Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

So, the quantum path integral is a generalization of the classical principle of least action- but here we know that all paths contribute something finite to the probability density. What confuses me is that this doesn't seem to involve the quantization of energy (action) at all. How does this come into play? Is it a separate assumption, only coming into play when we worry about the non-commutativity of certain operators?

share|cite|improve this question
afaik there is nothing inherently quantized in quantum mechanics, it only occurs when you solve for wavefunctions with certain boundary conditions – Lagerbaer Aug 23 '11 at 3:39
up vote 5 down vote accepted

The path integral in quantum mechanics computes the evolution kernel, which is the matrix element of the evolution operator: $\mathrm{exp}(iH t) $, ($H$ is the Hamiltonian), between two position eigenstates.

The path integral expresses the evolution kernel as a sum over paths:

$U(x,t, x_0) = \int_{x(0)=x_0}^{x(t)=x} \mathrm{exp}(\frac{iS}{\hbar}) \mathcal{D}x$.

where $U$ is the evolution kernel, $S$ is the classical action.

On the other hand, the evolution operator has an expansion as a sum over the energy eigenstates:

$U(x,t, x_0) =\sum_n \mathrm{exp}(\frac{-iE_n t}{\hbar})\psi_n(x) \psi_n^{*}(x_0)$

where, $\psi_n(x)$ are the energy eigenstates.

From this expression, it is clear that the evolution kernel has a discrete spectrum, whenever the energies are quantized.

In other words, in the case of quantized values of the energy, the Fourier transform of the evolution kernel:

$\hat{U}(x,\omega, x_0)\equiv \int_{-\infty}^{\infty} U(x,t, x_0) \mathrm{exp}(i\omega t) dt$

will be a sum of Dirac delta functions centered at the frequencies:

$\omega_n = \frac{E_n}{\hbar}$

Please observe that the weight of each Dirac delta function is just the projection operator on the corresponding discrete eigenstate.

share|cite|improve this answer
Hamiltonian can have both discrete and continuous spectrum (or only one of them - for the second consider a non-bounding potential of a shallow potential well, or even better - a free particle). You can write $U(x,t,x_0)$ as a sum (over a countable set) only for the purely discrete spectrum. – Piotr Migdal Aug 23 '11 at 9:47
@Piotr The form in which the "sum over states" is written is just illustrative, of course, if continuous spectrum is present, we should write an integral over the spectral projectors. What I wanted to emphasize is that we can detect the presence of discrete spectrum by observing the presence of Dirac "peaks" in the Fourier transform of the evolution kernel computed by means of a path integral. – David Bar Moshe Aug 23 '11 at 10:09
@DavidBarMoshe Sorry for bothering you, I know that this comment is not strictly related to your answer. Do you know any paper or book where I can read about the path integral measure, how it encodes ordering operator issues, how its definition is connected with the ordering prescription (e.g, Weyl), etc? I tried to read Glimm & Jaffe's book but it was too formal for me. I am looking for some reference somehow between the conventional path integral derivation of QM and QFT books and Glimm & Jaffe's. A link to some question in this site would be welcome as well. Thank you. – drake Mar 8 '13 at 19:05
@drake Please give me a few days, I'll prepare a list. – David Bar Moshe Mar 10 '13 at 7:14
@drake I have placed a list in a file exchange server:… – David Bar Moshe Apr 8 '13 at 12:38

In general, path integral does not imply quantization of anything. First, you can use path integrals for classical wave optics. Second, even in quantum mechanics there are non-bonding potentials (take the free particle potential $V(x)=0$ for the simplest case).

However, sometimes from the path integral point of view it can be argued that some things are quantized. Perhaps the best known example is the Aharonov-Bohm effect where to have non-ambiguous solution phase $e^{i S/\hbar}$ can not change when a particle makes a full circle around a flux of magnetic field.

share|cite|improve this answer

The quantization is not necessary in quantum mechanics. It is essential, but not necessary. Energy of a free particle is not quantized. Even for oscillator you are free to construct any state and energy.

Quantization appears at the next step. If you want for instance study stable states which do not change with time. It is not separate assumption. It is different problem.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.