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[tex][a, a^\dagger] = [b, b^\dagger] = 1[/tex]

then there is a unitary operator U such that

[tex]b = UaU^\dagger[/tex]

?

- Thread starter PBRMEASAP
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- #1

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[tex][a, a^\dagger] = [b, b^\dagger] = 1[/tex]

then there is a unitary operator U such that

[tex]b = UaU^\dagger[/tex]

?

- #2

MalleusScientiarum

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What do you mean by "constitute isomorphisms to each other"?

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Daniel.

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Meir Achuz

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- #7

MalleusScientiarum

Certainly the unitarity will preserve the commutation relations, but more generally I could have those operators acting on different vector spaces entirely, such as second quantized operators for a phonon-electron system. In that case no unitary transformation can change which vector space is being acted upon.

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If you have a collection of more than two objects, with a correspondence between each pair, then that correspondence is not one to one. I still don't get what you mean. It can't be that every creation operator is isomorphic to every other creation operator. If you define a relationMalleusScientiarum said:When I say that there's an isomorphism that means that I can put the operators in a one-to-one correspondence with each other.

a ~ b if [a,a*] = [b,b*] = 1,

then the relation is many-to-one. If that's not what you were trying to say, please correct me.

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reilly

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The eigenvalues of a, a destruction operator are all complex numbers, each one associated with a coherent state --- = exp (z a*)|0> -- where z is a complex number, and a* is a creation operator. So all destruction and creation ops have the same spectrum. And, the number operators a* a all have the same spectrum. I'm pretty sure that the unitary operator constructed above will work in general -- because the states for either set of ops are in 1-1 correspondence. (Smacks of functional analysis, a subject about which I'm a bit rusty.)

That's all she wrote.

Regards,

Reilly Atkinson

- #10

MalleusScientiarum

I'm certain what you're saying is correct, but I can't get past the fact that the operators are acting on number states of different particles, which I would guess would have to be in different vector spaces. I'm not as up to par with the specifics of this as I should be, maybe you can explain?

- #11

Haelfix

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Am I the only one here who is bothered by the fact that he hasn't written down the entirety of the CCR?

You want to include [A(i), A(j)] = 0 = [A*(i), A*(j)] or I think (prolly would have to think about it a little more) all bets are off (the converse is trivially true) as you are no longer in the Fock representation.

You want to include [A(i), A(j)] = 0 = [A*(i), A*(j)] or I think (prolly would have to think about it a little more) all bets are off (the converse is trivially true) as you are no longer in the Fock representation.

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- #12

reilly

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MalleusScientiarum said:

I'm certain what you're saying is correct, but I can't get past the fact that the operators are acting on number states of different particles, which I would guess would have to be in different vector spaces. I'm not as up to par with the specifics of this as I should be, maybe you can explain?

Here's a case in which the formidable power of mathematics is made plain. In simple terms, all particles exist and do their dances in the same space-time arena -- they all share the same x-y-z-t coordinates. Both spin and, say isotopic spin, extend the space-time description to include discrete-spectra operators. Technically, we are talking direct products of Hilbert Spaces --

|p> * |m> == |p,m>, where p and m are momenta and Sz eigenvalues.

So, we can always invent an operator, PT,with eigenvalues 0 for photons, 1 for electrons, 2 for positrons, 3 for protons, 4 for pi0s, ........ Thus, in an admittedly unelegant way, any state can be expressed as a direct product of the usual space-time-spin Hilbert Space and PT eigenstates.

And, think about the description of states with creation and destruction operators -- they create a state that may have as many particles of as many types as you want, all in one space -- the direct product of infinite dimensional Fock Spaces. (This approach says that all particles share the same vacuum state. Hmmmm?)

Regards,

Reilly

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- #13

MalleusScientiarum

That's much more clear now. Thank you.

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