Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Can we view the normal, non-relativistic quantum mechanics as a classical fields?

I know, that one can derive the Schrödinger equation from the Lagrangian density

$${\cal L} ~=~ \frac{i\hbar}{2} (\psi^* \dot\psi - \dot\psi^* \psi) - \frac{\hbar^2}{2m}\nabla\psi^* \cdot \nabla \psi - V({\rm r},t) \psi^*\psi$$

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.

share|improve this question
    
These are quantum field theory notes, but they discuss your question damtp.cam.ac.uk/user/tong/qft.html. –  Kyle Feb 20 '12 at 20:26
    
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. –  Thomas Feb 20 '12 at 21:03

5 Answers 5

up vote 6 down vote accepted

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.

share|improve this answer

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.

share|improve this answer

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.

share|improve this answer
    
You have to say specifically that this fails for two electrons. That's the whole thing that makes quantum mechanics different from classical field theory. –  Ron Maimon Nov 4 '12 at 1:23
    
As I know your statement that Dirac equation has negative probability density is totaly wrong, it is actually positive, but it is true that Klein-Gordon equation had negative one, and one of Dirac's goals was to solve that, but the problem Dirac had is negative energies, and that was solved by considering antimatter, not the exclusion principle. –  TMS Oct 7 at 22:55

Indeed, the true Lagrangian for the Schrödinger equation takes this from

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

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.

share|improve this answer

Tong's lecture notes cover this in detail. In particular, your lagrangian is easy to obtain from a complex scalar field that obeys the Klein-Gordon equation in the non-relativistic limit. This lagrangian does lead to a conserved Nöther current of the same form as that of Schrödinger's, but this does not have the interpretation of conserved probability, because $\psi \psi^*$ is not the probability density for a particle to be in specific position at specified time. In order to obtain QM, you'll need to quantize the field, construct the wave-function as the superposition of single-particle position states, then obtain the hamiltonian and the Schrödinger equation.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.