Preliminary Analysis

Suppose that at $t=0$ the state of the system is represented by

$$\psi(0) = \sum_m c_m \psi_m,$$ (1019)

where the $c_m$ 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 simply (see Sect. 4.12)

$$\psi(t) = \sum_m c_m \exp\left(-{\rm i} E_m t/\hbar\right)\psi_m.$$ (1020)

Now, the probability of finding the system in state $n$ at time $t$ is

$$P_n(t) = \vert\langle \psi_n\vert\psi\rangle\vert^2 = \left\... ...,E_n t/\hbar\right)\right\vert^2 = \vert c_n\vert^2 = P_n(0),$$ (1021)

since the unperturbed eigenstates are assummed to be orthonormal: i.e.,

$$\langle n\vert m\rangle =\delta_{nm}.$$ (1022)

Clearly, with $H_1=0$, the probability of finding the system in state $\psi_n$ 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 $P_n$--and, hence, $c_n$--to vary with time. Thus, we can write

$$\psi(t) = \sum_m c_m(t) \exp\left(-{\rm i} E_m t/\hbar\right)\psi_m,$$ (1023)

where $P_n(t)=\vert c_n(t)\vert^2$. Here, we have carefully separated the fast phase oscillation of the eigenstates, which depends on the unperturbed Hamiltonian, from the slow variation of the amplitudes $c_n(t)$, which depends entirely on the perturbation (i.e., $c_n$ is constant in time if $H_1=0$). Note that in Eq. (1023) the eigenstates $\psi_m$ are time-independent (they are actually the eigenstates of $H_0$ evaluated at the initial time, $t=0$).

The time-dependent Schrödinger equation [see Eq. (199)] yields

$${\rm i} \hbar \frac{\partial\psi(t)}{\partial t} = H(t) \psi(t) = [H_0+H_1(t)] \psi(t).$$ (1024)

Now, it follows from Eq. (1023) that

$$(H_0+H_1) \psi = \sum_m c_m \exp\left(-{\rm i} E_m t/\hbar\right)(E_m+H_1) \psi_m.$$ (1025)

We also have

$${\rm i} \hbar \frac{\partial\psi}{\partial t} = \sum_m\le... ...c_m E_m\right) \exp\left(-{\rm i} E_m t/\hbar\right)\psi_m,$$ (1026)

since the $\psi_m$ are time-independent. According to Eq. (1024), we can equate the right-hand sides of the previous two equations to obtain

$$\sum_m {\rm i} \hbar \frac{dc_m}{dt} \exp\left(-{\rm i} ... ...\sum_m c_m exp\left(-{\rm i} E_m t/\hbar\right)H_1 \psi_m.$$ (1027)

Projecting out the component of the above equation which is proportional to $\psi_n$, using Eq. (1022), we obtain

$${\rm i} \hbar \frac{dc_n(t)}{dt} = \sum_m H_{nm}(t) \exp\left( {\rm i} \omega_{nm} t\right)c_m(t),$$ (1028)

where

$$H_{nm}(t) = \langle n\vert H_1(t)\vert m\rangle,$$ (1029)

and

$$\omega_{nm} = \frac{E_n-E_m}{\hbar}.$$ (1030)

Suppose that there are $N$ linearly independent eigenstates of the unperturbed Hamiltonian. According to Eqs. (1028), the time-dependence of the set of $N$ coefficients $c_n$, which specify the probabilities of finding the system in these eigenstates at time $t$, is determined by $N$ coupled first-order differential equations. Note that Eqs. (1028) are exact--we have made no approximations at this stage. Unfortunately, we cannot generally find exact solutions to these equations. Instead, we have to obtain approximate solutions via suitable expansions in small quantities. However, for the particuilarly simple case of a two-state system (i.e., $N=2$), it is actually possible to solve Eqs. (1028) without approximation. This solution is of great practical importance.