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Quantum mechanics as classical field theory

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Can we view the normal, nonrelativistic quantum mechanics as a classical fields?

I know, that one can derive the Schrödinger equation from the Langrangian $L = \frac{i}{2} (\psi^* \dot\psi - \dot\psi^* \psi) - \Delta\psi^* \Delta \psi - m \psi^*\psi$ and the principle of least action. But I also heard, that this is not the true quantum mechanical picture as one has no probabilistic interpretation.

I hope the answer is not to obvious, but the topic is very hard to google (as I get always results for QFT :)). So any pointers to the literature are welcome.

This post has been migrated from (A51.SE)
asked Feb 20, 2012 in Theoretical Physics by altertoby (35 points) [ no revision ]
The Schroedinger lagrangian defines a non-relativistic quantum field theory, which is equivalent to many-body quantum mechanics. You may ask whether this field theory has a classical limit. It does, under the usual condition (the states have to be highly occupied). This means that you should look to Bosonic states, for example the case that $\psi$ describes an integer spin atom.

This post has been migrated from (A51.SE)
These are quantum field theory notes, but they discuss your question http://www.damtp.cam.ac.uk/user/tong/qft.html.

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4 Answers

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You certainly couldn't recover quantum effects with a classical treatment of that Lagrangian. If you wanted to recover quantum mechanics from the field Lagrangian you've written, you could either restrict your focus to the single particle sector of Fock space or consider a worldline treatment. To read more about the latter, look up Siegel's online QFT book "Fields" [hep-th/9912205] or Strassler's "Field Theory without Feynman Diagrams" [hep-ph/9205205] for applications of the technique.

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answered Feb 20, 2012 by josh (195 points) [ no revision ]
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Indeed, the true Lagrangian for the Schrödinger equation takes this from

$${\cal L}=i\psi^*\dot\psi-\frac{\hbar^2}{2m}|\nabla\psi|^2-V({\bf x},t)\psi^*\psi$$

and the action becomes

$$S=\int dtd^3x{\cal L}.$$

A Lagrangian for the Schrödinger equation has a meaning only in a quantum field theory context when you do a second quantization on the Schrödinger wavefunction. This applies in a lot fields and mostly in condensed matter and, generally speaking, to many-body physics. In this case, you have to generalize this equation to the Pauli equation and work with spinor and anticommutation rules to describe ordinary matter.

Then, the probabilistic interpretation applies to the states in the Fock space for the many-body problem you are considering.

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answered Feb 20, 2012 by JonLester (376 points) [ no revision ]
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Can we view the normal, nonrelativistic quantum mechanics as a classical fields?

Yes, you can view the wave function $\psi(x,t)$ as an ordinary complex-valued field in the spirit of, say, classical electrodynamics. This field describes the quantum mechanics of a single electron, but it is classical in the sense that it's an ordinary function $\psi(x,t) : \mathbb R^3\times\mathbb R \to \mathbb C$.

As usual, you can have a lot of fun with the Lagrangian. For instance, you can note that there it has a (global) $U(1)$ symmetry $\psi \mapsto e^{iθ}\psi$ and apply the Noether theorem. You will find a continuity equation for the quantity $\psi^*\psi$, which we commonly interpret as the probability density.

Of course, the Schrödinger equation is limited to non-relativistic physics, so people started to look for a relativistic equivalent. Dirac's eponymous equation was intended to be precisely that: an equation for a classical field that describes a quantum mechanical electron in a Lorentz-covariant way. Of course, there should be an equivalent of the probability density $\psi^*\psi$, which is always positive, but no matter how you spin it, this just didn't work out, even for the Dirac equation.

The solution to this problem is that electrons don't live in isolation, they are identical particles and linked together via the Pauli exclusion principle. Dirac could only make sense of his equation by considering a variable number of electrons. This is where the classical field $\psi$ has to be promoted to a quantum field $\Psi$, a process known as second quantization. ("First quantization" refers to the fact that the classical field $\psi$ already describes a quantum mechanical particle.)

It turns out that second quantization is also necessary to explain certain corrections to the ordinary Schrödinger equation. In this light, the classical field $\psi$ is really an approximation as it does neglect the influence of a variable number of particles.

The process of considering a variable number of particles is actually quite neat. If you go from one to two particles, you would have to consider a classical field $\psi(x_1,x_2)$ that depends on two variables, the particle positions. Going to $N$ particles, you would have a field $\psi(x_1,\dots,x_N)$ depending on that many variables. You can get all particles at once by considering an operator valued field $\Psi(x)$ instead, which creates a particle at position $x$. It turns out that you can just replace $\psi$ by $\Psi$ in the Lagrangian to get the right equations of motion for all particles at once.

Alas, I have to stop here, further details on second quantization and quantum field theory are beyond the scope of this answer.

Concerning literature, I found Sakurai's Advanced Quantum Mechanics to be a very clear if somewhat long-winded introduction to quantum field theory that starts where the Schrödinger equation left off.

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answered Feb 28, 2012 by Greg Graviton (575 points) [ no revision ]
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I am no expert in classical fields, but I guess you have no entanglement there, that is, there is no difference with a single particle, but there is a big difference with more than one.

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answered Feb 20, 2012 by user667 (20 points) [ no revision ]

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