First Order Perturbation Theory

This is achieved by doing the first iteration in

$$\frac{d}{dt}\vert\Psi_I(t)\rangle = -i V_I(t) \vert\Psi_I(t)\rangle$$

$$\leadsto \vert\Psi_I(t)\rangle = \Psi_I(t_0)\rangle -i \int_{t_0}^t dt' V_I(t') \vert\Psi_I(t')\rangle$$

$$= \vert\Psi_I(t_0)\rangle -i \int_{t_0}^t dt' V_I(t') \vert\Psi_I(t_0)\rangle +O(V_I^2).$$ (3.6)

We take $t_0=0$ for simplicity and the initial state therefore is $\vert\Psi_I(0)\rangle =\vert\Psi(0)\rangle$,

$$\vert\Psi_I(t)\rangle = \vert\Psi(0)\rangle -i \int_{0}^t dt' V_I(t') \vert\Psi(0)\rangle +O(V_I^2).$$ (3.7)

This can be worked out in some more detail by assuming a basis $\vert n\rangle$ of eigenstates of $H_0$,

$$H_0\vert n\rangle = \varepsilon_n \vert n\rangle,\quad \hat{1}=\sum_{n'}\vert n'\rangle \langle n'\vert\leadsto$$ (3.8)

$$\langle n\vert\Psi_I(t)\rangle = \langle n\vert\Psi(0)\rangle -i\sum_{n'} \int_{0}^t dt' \langle n\vert V_I(t')\vert n'\rangle \langle n' \vert\Psi(0)\rangle +O(V_I^2).$$

Let us assume the initial state $\vert\Psi(0)\rangle = \vert m\rangle$ is an eigenstate of $H_0$, then

$$\langle n\vert\Psi_I(t)\rangle = \delta_{nm} -i \int_{0}^t dt' \langle n\vert V_I(t')\vert m\rangle +O(V_I^2).$$ (3.9)

The probability to find the system in state $\vert n\rangle$ after time $t$ is then a transition probability. Use $\vert\langle n\vert\Psi(t)\rangle\vert^2 = \vert\langle n\vert\Psi_I(t)\rangle\vert^2$ (EXERCISE:CHECK!) to find within first order perturbation theory

$$P_{m\to n}(t) = \vert\langle n\vert\Psi(t)\rangle\vert^2 = \left\vert\int_{0}^t dt' \langle n\vert V_I(t')\vert m\rangle\right\vert^2$$

$$\vert\Psi(0)\rangle \equiv \vert m\rangle \ne \vert n\rangle.$$ (3.10)