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Feynman’s path integral formulation of quantum mechanics is based on the following two postulates:

  1. If an ideal measurement is performed to determine whether a particle has a path lying in a region of spacetime, the probability that the result will be affirmative is the absolute square of a sum of complex contributions, one from each path in the region.

  2. The paths contribute equally in magnitude but the phase of their contribution is the classical action, i.e. the time integral of the Lagrangian taken along the path.

Suppose that those postulates are as natural as it can be, i.e. they aren't a distilled version of something more elementary. If this is the case, then how to explain, in layman's terms, that how the discrete energy levels in, e.g. the harmonic oscillator arise from those two postulates? Are there qualitative answers?

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    $\begingroup$ These postulates do not imply that a system has discrete energy levels. Some quantum systems have a discrete energy spectrum (e.g. the hydrogen atom) others have a continuous spectrum (e.g. a free particle). Both cases can be described in the path integral formalism, so you need more information about the system (in particular some details of the Lagrangian or Hamiltonian) to determine which case you are dealing with. $\endgroup$ – By Symmetry Mar 29 '18 at 11:26
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    $\begingroup$ Further to the comment above, the simplest example of using he path integral formulation for a bound system with discrete energy levels is the harmonic oscillator. A Google will find you many articles explaining how the path integral formalism is applied to the harmonic oscillator. $\endgroup$ – John Rennie Mar 29 '18 at 11:30
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    $\begingroup$ Related: physics.stackexchange.com/q/39208/2451 and links therein. $\endgroup$ – Qmechanic Mar 29 '18 at 11:31
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    $\begingroup$ I don't think any of the comments above address the question, which is a good one. Of course we know the path integral is equivalent to other methods, which have quantized energy. This question is about seeing that directly from the path integral itself. $\endgroup$ – knzhou Mar 29 '18 at 11:44
  • $\begingroup$ Postulate 1 is wrong, or at least inaccurate: no measurement will give you the path of a quantum particle; there is no such path. $\endgroup$ – Stéphane Rollandin Mar 29 '18 at 16:21
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I'm not sure what you expect for qualitative examples in "layman's terms". The time-evolution amplitude from one point to another, $$\langle x_f|U(T)|x_i\rangle=K(x_f,x_i;T)~,$$ evaluated from the path integral is the propagator, which, for the oscillator, happens to be the celebrated 1866 Mehler kernel: a causal Green's function of the oscillator equation. Not coincidentally, this was available 60 years before QM. The point being that the path integral propagator is mostly classical. The quantization of energy levels is actually a feature of the compact time domain, as @Qmechanic comments.

Specifically, the classical action for the oscillator, from the above WP reference, amounts to $$ \begin{align} S_\text{cl} & = \int_{t_i}^{t_f} \mathcal{L} \,dt = \int_{t_i}^{t_f} \left(\tfrac12 m\dot{x}^2 - \tfrac12 m\omega^2 x^2 \right) \,dt \\[6pt] & = \frac 1 2 m\omega \left( \frac{(x_i^2 + x_f^2) \cos\omega(t_f - t_i) - 2 x_i x_f}{\sin\omega(t_f - t_i)} \right)~. \end{align} $$

The propagator, then, the above amplitude, can be evaluated from the functional integral as $$K(x_f, x_i;T) = \Large e^\frac{i S_{cl}}{\hbar} \small ~ \sqrt{ \frac{m\omega}{2\pi i \hbar \sin\omega(t_f - t_i)}}~~, $$ where $T=t_f-t_i$. Effectively, it is the exponential of the classical action, with a minor normalization correction due to quantum fluctuations, which, nevertheless is not that important.

  • The quantized energy levels are already in the discrete harmonics predicated by the periodicity of the classical action--themselves oblivious of $\hbar$.

This expression also equals to the conventional Hilbert space propagator in terms of Hermite functions,
$$ \begin{align} K(x_f, x_i;T ) & = \left( \frac{m \omega}{2 \pi i \hbar \sin\omega T } \right)^\frac12 \exp{ \left( \frac{i}{2\hbar} m \omega \frac{ (x_i^2 + x_f^2) \cos \omega T - 2 x_i x_f }{ \sin \omega T } \right) } \\[6pt] & = \sum_{n = 0}^\infty \exp{ \left( - \frac{i E_n T}{\hbar} \right) } \psi_n(x_f) ~\psi_n(x_i)^{*}~, \end{align} $$ with which you'd propagate your ket to the bra of the amplitude in conventional Hilbert space.(This is the expression Mehler summed in 1866.) But you might wish to pretend you are a Martian, unaware of that formulation, or Mehler's wonderful, "prescient", formula.

Rewrite this as $$ = \left( \frac{m \omega}{\pi \hbar} \right)^\frac12 e^\frac{-i \omega T} 2 \left( 1 - e^{-2 i \omega T} \right)^{-\frac12} \exp{ \left( - \frac{m \omega}{2 \hbar} \left( \left(x_i^2 + x_f^2\right) \frac{ 1 + e^{-2 i \omega T} }{ 1 - e^{- 2 i \omega T}} - \frac{4 x_i x_f e^{-i \omega T}}{1 - e^{ - 2 i \omega T} }\right) \right) }\\ \equiv \left( \frac{m \omega}{\pi \hbar} \right)^\frac12 e^\frac{-i \omega T } 2 ~ R(T)~. $$

The $e^{-in\omega T }$ Fourier modes of R(T) then multiplying this 0-point energy prefactor may be compared to the standard Hilbert space eigenstate expansion of the resolvent, to reassure you of the standard quantized spectrum of the quantum oscillator, $E_n = \left( n + \tfrac12 \right) \hbar \omega~. $

Here, faced with discreteness, you only need appreciate the essential periodicity of the system, the compactness that forces you to a harmonic structure: the waviness of the system; and that most of it is traceable to the classical action in this (somewhat exceptional) quadratic hamiltonian paradigm.

In their elementary textbook, Feynman and Hibbs work it out nicely in Probs 2-2, 3-8, (eqns 2-9,3-59) and "spike" in eqns (8-12), (8-13). (They even go ludicrously further than that, trying to make you "see" the Mehler kernel deconstruct itself to Hermite polynomials, taking requests of your type too far, in my opinion.) In any case, following the simple math is actually less obscure than summarizing it in "code" verbiage.

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  • $\begingroup$ I know I said "for example, the harmonic oscillator" in the question, but I was hoping for something that held for all 1D bound states, at least. It's not true that the classical action is periodic in those cases, right? I'm having trouble seeing why it should have a discrete spectrum in general -- the calculations get quite messy. $\endgroup$ – knzhou Apr 6 '18 at 17:25
  • $\begingroup$ Ach... that's where I did not want to go... WKB... A classical particle bouncing back and forth between the turning points of the potential it is trapped in will reflect that periodic motion in the classical action... but the systematics is messy, even though known among serious experts... I thought you just wanted some motivational illustration... $\endgroup$ – Cosmas Zachos Apr 6 '18 at 18:43
  • $\begingroup$ Mueller-Kirsten Ch 21 has stuff, but it is already technical.... $\endgroup$ – Cosmas Zachos Apr 6 '18 at 19:04
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Because this is a conceptual question, perhaps it needs a simple, conceptual answer:

Feynman's path integral can be represented as the integral of $e^{iS/ħ}$ over all possible paths. Note the $i$ in the exponent, which forces the integrand to be periodic with respect to the value of S (which is real-valued). Setting variation of the integral equal to zero selects out discrete sets of paths. That, I believe, is the simplest way to understand the discreteness of energy states in the oscillator (whose energy is closely tied to S).

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