New answers tagged

1

In the first summation for $U$: $$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}(it\Omega \frac{H_{2}}{\Omega})^{2n}=\cos(\Omega t)\begin{pmatrix} 0&0&0\\0&1&0\\0&0&1\end{pmatrix}(\text{This is wrong})$$ $$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}(it\Omega \frac{H_{2}}{\Omega})^{2n}=\begin{pmatrix} 1&0&0\\0&0&0\\0&0&0\...


4

You seem to be overcomplicating this somewhat! You are right to split it up into the two terms $H_1$ and $H_2$. So, we have $$ e^{-i(H_1+H_2)t}=e^{-iH_1t}e^{-iH_2t}. $$ Now, straightforwardly, $$ e^{-iH_1t}=I+(e^{-i\delta t}-1)|00\rangle\langle 00|. $$ Next, we need to think about the $e^{-iH_2t}$ term. Of course, it maps $|00\rangle$ to $|00\rangle$. So, ...


Top 50 recent answers are included