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Physics 471 Final Exam Solutions Fall 2006
- (a) (1 pt)The commutator [ˆx, pˆ] is
[ˆx, pˆ] = −i ¯h 2 [a + a†, a − a†] = −i ¯h 2
( [a†, a] − [a, a†]
) = i¯h
(b) (2 pt)
V (ˆx) = 12 mω^2 ˆx^2 = 12 mω^2 ¯h 2 mω
( a + a†
) 2 = 14 ¯hω
( a^2 + aa†^ + a†a + a†^2
)
(c) (2 pt)
〈n|V (ˆx)|n〉 = 14 hω¯ 〈n|
( a^2 + aa†^ + a†a + a†^2
) |n〉
= 14 hω¯
n〈n|a|n − 1 〉 +
n + 1〈n|a|n + 1〉
n〈n|a†|n − 1 〉 +
n + 1〈n|a†|n + 1〉
)
= 14 hω¯
(√ n(n − 1)〈n|n − 2 〉 + (n + 1)〈n|n〉 + n〈n|n〉
√ (n + 1)(n + 2)〈n|n + 2〉
) = 12 ¯hω
( n + (^12)
) .
(d) (2 pt) If |ψ〉 is expanded in energy eigenstates, the expansion coefficients are Cn = 〈n|ψ〉, so
Cn =
√ 2 mω 3¯h
√ ¯h 2 mω
〈n|
( a + a†
) | 1 〉 =
( 〈n| 0 〉 +
2 〈n| 2 〉
) .
Hence, there are only two nonzero Cn’s, C 0 =
√ 1 /3 and C 2 =
√ 2 /3. The possible outcomes of a measurement of the energy are 12 ¯hω with probability 1/3 and 52 ¯hω with probability 2/ 3
- (a) (2 pt) The desired probability is
P =
8 πa^3
∫ (^) ∞ 0 drr^2 e−^
r a
∫ (^) π/ 3 0 dθ sin θ cos^2 θ
∫ (^2) π 0 dφ
=
8 πa^3 2!a^3
( 1 − cos^3 (π/3)
) 2 π =
(b) (2 pt) Since the given wave function has a factor of Y 10 (θ, φ) it has = 1, m = 0. This means that the the nonzero Cnm’s must have the form Cn 10 with n ≥ 2. (c) (2 pt) The possible outcomes of a measurement of the energy are En = − 13 .6 eV/n^2 , with n = 2, 3 , · · ·.
(d) (2 pt) The probability of measuring E 2 is |C 210 |^2 , with
C 210 =
∫ d^3 rR 21 (r)Y 10 ∗(θ, φ)ψ(~r)
=
√ 3 8 πa^3
√ 1 24 a^5
√ 3 4 π
∫ (^) ∞ 0
drr^3 e−^
r a
∫ (^) π 0
dθ sin θ cos^2 θ
∫ (^2) π 0
dφ =
16 πa^4 3!a^4 4 π 3
and the probability is P (E 2 ) = 0.75.
- (a) (3 pt) The solution to the Schr¨odinger equation for the spherical well is
u(r) =
{ AI sin(κr) + BI cos(κr) 0 ≤ r ≤ a AII e−kr^ + BII ekr^ r > a
with κ =
√ 2 m(E)/h¯^2 + λ^2 /a^2 and k =
√ − 2 mE/¯h^2. (b) (1 pt) The boundary condition at r → 0 is u(r) → Cr and the boundary condition at r → ∞ is u(r) → 0. (c) (1 pt) The linear behavior in r as r → 0 implies BI = 0, and the vanishing of u(r) as r → ∞ implies BII = 0. (d) (2 pt) Matching the boundary conditions at r = a gives
AI sin(κa) = AII e−ka κAI cos(κa) = −kAII e−ka^ .,
and the ratio of these equations is
κa cot(κa) = −ka = −
λ^2 − κ^2 a^2.
(e) (1pt) As a function of κa, κa cot(κa) is positive until κa > π/2. This means that unless λ > π/2 the curve of −
λ^2 − κ^2 a^2 can never cross the κa cot(κa) curve.
- (a) (2 pt) The energy − 1 .51 eV corresponds to n = 3, which means that only the coeffi- cients of ψ 300 and ψ 311 contribute, giving
P (− 1 .51 eV) =
∣∣ ∣∣^4 10
∣∣ ∣∣
2
∣∣ ∣∣^ i 10
∣∣ ∣∣
2 = 17%.
(b) (1 pt) As a function of the nuclear charge, the binding energy of a single electron varies as Z^2 , so the binding energy of singly ionized He is − 54 .4 eV. (c) (2 pt) Since n − ` − 1 = 1, the polynomial v(ρ) is first order in ρ and the summation over k gives one condition, 6a 1 + 2a 0 = 0. Then, v(ρ) = a 0 (1 − ρ/3).
