Matrix Transformation: U^\dagger vs U

[Niles]

Hi guys

Ok, let's say I have a matrix given by

<br /> M = \sum_{ij}M_{ij}a_i^\dagger a_j,<br />

and I wish to transform it. Now in some books I have read they write the transformation as

<br /> s = \sum_{j}U_{ij}a_j,<br />

while in some notes I have read they write it as

<br /> s = \sum_{j}U^\dagger_{ij}a_j.<br />

What is the deal here? What is the proper way of doing this?

Kind regards,
Niles.

 

[peteratcam]

Suppose you have a Hamiltonian which can be represented as a quadratic form between boson operators:
<br /> {\cal H} = \sum_{ij}a^\dagger_iH_{ij}a_j = \mathbf a^\dagger\mathsf H \mathbf a<br />
where the matrix H_{ij} is hermitian.
Suppose you find the eigenvectors of H_{ij} and put them as normalised columns in a matrix U_{ij}, which is unitary.
Then a natural way to rewrite the hamiltonian is:
<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 />
where \mathsf D = \mathsf U^\dagger\mathsf H \mathsf U is diagonal, and the vector of boson operators is transformed as \mathbf b = U^\dagger\mathbf a.

Of course, I could equally well have said put the eigenvectors as normalised columns of the unitary matrix U^\dagger, and it would all be the other way round!

By the way, your first line is confusing: you have two matrices there. M acts on fock space as an operator (and is infinite dimensional for bosons), but M_ij is only as large as the number of sites.

 

[Niles]

Suppose you have a Hamiltonian which can be represented as a quadratic form between boson operators:
<br /> {\cal H} = \sum_{ij}a^\dagger_iH_{ij}a_j = \mathbf a^\dagger\mathsf H \mathbf a<br />
where the matrix H_{ij} is hermitian.
Suppose you find the eigenvectors of H_{ij} and put them as normalised columns in a matrix U_{ij}, which is unitary.
Then a natural way to rewrite the hamiltonian is:
<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 />
where \mathsf D = \mathsf U^\dagger\mathsf H \mathsf U is diagonal, and the vector of boson operators is transformed as \mathbf b = U^\dagger\mathbf a.

Of course, I could equally well have said put the eigenvectors as normalised columns of the unitary matrix U^\dagger, and it would all be the other way round!

By the way, your first line is confusing: you have two matrices there. M acts on fock space as an operator (and is infinite dimensional for bosons), but M_ij is only as large as the number of sites.

Just to be absolutely clear: When you say "other way around", then you mean that we go from

<br /> \mathbf b = U^\dagger\mathbf a<br />

to

<br /> \mathbf b = U\mathbf a<br />

?

 

[peteratcam]

Yes, U is a unitary matrix, which implies that U^\dagger is also unitary. Whether you attach a dagger doesn't really matter, as long as you are consistent with the definitions you choose.

In the context of diagonalising hamiltonians like my example, the way I have written it is probably more conventional. In another situations, the other way might be more suitable.

Check out active/passive transformations, it might help.