ˆ ˆ ˆ ˆ ˆ Aˆ − Aˆ B ˆ ) = i(AˆB ˆ −B ˆ A), ˆ so Hermitian. (a) i(AB − B A) = −i(B †
†
†
†
ˆ† = (b) Same, except for no factor of i, so it’s anti-Hermitian (O (c)
1 ˆˆ ( 2 AB
ˆ −O).
ˆ A) ˆ † = 1 (B ˆ † Aˆ† + Aˆ† B ˆ † ) = 1 (AˆB ˆ +B ˆ A), ˆ so Hermitian. +B 2 2
ˆ † = Aˆ† Aˆ†† = Aˆ† A, ˆ so Hermitian. (d) (Aˆ† A) (e)
1 ˆˆ (AB 2
ˆ A). ˆ +B
4.14 First, by the properties of the inner product, ˆ = Aϕ ˆ |ϕ ϕ|Aϕ
∗
ˆ Then, by the definition of the Hermitian adjoint, and the hermiticity of A, ˆ = Aˆ ϕ|ϕ = Aϕ ˆ |ϕ ϕ|Aϕ ˆ |ϕ = A , A = A . So, since Aϕ ˆ = b ϕ|ϕ = 4.16 If Bˆ |ϕ = b |ϕ, then ϕ|Bϕ b ϕ|ϕ = b ϕ|ϕ = b = b . So Bˆ = B.ˆ †
∗
∗
∗
1
1
∗
1
∗
1
1
†
1
ˆ ϕ = b1 . But Bϕ
|
4.35 (a) Normalize by requiring
|ψ| = A dx = A a = 1. So A = 1/√a. 2
2
a
0
2
(b) Note that ψ is antisymmetric about x = a/2. This means n ψ = 2/a ψ sin(nπx/a) = 0 for odd n, because those basis functions are symmetric about x = a/2. So the first non-zero energy component of ψ is n = 2.
Phys 580 HW# 2 Solutions
|
2
4.36 (a) ˆ = Hψ
−
¯h2 ∂ 2 ψ ¯ 2 k2 h = ψ 2m ∂x 2 2m
So it’s an energy eigenstate. (b) No; it’s a sum of p = (c) p =
±¯hk states.
±h¯ k can be found; in terms of momentum eigenstates |ψ = A| + ¯hk + √A2 | − h¯ k
