I offered a bounty on this question for a simple way to see that the Feynman path integral yields discrete energy levels for bound states, in one dimensional quantum mechanics. As shown there, there's in theory a simple explanation. The path integral computes the propagator by $$K(x_i, x_f, t) \equiv \langle x_f | e^{- i H t} | x_i \rangle = \int_{x(0) = x_i, x(t) = x_f} \mathcal{D}x(t)\, e^{iS[x(t)]/\hbar}.$$ On the other hand, applying a semiclassical approximation, the path integral is just $$K(x_i, x_f, t) \sim \bigg| \frac{\partial^2 S_0(x_i, x_f, t)}{\partial x_i \partial x_f}\bigg|\, e^{i S_0(x_i, x_f, t)/\hbar}$$ where $S_0$ is the on-shell action, i.e. the action for the classical path that goes from $x_i$ to $x_f$ in time $t$, where I'm ignoring issues about the existence and uniqueness of such a path. This makes perfect sense because in the semiclassical approximation we just expand about that classical path, and all the path integral does is provide the extra factor out in front, reflecting how much the nearby paths in the path integral amplify or suppress the classical one.

On the other hand, working in the energy eigenbasis, we have $$K(x_i, x_f, t) = \sum_{n,m} \langle x_f | n \rangle \langle n| \ e^{-iHt} |m \rangle \langle m | x_i \rangle = \sum_n \langle x_f | n \rangle \langle n | x_i \rangle e^{-i E_n t/\hbar}$$ so we get discrete energy if the Fourier transform of $K(x_i, x_f, t)$ in time has discrete support, which is equivalent to the same being true for $S_0(x_i, x_f, t)$. That means we can read off the energy discretization directly from the classical action. Indeed, for the case of the harmonic oscillator, $S_0(t)$ is a periodic function, reflecting the fact that the quantum energy levels are evenly spaced.

My issue is that I can't see how this works for a general potential well. I've tried calculating $S_0(t)$ for situations besides the harmonic oscillator, and it doesn't seem to have discrete spectrum at all. Is there a direct way to see this result, if it's true?


1 Answer 1


Briefly, to see the energy quantization in the path integral $K(q_f,q_i,T)$ one should:

  1. put $q_f=q_i$, i.e. consider only periodic virtual paths.

  2. integrate over $q_f=q_i$.

  3. Laplace transform $T\to E$.

Altogether the path integral becomes: $$G(E)~\propto~\int_0^{\infty} \!dT ~e^{\frac{iET}{\hbar}}\int_{\mathbb{R}}\! dq_f ~ K(q_f,q_f,T).$$ At the classical level, the Laplace transform induces a Legendre transformation from the action to the abbreviated action.

One may show that the discrete energy levels $E_n$ shows up as poles in $G(E)$, cf. Ref. 1. See also Gutzwiller trace formula, cf. Ref. 2.


  1. R. Rajaraman, Solitons and Instantons: An Intro to Solitons and Instantons in QFT, 1987; Section 6.3.

  2. P. Cvitanovic et. al., Chaos: Classical and Quantum, 2013; Chapters 35, 37 & 38. The pdf file is available at www.chaosbook.org.

  • $\begingroup$ Correction to answer (v1): Chapters in Ref. 2 have shifted 2 numbers in newer versions. $\endgroup$
    – Qmechanic
    Jul 29, 2019 at 6:32

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