Symmetry and conserved probability current for wave functions

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Noether's theorem connects symmetries to conserved quantities, with conservation of momentum arising from spatial translation invariance in classical mechanics. In quantum mechanics, the conservation of probability is tied to symmetries represented by unitary operators in the Hilbert space, as established by Wigner's theorem. This implies that probabilities must be conserved under inertial transformations, but there is no specific symmetry directly linked to probability conservation. The discussion highlights the distinction between classical field theory and quantum mechanics, emphasizing that the Lagrangian's role is less significant in quantum contexts. The conversation suggests that understanding quantum mechanics and field theory is best approached through texts like Weinberg's, which clarify these connections.
  • #31
Micha said:
If a particle has U(1) symmetry, it is charged and its antiparticle has opposite charge. If your theory allows for creation of particle and antiparticle pairs, conservation of probability generalizes to conservation of charge. If you have a theory with no particle/antiparticle creation, you must have conservation of probability.

I do not think that conservation of probability and conservation of charge are related in this way. The 0-compenent of the Dirac conserved current is positive definite. It does not represent charge since the Dirac equation describes both positrons and electrons. On the other, each of the four components of the Dirac spinor individually satisfy the Klein-Gordon equation and so also satisfy the Klein-Gordon conserved current, which does seem to be related to conservation of charge.

The Klein-Gordon pdf is what really interests me. I totally buy the fact that \phi\partial_{\mu}\phi^\dag-\phi^\dag\partial_{\mu}\phi is the charge current. But then what is the pdf? Can't we still ask, what is the probability of observing a particle in such-a-such region? But you know what? The Klein-Gordon pdf is still an open question and people are still publishing papers on it.
 
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  • #32
pellman said:
I would say the significance is something like this: if |\psi(x)|^2 is to be interpreted as a pdf, then it must satisfy a conservation law (so that it's normalization over all space is equal to 1 for all times).

Yes, in quantum mechanics this is guaranteed by the fact that the Hamiltonian H is Hermitean and that the time evolution operator U(t) = \exp(\frac{i}{\hbar} Ht) is unitary.

pellman said:
We also have the interpretation that the overall phase of \psi(x) is physically unimportant.

Yes, in quantum mechanics this is guaranteed by the interpretation of states as rays of vectors in the Hilbert space, rather than individual vectors.

There is no need to introduce Lagrangian to "prove" these things. I still believe that Lagrangians are out of place in quantum mechanics. Yes, you can formally introduce some action, Lagrangian, etc. in quantum mechanics and "derive" the Schroedinger equation and "conservation of probabilities" from them. But this will not tell you anything new that you didn't know already from the basic laws of QM.
 
  • #33
pellman said:
The Klein-Gordon pdf is what really interests me. I totally buy the fact that \phi\partial_{\mu}\phi^\dag-\phi^\dag\partial_{\mu}\phi is the charge current. But then what is the pdf? Can't we still ask, what is the probability of observing a particle in such-a-such region? But you know what? The Klein-Gordon pdf is still an open question and people are still publishing papers on it.

You can find more discussions of the Klein-Gordon equation and the probabilistic interpretation of its solutions in a very long thread

https://www.physicsforums.com/showthread.php?t=175155

Eugene.
 
  • #34
pellman said:
But there is something deep to the fact the time-translation invariance of the Lagrangian demands conservation of energy.

I would say the significance is something like this: if |\psi(x)|^2 is to be interpreted as a pdf, then it must satisfy a conservation law (so that it's normalization over all space is equal to 1 for all times). We also have the interpretation that the overall phase of \psi(x) is physically unimportant. And here we see that the two things are intimately tied: if the overall phase of \psi(x) did change the Lagrangian, then we would not have conservation of the quantity which we interpret as probability. The two things must go together.

I think, your idea is not bad. You should keep in mind though, that conservation of energy is a very general concept, which is problematic only in the context of general relativity.

Conservation of probability (which is also called unitarity), is also a very general concept. But the way, you present it here mathematically, as the norm of a single particle wave function, it is not very general. If you don't have a free theory, but allow for some interaction like electromagnetism, particles can be created out of energy.
So the fact, that the Schroedinger equation maintains the norm of the single particle wave function during time evolution, becomes a problem to fix for quantum field theory.
 

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