# Are Classical Field Theory and Quantum Mechanics of a single particle (nonrelativistic or “classical”) limits of Quantum Field Theory?

Recently I talked about QFT with another physicist and mentioned that the Quantum Field Theory of a fermion is a quantisation of its one-particle quantum mechanical theory. He denied this and responded that he rather sees the single particle QM as the non-relativistic limit of a QFT. He elaborated that the energies encountered are all much smaller than the particles mass, so we can ignore all multi-particle excitations in Fock Space and get an effective Hilbert space consisting of all single-particle excitations.

In turn, I asked what the corresponding limit of QED quantum mechanics of the massless photon is, and he of course responded that there can't be a nonrelativistic limit of QED exactly because of the masslessness. But there is classical ED, the classical limit of QED.

So is taking the classical or the nonrelativistic limit the same, or does one include the other, or is there some deep difference?

The question What does a Field Theory mean? has something to do with it, but does not fully answer my question.

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Check out this link, damtp.cam.ac.uk/user/tong/qft/two.pdf starting at the bottom of page 43. –  David Santo Pietro Mar 12 '12 at 7:19
Thanks, that's helpful. –  Turion Mar 29 '12 at 12:51
More on reduction from QFT to QM: physics.stackexchange.com/q/26960/2451 –  Qmechanic Sep 23 '12 at 16:51

Dear Turion, the Dirac quantum field may be formally obtained by quantizing the Dirac equation which is a relativistic but single-particle quantum mechanical equation.

The non-relativistic limit of the single-particle Dirac equation is the Pauli equation which is essentially the non-relativistic Schrödinger equation for a wave function with an extra 2-fold degeneracy describing the spin.

To get from the non-relativistic Schrödinger equation for an electron to a quantum field theory with a quantized Dirac field, you therefore

1. Have to add the spin and going to the Pauli equation - easy
2. Guess the right relativistic generalization of the Pauli equation - it's the single-particle Dirac equation that also has negative-energy solutions
3. Realize that the negative-energy solutions are inconsistent in the single-particle framework, so you have to second-quantize the wave function and obtain the Dirac quantum field

This sequence of steps is formal. One can't really "deduce" things in this order, at least not in a straightforward way. After all, the step 1 needed a creative genius of Pauli's caliber, the step 2 needed a creative genius of Dirac's caliber, and the step 3 needed a collaboration of dozens of top physicists who developed quantum field theory. Quite on the contrary, as you were correctly told, the meaningful well-defined operations go exactly in the opposite order - but it doesn't follow the steps above. You begin with the quantum field theory, including the Dirac field, which is the right full theory, and you may take various limits of it.

The non-relativistic limit is of course something totally different than the classical limit. The nonrelativistic limit is still a quantum theory, with probabilities etc. - but it doesn't respect the special role of the speed of light. On the other hand, the classical limit is something totally different - a classical deterministic theory that respects the Lorentz symmetry, and so on. Let us discuss the limits of quantum electrodynamics separately.

Classical limits

The classical, $\hbar\to 0$, limit of QED acts differently on fermions and bosons. The bosons like to occupy the same state. However, to "actually" send $\hbar$ to zero, you need quantities with the same units that are much larger than $\hbar$: $\hbar$ goes to zero relatively to them. What quantities may you find? Well, the electromagnetic field may carry a lot of energy in strong fields.

So you get a classical limit by having many photons in the same state. They combine into classical electromagnetic fields that are governed by classical Maxwell's equations; note that classical Maxwell's equations are "already" relativistic although people before Einstein hadn't fully appreciated this symmetry (although Lorentz wrote the "redefinition" down without realizing its relationship with the different inertial frames or symmetry groups, for that matter). You just erase hats from the similar Heisenberg equations for the electromagnetic fields.

Well, for extremely high frequencies, the number of photons won't be large because they carry a huge energy. So for high frequencies, you may also derive another classical limit - based on "pointlike particles", the photons.

Fermions, e.g. electrons described by the Dirac equation, obey the exclusion principle. So you can't have many of them. There is at most one particle per state. In the quantum mechanical theories, it has an approximate position and momentum that don't commute. The classical limit is where they do commute. So the classical limit must inevitably produce mechanics for the fermions - with positions and momenta of individual particles. As I mentioned, this picture may be relevant for high-energy bosons, too.

Nonrelativistic limit

The non-relativistic, $c\to\infty$, limit of QED is something completely different. It is still a quantum theory. Because the photons propagate by the speed of light and the speed is sent to infinity, the electromagnetic waves propagate infinitely quickly in the non-relativistic limit. That means that the charged (and spinning or moving charged) objects instantly influence each other by electric (and magnetic) fields.

When it comes to fermions, you undo one of the steps at the beginning: you reduce the speed of the electrons. Assuming there are no positrons for a while, the non-relativistic limit where the velocities are small will prevent you from the creation of fermion-antifermion pairs. So the number of particles will be conserved.

So it makes sense to decompose the Hilbert space into sectors with $N$ particles for various values of $N$ and you're back in multi-body quantum mechanics. They will also have the spin, as in the Pauli equation, and they will interact via instantaneous interactions - the Coulomb interaction and its magnetic counterparts (combine the Ampére and Biot-Savart laws for $B$ induced by currents with the usual magnetic forces acting on moving charges and spins). You will get the usual non-relativistic quantum mechanical Hamiltonian used for atomic physics.

There will be no waves because they move by the infinite speed. You won't be able to see them. But they won't destroy the conservation of energy etc. because in the non-relativistic limit, the power emitted by accelerating charges goes to zero because it contains $1/c^3$ or another negative power.

So in the non-relativistic limit, the photons just disappear from the picture, and their only trace will be the instantaneous interactions of the Coulomb type.

Classical non-relativistic limit

Of course, you may apply both limiting procedures at the same moment. Then you get non-relativistic point-like classical electrons interacting via the Coulomb and similar instantaneous interactions.

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I forgot to say: if you want the non-relativistic limit with positrons, it's dangerous. The limit works "with electrons only" or "with positrons only" but if you have states with both electrons and positrons, they may begin at low speeds, but they may annihilate and produce a huge energy that may, for example, accelerate nearby electrons and positrons to relativistic speeds, at least for a sufficient number of annihilation. So the nonrelativistic limit only exists if you carefully avoid states that are able to release a huge energy - e.g. by annihilation (or nuclear processes). –  Luboš Motl Jan 29 '11 at 16:06
Yet another great answer. If I fail to compliment them all I hope you will understand. –  Carl Brannen Jan 29 '11 at 22:31
Dear Carl, your pleasure is more important than compliments. ;-) Cheers, LM –  Luboš Motl Jan 30 '11 at 17:11

Some quick remarks (I hope that these are helpful to formulate more precise versions of your question):

• "Recently I talked about QFT with another physicist and mentioned that the Quantum Field Theory of a fermion is a quantisation of its one-particle quantum mechanical theory."

I think you meant to say that the Hilbert space of the QFT is constructed as the fermionic Fock space from the one particle Hilbert space, which is usually called second quantization, right?

• "He elaborated that the energies encountered are all much smaller than the particles mass, so we can ignore all multi-particle excitations in Fock Space and get an effective Hilbert space consisting of all single-particle excitations."

Particle creation is only one aspect of special relativity that is melded into quantum mechanics by quantum field theory. You also have to handle the non-zero probability of a particle described by nonrelativistic quantum mechanics to propagate with a velocity greater than c (the vacuum velocity of light). Take a free massive spin-zero boson in one dimension localized at a point $x$ at a time $t=0$, the time evolution is described by a Schrödinger equation. At an arbitrary small time $t \gt 0$ the wavefunction will be a Gaussian, which means there is a non-zero probability to find the particle at an arbitrary distance from $x$.

Therefore any limit $\lim_{c \to \infty}$ has to look at the dynamics as well, not only at the state space. In what sense the Schrödinger equation describes the dynamics of elementary particles with nonzero mass in the limit $\lim_{c \to \infty}$ is explained in every textbook of QFT that I know of.

• "...he of course responded that there can't be a nonrelativistic limit of QED exactly because of the masslessness."

Yes, that's just true, because the classical theory (Maxwell's equations) is already relativistic.

• "So is taking the classical or the nonrelativistic limit the same, or does one include the other, or is there some deep difference?"

I'm sorry but I don't understand this question. From a pure formal point of view, the classical limit is $\lim_{h \to 0}$, and the nonrelativistic limit is $\lim_{c \to \infty}$, and the examples that you cite yourself make already clear that this is not the same.

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Classical field theory, yes (extension of Ehrenfest's Theorem), but quantum mechanics, no, not in the strict sense. If you assume a free field theory in a number (Fock) state, in the nonrelativistic limit the quanta should obey Schroedinger's equation. However, a TRULY noninteracting field is unobservable and one could never even demonstrate that it exists! The ones we actually know about can thus only "approximately" reduce to QM in the non-relativistic limit.

I highly recommend a paper called "Quantum mechanics: myths and facts" by Hrvoje Nikolic

. Section 9 discusses at some length the fact that, strictly speaking, QM is definitely not the nonrelativistic limit of QFT in the interacting case...

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