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Why does the Bohr-Sommerfeld rule for quantization give the exact energy-levels for a simple harmonic oscillator?

Qmechanic
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MKR
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    Why shouldn't it? (Heuristically, you can see the Bohr-Sommerfeld quantization as arising from a WKB approximation that neglects all quantum effects of higher order than $\mathcal{O}(\hbar)$. It is then a "happy accident" that the quantum harmonic oscillator exhibits no quantum effects of higher orders, and exactly obeys the Bohr-Sommerfeld condition.) – ACuriousMind Jan 30 '15 at 19:40
  • Related: http://physics.stackexchange.com/q/129602/2451 – Qmechanic Jan 30 '15 at 20:07
  • I know, but one of my professors keeps saying that there is a physical reason behind that. It's not just an accident – MKR Jan 31 '15 at 03:55

2 Answers2

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One can use various kinds of supersymmetry to argue that the WKB approximation for the quantum harmonic oscillator is exact. One method uses localization of path integrals, cf. e.g. Ref. 1. Another method uses supersymmetric quantum mechanics, cf. e.g. Ref. 2.

References:

  1. R.J. Szabo, Equivariant Localization of Path Integrals, hep-th/9608068.

  2. F. Cooper, A. Khare, and U. Sukhatme, Supersymmetry and Quantum Mechanics, Phys. Rept. 251 (1995) 267, arXiv:hep-th/9405029.

Qmechanic
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It is one of rare cases when an approximate formula coincides with the exact one. Another example is the Plank formula obtained originally as a bridging of two experimental asymptotes. A fluke. There are some other cases of happy coincidents.

Vladimir Kalitvianski
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