# Formal integration of Heissenberg equation?

1. Feb 8, 2013

### climbon

Hi,

I am reading a paper and can't get the same answer they do, they have;

$$H= -i\hbar \sum_{\lambda} \int dK (a_{(k,\lambda)} g_{(1;K,\lambda)} \sigma_{1}^{+} e^{-i(\omega_{k}- \omega)t} - H.C )$$

Then states they do a formal integration of the Heissenberg equation for $$a_{k,\lambda}(t)$$ using the above Hamiltonian, they get;

$$a_{k,\lambda}(t) = a_{k,\lambda}(t_{0}) + \int_{t_{0}}^{t} dt^{'} g^{*} \sigma_{1}^{-}(t^{'}) e^{-i(\omega_{k}- \omega)t^{'}}$$

Could anyone help me as to how they get this answer from the Heissenberg equation...you be greatly appreciated!!

Thanks.

2. Feb 9, 2013

### Einj

Does $a_{(k,\lambda)}$ and $a^\dagger_{(k,\lambda)}$ follow some kind of orthogonality relation?
In fact, you get the correct result if $a_{(k,\lambda)}\cdot a_{(q,\lambda')}=0$ and $a_{(k,\lambda)}\cdot a^\dagger_{(q,\lambda')}=\delta(k-q)\delta_{\lambda\lambda'}$.

However, I'm just guessing.

3. Feb 9, 2013

### vanhees71

It would be much easier to help you, if you'd give the reference to the paper!