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Unformatted text preview: Homework Solutions # 7 (Liboﬀ Ch. 8)
8.7 The eigenfunctions of the semi-inﬁnite potential well are just those for
the ﬁnite one, with the extra boundary condition that φ(0) = 0. This is satisﬁed by the x > 0 halves of the odd-parity solutions to the ﬁnite well, √ now multiplied by 2 to make the normalization come out to 1 rather than 1 . Sketched, these look like the right half of Figure 8.4b, and other right 2 halves of sine functions with one and two zeros in 0 < x < a. The semiinﬁnite well, lacking the even parity solution, will have a higher ground state energy. The eigenstates you sketched are not, of course, eigenstates for the ﬁnite well—you’ve set ψ = 0 for x < 0. 8.16 The b → ∞ limit of equation 8.55a is, with cosh κb → ∞,
cos k1 a −
2 k1 − κ2 sin k1 a = 0 2k1 κ For E V , κ → k1 (from equations 8.54 and 8.55b), so we’re left with cos k1 a = 0, hence k1 a = nπ . The curves in Figure 8.13b all become ﬂat lines. 8.22 The standing-wave eigenfunctions are either odd or even. So their
x-derivative will have the opposite parity, and p = 0 because
∞ −ih ¯
−∞ dx ψ ∗ ∂ψ =0 ∂x as the integrand is an (even × odd) = (odd) function. 8.26 6 eV corresponds to ν = E/h = 1.45 × 1015 Hz. This is the near
ultraviolet. Phys 580 HW# 7 Solutions 1 ...
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