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Unformatted text preview: Homework Solutions # 5 (Liboﬀ Chapter 7)
7.5 Using the deﬁnition of a in terms of x and p, ˆ ˆ ˆ
[ˆ, a† ] = aˆ mω0 2i 1 [ˆ, p] xˆ [x, x] − ˆˆ [ˆ, p] + 2 2 [ˆ, p] = xˆ pˆ =1 2¯ h mω0 m ω0 ih ¯ 7.9 Say a† |n = Cn |n + 1 . Then the norm of this state is ˆ
ˆ n|aa† |n = n|(ˆ† a + 1)|n = n|(N + 1)|n = n + 1 ˆˆ aˆ √ ˆ Therefore, Cn = n + 1. Applying a† successively, we get 1 |n = √ a†n |0 ˆ n! This applies to the x-representation φn as well. You deal with a the same way. say a|n = Kn |n − 1 . Then, ˆ ˆ ˆ n|a† a|n = n|N |n = n ˆˆ So Kn = √ n. 7.15 Note that since x ∝ a + a† , m|x|n = 0 only for m = n ± 1. Since |ψ ˆˆˆ ˆ
is a superposition of the 0th and 3rd levels, and 3 = 0 ± 1, x = 0. The time dependence will just throw in some e−iωt factors, not aﬀecting the result. 7.18 Work with energy eigenstates. Since all states can be written as a
superposition of energy eigenstates, ﬁnding the minimum energy for energy eigenstates will give you the minimum for an arbitrary state as well. Since (see previous problem) x = p = 0 in all energy eigenstates, ∆x = ∆p = x2 − x p2 − p
2 = = x2 p2 2 In recitation problem 7.10 we found that V = T , so 1 12 k x2 = p 2 2m Phys 580 HW# 5 Solutions 1 √ meaning p2 = mk x2 and hence ∆p = mk ∆x. Now, the ground state energy is the sum of its kinetic and potential parts, h ¯ k ∆x∆p ≥ ω0 E0 = k x2 = k (∆x)2 = √ 2 mk So the minimum energy is at least 1 hω0 , which is exactly what the ground ¯ 2 state energy is.
∂ 7.30 With x → −i ∂k , ˆ −i ∂ ϕx (k ) = xϕx (k ) ∂k It’s obvious that solutions to this are the familiar plane waves, ϕx (k ) = eikx where k = p/h. ¯ Phys 580 HW# 5 Solutions 2 ...
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