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I was reading this, where it is mentioned that some quantum theories can have no classical limit or even more than one classical limit.

A possible example might be quantum spin, which doesn't have a classical analog. What is the precise condition (does it arise solely on imposing $\hbar \to 0$, which I am not able to see for quantum spin) that determines the number of classical limits of a quantum system?

What are some other examples of this, especially of systems having more than one classical limit.

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  • $\begingroup$ I am not sure, but I have a feeling that you should consult conformal mapping theory. Lagrangian (Hamiltonian) is a mapping between a physical symmetries of the system and the observables. I cannot answer your question, but I am almost positive that this kind of thinking will lead you to the answers. $\endgroup$ – MsTais Aug 24 '16 at 2:14
  • $\begingroup$ This is an interesting question, I will answer later when I have a bit of time. $\endgroup$ – yuggib Aug 24 '16 at 6:45
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First of all, I would like to specify what I mean for classical limit $\hslash\to 0$. It is useful to distinguish two cases: a) bosons and standard QM particles, b) fermions.

a) The classical limit means that we consider the behavior of a quantum system in the regime where the non-commutativity of canonical observables becomes negligible. The aim is to find an effective simpler description in this regime. Since the commutator between the canonical observables is proportional to Planck's constant $\hslash$, the classical limit is often written as $\hslash\to 0$.

b) For fermions the situation is more complicated, since their statistics is purely quantum and has no immediate classical analogue. The best way to recover an effective classical behavior for fermions is to take the coupled limit (following the considerations above) $\hslash\to 0$ and $N\to\infty$, where $N$ is the number of fermions. In this way it is possible to describe many fermions as a collective classical "plasma" (that obeys Vlasov evolution equation).

In the above, spin was (indirectly) neglected. Even if spin has no direct classical analogue, it does not "mess up" things to much: it may modify the behavior, but it does not prevent the possibility of having an effective description as discussed above.

Now, let's come to OP's questions:

  1. The first is whether there are systems that do not admit a classical limit. The answer is that, as far as I know, there is no "no-go" result that tells us that for some particular system it is impossible to obtain an effective description of a suitable kind. So I would say that, even if we are not always able to obtain (derive) the effective description, this should be seen more as a technical limitation than as a fundamental feature of some special physical system. Also, it is true that the effective descriptions, especially for fermions, may be quite far from what we would intuitively describe as a classical behavior, nevertheless it is classical in the sense that it is described, roughly speaking, by a classical commutative probability theory and not by a non-commutative probability theory of Hilbert-space operators.

  2. The second question is whether there are systems that admit more than one classical limit. If we interpret the classical limit as above in a) (depending on just one parameter), the classical description we obtain is unique. There are families of $\hbar$-dependent quantum states that would give different limits depending in which way the limit $\hbar\to 0$ is taken (subsequence extraction), but the resulting classical theory is always the same. On the other hand, if we consider a limit where there is more than one parameter, and those parameters may have a correlation between each other, then it is possible to obtain different limiting behaviors. As an example, in b) we may or not obtain a different limit if $N\sim O(\hslash^{-1})$ or $N\sim O(\hslash^{-5})$, and one of the two limits may be trivial. There is not, however, a general rule to see that, and we have to study systems case by case.

I would also like to conclude with a remark that is not completely related, but interesting. Since a quantum theory is more fundamental than the corresponding classical effective theory, the classical limit procedure is often solving ambiguities rather than being ambiguous itself. For example, there are classical motions that are not well-defined (e.g. they lack uniqueness) while the corresponding quantum motion is well-defined. In that situation, if we take the classical limit properly, we obtain a well-defined effective dynamics; and thus we have a prescription on how to treat the classical problem in the right way, that takes into account the underlying quantum theory and is properly defined. If you are interested in the (very technical) details on that type of situation, you may see this very recent paper.

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  • $\begingroup$ In the Wigner-function approach to QM (which is semi-classical in nature as it retains some non-classical correlations) the semi-classical limit of spin systems is reached when $1/S\to 0$ since, in this limit, the Moyal bracket collapses to the (classical) Poisson bracket on the sphere. See for instance Sec. 2.4 of iopscience.iop.org/article/10.1088/1751-8121/50/32/323001/meta $\endgroup$ – ZeroTheHero Apr 29 '18 at 16:56
  • $\begingroup$ My professor recently remarked that it is expected that Fermionic theories don't admit a classical limit because we simply can't put more than one particles in the same state and the classical limit is experimentally realized when we have a large number of particles in a single state. Could you comment on this argument? [...] $\endgroup$ – Dvij Mankad Mar 6 at 7:41
  • $\begingroup$ [...] Also, I don't see why he seemed to expect that the classical limit should correspond to a large number of particles in a single state. In my understanding, for example, in Bosons, the classical limit simply corresponds to $\hbar\to0$ and I don't see how that relates to a large number of particles in a single state. If you can suggest a review-type resource for all these related issues then it would be even fantastic. Apologies for the naivety of the questions. Thanks! :) $\endgroup$ – Dvij Mankad Mar 6 at 7:41

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