Assignment 11, solutions
Problem 1:
A system makes transitions between eigenstates of H0 under
the action of the time dependent Hamiltonian H0 +
Wcoswt, W << H0. Assume that at t =
0 the system is in the state |yi>.
(a) Find an expression for the probability of transition from |yi>
to |yf>, where |yi>
and |yf> are eigenstates of H0
with eigenvalues Ei and Ef.
(b) Find an expression for the probability of transition from |yi>
to |yf>,
for a constant perturbation W.
Solution:
 | Concepts, principles, relations
that apply to the problem:
Time-dependent perturbation
theory |
| Why do they apply?
For small perturbations time-dependent
perturbation theory can be used to calculate transition probabilities. |
| How do they apply?
(a) The probability of finding the system in the state |yf>
(f ¹ i) at time t is
Pif(t) = (1/h2)|ò0texp(iwfit')Wfi(t')dt'|2,
with wfi
= (Ef - Ei)/h
and Wfi(t) = <yf|W(t)|yi>,
with W(t) = Wcoswt.
We have to consider a sinusoidal perturbation starting at t = 0.
ò0t
exp(iwfit')Wfi(t')dt'
= (Wfi/2)ò0t
[exp(i(wfi+w)t'
+ exp(i(wfi-w)t']dt'
= (iWfi/2)[(1-exp(i(wfi+w)t)/(wfi+w)
+ (1-exp(i(wfi-w)t)/(wfi-w)].
Pif(t) =[|Wfi|2/(4h2)]|(1-exp(i(wfi+w)t)/(wfi+w)
+ (1-exp(i(wfi-w)t)/(wfi-w)|2,
with
Wfi = <yf|W|yi>.
The expression
(1-exp(i(wfi+w)t)/(wfi+w)
+ (1-exp(i(wfi-w)t)/(wfi-w)
has an appreciable amplitude only if the denominator of one of the two terms
is approximately zero,
i.e. if wfi
= ±w, or Ef = Ei
±
hw.
The first order effect of a perturbation that varies sinusoidally with time is
to receive from or transfer to the system a quantum of energy
hw.
If the system is initially in the
ground state, then Ef > Ei, and only the second term
needs to be considered. Then
Pif(t) = [|Wfi|2/(4h2)]
sin2((wfi
- w)t/2)/((wfi
- w)/2)2. |
| Details of the calculation:
(b) For a constant perturbation W we have
Pif(t) = (1/h2)|ò0texp(iwfit')Wfidt'|2
= (1/h2)|Wfiò0texp(iwfit')Wfidt'|2
(1/h2)|Wfi(1-exp(i(wfi)t)/wfi|2
= [|Wfi|2/h2]sin2(wfit/2)/(wfi/2)2. |
Problem 2:
Consider a particle in the ground state of a one-dimensional square well
of with a and depth V0.
Assume that the well is very deep and
k/k0 = (2m(E+V0)/(2mV0))1/2 << 1
for the ground state, so that the ground state wavefunction is nearly identical
to that of the infinite square well.
At t = 0, a time dependent perturbation W(t) = Wcoswt
is turned on. What is the minimum frequency necessary to free the particle
from the well? For frequencies greater than this minimum frequency, use
perturbation theory to find the transition rate.
Solution:
| Concepts, principles, relations
that apply to the problem:
Fermi's golden rule, the infinite square well |
| Why do they apply?
The initial wavefunction of the particle look very much like the
wavefunction of a particle in an infinite square well. The final state
of the particle is a continuum state. |
| How do they apply?
Let H0 be the Hamiltonian of the infinite square well.
Eigenfunctions of H0: fn(x) =
(2/a)1/2sin(npx/a)
with eigenvalues En =
n2p2h2/(2ma2).
the ground state energy is E1
= p2h2/(2ma2).
In a very deep well the ground state energy is approximately equal to -V0
+ E1 @
-V0.
The minimum frequency required to free the particle from the well therefore is wmin
= V0/h.
To find the transition rate we use Fermi's golden rule.
If W(t) = Wcos(wt),
then the transition probability per unit time is given by
w(i,E) = (p/(2h))r(E)|WEi|2dE-Ei,hw,
where WEi = <fE|W|fi>. |
| Details of the calculation:
Assume that the ejected electron is a free electron confined to a
region of width L >> a. Use periodic boundary conditions.
Then
fk(x) = L-1exp(ikx),
with k = 2pn/L,
n = 0, ±1, ±2, ... .
k2 = 2mE/h2 = 2m(w
- wmin)/h.
The number of states with wave vectors whose magnitudes
lie between k and k + dk is
dN = 2 dk/(2p/L) = Ldk/p.
(The particle can move towards the left or to the right.)
dN/dk = L/p.
The density of states is dN/dE = (dN/dk)(dk/dE).
With E = h2k2/(2m) we
have r(E)
= dN/dE = Lm/(pkh2).
WEi = (2/La)1/2Wò0adxeikxsin(px/a)
= (2/La)1/2(W/2i)ò0adxeikx(exp(ipx/a)
- exp(-ipx/a))
= (2/La)1/2(W/2)(eika+1)2pa/(p2
- k2a2)
= W(2/La)1/2eikacos(ka/2)2pa/(p2
- k2a2).
w(i,E) = (p/(2h))Lm/(pkh2)W2(2/La)cos2(ka/2)4p2a2/(p2
- k2a2)2
= W2cos2(ka/2)4p2am/[kh3(p2
- k2a2)2].
This is the transition probability per unit time. |
Problem 3:
An experimenter has carefully prepared a particle of
mass m in the first excited state of a one dimensional harmonic
oscillator, when he sneezes and knocks the center of the potential a small
distance a to one side. It takes him a time T to blow his nose.
and when he has done so he immediately puts the center back where it was. Find,
to lowest order in a, the probabilities P0 and P2 that the oscillator will now be in its ground state and its second excited
state.
Solution:
| Concepts, principles, relations
that apply to the problem:
Time dependent perturbation theory, the harmonic oscillator |
| Why do they apply?
For small perturbations time-dependent
perturbation theory can be used to calculate transition probabilities. |
| How do they apply?
For t < 0 and t > T the potential energy is U = (1/2)mw2x2,
H = H0, H0|fn> =
(n+(1/2))hw|fi>.
For 0 < t < T the potential energy is
U = (1/2)mw2(x-a)2 = (1/2)mw2x2
+ (1/2)mw2a2 - mw2ax
H = H0 + W, W = (1/2)mw2a2
- mw2ax.
The probability of finding the system in the state |yf>
(f ¹ i) at time t > T is
Pif(t) = (1/h2)|ò0Texp(iwfit')Wfidt'|2,
Let f = 0 and i = 1.
W01 = <f0|W|f1>
= - mw2a<f0|x|f1>.
[The stationary states of H0
fn(x) = NnHn(ax)exp(-a2x2/2)
with Nn = [a/(p1/22nn!)]1/2,
a = (mw/h)1/2 and Hn(ax) a Hermite polynomial.
Recursion relation: Hn+1(ax) = 2axHn(ax)
- 2nHn-1(ax).
H0(ax) = 1, H1(ax)
= 2ax, H2(ax)
= 4(ax)2 - 2.]
<f0(x)|x|f1(x)> = N0N1ò-¥¥dx
H0(ax) x H1(ax)
exp(-a2x2)
= N0N1(1/2a)ò-¥¥dx
H0(ax)(H2(ax) + 2H0(ax))exp(-a2x2)
=
N0N1(1/a)ò-¥¥dx
H02(ax)
exp(-a2x2)
=
N1/(N0a).
W01 = -(hmw3a2/2)1/2.
P10(t) = [mw3a2/(2h)](4/w2)sin2(wT/2)
= [2maw2/h]sin2(wT/2). |
| Details of the calculation:
Let f = 2 and i = 1.
W21 = <f2|W|f1>
= - mw2a<f2|x|f1>.
<f2(x)|x|f1(x)> = N2N1ò-¥¥dx
H2(ax) x H1(ax)
exp(-a2x2)
= N2N1(1/2a)ò-¥¥dx
(H3(ax) + 4H1(ax))H1(ax)exp(-a2x2)
=
N2N1(2/a)ò-¥¥dx
H12(ax)
exp(-a2x2)
= 2N2/(N1a).
W21 = -(hmw2a2)1/2.
P12(t) = [mw2a2/h](4/w2)sin2(wT/2)
= [4ma2/h]sin2(wT/2).
In first order perturbation theory
P1n(t)(n>2) = 0, since <fn(x)|x|f1(x)>
= 0 for n > 2. |
Problem 4:
Consider a one-dimensional oscillator in its ground state.
The unperturbed
Hamiltonian is H0 = p2/(2m) + (1/2)mw02x2.
At t = 0 the Hamiltonian becomes H = H0 + H1, where H1 = (1/2)mw12x2cosft,
w1 << w0.
Calculate the transition probability to the second excited state.
Can
there be a transition to any other excited state?
Solution:
| Concepts, principles, relations
that apply to the problem:
Time-dependent perturbation theory, the 1D harmonic oscillator |
| Why do they apply?
H = H0 + H1, H1 is a small, periodic
perturbation. We use time-dependent perturbation theory to calculate the
probability of transitions between stationary states of H0. |
|
How do they apply?
We have a periodic perturbation
W(t) = Wcosft.
If Ef > Ei,
then
Pif(t) = [|Wfi|2/(4h2)]
sin2((wfi
-
f)t/2)/((wfi
- f)/2)2.
Here
P02(t) = [((1/2)mw12)2/(4h2)]|<0|x2|2>|2
sin2((2w0
- f)t/2)/((2w0
- f)/2)2,
since wfi
= 2w0.
P02(t) is the probability that at time t we will find the
oscillator in its second excited state. |
| Details of the calculation:
The stationary states of H0 = p2/(2m) +
mw02x2/2 are
un(x) = NnHn(ax)exp(-a2x2/2)
with Nn = [a/(p1/22nn!)]1/2,
a = (mw0/h)1/2
and Hn(ax) a Hermite polynomial.
Recursion relation: Hn+1(ax) = 2axHn(ax)
- 2nHn-1(ax).
H0(ax) = 1, H1(ax)
= 2ax, H2(ax)
= 4(ax)2 - 2.
<0|x2|2> =
N0N2ò-¥¥dx
H0(ax) x2 H2(ax)
exp(-a2x2)
= N0N2ò-¥¥dx
xH0(ax) xH2(ax)
exp(-a2x2)
= N0N2(1/2a)ò-¥¥dx
H1(ax) xH2(ax)
exp(-a2x2)
= N0N2(1/a2)ò-¥¥dx
H12(ax)exp(-a2x2)
= [N0N2/(N12a2)]
= h/(21/2mw0).
P02(t) = [((1/2)mw12)2/(4h2)](h2/(2m2w02))
sin2((2w0
- f)t/2)/((2w0
- f)/2)2
= [w14/(32w02)]sin2((2w0
- f)t/2)/((2w0
- f)/2)2.
In first-order, time-dependent perturbation theory no other transitions are
allowed. |
Problem 5:
The electron in a hydrogen
atom is in a 3d state. Neglect the fine structure.
(a) To what state or states (i.e. 1s etc.) can it go by radiating a photon in
an allowed transition?
(b) What is the degeneracy of the electron (include spin, but ignore spin-orbit
interaction) in a 3d state?
Solution:
| Concepts, principles, relations
that apply to the problem:
Selection rules Dl
= 1, degeneracy of the energy level of the hydrogen atom |
| Why do they apply?
Optical transitions are only "allowed" if
Dl = 1. |
| How do they apply?
(a) Selection rules: Dl = 1.
An electron can only go to a 2p state in an allowed transition involving
photon emission.
(b) l = 2 m = -2, -1, 0, 1, 2, s = + ½, - ½
Degeneracy = 10. |
| Details of the calculation:
None |
Problem 6:
Consider a 2-dimensional system containing a gas of
electrons completely free to move in the x direction but confined by a
square-well potential of infinite depth and total width w in the y
direction. In such a system, the electrons can often be approximated as
non-interacting provided that the mass of the electron is replaced by an
effective mass. Assume for this problem that the effective mass of the
electrons is about 1/10 the mass of free electrons.
(a) This system contains states that involve quantum-mechanical motion in both
the x and y directions; describe qualitatively the nature of the
spectrum that you expect.
(b) Write or derive a formula for the discrete levels expected for quantized
motion in the y direction in terms of the width w.
(c) How small does the width w have to be before the transition energy between
the first two discrete levels found in b is larger than the average energy
available from thermal excitation at room temperature?
Solution:
| Concepts, principles, relations that apply to the
problem:
Separation of variable, the infinite square well |
| Why do they apply?
This is a two-dimensional problem, with confinement in one dimensions. |
| How do they apply?
(a) H = (-h2/(2m))[¶2/¶x2
+ ¶2/¶y2]
+ U(y), H y(x,y) = E y(x,y).
Separation of variables is possible.
y(x,y) =
c(x)
f(y). E = Ex + Ey.
(-h2/(2m))¶2c(x)/¶x2
= Exc(x).
Ex is a continuous eigenvalue. Ex =
hk2/(2m)
(-h2/(2m))¶2
f(y)/¶y2
+ U(y) f(y) = Eyf(y).
Ey is quantized. The Eyn are the eigenenergies of the
infinite square well.
E = hk2/(2m)
+ Eyn, k = (2m(E – Eyn)/h2)1/2.
The density of states dN/dE
µ dk/dE = [m/(2h2(E – Eyn))]1/2
will have spikes when E = Eyn, and the absorption spectrum, for
example, should reflect those spikes. (Fermi’s golden rule.)
E = hk2/(2m)
+ Eyn.
(b) Eyn = n2p2h2/(2mw2)
(c) 3p2h2/(2mw2)
> kT, w < 20 nm. |
| Details of the calculation:
None |
Problem 7:
The correctly normalized hydrogen ground state wavefunction
in 3D is given by
y0(r)
= (pa03)-1/2exp(-r/a0),
where a0 =
h2/(mee2)
is the Bohr radius, which is numerically ~0.529Ǻ.
(a) Confirm that this does
indeed satisfy the radial Schroedinger equation for hydrogen, and that the
wavefunction is normalized to ò d3r
|y(r)|2 = 1
(b) Two identical ions are introduced on the z-axis at locations z = +d and -d.
Assuming that the effect of each ion on the electron can be treated as a point
interaction,
Ue – ion
= n0
d(r – rion),
calculate the change in the hydrogen atom's
ground state energy using first order perturbation theory.Solution:
| Concepts, principles, relations
that apply to the problem:
The Schroedinger equation, stationary perturbation theory |
| Why do they apply?
We are asked to use first order perturbation
theory to calculate the perturbed atom's ground state energy. |
|
How do they apply?
(a) The time-independent Schroedinger equation for the hydrogen atom
is
 ,
where m = me for infinite
proton mass..
Writing fnlm(r) =
Rnl(r)Ylm(q,f)
= [unl(r)/r]Ylm(q,f)
we find -[h2/(2m)][(¶2/¶r2)
+ 2ml(l+1)/h2]ulm
(r) – [e2/r]ulm0(r) = Elmulm
(r).Since f100(r)
= (pa03)-1/2
exp(-r/a0), we have u10(r) = 2 a0-3/2
r exp(-r/a0).
We show that u10(r) satisfies
-[h2/(2m)][¶2/¶r2]
u10(r) – [e2/r] u10(r) = E10u10(r)
by simply differentiating.
(¶2/¶r2)rexp(-r/a0)
= [-(2/a0) + (r/a02)]exp(-r/a0)
Therefore [h2/(a0m)]
- [rh2/(2a02m)]
– e2 = Er.
This equation is satisfied if a0 =
h2/(me2) and E = -h2/(2a02m)
= - me4/(2h2).
Normalization: ò d3r |y(r)|2
= 4pò0¥
r2dr (pa03)-1
exp(-2r/a0) = ½ ò0¥
r’2dr’exp(-r’) = 1.
(b) The ground state of the hydrogen atom is
non-degenerate, we use first-order stationary perturbation theory for
non-degenerate energy eigenvalues.
The first order correction to the ground state energy is E11
= <f100|W|f100>.
E11 =
ò-11dcosq
ò0¥ r2dr
|R10(r)|2|Y00(q,f)|2
n0 d(r-d)
d(cosq-1)/r2
+ ò-11dcosq
ò0¥ r2dr
|R10(r)|2|Y00(q,f)|2
n0 d(r-d)
d(cosq+1)/r2
= n0 |R10(d)|2|Y00(q
= 0)|2 + n0 |R10(d)|2|Y00(q
= p)|2 = 2n0
|y100(d)|2 = 2n0exp(-2d/a0)/(pa3).
Alternatively:
E11 =
ò0¥dxò0¥dyò0¥dz
|y100 (r)|2n0
d(x)d(y)
(d(z-d) +
d(z+d)) = 2n0 |y100(d)|2. |
| Details of the calculation:
None |