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Matrix Transformations

  1. Feb 18, 2010 #1
    Hi guys

    Ok, lets say I have a matrix given by

    [tex]
    M = \sum_{ij}M_{ij}a_i^\dagger a_j,
    [/tex]

    and I wish to transform it. Now in some books I have read they write the transformation as

    [tex]
    s = \sum_{j}U_{ij}a_j,
    [/tex]

    while in some notes I have read they write it as

    [tex]
    s = \sum_{j}U^\dagger_{ij}a_j.
    [/tex]

    What is the deal here? What is the proper way of doing this? :confused:

    Kind regards,
    Niles.
     
  2. jcsd
  3. Feb 18, 2010 #2
    Re: Transformations

    Suppose you have a Hamiltonian which can be represented as a quadratic form between boson operators:
    [tex]
    {\cal H} = \sum_{ij}a^\dagger_iH_{ij}a_j = \mathbf a^\dagger\mathsf H \mathbf a
    [/tex]
    where the matrix [tex]H_{ij}[/tex] is hermitian.
    Suppose you find the eigenvectors of [tex]H_{ij}[/tex] and put them as normalised columns in a matrix [tex]U_{ij}[/tex], which is unitary.
    Then a natural way to rewrite the hamiltonian is:
    [tex]
    {\cal H} = \mathbf a^\dagger\mathsf H \mathbf a =
    \mathbf a^\dagger \mathsf U\mathsf U^\dagger\mathsf H \mathsf U\mathsf U^\dagger\mathbf a = \mathbf b^\dagger \mathsf D \mathbf b
    [/tex]
    where [tex]\mathsf D = \mathsf U^\dagger\mathsf H \mathsf U[/tex] is diagonal, and the vector of boson operators is transformed as [tex]\mathbf b = U^\dagger\mathbf a[/tex].

    Of course, I could equally well have said put the eigenvectors as normalised columns of the unitary matrix [tex]U^\dagger[/tex], and it would all be the other way round!

    By the way, your first line is confusing: you have two matrices there. M acts on fock space as an operator (and is infinite dimensional for bosons), but M_ij is only as large as the number of sites.
     
  4. Feb 18, 2010 #3
    Re: Transformations

    Just to be absolutely clear: When you say "other way around", then you mean that we go from

    [tex]
    \mathbf b = U^\dagger\mathbf a
    [/tex]

    to

    [tex]
    \mathbf b = U\mathbf a
    [/tex]

    ?
     
  5. Feb 18, 2010 #4
    Re: Transformations

    Yes, [tex]U[/tex] is a unitary matrix, which implies that [tex]U^\dagger[/tex] is also unitary. Whether you attach a dagger doesn't really matter, as long as you are consistent with the definitions you choose.

    In the context of diagonalising hamiltonians like my example, the way I have written it is probably more conventional. In another situations, the other way might be more suitable.

    Check out active/passive transformations, it might help.
     
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