Representations and change of basis

[Niles]

Hi guys

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

<br /> <br /> {\cal H} = \mathbf a^\dagger\mathsf H \mathbf a =<br /> \mathbf a^\dagger \mathsf U\mathsf U^\dagger\mathsf H \mathsf U\mathsf U^\dagger\mathbf a = \mathbf b^\dagger \mathsf D \mathbf b<br /> <br />

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

<br /> a(\mathbf k) = \sum_\nu &lt;\mathbf k | \psi_\nu&gt;a_\nu.<br />

In my book they write this as

<br /> a(\mathbf r) = \sum_{\mathbf k}{e^{i\mathbf k\cdot r}} a(\mathbf k).<br />

How do I show this rigorously?

 

[peteratcam]

The two situations are identical.
Consider a unitary matrix, U_{ij}
Definition of unitary is that \sum_k U_{ik}U^*_{jk}=\delta_{ij}
Unitary transformation of a vector is a_i = \sum_j U_{ij}a&#039;_j
Now change the notation a bit, so instead of writing U_{ij} we write U(r,k)

Consider a unitary matrix, U(r,k)
Definition of unitary is that \sum_q U(r,q)U^*(r&#039;,q)=\delta(r,r&#039;)
Unitary transformation of a vector is a(r) = \sum_k U(r,k)\tilde a(k)
You can check that, with the appropriate normalisations (I think its 1/sqrt(N)), the 'Fourier transform matrix', U(r,k) \propto e^{ikr} 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.