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Physics 471 Solutions Exam 2 Fall 2006
- (a) (2 pt) The eigenvalues of H are given by
det(H − E I) = det
2 a − E 0 a 0 2 a − E 0 a 0 2 a − E
= (2a − E)
( (2a − E)^2 − a^2 )
) = 0 ⇒ E = a, 2 a, 3 a.
(b) (2 pt) The eigenvectors satisfy
He 1 = 2ae 1 He 2 = 3ae 2 He 3 = ae 3.
(c) (2 pt) ψ(0) can be expanded in eigenstates of H as
ψ(0) = e† 1 ψ(0)e 1 + e† 2 ψ(0)e 2 + e† 3 ψ(0)e 3 ψ(0) = √^12 e 1 + 12 e 2 + 12 e 3.
Since the expansion of ψ(0) involves every eigenvector of H, the possible outcomes of a measurement of the energy are a, 2 a, 3 a. (d) (2 pt) The probabilities are
P (a) =
P (2a) =
P (3a) =
- (a) (1 pt) Since the δ-function only affects the solution at x = a, the potential term can be ignored elsewhere, giving, effectively,
¯h^2 2 m
d^2 ψ dx^2 = Eψ.
Substitution of any of the terms in the given solution then yields
E = −
¯h^2 k^2 2 m
(b) (2 pt) The vanishing of ψ at x = 0 gives BI = 0 and the vanishing of ψ as x → ∞ gives BII = 0.
(c) (3 pt) At x = a, the equality of ψI (a) and ψII (a) together with the derivative condition give
AI sinh(ka) = AII e−ka −kAII e−ka^ − kAI cosh(ka) = − λ a AII e−ka
and eliminating AI in favor of AII results in the condition
ka (1 + coth(ka)) = λ.
1 2 3 4 5 ka
2
4
6
8
10
ka + kacothka
(d) (1 pt) From the figure, λ > 1 for a bound state solution to exist.
- (a) (1 pt) Using the formula on the cover, a†| 4 〉 =
5 | 5 〉, so 〈 5 |a†| 4 〉 =
(b) (1 pt) This basis is not orthonormal since, for example, 〈f 0 |f 2 〉 6 = 0. (c) (1 pt) Since x = −1 is in the region of integration, the integral is ( (−1)^2 − 3(−1) + 5)
) | − 1 | = 9
(d) (2 pt) The probability that p lies between p and p + dp is ∣∣ ∣∣ ∣
√ (^) a π¯h
(1 + ipa/¯h)
∣∣ ∣∣ ∣
2 dp = a dp π¯h(1 + p^2 a^2 /h¯^2 )
