3 =
-
1
2
3
2
×2
×1


2
!
l = l1 + l2 = 1


0
The ×n notation is simply the multiplicity of the given value, also
S, P, D, F, ... = (l = 0, 1, 2, 3, 4, ...) respectivily
31

3



2 ×2
+ l3 =

1 ×3



0
and the table is given by
s!
3/2
3/2
3/2
3/2
1/2
1/2
1/2
1/2
l!
3
2
1
0
3
2
1
0
2s! +1 L
j!
9/2,7/2,5/2,3/2
7/2,5/2,3/2,1/2
5/2,3/2,1/2
3/2
7/2,5/2
5/2,3/2
3/2,1/2
1/2
j!
4F
9/2,7/2,5/2,3/2
4D
7/2,5/2,3/2,1/2
4P
5/2,3/2,1/2
4S
3/2
2F
7/2,5/2
2D
5/2,3/2
2P
3/2,1/2
2S
1/2
m!j
28×1
20×2
12×3
4×1
14×2
10×4
6×6
2×2
and when we add all the possible number of states we find a total of 216 possible states.
Problem # 11
Find all the spectral terms allowed by the Pauli Exclusion Principle that arises from the
Russell-Saunders coupling of 8 equivalent d 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]
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 10 states allowed in the D shell and the
second electron can only go into any of the 9 other states. thus the number of possible states is
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
= 45 allowed states
2!
we know
s1 = s2 =
1
s! = s1 ± s2 =
0
1
2
l1 = l2 = 2


4





3
!
l = l1 ± l2 = 2



1



0
symmetric
anti-symmetric
symmetric
anti-symmetric
symmetric
anti-symmetric
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
32
s! = 1 l ! = 3, 1
s! = 0 l ! = 4, 2, 0
and the table is given by
s!
1
1
0
0
0
l!
3
1
4
2
0
2s! +1 L
j!
4,3,2
2,1,0
4
2
0
m!j
21
9
9
5
1
j!
3F
4,3,2
3P
2,1,0
1G
4
1D
2
1S
0
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+++ 4f3 5s2 p6
Gd+++ 4f7 5s2 p6
Ho+++ 4f10 5s2 p6
thus, the 5sp subshells are full, and all the action arises from the 4f subshells.
Hund’s rules are given as
1. Full shells and subshells do not contribute to total S, the total spin angular momentum and L, the
total 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 of L has the lowest energy.
4. For atoms with less then half-filled shells, the level with the lowest value of J 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 Hund’s rules to obtain the spectral terms for each ion in the ground state.
for Nd+++ we find
ml
ms
+ 21
− 21
3
↑
2
↑
1
↑
0
33
-1
-2
-3
therefore
3
2
= 6
maximum S! =
maximum L!
and since this shell is less than half-filled we find the lowest value of J !
minimum J ! = |L! − S! | =
9
2
therefore the spectral term is given by
4
I9/2
for Gd+++ we find
ml
ms
+ 21
− 21
3
↑
2
↑
1
↑
0
↑
-1
↑
-2
↑
-3
↑
-1
↑
-2
↑
-3
↑
therefore
7
2
= 0
maximum S! =
maximum L!
and since this shell is half-filled we find J ! to be
J ! = S! =
7
2
therefore the spectral term is given by
8
S7/2
for Ho+++ we find
ml
ms
+ 21
− 21
3
↑
↓
2
↑
↓
1
↑
↓
0
↑
therefore
maximum S! = 2
maximum L! = 6
34
and since this shell is more than half-filled we find the highest value of J !
maximum J ! = |L! + S! | = 8
therefore the spectral term is given by
5
I8
(ii). Calculate the g-values and effective magnetic moments for these ions
The g-values and effective magnetic moments are given by
!
"
3
j( j + 1) + s(s + 1) − l(l + 1)
g = 1+
and #µi " = −gµB j( j + 1)
2 j( j + 1)
µB =
e!
2m
for Nd+++ we find
s=
3
2
l=6
j=
9
2
so we find
)
g = 1+
3 5
9 11
2 ( 2 ) + 2 ( 2 ) − 6(7)
2 92 ( 11
2)
*
=
8
11
#µi " = −
6 e! √
11
11 m
for Gd+++ we find
s=
7
2
l=0
j=
7
2
so we find
)
g = 1+
7 7
7 7
2(2) + 2(2)
2 72 ( 72 )
*
= 2
#µi " = −
3 e! √
7
2m
for Ho+++ we find
s=2
l=6
j=8
so we find
"
!
5
72 + 6 − 42
g = 1+
=
144
4
#µi " = −
15 e! √
2
4 m
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.
35
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
5
E1
3
10
2k( j + 2) =
k( j + 1)
3
1
jmin =
2
E2 =
therefore
36
jmax =
7
2
1
2
jmin =
thus
jmax = l + s =
7
2
jmin = |l − s| =
1
2
and we find
3
7 5 3 1
j ! = , , , l ! = 2 s! =
2 2 2 2
2
Problem # 14
Consider a single d electron (l = 2). Calculate the allowed values of l, j, m j and g, presenting your
results 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 m j and the transitions in terms of Δm j . [Relavent selection
rules: Δl = ±1, Δ j = 0, ±1, Δm j = 0, ±1]
We know that
j = |l ± s| thus l = 2 s =
1
2
5 3
j= ,
2 2
and the table is given by
l
j
mj
g
2
5
2
5
2
5
2
5
2
5
2
5
2
3
2
3
2
3
2
3
2
5
2
3
2
1
2
- 21
- 22
- 25
3
2
1
2
- 21
- 23
6
5
6
5
6
5
6
5
6
5
6
5
4
5
4
5
4
5
4
5
2
2
2
2
2
2
2
2
2
37
Figure 1: Possible transitions from D to P shells
Problem # 15
Consider a single p electron. Obtain all the possible kets | j, m j ", in terms of the kets |ml , ms " with the
appropriate Clebsch-Gordan coefficients.
since we are considering an electron in a p shell we know that
3 1
l=1 j= ,
2 2
and the m j values are given as
3
3 1 1 3
mj = , ,− ,−
2
2 2 2 2
1
1 1
j =
mj = ,−
2
2 2
j =
and the eigenkets are given as
| j, m j " = |ml, ms "
we will need to use
B
B
C
C
B
B3 3
1
B ,±
= BB±1, ±
B2 2
2
B
B
C
CB
C
B3 1
B
B
1
1
B ,±
B0, ±
= BB±1, ∓
B2 2
2 B
2
B
B
C
CB
C
B
B
B1 1
1
1
B0, ±
B ,±
= BB±1, ∓
B2 2
2 B
2
| j, m j "! =
∑
m j=ml +ms
|ml , ms"#ml , ms | j, m j "!
38
where #ml , ms | j, m j " is the Clebsh-Gordon coefficient, to find our C-G coefficients. We will also need
to use raising and lowering operators.
#
J± | j, m j "! = ! j( j + 1) − m j (m j ± 1)| j, m j ± 1"!
3
L± |l, ml "! = ! l(l + 1) − ml (ml ± 1)|l, ml ± 1"!
3
S± |s, ms"! = ! s(s + 1) − ms(ms ± 1)|s, ms ± 1"!
and we also know that
J± = L± + S±
J− | j, m j " = L− |l, ml " + S−|s, ms "
we know that the C-G coefficients for
B
C
B3 3
B ,±
B2 2 =
B
C
B
B±1, ± 1
B
2
is 1, now for the rest of the eigenkets we find
B
B
B
C
C
C
B3 3
B 1
B 1
J− BB ,
= L− BB1,
+ S− BB1,
2 2
2
2
using the raising and lowering relationships we find
B
B
B
C
C
C
√ B 1
√ B3 1
B
1
B
B
B
= 2 B0,
+ 1 B1, −
3B ,
2 2
2
2
therefore
and next we use
we find
therefore
B
B
C ( B
C
C
B3 1
1
2 BB 1
1 BB
B ,
+ √ B1, −
0,
B2 2 =
3B 2
2
3
B
B
B
C
C
C
B3 3
B
B
1
1
B
B
B
J+ B , −
= L+ B−1, −
+ S+ B−1, −
2 2
2
2
B
B
B
C
C
C
√ B
√ B3 1
B
1
1
B
B
B
= 2 B0, −
+ 1 B−1,
3B ,−
2 2
2
2
B
B
C ( B
C
C
B
B3 1
B
1
2
1
1
B ,−
B0, −
+ √ BB−1,
B2 2 =
3B
2
2
3
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.
39
B
B
B
C
C
C
B1 1
B
B 1
1
B ,
= α BB1, −
+ β BB0,
B2 2
2
2
B
B
B
C
C
C
B
B
B1 1
1
1
B
B
B ,−
= δ B−1,
+ γ B0, −
B2 2
2
2
B
B
G
G
we can make B 21 , 12 orthogonal to B 32 , 12 by saying
BB
H
C
1 1 BB BB 3 1
=0
,
,
2 2BB2 2
thus we find
B
B
B" )( B
H
C
C*
! H
1
1 BB
2 BB 1
1 BB
1 BB
0,
+ √ B1, −
=0
α 1, − B + β 0, B
2
2
3B 2
2
3
(
(
1
1
2
2 1 1
1 1
1
1
α
1
α#1, − |0, " + β √ #0, |1, − " + √ #1, − |1, − " + β
#0, |0, " = 0
3
2 2
2
2
2
3 2 2
3 2
3
we can see that the first two → 0 due to orthonormality and so
(
α
2
√ = −β
and we know |α|2 + |β|2 = 1
3
3
therefore
(
α=−
2
3
1
β= √
3
and so we find
( B
B
B
C
C
C
B 1
B1 1
B
2
1
1
B
B ,
B
B 2 2 = − 3 B1, − 2 + √3 B0, 2
to find the last C-G coefficients
H
thus we find
BB
C
1 1 BB BB 3 1
=0
,−
,−
2 2BB2 2
B
B
B" )( B
H
C
C*
! H
1
2 BB
1 BB
1
1 BB
1 BB
0, −
=0
+ √ B−1,
δ −1, B + γ 0, − B
2
2
3B
2
2
3
1
δ
1
√ #−1, | − 1, " + γ
2
2
3
(
(
1
2
2
1
1
1
1
γ
1
#0, − |0, − " + δ
#−1, |0, − " + √ #0, − |−, " = 0
3
2
2
3
2
2
2
2
3
40
we can see that the last two → 0 due to orthonormality and so
(
δ
2
√ = −γ
and we know |δ|2 + |γ|2 = 1
3
3
therefore
(
δ=−
2
3
1
γ= √
3
and so we find
( B
B
B
C
C
C
B1 1
1
1 BB
2 BB
1
B ,−
B 2 2 = − 3 B−1, 2 + √3 B0, − 2
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
'
&
Z
1 x2 3
1
π
ΔΦ =
2m[E −V (x)]dx = n +
! x1
2
and the tunneling rate can be found using
2
ln Tn = −
!
Z x2 3
x1
2m[V (x) − E]dx
Problems
Problem # 1
A particle of mass m and electric charge q is confined within a one-dimensional well, with
−V0 |x| < a/2
V (x) =
0
|x| > a/2
assume that V0 1
π2 !2
,
ma2
i.e. the well is deep.
41
(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 ε0 (in the x-direction) is applied to the system. Choose the zero of the
electrostatic potential φ so that (φ(x = 0) = 0. Assume that V0 1 qε0 a.
(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
nπ! =
Z 3
2m(E −V (x)dx =
and solving for the energy we find
Z a/2 3
−a/2
2m(E +V0 )dx =
1
E = −V0 +
2
&
nπ!
a
3
2m(E +V0 )a, n = 1, 2, 3, ...
'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 energy
bound state corresponds to
&
'
1 nbound π! 2
=0
−V0 +
2m
a
solving for nbound , we find
a3
nbound =
2mV0
π!
(c). Now the bound states are no longer stable. Explain why.
Turning on the electric field adds a term −qε0 x 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 its
energy 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 −V0
Since V0 1
π2 !2
,
ma2
we approximate (as specified) the ground state energy as
& '
1 π! 2
≈ −V0
E " −V0 +
2m a
The tunneling factor is given by e−2γ , where
γ=
1
!
Z x1 3
x0
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 ⇒ −qε0 x1 = −V0 ⇒ x1 =
42
V0
qε0
we now evaluate
1
γ=
!
√
√
2m
2
1 2 2m
(0 − (−qεx0 +V0 )3/2 ≈
(V0 )3/2
2m(−qε0 x +V0 )dx =
! 3 −qε0
3!qε0
Z x1 3
x0
where we use V0 1 qε0 a in the approximation. Our tunneling probability is then
J
√
4 2m
(V0 )3/2
T 2 exp −
3!qε0
I
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 one
dimensional 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 infinite
potential barrier in this 1-D problem, so the phase change is given by ΔΦ = (n + 34 )π instead of (n + 21 )π.
Hint:
√
Z ∞
√ −x
π
xe dx =
2
0
We know that
&
'
Z
3
1 r0 3
2m(E −V (r))dr = n +
ΔΦ =
π n = 0, 1, 2, ...
! 0
4
Z
1 r0 3
2m(E −V0 ln(r/a))dr
=
! 0
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
!
If we let
Z r0 3
0
1 r0 3
2m(E −V0 ln(r/a))dr =
2mV0 (ln(r0 /a) − ln(r/a))dr
! 0
Z r0 3
13
=
2mV0
ln(r0 /r)dr
!
0
Z
x = ln(r0 /r)
r = r0 e−x
dr = −r0 e−x dx
so we find
13
2mV0
!
√ √
'
&
Z ∞
√ −x
3
π 2mV0 E/V0
pi
xe dx =
ae
= n+
0
2
43
!
4
thus
E/V0
e
√
2 π!(n + 3/4)
√
=
a 2mV0
and finally
&
"
'"
'
! (
! (
&
2π
3
! 2π
!
3
n+
= V0 ln n +
+V0 ln
En = V0 ln
a mV0
4
4
a mV0
Problem # 3
A pareticle of mass m moves in the potential V (x) = β|x|, where β is a constant. Use the WKB
approximation to find the energy levels, En .
the first thing we know is that
β|x| = E thus x = ±
E
β
x1 = −
E
β
x2 =
E
β
and we know that the energies can be found using
1
ΔΦ =
!
thus we know
so we find
1
!
Z x2 √
x1
Z E/β √
−E/β
2
!
2m(E −V (x))
2m(E − β|x|)
Z E/β √
0
1/2
1/2
'
&
1
π
dx = n +
2
&
'
1
dx = n +
π
2
2m(E − βx)1/2 dx
44
if we let
u = E − βx
du = −βdx
du
−
= dx
β
thus
2
−
!β
*0
√
√
4
2m
2m 3/2
4
2m(u)1/2 du = −
u3/2 =
E
3!β
3!β
Z 0√
E
)
E
and so we find the allowed energies to be
√
&
'
1
4 2m 3/2
En = n +
π
3!β
2
⇒
&
' "2/3
1
3!β
n+
En = √
π
2
4 2m
!
Problem # 4
Use the WKB approximation to find the allowed energies En of an infinite square well with a “shelf”,
of height V0 , extending half-way across:


if 0 < x < a/2
V0 ,
V (x) = 0,
if a/2 < x < a


∞
otherwise
Express your answer in terms of V0 and En0 ≡ (nπ!)2/2ma2 (the nth allowed energy for the “unperturbed” infinite square well, with no shelf). Assume that E10 > V0 , but do not assume that En 1 V0 .
45
the allowed energies can be found using
1
ΔΦ =
!
Z x2 √
x1
2m(E −V (x))
1/2
&
'
1
dx = n +
π
2
since we know that “n” is proportional to “n+1/2” we can just write this as
Z
1 x2 √
ΔΦ =
2m(E −V (x))1/2 dx = nπ
! x1
to find what the integral is we can split it up into to pieces i.e
√ !Z
√
"
Z a
,
a/2
2m
2m a +
1/2
1/2
dx =
(E −V0 )1/2 + E 1/2 = nπ
(E −V0 ) dx +
E
!
! 2
0
a/2
thus
√
and so we find
foiling the left side we find
,
2m a +
1/2
1/2
(E −V0 ) + E
= nπ
! 2
+
,2
n2 π 2 !2
= 4En0
(E −V0 )1/2 + E 1/2 = 4
2ma2
3
2E + 2 E(E −V0 ) −V0 = 4En0
if we rearrange this equation and square both sides we find
and this simplifies to
3
(2 E(E −V0 )2 = (4E +V0 − 2E)2
4E 2 − 4EV0 = 16(E 0n )2 + 4En0V0 − 8En0 E + 4En0V0 +V02 − 4EV0 − 8EEn0 + 4E 2
thus we find
16EEn0 = 16(En0 )2 + 8En0V0 +V02
and so we find
En = En0 +
V2
V0
+ 00
2 16En
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 ε. The various energies and potentials are indicated in the figure. The
surface of the metal is the plane x = 0, and the potential for x > 0 (vacuum) is:
V (x) = ϕ + EF − eεx
where EF is the Fermi level of the electron gas in the metal and ϕ is the work function. Use WKB
approximation to show that the transmission coefficient is given by
)
*
4 √ ϕ3/2
T = exp −
2m
3
!eε
Cold emmision is the principle underlying the field ion microscope.
46
since we know that the transmission coefficient can be found using
√
2 R x2 2m(V (x)−E)1/2 dx
x1
T = e− !
we can write this as
ln T = −
2
!
Z x2 √
x1
2m(V (x) − E)1/2 dx
and since we know what the potential is, we get
2
−
!
Z Φ/eε √
0
2m(ϕ + EF − eεx − EF )
1/2
2
dx = −
!
Z Φ/eε √
0
2m(ϕ − eεx)1/2 dx
if we let
u = ϕ − eεx
du = −eεdx
du
−
= dx
eε
substituting this into the above equation we get
*0
) √
√ Z
√
2 2m 0 1/2
4 2m 3/2
4 2m 3/2
=−
(u) du =
u
ϕ
!eε ϕ
3!eε
3!eε
ϕ
thus we find the transmission cofficient to be
√
4 2m 3/2
ϕ
ln T = −
3!eε
and solving for T we find
)
4 √ ϕ3/2
T = exp −
2m
3
!eε
47
*
Problem # 6
Use the WKB approximation in the form
Z r2
r1
p(r)dr = (n − 1/2)π!
to estimate the bound state energies for hydrogen. Don’t forget the centrifugal term in the effective
potential. The following integral may help:
Z b 3
√
π √
1
(x − a)(b − x)dx = ( b − a)2
2
a x
Note that you recover the Bohr energy levels when n 1 1 and n 1 1/2. answer
Enl ≈
−13.6 eV
3
[n − (1/2) + l(l + 1)]2
Chapter 6: Time-Dependent Pertubation Theory
6.1 General Formalism
this is an exact solution but totally useless
1
i!ȧk = ∑ an(t)#k|H ! |n"eiωkn t where ωkn = [Ek0 − En0]
!
the problem is that ȧk depends on the time dependence of all an (t). We can find ak by using
ak (t) = −
where H
state.
! are
1
!
Z t
0
!
H k!j eiωk jt dt !
the appropriate matrix elements and k is the final state of the system and j is the initial
6.2 Harmonic Pertubations
The transition probability/unit time to any one of the final states |k" is given by
W=
1
|ak (t)|2
t∑
k
R
if we assume many states we can write this as an integral ∑k → ρ(Ek )dEk
1
W=
ak (t)ρ(Ek)dEk where dEk = !d(Δω)
t
Fermi’s Golden Rule #1 is given by
Z
W ∼ |#k|α2|i"#i|α1| j"|2
Fermi’s Golden Rule #2 is given by
W=
2π
|#k|α| j"|2ρ(Ek )
!
48
6.3 Interaction of Atomic Electron With E-M Radiation
The Hamiltonian for this interaction is given by
and the transition rates are given by
H k!j = eE!0 #k|!r| j" cos ωt
Wk j =
e2 π
|#k|!r| j"|2ρ(ωk j )
3ε0 !2
6.4 Selection Rules for Electric Dipole Transitions
The selection rules are as follows
Δ" = ±1 for x, y, andz axis
Δm" = 0, ±1 0 for the z axis and ± 1 for the x and y axis
ΔS = 0
The spin cannot flip in an electric dipole interaction.
The polarization of an E-M wave
i.e. !E field is along z-axis, and ,
Δm" = 0 only if #k|z| j" =
$ 0
Δm" = ±1 only if #k|x| j" = #k|y| j" =
$ 0
also, the component of !E-field are out of phase by π/2.
The Weak Field Limit: !S ·!L Coupling
The quantum numbers n, ", j, m j can be expressed
|n " j m j " in terms of |n " m" ms "
will give us a new set of selection rules
Δ"
Δj
Δm j
ΔS
=
=
=
=
±1 for x, y, andz
0, ±1 (cannot go from 0 → 0)
0, ±1 (not from 0 → 0 if Δ j = 0)
0
6.5 Spontaneous Emmision: Einstein A and B coefficients
The transmission and absorption rates are given by
RT = [A + Bρ(ω)]N2 transmission rate
RA = Bρ(ω)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.
+
,
e2 ω3
|#1|!r|2"|2
A = Bρ(ω) e!ω/kb T − 1 =
3
3π!c ε0
49
6.6 Higher Order Transitions
We have the following pertubation for transitions for an E-M field given by
H ! = −eE0 #k|!reik·r | j" cos ωt
and if
eik·r = 1
k·r * 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
1
(ik · r)2 + ..]| j"
2!
where 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.
#k|!reik·r | j" = #k|!r[1 + ik · r +
6.7 Life Times and Line Widths
Natural Broadening
The transition rate is defined as
RT =
1
τ
where τ is the lifetime
τ ∼ 10−9 s electric dipole transitions
τ ∼ 106 s electric quadrupole transition
The uncertanty principle say
!
τ
this means that the emitted photon is not sharp but is spread out
ΔE · τ ≈ !
⇒ ΔE ≈
!ω has a width
1
τ
Collision
The path length is given by
1
4σn
where σ 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
l
τc =
v̄ is the average velocity
v̄
and the rate is given by
1
RT =
τc
The equipartition theorem states
(
3
|v̄| =
kb T
2
where kb is the Boltzman constant and T is the temperature.
l=
50
Problems
Problem # 1
An infinite one-dimensional square well [V=0, for |x|≤ 2a , V=∞ elsewhere] contains a particle of mass
m in the ground state. At time t = 0 a pertubation H ! = Aδ(x) is turned on: δ(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 appropriate matrix elements. Look carefulle at the parity of the two matrix elements before evaluating them!
“Initial” means ωt * 1, where !ω is the relavent energy differece.
We are given
H ! = Aδ(x)
ΔE = !ω
and the wave functions for the one-dimensional infinite square well are given by
(
(
$ nπx %
2 $ nπx %
2
sin
ψeven =
cos
ψodd =
a
a
a
a
The integral expression derived in class is given by
F
1E
i t ! iωk j t ! !
dt
ωk j = Ek0 − En0
H k je
ak (t) = −
! 0
!
where k is the final state of the system and j is the initial state.
Z
We need to find the matrix elements H k!j . (we also know that n = 1 is an even wave function and n = 2
is an odd wavefunction)
For the n : 1 → 2 we find
H 21!
2A
= #2|Aδ(x)|1" =
a
Z a/2
−a/2
cos
thus we know that
a
'
2A
2πx
sin
δ(x)dx =
cos(0) · sin(0) = 0
a
a
Z t
H 21! eiω21t dt ! = 0
$ πx %
i
a2 (t) = −
!
0
&
!
For the n : 1 → 3 we find
H 31!
2A
= #3|Aδ(x)|1" =
a
we know what ω31 is
Z a/2
−a/2
cos
$ πx %
a
&
'
2A
3πx
2A
cos
δ(x)dx =
cos(0) · cos(0) =
a
a
a
F 4π2 !
1E 0
E3 − E10 =
!
ma2
we will need to use the following identity for the next part
ω31 =
2i sin(ω31t/2) = eiω31t/2 − e−iω31t/2
we find the coefficient to be
51
!
"
2A eiω31t − 1
e
dt = −
t
!a
ω31t
0
0
*
)
!
"
2A iω31t/2 eiω31t/2 − e−iω31t/2
2A iωt/2 i sin(ω31t/2)
= − e
t =− e
t
!a
ω31t
!a
ω31t/2
i
a3 (t) = −
!
Z t
!
H 31! eiω31t dt !
2Ai
=−
!a
Z t
iω31 t !
!
so we get
!
"
2A iω31t/2 i sin(ω31t/2)
t
a3 (t) = − e
!a
ω31 t/2
The rate of transition is given by
|a3 (t)|2
=
t
&
2A
t
!a
'2 &
sin(ω31t/2)
ω31t/2
'2
since ω31t * 1
sin(ω31t/2)
⇒1
ω31 t/2
we get
|a3 (t)|2
4A2
= 2 2t
t
! a
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 !ε =!ε0 e−t/τ is
applied to the atom. The perturbing Hamiltonian is thus
H ! = −e!r ·!ε0 e−t/τ = −ezε0 e−t/τ and since z is given by z = r cos(θ)
we find the Hamiltonian to be
H ! = −er cos(θ)ε0 e−t/τ
and the three hydrogen wave functions needed for this problem are given as
&
'
r
1
1
1
1
1
−r/a0
1−
e−r/2a0 ψ210 = √
cos(θ)e−r/2a0
ψ100 = #
e
ψ200 = √
5/2
2a0
2πa0 2a0
2π 4a
πa30
0
(a) Show that the electron has zero probability of being excited into the 2s state Ψnlm = Ψ200
what we are looking for is |a200 (t)|2 so we must use the formula
i
a200 (t) = −
!
Z t
0
!
H k!j eiωk jt dt !
where
!
H 200,100
= #200| − er cos(θ)ε0e−t/τ |100" = −eε0 e−t/τ #200|r cos(θ)|100"
eε0 e−t/τ
= −√
8πa30
Z 2π
0
dφ
Z π
0
cos(θ) sin(θ)dθ
52
Z ∞
0
'
&
r
e−3r/2a0 dr
r 1−
2a0
3
but since we know that
Z π
0
cos(θ) sin(θ)dθ = 0
causes
!
H 200,100
=0
and thus
i t ! iωk jt ! !
H e
dt = 0
a200 (t) = −
! 0 kj
and if you square this number, you still get 0 thus the probability of this transition happening is nill.
Z
(b) Show that after time t 1 τ, the probability that the atom is in the 2p state Ψnlm = Ψ210 is
|a210 (t)|2 =
e2 ε20 a20
215
310 !2 (ω2 + 1/τ2 )
!
we must first find the matrix element H 210,100
which is given by
!
H 210,100
= #210| − er cos(θ)ε0e−t/τ |100" = −eε0 e−t/τ #210|r cos(θ)|100"
eε0 e−t/τ
= −√
8πa40
Z 2π
0
dφ
Z π
0
2
cos (θ) sin(θ)dθ
Z ∞
0
r4 e−3r/2a0 dr
and so we find the integrals to be
Z 2π
0
thus we find
dφ
Z π
0
cos2 (θ) sin(θ)dθ
Z 2π
0
dφ
Z π
0
letting u = cos(θ) du = − sin(θ)dθ
2
cos (θ) sin(θ)dθ = 2π
Z −1
1
u2 du =
4π
3
and for the other integral we find
Z ∞
0
4 −3r/2a0
r e
&
2a0
dr = 4!
3
putting this all together we get
!
H 210,100
'5
'
&
eε0 e−t/τ 4π
28 a0
2a0 5
√ eε0 e−t/τ
=− √
=
−
4!
3
8πa40 3
35 2
now that we have this solution we can find the coefficient. Plugging this into the equation derived in
class we find
i t ! iωk jt ! !
a210 (t) = −
H e
dt
! 0 kj
Z
ieε0 a0 28 t t ! (iω−1/τ) !
√ 5
e
dt
=
!
23 0
Z t
+
,
1
t(iω−1/τ)
t ! (iω−1/τ) !
e
−1
e
dt =
iω − 1/τ
0
Z
53
but since we know that
et(iω−1/τ) → 0
for t 1 τ
and so we find the coefficient to be given as
ieε0
a210 (t) = −
!
&
a
√0
2
'
28
35
&
1
iω − 1/τ
'
and the probability is given by
|a210 (t)|2 = a210 (t)a210 (t)∗ =
e2 ε20 a20
215
310 !2 (ω2 + 1/τ2 )
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
p2 1 2
+ kx
H0 =
2m 2
A homogenius electric field ε(t) is switched on at t = 0, so that the system is perturbed by the interaction
H ! = −qxε(t)
if ε(t) has the form
ε(t) = ε0 e−t/τ
where ε0 and τ are constants, and if the oscillator is in the grounds state for t ≤ 0, find the probability
that 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 light
that 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
2
2
|!E(t)| = E0 cos(ωt)e−t /T
with E0 , ω, and T all constants. The frequency ω satisfies
8
!ω = (13.6 eV)
9
The constant T is large compared to all other timescales in the problem. At t → ∞, the atom is, in
general, 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) = H 0 (t) + λH ! (t)
54
the Adiabatic approximation is used if H (t) varies very slowly
i(φn −φk )
ȧk = − ∑ an #ψk |ψ̇n "e
n
and if k $= n we find
#ψk |ψ̇n " =
the Bohr angular frequency is given by
ωkn =
1
φn = −
!
Z t
0
En (τ)dτ
#ψk |H˙ |ψn "
En − Ek
Ek − En
!
and
Rt
an
#ψk |H˙ |ψn "ei 0 ωk j (τ)dτ
n$=k !ωk j
ȧk = − ∑
6.9 Sudden Approximation
if a system is in specific eigenstate at t = 0 and H 0 → H 1 “instantaneously” we use sudden approximation
method. Suppose that the eigenfuntion
|Ψ" → |Φ"
then
t < 0 we find |Ψ(!r,t)" = ∑ an |ψn "e−iEnt/!
n
t > 0 we find ||Φ(!r,t)" = ∑ bη |φ
η
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
ˆ y + k̂Bz is a constant and !L is the angular momentum operator.
Hamiltonian !B ·!L cos ωt, where !B = îBx + jB
By considering appropriate matrix elements, show that the selection rules for magnetic dipole transitions are Δ" = 0, Δm" = 0, ±1. Which of these apply when !L is (i) parallel, and (ii) perpendicular to the
z-axis? [Hint: Express Lx and Ly in terms of L+ and L− .]
We know that
H ! = !B ·!L cos ωt = (Bx Lx + By Ly + Bz Lz ) cos ωt
thus we know that
#k|H ! | j" = cos ωt[Bx #k|Lx | j" + By#k|Ly | j" + Bz #k|Lz | j"]
55
for !L parallel to the z-axis we find
Bx Lx = By Ly = 0
and
#k|H ! | j" = Bz #k|Lz | j" cos ωt = Bz #l ! l|Lz |m! l ml " = Bz cos ωt !mδl ! l δm! l ml
thus the selection rules when !L is parallel to the z-axis
Δl = 0
Δm = 0
if !L is perpendicular to the z-axis we find
Bz Lz = 0
and
#k|H ! | j" = cos ωt[Bx#k|Lx | j" + By #k|Ly | j"]
we know that Lx and Ly can be expressed by lowering and raising operators
Ly =
1
(L+ − L− )
2i
1
Lx = (L+ + L− )
2
since we know that the relationship for finding expectation values using lowering and raising operatots
is given by
#l ! m! |L± |lm" =
thus
and
#l ! m! |L+ |lm" =
#l ! m! |L− |lm" =
3
3
3
l(l + 1) − ml (ml ± 1)!δl ! l δm! l ml ±1
l(l + 1) − ml (ml + 1)!δl ! l δm! l ml +1
l(l + 1) − ml (ml − 1)!δl ! l δm! l ml −1
thus the selection rules when !L is perpendicular to the z-axis are
Δl = 0
Δm = ±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/τ)2] where E0 and τ are constants.
Thus, the perturbing Hamiltonian is -qE(t)x. Assuming that dE(t)/dt is small, and that at t=−∞ the oscillator 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 in
terms 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
if we let
p2 1
+ mω2 x2 − qE(t)x
H =
2m 2
k = mω2
56
we find
H =
p2 1 2
+ kx − qE(t)x
2m 2
we can complete the square and we find
)&
' &
' *
qE(t) 2
qE(t) 2
p2 1
+ k x−
−
H =
2m 2
k
k
we get
H =
now we need to find
letting x0 =
qE(t)
k
F
p2 1 E
+ k (x − x0 )2 − x20
2m 2
(
!
∂H
= −kx˙0 (x − x0 ) − kx˙0 x = −kx˙0 x = −
mω2 x˙0 (a† + a)
∂t
2mω
and so
∂H
=
∂t
and
#k|H˙ |0" =
and we also know
(
(
2
! 2qt
E0 e−(t/τ) (a† + a)
2
2mω τ
2
! 2qt
E0 e−(t/τ) #k|(a† + a)|0"
2
2mω τ
E1 − E0 !ω(1 + 1/2) − !ω(1/2)
=
=ω
!
!
this tells us that the only allowed transition is from n = 0 → n = 1. So the probabilty transition is found
by using
B
B
B 1 Z t ∂H iωt ! ! B2
2
B
P10 = |a1 (t)| = B
e dt BB
!ω 0 ∂t !
ω10 =
so we can find
1
a1 (t) =
!ω
(
! 2q
E0
2mω τ2
if we let
1
K=
!ω
we find
a10 (t) = K
(
Z ∞
−∞
Z ∞
2
−∞
te−(t/τ) eiω10t dt
! 2q
E0
2mω τ2
2 −iω t)
10
te−((t/τ)
dt
integrals of this form can be solved by completing the square, i.e
Z ∞
−∞
2 +bx)
e−(ax
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
τ2
b = −iω10
57
we get
Z ∞
(yτ − iωτ2 /2)e−y
Z ∞
yτe−y dy + e−(ωτ
a10 (t) = Kτ
−∞
which gives us
a10 (t) = −e−(ωτ
2 /4)
Kτ
2
−∞
but we know that
Z ∞
−∞
so the integral simplyfies to
−(ωτ2 /4)
a10 (t) = e
2 −(ωτ2 /4)
2 /4)
Kτ
dy
Z ∞
iωτ2 −y2
e dy
−∞
2
2
yτe−y dy = 0
√
Z ∞
π
iω −y2
−(ωτ2 /4)
3
e dy = e
Kτ iω
Kτ
3
2
−∞
2
thus the probability is given by
|a10 (t)|2 = e−(ω10 τ
=
2 /2)
K 2 τ6 ω2
π
4
q2 E02 τ2 π −(ω10 τ2 /2)
e
2m!ω
Problem # 3
The Hydrogen nucleus of mass 3 (tritium) is radioactive, and decays into a helium nucleus of mass 3
by β-decay. (The helium atom is thus singly ionized.) Suppose the hydrogen atom is initially in its ground
state. 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
3
3
1H → 2H
by β decay n → p+ + e− + ν¯e
the states are given by


1s
state =
2s


2p
n = 1 l = 0 ml = 0
n = 1 l = 0 ml = 0
n = 2 l = 1 ml = ±1, 0
For helium, spherical harmonics are unchanged (µ ≈ me ) but the radial wave functions are changed
because they depend on the potential, which depend on z. (z hydrogen=1, z helium=2)
ψ100 = √
ψ210
1
3/2
e
πa0
1
= √
4 2π
−r/a0
&
z
a0
'3/2
zr −zr/2a0
e
cos θ
a0
&
'3/2 !
"
zr −zr/2a0
2−
e
a0
& '3/2
z
zr −zr/2a0
1
e
ψ21±1 = √
sin θe±iφ
a0
8 π a0
1
ψ200 = √
4 2π
z
a0
the probability that 1s → 2s is given by
P = |#200|100"|2
58
so we first need to find
#200|100" =
Z ∞
0
Z ∞!
1
ψ∗200 ψ100 r drdΩ = 3 √
a0 2
2
this can be solve using
Z ∞
"
2r −2r/a0 2
2−
e
r dr
a0
0
xn e−x/a = n!an+1
0
so we find
"
! $ %
1
a0 3 6 $ a0 %4
4
−
=−
#200|100" = 3 √ 2 ·
2
a0 2
2
a0 2
and the probability is given by
1
4
|#200|100"|2 =
and the probability that 1s → 2p is not allowed, because
|#210|100"| =
Z ∞
0
2
ψ∗210 ψ100 r dr
Z π
0
Z 2π
0
dφ
Z π
0
sin θ cos θdθ
sin θ cos θdθ = 0
therefore
|#210|100"|2 = 0
and we also find
|#21 ± 1|100"| =
Z ∞
0
ψ∗21±1 ψ100 r2 dr
Z π
0
Z 2π
0
dφ
Z π
0
sin2 θdθ
sin2 θdθ = 0
therefore
|#21 ± 1|100"|2 = 0
Problem # 4
An electron is held in a spin up state along the z−axis by a magnetic filed, !B = B0 ẑ. 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?
59
(c). Now suppose the weak permanent magnetic field described in (b) is along the x−direction with
a magnitude B1 * B0 .The experimenter slowly reverses the magnetic field along the z−direction in the
following manner:
t
Bz (t) = −B0 for − T ≤ 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 electric
field ε is turned on in an amount of time much less than ω−1 , where ω is the angular frequency of the
oscillator. 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 E the scattering cross-section
is given by
prob
S
where S =
and I = prob flux = v|ψi |2
σ=
I
sec
and the differential cross section is given as
D(θ) =
S(θ)
dσ
=
dΩ
I
where S(θ) is the probability/second that particle scatters into a unit solid angle at angle θ. We are
assuming that scattering center has all the mass. We also know that the total cross section is given as
σ≡
Z
D(θ)dΩ
7.1 Born Approximation
If we take a cube of side L, the density of the states is given by
ρθ (k f ) =
mL3 k f
8π3!2
we can write the initial and final wavefunctions as
|ψi " =
1
L3/2
!
eiki!r
|ψ f " =
1
L3/2
!
eik f!r
and we will need to use Fermi’s golden rule # 2, and we find
thus
$ m %2
dσ
S(θ) Wi f (θ)
=
=
= |Vk f ·ki |2
dΩ
I
I
2π!2
$ m %2
dσ
= |Vk f ·ki |2
dΩ
2π!2
60
where
|Vk f ·ki | =
if we let
Z
V (r)ei(ki −k f )·r dV
! =!ki −!k f
K
|!ki | = |!k f |
than we find for K
K = 2k sin(θ/2)
thus we can write
4π
Vk f ·ki =
K
Z
V (r) sin(Kr)rdr
Problems
Problem # 1
Use the Born approximation to obtain the differential and total cross section for scattering by the
exponential potential V (r) = V0 e−αr .
The differential cross section (in terms of Vk!f !ki ) is given as
$ m %2
dσ
= |Vk!f !ki |2
dΩ
2π!2
where
4π ∞
V (r) sin(Kr)rdr
Vk!f !ki =
K 0
The potential for this system is given as
Z
V (r) is the potential of the system.
V (r) = V0 e−αr
so we have
Vk!f !ki =
4π
K
Z ∞
0
V0 e−αr sin(Kr)rdr
but we know that
eiKr − e−iKr
2i
Note: This integral has limits that are imaginary, this is not a problem because the function has no
poles and we may choose our contour along the real line where all other paths vanish. Thus our effective
limits are from 0 → ∞. Thus
sin(Kr) =
2πV0
Vk!f !ki =
Ki
and using
Z ∞
0
−αr
e
−e
2πV0
)rdr =
Ki
Z ∞
xn e−x/a dx = n!an+1
iKr
(e
−iKr
0
Z
(e−(α−iK)r − e−(α+iK)r )rdr
we find
!
"
1
1
8παV0
2πV0
−
= 2
Vk!f !ki =
2
2
Ki (α − iK)
(α + iK)
(α + K 2 )2
61
and
|V k!f !ki |2 =
(α8πV0 )2
(α2 + K 2 )4
K = 2k sin(θ/2)
and the differential cross section is
dσ
=
dΩ
&
&
'
'
mV0 2
16α2
(α8πV0 )2 $ m %2
=
(α2 + k2 )4
2π!2
(α2 + 4k2 sin2 (θ/2))4 !2
and the total cross section can be found by integrating over the solid angle
16α2
dσ =
(α2 + 4k2 sin2 (θ/2))4
thus
σ=
&
mV0
!2
&
'2
mV0
!2
16α
'2
2
dΩ =
Z 2π
0
dφ
&
mV0
!2
Z π
0
'2
16α2
Z
1
dΩ
(α2 + 4k2 sin2 (θ/2))4
sin(θ)
dθ
(α2 + 4k2 sin2 (θ/2))4
we know that
1
sin2 (θ/2) = [1 − cos(θ)] let u = 1 − cos(θ) du = sin(θ)dθ
2
so the integral now becomes
σ=
&
mV0
!2
'2
2
16α · 2π
Z 2
0
1
(α2 + 2k2 u)4
du let w = α2 + 2k2 u du =
dw
2k2
thus we find
σ=
&
mV0
!2
'2
π
16α · 2
k
2
Z α2 +4k2
α2
16πα2
w dw =
3k2
−4
&
mV0
!2
'2 !
1
1
− 2
6
α
(α + 4k2 )3
"
Problem # 2
2
A particle of mass m is scattered by an attractive Gaussian potential of the form V (r) = −V0 e−(r/R) ,
where V0 and R are constants.
(a). Sketch the form of the potential.
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(b). Using the first Born approximation, calculate the differential cross section as a function of the
incident k-value and the angle θ through which the particle is scattered.
The differential cross section is given by
$ m %2
dσ
= |Vk!f !ki |2
dΩ
2π!2
where
4π ∞
4πV0
Vk!f !ki =
V (r) sin(Kr)rdr = −
K 0
k
as before, this can be written
Z
2πV0
Vk!f !ki = −
Ki
Z ∞
−((r/R)2 −iKr)
re
0
Z ∞
0
2πV0
dr +
ki
2
e−(r/R) sin(Kr)rdr
Z ∞
0
2 +iKr)
re−((r/R)
)dr
integrals of this form can be solved by completing the square, i.e
Z ∞
−∞
e−(ar
2 +br)
dr let y ≡
√
a[r + (b/2a)] then (ar2 + br) = y2 − (b2 /4a)
if we apply this to our problem we know that
a=
1
R2
b = ±iK
r = Ry −
R2 b
dr = Rdy
2
so the integral now becomes
−R2 K 2 /4 2πV0
Vk!f !ki = −e
Ki
!Z
0
∞
−y2
e
R3 iK
)dy −
(R y +
2
2
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Z ∞
0
−y2
e
' "
&
R3 iK
2
dy
R y−
2
which symplifies to
−R2 K 2 /4
Vk!f !ki = −e
2πV0R
3
Z ∞
0
2
e−y dy
and using
Z ∞
0
2 /a2
x2n e−x
dx =
this becomes
−R2 k2 /4
Vk!f !ki = −e
2πV0 R
3
Z ∞
0
√ (2n!) $ a %2n+1
π
n!
2
2
e−y dy = −(π)3/2V0 R3 e−R
2 k2 /4
so the solution is given by
|Vk!f !ki |2 = (π)3V02 R6 e−R
2 K 2 /2
but we know K = 2k sin(θ/2)
and the differential cross section is given by
dσ $ m %2 3 2 6 −2kR2 sin2 (θ/2)
πm2 2 6 −2k2 R2 sin2 (θ/2)
(π) V0 R e
V0 R e
=
=
dΩ
2π!2
4!4
(c). Sketch the form of the differential cross section versus θ , and comment on its relationship to the
scattering potential.
The form of the differential cross section has a Gaussian form just like the the scattering potential ,
one is increasing and the other one is decreasing.
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(d). Write down the angle θ! at which the differential cross section falls of to 1/e of its value at θ = 0.
we can solve this by letting the exponent be
2kR2 sin2 (θ! /2) = 1
sin(θ! /2) = √
θ
!
1
2kR
−1
2 sin
=
&
1
√
2kR
'
(e). Now consider the high-energy limit kR1 1. Using your result in (d), find an approximate value
for θ! , and show that θ! * 1.
For the high energy limit we know that
sin(θ! /2) ≈
thus
θ!
2
for very small angles
1
θ!
≈√
2
2kR
so we find
!
θ ≈
√
2
kR
Problem # 3
Consider a particle of mass m scaterring off a delta-shelf potential in 3D
V (r) = V0 bδ(r − b)
(a). What is the scattering amplitude of a particle with energy !2 k2 /2m in the first Born approximation.
(b). What is the differential cross section in the first Born approximation?
(c). What is the total cross section in the limit k → 0?
Problem # 4
Calculate the differential and total cross-sections for scattering from a Yukawa potential
V (r) = β
e−µr
r
where β and µ are constants in the Born approximation. Express your answers as a function of E.
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