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This is in most probability a super basic question but I can't find easy answers anywhere. In my early education, I learned that in QM systems, the classical expressions of momentum, energy etc don't make much sense, hence we define analogues for them to do calculations in QM.

However... when I am learning the Path integral formulation, for studying a free particle propagator, classical results are assumed to get the Kernel... so what gives?

Qmechanic
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  • Good question. It is indeed weird, and interesting, that a classical action is also capable of giving you quantum field theory, if you're willing to exponentiate it with iota and integrate over all field configurations. This way, you include not just the classical solution for the field (one that extremizes the classical action), but also quantum corrections. – Avantgarde Oct 22 '22 at 03:08
  • The path integral is a quantization procedure. It tells us how to go from a classical theory to a quantum theory that has the classical theory as a limiting case (in some sense). You are looking at the automatic translator algorithm, if you will. – FlatterMann Oct 22 '22 at 05:33
  • "when I am learning the Path integral formulation, for studying a free particle propagator, classical results are assumed to get the Kernel" Can you be more specific what "classical results" you are referring to here? – ACuriousMind Oct 22 '22 at 14:49
  • @ACuriousMind For example, expression of velocity, kinetic energy etc – tryst with freedom Oct 22 '22 at 22:25

3 Answers3

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OP's title question is quite broad and perhaps best addressed in the realm of deformation quantization.

It should stressed that the quantization of a classical quantity to a quantum operator is a non-unique/ambiguous procedure, which might not work for certain theories.

Conversely, a quantum operator corresponds to many different symbols/functions, e.g. Weyl or normal ordering symbol.

These symbols/functions are semiclassical power series in $\hbar$. The leading term at order ${\cal O}(\hbar^0)$ is the classical quantity itself. So in the limit $\hbar\to 0$, we in principle recover the classical theory.

So when we rewrite a correlation function from the operator formalism to the path integral formalism, we choose in principle a specific operator ordering prescription, which leads to a specific choice of symbols/functions in the path integral.

For more details, see e.g. this and this related Phys.SE posts.

Qmechanic
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It is not completely correct that classical results are used to construct the kernel of the quantum propagator in the path integral formulation.

In fact, first of all, the "integral" sums over all the paths independently of the fact that they satisfy or not the classical equation of motion.

Actually, an even stronger fact holds. The measure (in the Euclidean formulation) of the (correspondings of) classical paths is negligible in the space of all summed paths. (In the real time formulation an analogous fact is valid but there is no proper measure in that formulation).

The quantum nature of the path integral relies just upon the use of non-classical paths. Quantum interference arises by them.

  • If the path integral would have summed only over those paths that satisfy classical equations of motion, would the predictions of the theory have been different? – flippiefanus Oct 22 '22 at 13:05
  • I am not an expert, but the set of classical paths has zero measure so that you would obtain zero as the propagator! – Valter Moretti Oct 22 '22 at 13:48
  • The real line is of measure zero in the complex plane, but if one restricts the integration over the real line only the results are finite. So I expect something similar for functional integrals. – flippiefanus Oct 23 '22 at 02:48
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It helps to remember that the separation between quantum and classical theories are just the different ways we model nature. Nature out there doesn't care about our theories. It is the same nature that produces all the phenomena that we observe.

So, if we talk about a photon or about a classical optical beam, physically they represent the same thing. If our equations of motion work for one, it should also work for the other.

The thing that makes a difference is the interactions. We can basically add interactions to our classical theories and end up with quantum theories. When we take the quantum theory and make the interaction strength zero, we get back the classical theory. The quantization property is introduced by the properties of interactions.

In quantum physics, we often study quantum states, which are supposed to be different from classical states. However, if you look carefully, you'll see that the difference comes from consequences of interactions. It is the methods we use to prepare such states that make them different. For example, entanglement is a consequence of the interactions that are involved in the preparation process of such states.

Hope this helps. It is not an easy question to answer, but I hope that this understanding can give an idea of how to see the difference.


To respond to the comments "... free theories have quantum effects." No they don't.

Double slit interference: linear process. The process itself is not a quantum process. The same interference happens with classical light. The quantum aspect comes in with the observation of single photons/electrons at the screen. This happens because the observation requires an interaction with the screen.

Bell's inequality: it is an entanglement witness. As I explained, entanglement is a property that is obtained by the state preparation process which requires interactions.

Uncertainty principle: fundamentally a result of Fourier analysis which does not have anything to do with quantum mechanics per se. Quantum mechanics inherits the uncertainty principle due to the link between momentum and the wave vector which comes from the quantization associated with interactions.

In particle physics, it is generally accepted that a free-field theory (i.e., a theory without interactions) is equivalent to a classical theory.

Anything else?

flippiefanus
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    Interactions...? In what aspect of the Path integral is this idea encoded in? – tryst with freedom Oct 22 '22 at 03:13
  • Without the interactions that Path integral represents a classical theory. The interactions make it a quantum theory. – flippiefanus Oct 22 '22 at 03:14
  • But free theories have quantum effects. Like double slit interference, identical particle interference, Bell's inequality, uncertainty principle. How is a quantum theory classical without interactions? – Ryder Rude Oct 22 '22 at 07:52
  • @RyderRude The double slit doesn't show quantum effects. Only the emission and absorption processes do. – FlatterMann Oct 22 '22 at 10:02