Matrix Transformation: U^\dagger vs U

[Niles]

Hi guys

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

$$M = \sum_{ij}M_{ij}a_i^\dagger a_j,$$

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

$$s = \sum_{j}U_{ij}a_j,$$

while in some notes I have read they write it as

$$s = \sum_{j}U^\dagger_{ij}a_j.$$

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:
$${\cal H} = \sum_{ij}a^\dagger_iH_{ij}a_j = \mathbf a^\dagger\mathsf H \mathbf a$$
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:
$${\cal H} = \mathbf a^\dagger\mathsf H \mathbf a =<br /> \mathbf a^\dagger \mathsf U\mathsf U^\dagger\mathsf H \mathsf U\mathsf U^\dagger\mathbf a = \mathbf b^\dagger \mathsf D \mathbf b$$
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:
$${\cal H} = \sum_{ij}a^\dagger_iH_{ij}a_j = \mathbf a^\dagger\mathsf H \mathbf a$$
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:
$${\cal H} = \mathbf a^\dagger\mathsf H \mathbf a =<br /> \mathbf a^\dagger \mathsf U\mathsf U^\dagger\mathsf H \mathsf U\mathsf U^\dagger\mathbf a = \mathbf b^\dagger \mathsf D \mathbf b$$
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

$$\mathbf b = U^\dagger\mathbf a$$

to

$$\mathbf b = U\mathbf a$$

?

 

[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.