Time-dependent perturbation theory

Suppose that the Hamiltonian of the system under consideration can be written

$$H = H_0 + H_1(t),$$ (730)

where $H_0$ does not contain time explicitly, and $H_1$ is a small time-dependent perturbation. It is assumed that we are able to calculate the eigenkets of the unperturbed Hamiltonian:

$$H_0 \vert n\rangle = E_n \vert n\rangle.$$ (731)

We know that if the system is in one of the eigenstates of $H_0$ then, in the absence of the external perturbation, it remains in this state for ever. However, the presence of a small time-dependent perturbation can, in principle, give rise to a finite probability that a system initially in some eigenstate $\vert i\rangle$ of the unperturbed Hamiltonian is found in some other eigenstate at a subsequent time (since $\vert i\rangle$ is no longer an exact eigenstate of the total Hamiltonian), In other words, a time-dependent perturbation causes the system to make transitions between its unperturbed energy eigenstates. Let us investigate this effect.

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

$$\vert A\rangle = \sum_n c_n \vert n\rangle,$$ (732)

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

$$\vert A, t_0, t\rangle = \sum_n c_n \exp([-{\rm i} E_n (t-t_0)/\hbar] \vert n\rangle.$$ (733)

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

$$P_n(t) = \vert c_n \exp[-{\rm i} E_n (t-t_0)/\hbar]\vert^2 = \vert c_n\vert^2 = P_n(t_0).$$ (734)

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

$$\vert A, t_0, t\rangle = \sum_n c_n(t) \exp[-{\rm i} E_n (t-t_0)/\hbar] \vert n\rangle,$$ (735)

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 (i.e., $c_n$ is constant if $H_1= 0$). Note that in Eq. (735) the eigenkets $\vert n\rangle$ are time-independent (they are actually the eigenkets of $H_0$ evaluated at the time $t_0$).

Schrödinger's time evolution equation yields

$${\rm i} \hbar \frac{\partial}{\partial t}\vert A, t_0, t\ra... ... = H \vert A,t_0,t\rangle= (H_0+H_1) \vert A,t_0,t\rangle.$$ (736)

It follows from Eq. (735) that

$$(H_0+H_1) \vert A,t_0,t\rangle = \sum_m c_m(t) \exp[-{\rm i} E_m (t-t_0)/\hbar] (E_m + H_1) \vert m\rangle.$$ (737)

We also have

$${\rm i} \hbar \frac{\partial}{\partial t}\vert A,t_0,t\rang... ..._m\right) \exp[-{\rm i} E_m (t-t_0)/\hbar] \vert m\rangle,$$ (738)

where use has been made of the time-independence of the kets $\vert m\rangle$. According to Eq. (736), we can equate the right-hand sides of the previous two equations to obtain

$$\sum_m {\rm i} \hbar \frac{d c_m}{dt}\exp[-{\rm i} E_m \... ...m(t) \exp[-{\rm i} E_m (t-t_0)/\hbar] H_1 \vert m\rangle.$$ (739)

Left-multiplication by $\langle n\vert$ yields

$${\rm i} \hbar \frac{d c_n}{dt} = \sum_m H_{nm}(t) \exp[{\rm i} \omega_{nm} (t-t_0)] c_m(t),$$ (740)

where

$$H_{nm}(t) = \langle n \vert H_1(t)\vert m \rangle,$$ (741)

and

$$\omega_{nm} = \frac{E_n - E_m}{\hbar}.$$ (742)

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 Eq. (740), the time variation of the coefficients $c_n$, 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 Eq. (740) is exact--we have made no approximations at this stage. Unfortunately, we cannot generally find exact solutions to this equation, 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 Eq. (740) without approximation. This solution is of enormous practical importance.