Representations and change of basis

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SUMMARY

This discussion focuses on the transformation of Hamiltonians in quantum mechanics through unitary rotations in Hilbert space and their equivalence to momentum-space representations. The transformation is expressed as {\cal H} = \mathbf a^\dagger\mathsf H \mathbf a and involves diagonalization via a unitary matrix, U(r,k), which is shown to be analogous to a Fourier transform. The rigorous demonstration of the equivalence between position and momentum representations is explored, emphasizing that this method only diagonalizes translationally invariant Hamiltonians.

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Niles
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Hi guys

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

[tex] <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 /> [/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?
 
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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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