- #1

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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.

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- Thread starter domhal
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- #1

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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.

- #2

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- #3

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- #4

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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, [tex][e^{u\cdot\nabla},H]=0[/tex] for abritary vector u. From this it follows, that [tex][-i\hbar\nabla,H]=0[/tex], and with Shrodinger's equation, that [tex]\langle\Psi|-i\hbar\nabla|\Psi\rangle[/tex] is conserved in time. Analogously to this, you could argue that quantity [tex]\langle\Psi|\Psi\rangle[/tex] is conserved simply because [tex][1,H]=0[/tex], but that looks dumb. These don't look the same kind of conservation laws.I can't see how the situation is different from, for example, that of spatial translation giving conservation of momentum.

- #5

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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, [tex][e^{u\cdot\nabla},H]=0[/tex] for abritary vector u. From this it follows, that [tex][-i\hbar\nabla,H]=0[/tex], and with Shrodinger's equation, that [tex]\langle\Psi|-i\hbar\nabla|\Psi\rangle[/tex] is conserved in time. Analogously to this, you could argue that quantity [tex]\langle\Psi|\Psi\rangle[/tex] is conserved simply because [tex][1,H]=0[/tex], 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.

- #6

vanesch

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From a physical point of view, I would even say: the argument goes in the other way! Because the physical manifestation of a symmetry is, well, that there is no observational difference between the system as such, and the "symmetry candidate" applied to the system, and because all observables are essentially expectation values, which are essentially weighted probabilities, the necessary and sufficient condition for a symmetry candidate on a physical system to be actually a symmetry, is that all probabilities "before" and "after" are the same. Hence, symmetries NEED to conserve probabilities (and hence all observable phenomena) in order for them to merit the denomination of symmetry.

And from this condition follows mathematically that they must be unitary or anti-unitary representations of their group (because symmetries also always form a group, for the same physical reasons).

EDIT: we can, because of this, apply two kinds of approaches to setting up the quantum description of a system. We can make a list of symmetries, and try to find a good representation of them, which can then serve as a quantum description ; or we can have another way of finding the quantum description, and then go fishing for the different symmetry representations it contains.

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- #7

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"What symmetry gives probability conservation? Or, what symmetry does probability conservation give?"

should be answered: "Any symmetry. All symmetries".

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