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Is this a theorem?

  1. Jul 21, 2005 #1
    Is there a theorem that says that for linear operators a and b on some vector space, if

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

    where [A, B] denotes the commutator AB-BA, then there exists a unitary operator U such that

    [tex] b = UaU^\dagger [/tex]
     
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  3. Jul 21, 2005 #2

    Hurkyl

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    No.

    Notice that UaU* and a have the same eigenvalues:

    If av = λv, then if we set w = Uv, we have:

    a(U*w) = λ(U*w)
    (UaU*)w = λw

    So if we take a to be an operator with a discrete spectrum satisfying [a, a*], then we can choose a constant δ and set b = a + δ so that b and a have different spectra. Then:

    [b, b*] = [a + δ, a* + δ*] = [a, a*] = 1

    So [a, a*] = [b, b*] = 1, and we cannot write b = UaU*.
     
  4. Jul 21, 2005 #3
    An interesting thing to note is that [tex] a^{\dagger} [/tex] doesn't have eigenvectors at all. Just try it out.

    However, despite the unitary transformation absence, I should imagine that these operators instead form isomorphisms to each other, since they could be acting on different vector spaces. Cross reference many coupled harmonic oscillators in quantum mechanics, or more generally the method of second quantization.
     
    Last edited by a moderator: Jul 21, 2005
  5. Jul 21, 2005 #4
    Thank you, Hurkyl. That makes good sense.

    MalleusScientiarum,

    Could you explain what you mean by
    ?
     
  6. Jul 21, 2005 #5

    Hurkyl

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    I imagine you mean that there ought to be an isomorphism f of Hilbert spaces such that we have the equality [itex]\hat{a} = b[/itex] where [itex]\hat{a} := f^{-1} \circ a \circ f[/itex] is the the action of a carried over to the other vector space. (In other words, af(v) = f(bv))

    But the eigenvalues are an invariant: a and [itex]\hat{a}[/itex] have to have the same eigenvalues... so if a and b have different spectra, they cannot be "isomorphic".

    I suspect you can prove that there exists a U such that b = UaU* if and only if they have the same spectrum... at least if they have a complete set of eigenvectors. I imagine it would be straightforward to prove that the change of basis operation that takes an eigenbasis of a to an eigenbasis of b is a unitary transformation.
     
  7. Jul 21, 2005 #6
    In this case, [itex]a[/itex] has a continuous spectrum of eigenvalues--the complex numbers. I will think about it some more. Thanks.
     
  8. Jul 22, 2005 #7
    Can you give a hint as to where to begin? Does this fact depend only on the fact that a and b have the same spectrum, or does it also rely on the commutator relations they satisfy?

    Also, how do you write greek letters without using latex?
     
  9. Jul 22, 2005 #8

    Hurkyl

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    & alpha ; = α


    At least in the finite-dimensional case, if your linear operator has a complete set of eigenvalues, one nice thing you can do is to make an orthonormal basis out of eigenvectors. Then, you can describe the operator formally by its action on the basis elements.

    If you have two operators with the same eigenvalues, and we construct this formal representation for each, how do they compare?

    Finally, remember that all that "unitary transformation" means is that the norm of a vector before the transformation equals the norm of that vector after the transformation.
     
  10. Jul 23, 2005 #9
    When you say make an orthonormal basis out of eigenvectors, are you talking about defining an inner product? When one speaks of a unitary transformation, it is always with respect to a particular inner product, right? Or are you talking about Gram-Schmidt procedure?

    edit: α β γ δ ... that is neat. Thanks for showing me that!
     
    Last edited: Jul 23, 2005
  11. Jul 23, 2005 #10

    Hurkyl

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    I'm not talking about defining an inner product: your vector space should already come equipped with one. (Otherwise, it wouldn't make sense to ask about a unitary transformation, as you suspected!) The origin of your problem is that these are linear operators on a Hilbert space is it not? By definition, a Hilbert space comes equipped with an inner product.

    I have made a mistake, though: I forgot that you can only get the orthonormal eigenbasis if the transformation is symmetric. (Again, in the finite dimensional case)

    All right, so I've unconvinced myself. :smile: When rewriting everything in terms of an eigenbasis (which again I'm assuming exists), we need two things:

    (1) The action of the operator on each basis vector
    (2) The inner product of any pair of basis vectors.

    So now it's obvious how one would go about constructing a counterexample to my claim.

    However, I still conjecture that if a and b have the same spectrum, that there exists an invertible linear operator T such that aT = Tb, and T can be simply given in terms of a change of basis. (This is easy in the finite dimensional case) And that if a and b have orthonormal eigenbases, then T can be chosen to be unitary.
     
  12. Jul 23, 2005 #11
    You guessed it. The particular "a" that I'm talking about is the annihilation operator from quantum mechanics. So we are dealing with a Hilbert space, which comes equipped with an inner product. And in order for the relation [a, a*] = 1 to be possible, it has to be an infinite dimensional space. The eigenvectors of a are called coherent states, and they are not orthogonal. According to the quantum optics book I was reading when I encountered this problem, the coherent states are "overcomplete", but I don't know what that means. Apparently you can still write a vector as a superposition of these coherent states--otherwise they wouldn't be very useful.

    There is one more connection between a and b that I didn't mention, and it may hold the key to the problem:

    b = μa + νa*, where |μ|^2 - |ν|^2 = 1. I thought this was just to ensure that [b,b*] = 1. Does this also provide a unitary transformation between a and b? The author is not very clear about why such a unitary transformation exists. I'm trying to figure out exactly what is needed to make that conclusion. Either it's obvious and i'm not seeing it, or he was being vague on purpose to avoid going into details.
     
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