,FINAL example 1
SOLUTION GUIDE
SI-MKS
c = 2.99792458 10 m s
8 –1
Speed of light in free space
= 6.58211889 10
– 16
Planck’s constant eV s
= 1.054571596 10
– 34
Js
e = 1.602176462 10
– 19
Electron charge C
m 0 = 9.10938188 10
– 31
Electron mass kg
m n = 1.67492716 10
– 27
Neutron mass kg
m p = 1.67262158 10
– 27
Proton mass kg
k B = 1.3806503 10
–23 –1
Boltzmann constant JK
k B = 8.617342 10
–5 –1
eV K
0 = 8.8541878 10
– 12 –1
Permittivity of free space Fm
0 = 4 10
–7 –1
Permeability of free space Hm
Speed of light in free space c = 1 00
N A = 6.02214199 10
23 –1
Avagadro’s number mol
a B = 0.52917721 10 m
–10
Bohr radius
4 0
2
a B = ----------------
-
m0e 2
Inverse fine-structure constant –1 = 137.0359976
4 0 c
–1 = -----------------
-
e2
Applied quantum mechanics
,PROBLEM 1
The first four lowest energy states of a one-dimensional harmonic oscillator with characteristic fre-
quency 0 are subject to the perturbation
–1 –1
1 ------- 0 -------
2 2
W 00 W 01 W 02 W 03 –1
------- 2 0 0
W 10 W 11 W 12 W 13
W = = 0 2
W 20 W 21 W 22 W 23 1
0 0 --- 0
W 30 W 31 W 32 W 33 2
– 1
------- 0 0 0
2
where « 1 .
(a) Find the new eigenenergies to first-order in time-independent perturbation theory. (50%)
(b) Find the new eigenenergies to second-order in time-independent perturbation theory. (50%)
PROBLEM 1 SOLUTION:
(a) The eigenenergies of the unperturbed Hamiltonian are E n = 0 n + --- for n = 0 1 2 .
0 1
2
1
The first-order correction is E = W nn where W nn = nŴ n so that the new energy eigenvalues
0
to first-order are E n = E n + W nn .
E 0 = ---------0 + 0 = ---------0 1 + 2
2 2
3
E 1 = ------------0 + 2 0 = ---------0 3 + 4
2 2
5
E 2 = ------------0 + -------------0 = ---------0 5 +
2 2 2
7
E 3 = ------------0
2
(b) The new energy eigenvalues to second-order are given by
2
0 W nm
E n = E n + W nn + ----------------------
0
-
0
m n En – Em
and so
2
2 2 2
E 0 = ---------0 + 0 – ---------------0 – ---------------0 = ---------0 + 0 – ------------------0
2 2 6 2 3
2
3
E 1 = ------------0 + 2 0 + ---------------0
2 2
5
E 2 = ------------0 + -------------0
2 2
7
2
E 3 = ------------0 + ---------------0
2 6
, PROBLEM 2
In first-order time-dependent perturbation theory a particle initially in eigenstate n of the unper-
ˆ
turbed Hamiltonian scatters into state m with probability a m t after the perturbation W x t is
2
applied at time t = 0 .
(a) Derive the expression for the time dependent coefficient
t = t
1 i mn t
a m t = ----- W mn e dt
i
t = 0
ˆ
where the matrix element W mn = mW x t n and mn = E m – E n is the difference in eigenen-
ergies of the states m and n . (40%)
(b) An electron is initially in the ground state of a one-dimensional harmonic oscillator with
Hamiltonian H ˆ = bˆ † bˆ + 1 2 where is the oscillator’s characteristic frequency and the
m 0 1 2
operator bˆ = ----------
ip̂ x ˆ x t = V x 3 e –t is applied
x̂ + ---------- . At time t = 0 a perturbation W
2 m 0 0
where V0 and are constants. What are the allowed transitions? Calculate the probability of transi-
tion to each excited state of the system in the long time limit, t . (50%)
(c) What value of maximizes the transition probability? Explain your result. (10%)
PROBLEM 2 SOLUTION:
(a) Consider a quantum-mechanical system described by Hamiltonian H ˆ and for which we know
0
the solutions to the time-independent Schrödinger equation. That is,
Hˆ n = E n
0 n
are known. The time-independent eigenvalues are E n = n , and the orthonormal eigenfunctions
are n . The eigenfunction n evolves in time according to
– i t – i n t
n t = ne n = n x e
and satisfies
– i t ˆ ne –in t
i ne n = H 0
t
ˆ t the effect of which is to create
At time t = 0 we apply a time-dependent change in potential W
a new Hamiltonian:
Hˆ = Hˆ +W ˆ t
0
and state t , which evolves in time according to
i t = H
ˆ +W
0
ˆ t t
t
We seek solutions to the time-dependent Schrödinger equation in the form of a sum over the known
eigenstates of the unperturbed system,
– i n t
t = a n t ne
n
where a n t are time-dependent coefficients. Substituting gives
d – i t ˆ +Wˆ t a t ne –in t
i ----- a n t ne n = H 0 n
dt n n
Using the product rule for differentiation ((fg)' = (f 'g + fg')), one may rewrite the left-hand side as
Applied quantum mechanics