We know the physical relevance of the classical limit of quantum mechanics quite well. However, if I take the classical limit of a quantum field theory, the answer is not so clear.

Suppose I take the Hamiltonian for a free electron moving in one dimension, which is $\hat H=\hat p^2/2m$. The classical limit of this theory is the Hamiltonian $H=p^2/2m$, which is that of a point particle moving at a constant velocity.

However, suppose that I now take the Hamiltonian for $N$ free electrons, which is $\hat H=\int dx\,\psi^\dagger(x)(\hat p^2/2m)\psi(x)$. The classical limit of this theory is the Hamiltonian $H=\int dx\,\bar \psi(x)(-\hbar^2\partial^2/2m)\psi(x)$.

Shouldn't we just get $N$ point particles moving at a constant velocity? Instead, we get this weird one-dimensional wave...

  • $\begingroup$ How do you define the classical limit of quantum mechanics (by which you mean non-relativistic single particle theory?)? $\hbar \to 0$? That's not even defined. Take a hydrogen atom and vary $\hbar$. What happens? Do you get a classical piece of matter of just a hydrogen atom with a scaled energy spectrum and size? $\endgroup$
    – CuriousOne
    Apr 3, 2016 at 4:38
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    $\begingroup$ @CuriousOne The meaning of taking the classical limit of quantum theories is pretty standard and accepted, and it means considering the effective theory emerging from the quantum one when the quantum effects become negligible (or, as Bohr said, "in the limit of large quantum numbers"). It is necessary for a quantum theory to be reasonable that it reproduces the macroscopic classical theory, in a suitable effective regime. The proof of this fact is far from trivial, but we're recently starting to have a precise and quite satisfactory picture for many interesting quantum theories. $\endgroup$
    – yuggib
    Apr 3, 2016 at 8:49
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    $\begingroup$ @yuggib: Just because some handwaving is "pretty standard and accepted" and the same nonsensical argument can be found in the first chapter of every entry level quantum mechanics text in the world does NOT mean that one has to swallow it hook, line and sinker. $\hbar$ in the Schroedinger equation sets a scale and that is all it does. There is no limit procedure that can change that scale into classical behavior. Try it for yourself and see what happens. That's also not how nature transitions from QM to CM. It uses decoherence, instead. $\endgroup$
    – CuriousOne
    Apr 3, 2016 at 10:29
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    $\begingroup$ @CuriousOne I am not the one handwaving, see the references in the other comment below my answer. There is a huge mathematical and physical literature on the topic, and many things are now well understood in a rigorous and satisfactory fashion. And by the way, it is actually my topic of research, so yes I have tried it for myself and I could give you all the results you want. Ignorance is not an excuse for being rude, and wrong. $\endgroup$
    – yuggib
    Apr 3, 2016 at 11:15

1 Answer 1


It all depends on the scaling, i.e. on which parameter is taken to be small (large) in your effective description of the system.

It is customary to interpret the semiclassical parameter to be a quantity "equivalent" to $\hbar$, but going to zero. This is convenient, for in the classical scale of energies Planck's constant is comparatively very small. Equivalently, we may think of the semiclassical parameter as being representing the inverse of the "characteristic frequency" of the particle's wave (and therefore the classical limit is the limit of very high frequencies).

Another different parameter is the number of particles $N$. We may think of taking the limit $N\to\infty$ in a given $N$-particle system. It turns out that, mathematically, this is similar to taking the classical limit but the physical interpretation is quite different.

So, let's consider a system of $N$ free non-relativistic bosons of mass $1/2$. Their Hamiltonian can be written as $$H_N=\sum_{j=1}^N -\hbar^2\Delta_{x_j}\; ;$$ where $\Delta_x$ is the Laplacian, or equivalently in "second quantization" notation $$H_N = -\hbar^2\int_{\mathbb{R}^3}a^*(x)\Delta_x a(x)dx\; \Bigr\rvert_{L^2_s(\mathbb{R}^{3N})}\; ;$$ where the restriction to $L^2_s(\mathbb{R}^{3N})$ means that we are just considering the sector with $N$ particles (since in fact the number of particles is here conserved, and it's not so useful to consider the whole Fock space).

Now if you take the limit $N\to\infty$, you get indeed an energy functional (not an operator anymore, hence a "classical" infinite dimensional field theory) of the type $$E(u)=-\hbar^2\int_{\mathbb{R}^3}\bar{u}(x)\Delta_x u(x)dx\; ;$$ where $u\in L^2(\mathbb{R}^3)$ is the "classical" (more properly, mean field) variable corresponding to the annihilation operator valued distribution $a(x)$. The interpretation is of a free quantum mean-field theory : $u$ represent the effective wavefunction of a single particle under the effect of all the other particles combined (that in this case reduces to a free particle, since there were no interaction). With a two body weak (in the limit $N\to\infty$) interaction you would have got the Hartree energy functional and corresponding Hartree dynamics.

If you take the limit $\hbar\to 0$ instead, you get $N$ free classical particles , with energy function $$E(\vec{x},\vec{p})=\sum_{j=1}^N p_j^2\; ;$$ where $p_j\in\mathbb{R}^3$ is the momentum of the $j$-th particle.

As you see, the two limits have quite different physical intepretations, even if they're actually mathematically pretty similar. I remark that they can also be combined in a "commutative" way; in the end you would get a classical Vlasov-type evolution for infinitely many classical particles (both if you do before the $\hbar\to 0$ and then the $N\to\infty$ or vice-versa).

The situation is different if you consider a "true" QFT, where particles can be created or destroyed, e.g. photons in QED. There, the classical limit $\hbar\to 0$ directly yields, as expected, a classical field theory. On the other hand, the mean field is not so meaningful since there are quantum states with an undefined (possibly very large) number of particles; and since the number is not conserved, even if you start with a fixed number of particles, after evolution you get a state with a non-zero probability of having different numbers of particles.


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