Zewail City of Science and Technology–Fall 2016
PEU 430: Solution Solution to Quantum Quantum Mechanic Mechanicss 3–Midterm 3–Midterm 1
1
(1) [20 points] (a) A particle of mass m is initially in the ground state (E 1 ) of an infinite square well of width L. Starting at time t = 0, the system is subject to the perturbation H (t) = V 0 x2 e
t/τ
−
,
where V 0 and τ are constants. (i) Find the probability that the energy after time T is measured to be E 2 . (ii) Calculate the probability in the limit T
→ ∞.
Hint: Don’t evaluate the spatial integrals. (b) A hydrogen atom starts in the state n = 4, l = 3, m = 3, where we ignore the spin. What possible states will the atom go through as it decays spontaneously to the ground state.
Solution (a)(i) For a particle initially in the state 1 at t = 0, the probability of being found in the state 2 at t = T (transition probability) is:
|
|
P 1
2 (T )
→
= 2 ψ(T )
| |
2
| ,
where
|ψ(t) = c (t)e
E i t/
−
i
|i,
i=1
c1 (0) = 1, c j (0) = 0, j = 1.
∀
Hence, 2 (T )
P 1
→
= c2 (T ) 2,
|
|
where 1 c2 (T ) = i
T
0
V 0 dt = 2 x2 1 i
i(E 2 −E 1 )t/
2|H (t)|1e
|
iω21 t−t/τ
e | iω e
V 0 = 2 x2 1 i
0
2
0
iω21 T −T/τ
21
21
2
t/τ i(E 2 −E 1 )t/
−
e
e
dt
0
T
− 1/τ V 1 − = 2|x |1 , i iω − 1/τ where we defined ω = (E − E )/ . 21
| |
T
1
2
(1)
The transition probability now reads P 1
2
→
2
2
2
| V | |2|x |1| (T ) = |c (T )| = 1 + e ω + 1/τ 2
0 2
2
2 21
2T /τ
−
2
T /τ
−
− 2e
cos ω21
T
(ii) In the limit T
→ ∞, we have P 1
2
→
2
2
|V | |2|x |1| ( ∞) = 0 2
2
2 + 1/τ 2 ω21
(ii) The spontaneous transitions will occur subject to the selction rules ∆l =
±1,
∆m = 0 or
± 1.
Since the initial state is 433 , the sequence of transitions is
| |433−→|322−→|211−→|100 in all the above transitions we have ∆l = −1, ∆m = −1. (2) [20 points]
Calculate the life time of the first excited state of the hydrogen atom 210 when it undergoes a spontaneous transition (electric dipole transition) to the ground state.
|
Hint: You may use
28 a 100 z 210 = . 35 2
√
|| Solution
The life time of an excited state is given by τ =
1 , A
where A is the spontaneous emission rate A = 3
ω03 2 . 3π0 c3
|P|
Here is the matrix element of the electric dipole moment between the two relevant states
P
ˆ = 100|z |210k. ˆ ˆ 100|z |210k) P = q 100|r|210 = q (100|x|210ˆi + 100|y|210 j + where the vanishing of 100|x|210 and 100|y |210 is due to the m selection rule explained in the textbook. Hence,
|P|
2
q 2 215 a2 = , 310
3π0 c3 310 τ = q 2 ω03 215 a2
(3) [20 points] Consider a particle in the potential V (x) = λx 4 ,
λ > 0.
Numerical integration gives the following values for the ground state and first excited state energies, respectively E 0 = 1.060
2
2/3
2m
1/3
λ
,
E 1 = 3.800
2
2m
2/3
λ1/3
use the variational method to find approximate values (within 2% of the numerical values) of the energies of (a) the ground state, and (b) the first excited state.
Hint: You may need the integral ∞
x2ne
x2
−
dx =
(2n
− 1)(2n − 3)(2n − 5) ··· 5 × 3 × 1 √ π. 2n+1
0
Solution (a) For the ground state we chose the following trial wavefunction (ground state of the SHO) ψ0 (x) = Ae α x /2 , −
4
2
2
1/4
α2 π
where normalizability will lead to A =
.
We have
H = ψ |H |ψ 0
0
0
∞
α = π
√ α
α2 x2 /2
−
e
√ π
2
4
α2 x2 /2
−
2
−∞ ∞
=
2
− d + λx e dx 2m dx − (α x − α ) + λx e
α2 x2 /2
−
e
2
4 2
2
α2 x2
4
−
2m
−∞
dx
(2)
2
α2 3 λ = + . 2m 2 4 α4 We know that H E 0 , where E 0 is the smallest eigenvalue of H or the ground state energy, i.e., H 0 is bounded from below by E 0. We will chose α which minimizes H 0 d 6mλ 1/6 H = 0 = α = . 2 dα Substituting this value for α in the expression for H 0 we get
≥
⇒
E 0
≈ 1.082
2
2/3
λ1/3.
2m
This is already within 2% of the value obtained using numerical integration. (b) To find a trial wavefunction for the first excited state we pick a trial function orthogonal to the ground state wavefunction we used before. Since the ground state wavefunction is even, then any odd function will be orthogonal to it. In particular the first excited state of the SHO ψ1(x) = Bxe where normalizability gives B =
H = ψ |H |ψ 1
1
1
2α3 = π =
√ π
.
α2 x2 /2
−
xe
−∞ ∞
α2 x2 /2
−
e
−∞
,
1/4
4α6 π
∞
√ 2α 3
α2 x2 /2
−
−
−
2
d2 + λx4 xe 2 2m dx
2
2m
2 2
3α2 15 λ = + . 2m 2 2 α4 The best value of α satisfies
⇒ α = 5
dx
(α x − 3α x ) + λx e 4 4
2
d H 1 =0= dα
α2 x2 /2
−
1/6
10mλ 2
.
6
α2 x2
−
dx (3)
Substituting this value for α in the expression for H 1 we get
≈ 3.847
E 1
2
2/3
λ1/3.
2m
This is already within 1.5% of the value obtained using numerical integration.
(4) [20 points] For a particle in a constant potential well
−V V (x) = 0
|x| < a |x| > a,
−f (x) V (x) = 0
|x| < a |x| > a,
0
c
V 0 > 0
the system will always have at least one bound state ψb . Use this result and the variational principle to prove that there is at least one bound state for a well with arbitrary potential V 0 < f (x).
Solution For V =
−V , we know the system has a bound state ψ , i.e., a state with E < 0 (4) ψ|H |ψ = E , where H = − V . For V (x) = −f (x) and using ψ as a trial wavefunction we know that ψ|H |ψ ≤ E (5) where H = − V and E is the ground state energy of H . 0
b
0
0
p2 2m
b
b
0
b
0
p2 2m
0
0
Subtracting equations (4) and (5),
ψ|(V − f (x))|ψ ≤ E − E . 0
0
b
The left-hand side of this inequality is always negative since f (x) > V 0 . Since E b < 0 then E 0 < 0 and the system will always have at least one bound state. 6
