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(xi,xf,t)≡⟨xf|e−iHt|xi⟩=∫x(0)=xi,x(t)=xfDx(t)eiS[x(t)]/ℏ.
On the other hand, applying a semiclassical approximation, the path integral is just
K(xi,xf,t)∼|∂2S0(xi,xf,t)∂xi∂xf|eiS0(xi,xf,t)/ℏ
where
S0 is the on-shell action, i.e. the action for the classical path that goes from
xi to
xf 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(xi,xf,t)=∑n,m⟨xf|n⟩⟨n| e−iHt|m⟩⟨m|xi⟩=∑n⟨xf|n⟩⟨n|xi⟩e−iEnt/ℏ
so we get discrete energy if the Fourier transform of
K(xi,xf,t) in time has discrete support, which is equivalent to the same being true for
S0(xi,xf,t). That means we can read off the energy discretization directly from the classical action. Indeed, for the case of the harmonic oscillator,
S0(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 S0(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?
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