Transforming a Hamiltonian

1. Feb 14, 2010

Niles

Hi guys

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

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

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
$$a_i = \sum\limits_j {S_{ij} \gamma _j }.$$

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.

Best regards,
Niles.

2. Feb 14, 2010

The Hamiltonian can be written in a matrix form
$$\hat H = \mathbf{a}^{\dagger} \mathbf{H} \mathbf{a}.$$

Since the Hamiltonian matrix is Hermitian, it can be diagonalized by a unitary transformation, i.e.
$$\mathbf{H}= \mathbf{S} \Lambda \mathbf{S}^{\dagger},$$
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
$$\hat H = \gamma ^{\dagger} \Lambda \gamma = \sum_i \lambda_i \gamma_i^{\dagger} \gamma_i$$
and there would be no reason for further transformations. Here $$\Lambda=diag(\lambda_i)$$.

Last edited: Feb 14, 2010
3. Feb 14, 2010

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 $\gamma _i = \sum\limits_j {S_{ij} a_j }.$, what do we write on the right side of $\leftrightarrow$?

4. Feb 14, 2010

$$\mathbf{a} = \mathbf{S}^{\dagger} \mathbf{\gamma} \Leftrightarrow a_i = \sum_j (S^{\dagger})_{ij} \gamma_j = \sum_j S_{ji}^* \gamma_j,$$
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).

5. Feb 14, 2010

rkrsnan

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

6. Feb 14, 2010

Niles

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