# Representations and change of basis

1. Feb 22, 2010

### Niles

Hi guys

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

$${\cal H} = \mathbf a^\dagger\mathsf H \mathbf a = \mathbf a^\dagger \mathsf U\mathsf U^\dagger\mathsf H \mathsf U\mathsf U^\dagger\mathbf a = \mathbf b^\dagger \mathsf D \mathbf b$$

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

$$a(\mathbf k) = \sum_\nu <\mathbf k | \psi_\nu>a_\nu.$$

In my book they write this as

$$a(\mathbf r) = \sum_{\mathbf k}{e^{i\mathbf k\cdot r}} a(\mathbf k).$$

How do I show this rigorously?

2. Feb 22, 2010

### 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 $$a_i = \sum_j U_{ij}a'_j$$
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 $$\sum_q U(r,q)U^*(r',q)=\delta(r,r')$$
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.