I have been following chapter 9 of Peskin and Schroeder where in section 9.2 the functional integral formalism is used for the quantum theory of a real scalar field.

As far as I understand, the path integral operates over all intermediate field configurations from a specified start state to a specified end state.

In particular, there does not seem to be any constraint that the intermediate field configurations are quantized in any way, e.g. to have integral number of particles.


Are the intermediate field configurations constrained to have an integral number of particles?

If so, then where does this appear in the mathematics?

If not, then why does the physics not include terms where fractions of a particle appear?

  • $\begingroup$ They are never quantized. Note, however, that "value" of the field configuration has no real significance here. One way to see this is that a-numbers, which are required to describe fermions, have no absolute value or something of this sort. $\endgroup$ – user178876 Feb 18 '18 at 22:55
  • $\begingroup$ So is it true that the starting and end states can have real physical significance (e.g. representing starting and ending particles), but integrating over the unconstrained intermediate field configurations is more of a mathematical trick that ends up producing the right physics? $\endgroup$ – Peter de Rivaz Feb 18 '18 at 23:04
  • $\begingroup$ Not really. Your question is in the context of QFT, where it may not make too much sense because of what I wrote above. You may ask it in the context of QM, where instead of a field (operator) you deal with with wave functions. There, however, this question does not arise in this form. I also partly agree with this answer, but not on the part that the intermediate field configurations are "classical". There are no a-numbers in classical physics. $\endgroup$ – user178876 Feb 18 '18 at 23:36
  • $\begingroup$ @marmot True, fermions are more complicated. I would call a Grassman-valued field semiclassical, i.e. not fully quantum, but I guess that’s up for debate. $\endgroup$ – knzhou Feb 18 '18 at 23:46

When the path integral is used to define a quantum theory, the intermediate states are classical, so the question doesn't make sense.

You can see this in the path integral for quantum mechanics. In the usual formulation, we consider paths $x(t)$. Each path describes the motion of a classical particle; the quantumness comes from integrating over paths, not the paths themselves. Similarly, in the path integral for quantum field theory, the intermediate states are simply classical field configurations $\phi(x)$. The number of particles in a field is a quantum notion and does not apply.

But there's a closely related question that I think you wanted to ask: if we want to use the path integral to find the amplitude to go from one quantum state to another, how do we map the quantum states to classical boundary conditions?

For quantum mechanics we associate a classical particle with position $x$ with the quantum state $|x \rangle$. In quantum field theory we usually associate a classical field $\phi(x)$ with the field eigenstate $| \phi(x) \rangle$ which satisfies $$\hat{\phi}(x) | \phi(x) \rangle = \phi(x) |\phi(x) \rangle.$$ Since $\hat{\phi}(x)$ contains creation and annihilation operators, $|phi(x) \rangle$ can't have a definite number of particles, or even a bounded number (because if the highest number was $n$, acting with $\hat{\phi}(x)$ would give a part with $n+1$ particles). Instead, it's a superposition of all possible numbers of particles. This is pretty unwieldy for scattering problems, but as you've seen in P&S, we can compute $S$-matrix elements without thinking about the boundary conditions at all, which is why they spend so little time on the subject.

A third point is that, instead of using the path integral to define the quantum theory, you may instead want to start in the quantum theory and derive the path integral. In quantum mechanics, we do this by inserting resolutions of the identity in the position basis, $$1 = \int\, dx |x \rangle \langle x |.$$ For quantum field theory we instead use the field basis defined above, $$1 = \int\, \mathcal{D}\phi(x) \, |\phi(x)\rangle \langle \phi(x)|.$$ In this picture, the intermediate states are quantum. They don't contain a definite number of particles, but instead are superpositions of states that are.

I totally understand being confused about this, because P&S shove a whole lot under the rug. They run through everything I said in about one paragraph, leaving everything implicit, and elide the difference between the path integral as a definition of a quantum theory (where the intermediate states are classical) and the path integral as a derived consequence of a quantum theory (where the intermediate states are quantum), switching between the two in a single sentence. For more detail about path integrals, I recommend these lecture notes.

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    $\begingroup$ +1 because I overall agree, but I'd like to mention that the a-number vs. c-number issue is really more delicate. Imagine a pion, i.e. a bound state of two states. That is described by a product of two a-numbers, which is a c-number, which is not to be confused with an ordinary complex number. So inside the path integral these are just algebraic objects, not at all classical field configurations. Only the result of the path integral is a number. (That's why the question doesn't make too much sense to begin with, but as you point out the literature is very sketchy, so the OP can't be blamed.) $\endgroup$ – user178876 Feb 19 '18 at 2:59
  • $\begingroup$ Thanks, those lecture notes look very useful - it is surprising how much there is to understand even in the 0-dimensional case! I still don't understand how the path integral formalism can give rise to quantized phenomena (such as not being able to have half an electron) but maybe it will become clearer when I have absorbed more of the material... $\endgroup$ – Peter de Rivaz Feb 19 '18 at 21:46

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