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Representations and change of basis

  1. Feb 22, 2010 #1
    Hi guys

    1) We are looking at a Hamiltonian H. I make a rotation in Hilbert space by the transformation

    [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 D is diagonal.


    Now, is there a difference between a rotation of this kind and the transformation I perform when I go to momentum-representation?

    2) When I want to write my second quantization operators in momentum-space, I write them as

    [tex]
    a(\mathbf k) = \sum_\nu <\mathbf k | \psi_\nu>a_\nu.
    [/tex]

    In my book they write this as

    [tex]
    a(\mathbf r) = \sum_{\mathbf k}{e^{i\mathbf k\cdot r}} a(\mathbf k).
    [/tex]

    How do I show this rigorously?
     
  2. jcsd
  3. Feb 22, 2010 #2
    The two situations are identical.
    Consider a unitary matrix, [tex]U_{ij}[/tex]
    Definition of unitary is that [tex]\sum_k U_{ik}U^*_{jk}=\delta_{ij}[/tex]
    Unitary transformation of a vector is [tex]a_i = \sum_j U_{ij}a'_j[/tex]
    Now change the notation a bit, so instead of writing [tex]U_{ij}[/tex] we write [tex]U(r,k)[/tex]

    Consider a unitary matrix, [tex]U(r,k)[/tex]
    Definition of unitary is that [tex]\sum_q U(r,q)U^*(r',q)=\delta(r,r')[/tex]
    Unitary transformation of a vector is [tex]a(r) = \sum_k U(r,k)\tilde a(k)[/tex]
    You can check that, with the appropriate normalisations (I think its 1/sqrt(N)), the 'Fourier transform matrix', [tex]U(r,k) \propto e^{ikr}[/tex] works. You can think of fourier transformation as a big square matrix where position labels the columns, and momentum labels the rows. It is a change of basis, like any unitary transformation.

    This procedure, however, will only diagonalise translationally invariant Hamiltonians.
     
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