General Analysis

Suppose that at $t=t_0$ the state of the system is represented by

$$\vert A\rangle = \sum_n c_n\, \vert n\rangle,$$ (8.3)

where the $c_n$ are complex numbers. Thus, the initial state is some linear superposition of the unperturbed energy eigenstates. In the absence of the time-dependent perturbation, the time evolution of the system is given by

$$\vert A, t_0, t\rangle = \sum_n c_n \exp\left[\frac{-{\rm i}\,E_n \,(t-t_0)}{\hbar}\right]\vert n\rangle.$$ (8.4)

(See Section 3.5.) Here, $\vert A, t_0, t\rangle$ is the state ket of the system at time $t$ , given that the state ket at the initial time $t_0$ is $\vert A\rangle$ . Now, the probability of finding the system in state $\vert n\rangle$ at time $t$ is

$$P_n(t) = \left\vert c_n \exp\left[\frac{-{\rm i}\,E_n (t-t_0)}{\hbar}\right]\right\vert^{\,2} = \vert c_n\vert^{\,2} = P_n(t_0).$$ (8.5)

Clearly, with $H_1= 0$ , the probability of finding the system in state $\vert n\rangle$ at time $t$ is exactly the same as the probability of finding the system in this state at the initial time, $t_0$ . However, with $H_1\neq 0$ , we expect $P_n(t)$ to vary with time. Thus, we can write

$$\vert A, t_0, t\rangle = \sum_n c_n(t) \exp\left[\frac{-{\rm i}\,E_n\,(t-t_0)}{\hbar}\right]\vert n\rangle,$$ (8.6)

where $P_n(t) = \vert c_n(t)\vert^{\,2}$ . Here, we have carefully separated the fast phase oscillation of the eigenkets, which depends on the unperturbed Hamiltonian, from the slow variation of the amplitudes $c_n(t)$ , which depends entirely on the perturbation (because the $c_n$ are all constant if $H_1= 0$ ) [32]. Note that the eigenkets $\vert n\rangle$ , appearing in Equation (8.6), are time-independent (they are actually the eigenkets of $H_0$ evaluated at the time $t_0$ ).

Schrödinger's time evolution equation yields

$${\rm i}\,\hbar\, \frac{\partial}{\partial t}\,\vert A, t_0, t\rangle = H\,\vert A,t_0,t\rangle= (H_0+H_1) \,\vert A,t_0,t\rangle.$$ (8.7)

(See Section 3.2.) It follows from Equation (8.6) that

$$(H_0+H_1)\, \vert A,t_0,t\rangle = \sum_m c_m(t) \exp\left[\frac{-{\rm i}\,E_m\, (t-t_0)}{\hbar}\right] (E_m + H_1)\,\vert m\rangle.$$ (8.8)

We also have

$${\rm i}\,\hbar\, \frac{\partial}{\partial t}\,\vert A,t_0,t\rangl... ...m\right] \exp\left[\frac{-{\rm i}\,E_m \,(t-t_0)}{\hbar}\right] \vert m\rangle,$$ (8.9)

where use has been made of the time-independence of the kets $\vert m\rangle$ . According to Equation (8.7), we can equate the right-hand sides of the previous two equations to obtain

$$\begin{multline} \sum_m {\rm i}\,\hbar\, \frac{d c_m}{dt}\exp\left[\frac{-{\rm i... ...frac{-{\rm i}\,E_m \,(t-t_0)}{\hbar}\right] H_1\, \vert m\rangle. \end{multline}$$

Left-multiplication by $\langle n\vert$ yields [32]

$${\rm i}\,\hbar\, \frac{d c_n}{dt} = \sum_m H_{nm}(t)\, \exp[\,{\rm i}\,\omega_{nm}\, (t-t_0)]\, c_m(t),$$ (8.10)

where

$$H_{nm}(t) = \langle n \vert\,H_1(t)\,\vert m \rangle,$$ (8.11)

and

$$\omega_{nm} = \frac{E_n - E_m}{\hbar}.$$ (8.12)

Here, we have made use of the standard orthonormality result, $\langle n\vert m\rangle =\delta_{nm}$ . Suppose that there are $N$ linearly independent eigenkets of the unperturbed Hamiltonian. According to Equation (8.11), the time variation of the coefficients $c_n(t)$ , which specify the probability of finding the system in state $\vert n\rangle$ at time $t$ , is determined by $N$ coupled first-order differential equations. Note that the set of equations specified in Equation (8.11) is exact--we have made no approximations at this stage. Unfortunately, we cannot generally find exact solutions to this set, so we have to obtain approximate solutions via suitable expansions in small quantities. However, for the particularly simple case of a two-state system (i.e., $N=2$ ), it is actually possible to solve Equation (8.11) without approximation. This solution is of great practical importance.