Interaction Picture - Exercises

To do this, we first consider the definition of the interaction picture operator, which is $$\begin{equation} \chat_{\k\sigma,I}(t) = e^{i\hat{H}_0t/\hbar} \chat_{\k\sigma} e^{-i\hat{H}_0t/\hbar}. \end{equation}$$ We now apply the Baker-Campbell-Hausdorff formula to get $$\begin{equation} \chat_{\k\sigma,I}(t) = \chat_{\k\sigma} + \frac{it}{\hbar}[\hat{H}_0, \chat_{\k\sigma}] + \frac{1}{2} \left(\frac{it}{\hbar}\right)^2[\hat{H}_0,[\hat{H}_0,\chat_{\k\sigma}]] + \cdots \end{equation}$$ Let's first calculate the commutator \([\hat{H}_0,\chat_{\k\sigma}]\). $$\begin{align} [\hat{H}_0,\chat_{\k\sigma}] ={}& \left[\sum_{\k'\sigma'}\xi_{\k'} \chat^\dagger_{\k'\sigma'} \chat_{\k'\sigma'}, \chat_{\k\sigma}\right] \\ ={}& \sum_{\k'\sigma'}\xi_{\k'}\left[ \chat^\dagger_{\k'\sigma'} \chat_{\k'\sigma'}, \chat_{\k\sigma}\right] \\ ={}& \sum_{\k'\sigma'}\xi_{\k'}\left( \chat^\dagger_{\k'\sigma'} \chat_{\k'\sigma'} \chat_{\k\sigma} - \chat_{\k\sigma} \chat^\dagger_{\k'\sigma'} \chat_{\k'\sigma'}\right) \\ ={}& \sum_{\k'\sigma'}\xi_{\k'}\left(\chat^\dagger_{\k'\sigma'}\chat_{\k'\sigma'}\chat_{\k\sigma} - \left(-\chat^\dagger_{\k'\sigma'} \chat_{\k\sigma} + \delta_{\k\k'}\delta_{\sigma\sigma'}\right)\chat_{\k'\sigma'}\right)\\ ={}& \sum_{\k'\sigma'}\xi_{\k'}\left(\chat^\dagger_{\k'\sigma'}\left\{\chat_{\k'\sigma'}, \chat_{\k\sigma}\right\} - \delta_{\k\k'}\delta_{\sigma\sigma'}\chat_{\k'\sigma'}\right) \\ ={}& -\xi_{\k} \chat_{\k\sigma} \end{align}$$ This means it is actually easy to evaluate the series completely! This is so because if we have \(n\) nested commutators, for example, if \(n=2\), $$\begin{align} [\hat{H}_0, [\hat{H}_0, \chat_{\k\sigma}]] ={}& [\hat{H}_0, -\xi_{\k} \chat_{\k\sigma} ]\\ ={}& -\xi_{\k} [\hat{H}_0,\chat_{\k\sigma}] \\ ={}& (-\xi_{\k})^2 \chat_{\k\sigma} \end{align}$$ Thus, the series becomes $$\begin{align} \chat_{\k\sigma,I}(t) ={}& \chat_{\k\sigma} + \frac{it}{\hbar}(-\xi_{\k})\chat_{\k\sigma} + \frac{1}{2}\left(\frac{it}{\hbar}\right)^2 (-\xi_{\k})^2 \chat_{\k\sigma} + \cdots \\ ={}& \chat_{\k\sigma}\sum_{n=0}^\infty \frac{1}{n!}\left(\frac{-it\xi_{\k}}{\hbar}\right)^n \\ ={}& e^{-it\xi_{\k}/\hbar} \chat_{\k\sigma} \end{align}$$ Similarly, one can show that $$\begin{equation} \chat^\dagger_{\k\sigma,I}(t) = e^{it\xi_{\k}/\hbar} \chat^\dagger_{\k\sigma}, \end{equation}$$ either by taking the Hermitian conjugate of our previous result, or by direct computation.

To do this, we first consider the definition of the interaction picture operator, which is $$\begin{equation} \bhat_{\k,I}(t) = e^{i\Hhat_0 t/\hbar} \bhat_{\k} e^{-i\Hhat_0 t/\hbar} \end{equation}$$ We now apply the Baker-Campbell-Hausdorff formula to get $$\begin{equation} \bhat_{\k,I}(t) = \bhat_{\k} + \frac{it}{\hbar}[\Hhat_0,\bhat_{\k}] + \frac{1}{2}\left(\frac{it}{\hbar}\right)^2[\Hhat_0,[\Hhat_0,\bhat_{\k}]] + \cdots \end{equation}$$ Let's first compute the commutator \([\Hhat_0, \bhat_{\k}]\). We find $$\begin{align} [\Hhat_0, \bhat_{\k}] ={}& \left[\sum_{\k'} \xi_{\k'} \bhat^\dagger_{\k'}\bhat_{\k'}, \bhat_{\k}\right] \\ ={}& \sum_{\k'} \xi_{\k'} [\bhat^\dagger_{\k'}, \bhat_{\k}]\bhat_{\k'} \\ ={}& \sum_{\k'} \xi_{\k'} (-\delta_{\k\k'}) \bhat_{\k'} \\ ={}& -\xi_{\k} \bhat_{\k} \end{align}$$ Thus we see that $$\begin{align} \bhat_{\k,I}(t) ={}& \bhat_{\k} \frac{it\xi_{\k}}{\hbar}\bhat_{\k} + \frac{1}{2}\left(\frac{-\xi_{\k} it}{\hbar}\right)^2\bhat_{\k}+ \cdots \\ ={}& \sum_{n=0}^\infty \frac{1}{n!}\left(-\frac{it\xi_{\k}}{\hbar}\right)^n \bhat_{\k} \\ ={}& e^{-it\xi_{\k}/\hbar} \bhat_{\k} \end{align}$$ From this, we also see that $$\begin{equation} \bhat^\dagger_{\k,I}(t) = e^{it\xi_{\k}/\hbar} \bhat^\dagger_{\k}. \end{equation}$$