PBRMEASAP

Is there a theorem that says if

$$[a, a^\dagger] = [b, b^\dagger] = 1$$

then there is a unitary operator U such that

$$b = UaU^\dagger$$

?

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MalleusScientiarum

Well, you can show that if b satisfies that relation with a then the commutator is certainly true. I would imagine that such sets of operators would constitute isomorphisms to each other.

PBRMEASAP

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

dextercioby

Homework Helper
If there isn't a theorem, then u can build it. Give the statement and then formulate the proof. It's trivial to show the preservation of the commutation relations which define the unit Heisenberg algebra by a unitary transfrmation. Build the irred. rep. of the algebra and then show that $\hat{U}$ is an isometry of the irred. space.

Daniel.

PBRMEASAP

So the answer is yes? In your post, what is U? Are you assuming (or is it somehow obviously true) that there is a transformation b = UaU*, and then from there showing that U must be unitary? Also, how do you build an irreducible representation for a and a*? The most natural representation I can think of for them is in the number (Fock) state basis. And a* doesn't leave any finite subspace invariant, although I don't know if that matters.

Meir Achuz

Homework Helper
Gold Member
Of course, the converse is true. But two different annihilation operators need not be connected by a UT. They could destroy different particles and be unrelated.

MalleusScientiarum

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.

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.

PBRMEASAP

MalleusScientiarum 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.
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 relation

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.

reilly

For bosons with spin, there are creation and destruction operators for all combinations of spin and momentum. Clearly unitary transforms exist that will map p,m -> q,n where p and q are momenta and m and n are values of Sz.The next issue is connecting different particles, say photons and neutral pi mesons, pi0. Certainly photon and pion operators commute, and one can get a unitary xform that will do the trick; let p,m stand for photon operator eigenvalues as above, and let q be the momentum of a pi0. the unitary xform is SUM |p, m><q|. We'll let the sum go over all p and q, and fix m.

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

MalleusScientiarum

Reilly,

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?

Haelfix

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.

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reilly

MalleusScientiarum said:
Reilly,

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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MalleusScientiarum

That's much more clear now. Thank you.

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