Why Did My Teacher Make a Substitution in Transforming a Hamiltonian?

[Niles]

Hi guys

Say I have a Hamiltonian given by
<br /> H = \sum\limits_{i,j} {a_i^\dag H_{ij} a_j^{} }<br />

I wish to perform a transformation given by
<br /> \gamma _i = \sum\limits_j {S_{ij} a_j }.<br />

Now, what my teacher did was to make the substituion \gamma_i \rightarrow a_i and a_i \rightarrow \gamma_i, so we get the transformation
<br /> a_i = \sum\limits_j {S_{ij} \gamma _j }.<br />

This expression he then inserted in H to find H in the new basis, but I don't understand why he could just make a substituion in the transformation and then insert it? Is a_i = \sum\limits_j {S_{ij} \gamma _j } when we express the creation/annihilation operators in terms of the transformation or what?

I hope you will shed some light on this.Niles.

 

[saaskis]

The Hamiltonian can be written in a matrix form
<br /> \hat H = \mathbf{a}^{\dagger} \mathbf{H} \mathbf{a}.<br />

Since the Hamiltonian matrix is Hermitian, it can be diagonalized by a unitary transformation, i.e.
<br /> \mathbf{H}= \mathbf{S} \Lambda \mathbf{S}^{\dagger},<br />
where \mathbf{S}^{\dagger} = \mathbf{S}^{-1} and \mathbf{\Lambda} is a diagonal matrix.

So what do you get, when you make the change of basis to \mathbf{\gamma} = \mathbf{S} \mathbf{a}? This is equivalent to \mathbf{a} = \mathbf{S}^{\dagger} \mathbf{\gamma}, of course. I don't know if this answers your question, though. The original Hamilton operator remains unchanged in the transformation, but it is now simply expressed in a basis, in which it is diagonal. If you had calculated the matrix elements H_{ij} in this basis in the first place, we could directly write
<br /> \hat H = \gamma ^{\dagger} \Lambda \gamma = \sum_i \lambda_i \gamma_i^{\dagger} \gamma_i
and there would be no reason for further transformations. Here \Lambda=diag(\lambda_i).

 

[Niles]

Thanks, that made things clearer to me, but you say that we have
\mathbf{\gamma} = \mathbf{S} \mathbf{a} \quad \leftrightarrow \quad \mathbf{a} = \mathbf{S}^{\dagger} \mathbf{\gamma}.

In the case of <br /> <br /> \gamma _i = \sum\limits_j {S_{ij} a_j }.<br /> <br />, what do we write on the right side of \leftrightarrow?

 

[saaskis]

In the case of <br /> <br /> \gamma _i = \sum\limits_j {S_{ij} a_j }.<br /> <br />, what do we write on the right side of \leftrightarrow?

<br /> \mathbf{a} = \mathbf{S}^{\dagger} \mathbf{\gamma} \Leftrightarrow a_i = \sum_j (S^{\dagger})_{ij} \gamma_j = \sum_j S_{ji}^* \gamma_j,<br />
by the definition of the adjoint matrix. Hope this helps. And just to make the notation clear, above I defined \mathbf{a} = (a_1,a_2,\dots)^T (column vector) and \mathbf{a}^{\dagger} = (a_1^{\dagger},a_2^{\dagger},\dots) (row vector).

 

[rkrsnan]

<br /> \mathbf{\gamma} = \mathbf{S} \mathbf{a} \quad \leftrightarrow \quad \gamma _i = \sum\limits_j {S_{ij} a_j <br />
<br /> \mathbf{a} = \mathbf{S}^\dagger \mathbf{\gamma} \quad \leftrightarrow \quad a _i = \sum\limits_j {S^\dagger_{ij} \gamma_j \quad \leftrightarrow \quad a _i = \sum\limits_j {S^*_{ji} \gamma_j <br />

 

[Niles]

Thanks, I get it now. It's very kind of you to help me (both of you).