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Transforming a Hamiltonian

  1. Feb 14, 2010 #1
    Hi guys

    Say I have a Hamiltonian given by
    [tex]
    H = \sum\limits_{i,j} {a_i^\dag H_{ij} a_j^{} }
    [/tex]

    I wish to perform a transformation given by
    [tex]
    \gamma _i = \sum\limits_j {S_{ij} a_j }.
    [/tex]

    Now, what my teacher did was to make the substituion [itex]\gamma_i \rightarrow a_i[/itex] and [itex]a_i \rightarrow \gamma_i[/itex], so we get the transformation
    [tex]
    a_i = \sum\limits_j {S_{ij} \gamma _j }.
    [/tex]

    This expression he then inserted in H to find H in the new basis, but I don't understand why he could just make a substituion in the transformation and then insert it? Is [itex]a_i = \sum\limits_j {S_{ij} \gamma _j }[/itex] when we express the creation/annihilation operators in terms of the transformation or what?

    I hope you will shed some light on this.

    Best regards,
    Niles.
     
  2. jcsd
  3. Feb 14, 2010 #2
    The Hamiltonian can be written in a matrix form
    [tex]
    \hat H = \mathbf{a}^{\dagger} \mathbf{H} \mathbf{a}.
    [/tex]

    Since the Hamiltonian matrix is Hermitian, it can be diagonalized by a unitary transformation, i.e.
    [tex]
    \mathbf{H}= \mathbf{S} \Lambda \mathbf{S}^{\dagger},
    [/tex]
    where [tex] \mathbf{S}^{\dagger} = \mathbf{S}^{-1} [/tex] and [tex] \mathbf{\Lambda} [/tex] is a diagonal matrix.

    So what do you get, when you make the change of basis to [tex]\mathbf{\gamma} = \mathbf{S} \mathbf{a} [/tex]? This is equivalent to [tex] \mathbf{a} = \mathbf{S}^{\dagger} \mathbf{\gamma}[/tex], of course. I don't know if this answers your question, though. The original Hamilton operator remains unchanged in the transformation, but it is now simply expressed in a basis, in which it is diagonal. If you had calculated the matrix elements [tex]H_{ij}[/tex] in this basis in the first place, we could directly write
    [tex]
    \hat H = \gamma ^{\dagger} \Lambda \gamma = \sum_i \lambda_i \gamma_i^{\dagger} \gamma_i [/tex]
    and there would be no reason for further transformations. Here [tex] \Lambda=diag(\lambda_i) [/tex].
     
    Last edited: Feb 14, 2010
  4. Feb 14, 2010 #3
    Thanks, that made things clearer to me, but you say that we have
    [tex]\mathbf{\gamma} = \mathbf{S} \mathbf{a} \quad \leftrightarrow \quad \mathbf{a} = \mathbf{S}^{\dagger} \mathbf{\gamma} [/tex].

    In the case of [itex]

    \gamma _i = \sum\limits_j {S_{ij} a_j }.

    [/itex], what do we write on the right side of [itex]\leftrightarrow[/itex]?
     
  5. Feb 14, 2010 #4
    [tex]
    \mathbf{a} = \mathbf{S}^{\dagger} \mathbf{\gamma} \Leftrightarrow a_i = \sum_j (S^{\dagger})_{ij} \gamma_j = \sum_j S_{ji}^* \gamma_j,
    [/tex]
    by the definition of the adjoint matrix. Hope this helps. And just to make the notation clear, above I defined [tex]\mathbf{a} = (a_1,a_2,\dots)^T [/tex] (column vector) and [tex] \mathbf{a}^{\dagger} = (a_1^{\dagger},a_2^{\dagger},\dots) [/tex] (row vector).
     
  6. Feb 14, 2010 #5
    [tex]
    \mathbf{\gamma} = \mathbf{S} \mathbf{a} \quad \leftrightarrow \quad \gamma _i = \sum\limits_j {S_{ij} a_j
    [/tex]
    [tex]
    \mathbf{a} = \mathbf{S}^\dagger \mathbf{\gamma} \quad \leftrightarrow \quad a _i = \sum\limits_j {S^\dagger_{ij} \gamma_j \quad \leftrightarrow \quad a _i = \sum\limits_j {S^*_{ji} \gamma_j
    [/tex]
     
  7. Feb 14, 2010 #6
    Thanks, I get it now. It's very kind of you to help me (both of you).
     
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