# Probability conservation and symmetry

by domhal
Tags: conservation, probability, symmetry
 P: 4 What symmetry gives probability conservation? Or, what symmetry does probability conservation give? I have been trying (unsuccessfully) to find an answer to this question. I think the question makes sense. That is, I can't see how the situation is different from, for example, that of spatial translation giving conservation of momentum.
 P: 223 Unfortunately I don't have time to flip out a complete answer, but I would tell you to look at the Lagrangian formulation of the Schrodinger equation and to look up Noether's theorem. I seem to recall actually working this out at some point, but those notes are probably buried in a box somewhere.
 Sci Advisor HW Helper P: 11,896 Actually the theorem of Wigner insures probability conservation for any symmetry of the system. By that theorem, symmetry transformations are implemented in (rigged) Hilbert space language by either unitary or antiunitary operators which are known to conserve scalar products, hence probabilities.
P: 2,032
Probability conservation and symmetry

 I can't see how the situation is different from, for example, that of spatial translation giving conservation of momentum.
I think probablity conservation looks fundamentaly different from mometum conservation. I recall momentum conservation being derived by assuming that Hamilton's operator commutes with a translation operator. For one particle state, $$[e^{u\cdot\nabla},H]=0$$ for abritary vector u. From this it follows, that $$[-i\hbar\nabla,H]=0$$, and with Shrodinger's equation, that $$\langle\Psi|-i\hbar\nabla|\Psi\rangle$$ is conserved in time. Analogously to this, you could argue that quantity $$\langle\Psi|\Psi\rangle$$ is conserved simply because $$[1,H]=0$$, but that looks dumb. These don't look the same kind of conservation laws.
P: 223
 Quote by jostpuur I think probablity conservation looks fundamentaly different from mometum conservation. I recall momentum conservation being derived by assuming that Hamilton's operator commutes with a translation operator. For one particle state, $$[e^{u\cdot\nabla},H]=0$$ for abritary vector u. From this it follows, that $$[-i\hbar\nabla,H]=0$$, and with Shrodinger's equation, that $$\langle\Psi|-i\hbar\nabla|\Psi\rangle$$ is conserved in time. Analogously to this, you could argue that quantity $$\langle\Psi|\Psi\rangle$$ is conserved simply because $$[1,H]=0$$, but that looks dumb. These don't look the same kind of conservation laws.
They're actually very similar, in that they are related to Noether's Theorem and "conserved charges". This is why I was saying to look at the lagrangian.
Emeritus