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Ladder operator theorem?

  1. Jul 19, 2005 #1
    Is there a theorem that says if

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

    then there is a unitary operator U such that

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

  2. jcsd
  3. Jul 19, 2005 #2
    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.
  4. Jul 19, 2005 #3
    What do you mean by "constitute isomorphisms to each other"?
  5. Jul 29, 2005 #4


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    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 [itex] \hat{U} [/itex] is an isometry of the irred. space.

  6. Jul 29, 2005 #5
    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.
  7. Jul 29, 2005 #6

    Meir Achuz

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    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.
  8. Jul 29, 2005 #7
    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.
  9. Jul 29, 2005 #8
    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.
  10. Jul 29, 2005 #9


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    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.
    Reilly Atkinson
  11. Jul 29, 2005 #10

    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?
  12. Jul 30, 2005 #11


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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.
    Last edited: Jul 30, 2005
  13. Jul 30, 2005 #12


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    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?)

    Last edited: Jul 30, 2005
  14. Jul 30, 2005 #13
    That's much more clear now. Thank you.
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