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So I am dealing with the following hamiltonian, and the following perturbation: $$H=-\mu B_0\sigma_z$$ $$V=\mu B_1(\cos(\omega t)\hat x-\sin(\omega t)\hat y)\cdot{\bf \sigma}$$ I am asked for the probability that the spin will go from ground state at time 0 to the excited state at time t using first order time dependent perturbation theory.

I know that the ground state is $\begin{pmatrix}1\\0 \end{pmatrix}$ with energy $-\mu B_0$. This means the excited state is $\begin{pmatrix}0\\1 \end{pmatrix}$ with energy$+\mu B_0$. So the probability I want is:

$$P=|(0 1)U^{(1)}(t,0)\begin{pmatrix}0\\1\end{pmatrix}|^2$$ Where I have $$U^{(1)}=\frac{1}{i\hbar}\int^t_0 dt_1e^{-iH_f(t-t_1)/\hbar} V(t_1) e^{-iH_it_1/\hbar}$$ So I have 2 questions. The first is if this is the correct expression I want to use for finding the probability. And secondly, how do I treat the matrices? Seeing as they are a constant, can I just pull them out and then integrate the remaining pieces?

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Yes, the expression seems correct. To compute the probability, you have to compute $|\langle 1|U^{(1)}|0\rangle|^2$ . As $|0\rangle$ and $|1\rangle$ do not depend on time, you can just pull them inside the integral, and integrate the (1,0) component of the 2 x 2 matrix $V(t_1)$, denoted below by $V_{10}(t_1)$. So

\begin{align}p&=\left|\frac{1}{i\hbar}\int^t_0 dt_1e^{-iH_f(t-t_1)/\hbar} \langle 1|V(t_1)|0\rangle e^{-iH_it_1/\hbar}\right|^2\\ &=\frac{1}{\hbar^2}\left| \int^t_0 dt_1e^{-iH_f(t-t_1)/\hbar} V_{10}(t_1) e^{-iH_it_1/\hbar} \right|^2\\ &=\frac{1}{\hbar^2}\left| \int^t_0 dt_1e^{+iH_f t_1/\hbar} V_{10}(t_1) e^{-iH_it_1/\hbar} \right|^2\\ &=\frac{1}{\hbar^2}\left| \int^t_0 dt_1e^{+i(H_f-H_i)t_1/\hbar} V_{10}(t_1) \right|^2\\ &=\frac{1}{\hbar^2}\left| \int^t_0 dt_1e^{+2i \mu B_0 t_1/\hbar} V_{10}(t_1) \right|^2 \end{align}

Hope this helps.

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