Q1P Use the WKB approximation to fin... [FREE SOLUTION]

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Introduction to Quantum Mechanics

David J. Griffiths

Physics

2nd Edition 路 469 pages

ISBN: 9780131118928

Chapter 8: Q1P (page 319)

Use the WKB approximation to find the allowed energies $(E_{n})$ of an infinite square well with a 鈥渟helf,鈥� of height $V_{0}$ , extending half-way across
role="math" localid="1658403794484" $V (x) = {\begin{matrix} v_{0,} & (0 < x < a / 2) \\ 0, & (a / 2 < x < a) \\ 鈭�, & (otherwise) \end{matrix}$
Express your answer in terms ofrole="math" localid="1658403507865" $V_{0} and E_{n}^{0} 鈮� {(苍蟺魔)}^{2} / 2 {ma}^{2}$ (the nth allowed energy for the infinite square well with no shelf). Assume that, but do not assume that $E_{1}^{0} > V_{0}$ . Compare your result with what we got in Section 7.1.2, using first-order perturbation theory. Note that they are in agreement if either $V_{0}$ is very small (the perturbation theory regime) or n is very large (the WKB鈥攕emi-classical鈥攔egime).

Short Answer

Expert verified

The allowed energies of an infinite square well with a 鈥淪helf鈥� of height $V_{0} i s E_{n} = E_{n}^{0} + \frac{V_{0}}{2} + \frac{V_{0}^{2}}{16 E_{n}^{0}}$

Step by step solution

Parameters.

$V (x) = {\begin{matrix} v_{0,} & (0 < x < a / 2) \\ 0, & (a / 2 < x < a) \\ 鈭�, & (otherwise) \end{matrix}$

$E_{n}^{0} 鈮� {(n 蟺魔)}^{2} / 2 m a^{2}$

Finding allowed energies (En) of an infinite square well with a “Shelf” of height V0

$P (x) is real, so {鈭�}_{x}^{0} p (x) dx = 苍蟺魔, with n = 1, 2, 3, . . . and p (x) = 2 m [E - V (x)] {鈭�}_{x}^{0} p (x) dx = 苍蟺魔 {鈭�}_{x}^{a} p (x) dx = \sqrt{2 mE} (\frac{a}{2}) + \sqrt{2 m (E - V_{0})} (\frac{a}{2}) Here, = \sqrt{2 m} (\frac{a}{2}) (\sqrt{E} + \sqrt{E - V_{0}}) = 苍蟺魔鈬� E + E - V_{0} + 2 \sqrt{E (E - V_{0})} = \frac{4}{2 m} {(\frac{苍蟺魔}{a})}^{2} = 4 E_{n,}^{0 .} 2 \sqrt{E (E - V_{0})} = (4 E_{n,}^{0 .} - 2 E + V_{0}) Squaring again : 4 E (E - V_{0}) = 4 E^{2} - 4 {EV}_{0} = 16 E_{n,}^{0 .} + 4 E^{2} + V_{0}^{2} - 16 {EE}_{n,}^{0 .} + 8 E_{n,}^{0 .} V_{0} - 4 {EV}_{0} 鈬� 16 {EE}_{n,}^{0 .} = 16 E_{n,}^{0 .} + 8 E_{n,}^{0 .} V_{0} - 4 {EV}_{0} 鈬� E_{n} = E_{n,}^{0 .} + \frac{V_{0}}{2} + \frac{V_{0}^{2}}{16 E_{n,}^{0 .}}$

Perturbation theory gave

$E_{n} = E_{n}^{0} + \frac{V_{0}}{2}$

The extra term goes to zero for very $s m a l l V_{0,} o r (\sin c e E_{n}^{0} ~ n^{2})$ ,for large n.

Thus the allowed energies of an infinite square well with a 鈥淪helf鈥� of height $V_{0}$ is

$E_{n} = E_{n}^{0} + \frac{V_{0}}{2} + \frac{V_{0}^{2}}{16 E_{n}^{0}}$

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Parameters.

Finding allowed energies (En) of an infinite square well with a “Shelf” of height V0

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