In spectroscopy, we start with a system in some stationary state, expose it to electromagnetic radiation (light), and then observe whether the system has made a transition to another stationary state. The radiation produces a time-dependent potential-energy term in the Hamiltonian, so we must use the time-dependent Schrödinger equation. The most convenient approach here is an approximate one called time-dependent perturbation theory.
Let the system (atom or molecule) have the time-independent Hamiltonian $\hat{H}^{0}$ in the absence of the radiation (or other time-dependent perturbation), and let $\hat{H}^{\prime}(t)$ be the timedependent perturbation. The time-independent Schrödinger equation for the unperturbed problem is
\(
\begin{equation}
\hat{H}^{0} \psi{k}^{0}=E{k}^{0} \psi_{k}^{0} \tag{9.112}
\end{equation}
\)
where $E{k}^{0}$ and $\psi{k}^{0}$ are the stationary-state energies and wave functions. The time-dependent Schrödinger equation (7.97) in the presence of the radiation is
\(
\begin{equation}
-\frac{\hbar}{i} \frac{\partial \Psi}{\partial t}=\left(\hat{H}^{0}+\hat{H}^{\prime}\right) \Psi \tag{9.113}
\end{equation}
\)
where the state function $\Psi$ depends on the spatial and spin coordinates (symbolized by $q$ ) and on the time: $\Psi=\Psi(q, t)$. (See Chapter 10 for a discussion of spin coordinates.)
First suppose that $\hat{H}^{\prime}(t)$ is absent. The unperturbed time-dependent Schrödinger equation is
\(
\begin{equation}
-(\hbar / i) \partial \Psi^{0} / \partial t=\hat{H}^{0} \Psi^{0} \tag{9.114}
\end{equation}
\)
The system's possible stationary-state state functions are given by (7.99) as $\Psi{k}^{0}=$ $\exp \left(-i E{k}^{0} t / \hbar\right) \psi{k}^{0}$, where the $\psi{k}^{0}$ functions are the eigenfunctions of $\hat{H}^{0}$ [Eq. (9.112)]. Each $\Psi_{k}^{0}$ is a solution of (9.114). Moreover, the linear combination
\(
\begin{equation}
\Psi^{0}=\sum{k} c{k} \Psi{k}^{0}=\sum{k} c{k} \exp \left(-i E{k}^{0} t / \hbar\right) \psi_{k}^{0} \tag{9.115}
\end{equation}
\)
with the $c{k}$ 's being arbitrary time-independent constants, is a solution of the time-dependent Schrödinger equation (9.114), as proved in the discussion leading to Eq. (7.100). The functions $\Psi{k}^{0}$ form a complete set (since they are the eigenfunctions of the Hermitian operator $\hat{H}^{0}$ ), so any solution of (9.114) can be expressed in the form (9.115). Hence (9.115) is the general solution of the time-dependent Schrödinger equation (9.114), where $\hat{H}^{0}$ is independent of time.
Now suppose that $\hat{H}^{\prime}(t)$ is present. The function (9.115) is no longer a solution of the time-dependent Schrödinger equation. However, because the unperturbed functions $\Psi{k}^{0}$ form a complete set, the true state function $\Psi$ can at any instant of time be expanded as a linear combination of the $\Psi{k}^{0}$ functions according to $\Psi=\sum{k} b{k} \Psi{k}^{0}$. Because $\hat{H}$ is timedependent, $\Psi$ will change with time and the expansion coefficients $b{k}$ will change with time. Therefore,
\(
\begin{equation}
\Psi=\sum{k} b{k}(t) \exp \left(-i E{k}^{0} t / \hbar\right) \psi{k}^{0} \tag{9.116}
\end{equation}
\)
In the limit $\hat{H}^{\prime}(t) \rightarrow 0$, the expansion (9.116) reduces to (9.115).
Substitution of (9.116) into the time-dependent Schrödinger equation (9.113) and use of (9.112) gives
\(
\begin{aligned}
-\frac{\hbar}{i} \sum{k} \frac{d b{k}}{d t} \exp \left(-i E{k}^{0} t / \hbar\right) \psi{k}^{0}+ & \sum{k} E{k}^{0} b{k} \exp \left(-i E{k}^{0} t / \hbar\right) \psi{k}^{0} \
& =\sum{k} b{k} \exp \left(-i E{k}^{0} t / \hbar\right) E{k}^{0} \psi{k}^{0}+\sum{k} b{k} \exp \left(-i E{k}^{0} t / \hbar\right) \hat{H}^{\prime} \psi{k}^{0} \
-\frac{\hbar}{i} \sum{k} \frac{d b{k}}{d t} & \exp \left(-i E{k}^{0} t / \hbar\right) \psi{k}^{0}=\sum{k} b{k} \exp \left(-i E{k}^{0} t / \hbar\right) \hat{H}^{\prime} \psi{k}^{0}
\end{aligned}
\)
We now multiply by $\psi{m}^{0} *$ and integrate over the spatial and spin coordinates. Using the orthonormality equation $\left\langle\psi{m}^{0} \mid \psi{k}^{0}\right\rangle=\delta{m k}$, we get
\(
-\frac{\hbar}{i} \sum{k} \frac{d b{k}}{d t} \exp \left(-i E{k}^{0} t / \hbar\right) \delta{m k}=\sum{k} b{k} \exp \left(-i E{k}^{0} t / \hbar\right)\left\langle\psi{m}^{0}\right| \hat{H}^{\prime}\left|\psi_{k}^{0}\right\rangle
\)
Because of the $\delta{m k}$ factor, all terms but one in the sum on the left are zero, and the left side equals $-(\hbar / i)\left(d b{m} / d t\right) \exp \left(-i E_{m}^{0} t / \hbar\right)$. We get
\(
\begin{equation}
\frac{d b{m}}{d t}=-\frac{i}{\hbar} \sum{k} b{k} \exp \left[i\left(E{m}^{0}-E{k}^{0}\right) t / \hbar\right]\left\langle\psi{m}^{0}\right| \hat{H}^{\prime}\left|\psi_{k}^{0}\right\rangle \tag{9.117}
\end{equation}
\)
Let us suppose that the perturbation $\hat{H}^{\prime}(t)$ was applied at time $t=0$ and that before the perturbation was applied the system was in stationary state $n$ with energy $E{n}^{(0)}$. The state function at $t=0$ is therefore $\Psi=\exp \left(-i E{n}^{0} t / \hbar\right) \psi{n}^{0}$ [Eq. (7.99)], and the $t=0$ values of the expansion coefficients in (9.116) are thus $b{n}(0)=1$ and $b_{k}(0)=0$ for $k \neq n$ :
\(
\begin{equation}
b{k}(0)=\delta{k n} \tag{9.118}
\end{equation}
\)
We shall assume that the perturbation $\hat{H}^{\prime}$ is small and acts for only a short time. Under these conditions, the change in the expansion coefficients $b_{k}$ from their initial values at the time the perturbation is applied will be small. To a good approximation, we can replace the expansion coefficients on the right side of (9.117) by their initial values (9.118). This gives
\(
\begin{equation}
\frac{d b{m}}{d t} \approx-\frac{i}{\hbar} \exp \left[i\left(E{m}^{0}-E{n}^{0}\right) t / \hbar\right]\left\langle\psi{m}^{0}\right| \hat{H}^{\prime}\left|\psi_{n}^{0}\right\rangle \tag{9.119}
\end{equation}
\)
Let the perturbation $\hat{H}^{\prime}$ act from $t=0$ to $t=t^{\prime}$. Integrating from $t=0$ to $t^{\prime}$ and using (9.118), we get
\(
\begin{equation}
b{m}\left(t^{\prime}\right) \approx \delta{m n}-\frac{i}{\hbar} \int{0}^{t^{\prime}} \exp \left[i\left(E{m}^{0}-E{n}^{0}\right) t / \hbar\right]\left\langle\psi{m}^{0}\right| \hat{H}^{\prime}\left|\psi_{n}^{0}\right\rangle d t \tag{9.120}
\end{equation}
\)
Use of the approximate result (9.120) for the expansion coefficients in (9.116) gives the desired approximation to the state function at time $t^{\prime}$ for the case that the time-dependent perturbation $\hat{H}^{\prime}$ is applied at $t=0$ to a system in stationary state $n$. [As with time-independent perturbation theory, one can go to higher-order approximations (see Fong, pp. 234-244).]
For times after $t^{\prime}$, the perturbation has ceased to act, and $\hat{H}^{\prime}=0$. Equation (9.117) gives $d b{m} / d t=0$ for $t>t^{\prime}$, so $b{m}=b_{m}\left(t^{\prime}\right)$ for $t \geq t^{\prime}$. Therefore, for times after exposure to the perturbation, the state function $\Psi$ is [Eq. (9.116)]
\(
\begin{equation}
\Psi=\sum{m} b{m}\left(t^{\prime}\right) \exp \left(-i E{m}^{0} t / \hbar\right) \psi{m}^{0} \quad \text { for } t \geq t^{\prime} \tag{9.121}
\end{equation}
\)
where $b{m}\left(t^{\prime}\right)$ is given by (9.120). In (9.121), $\Psi$ is a superposition of the eigenfunctions $\psi{m}^{0}$ of the energy operator $\hat{H}^{0}$, the expansion coefficients being $b{m} \exp \left(-i E{m}^{0} t / \hbar\right)$. [Compare (9.121) and (7.66).] The work of Section 7.6 tells us that a measurement of the system's energy at a time after $t^{\prime}$ will give one of the eigenvalues $E{m}^{0}$ of the energy operator $\hat{H}^{0}$, and the probability of getting $E{m}^{0}$ equals the square of the absolute value of the expansion coefficient that multiplies $\psi{m}^{0}$; that is, it equals $\left|b{m}\left(t^{\prime}\right) \exp \left(-i E{m}^{0} t / \hbar\right)\right|^{2}=\left|b{m}\left(t^{\prime}\right)\right|^{2}$.
The time-dependent perturbation changes the system's state function from $\exp \left(-i E{n}^{0} t / \hbar\right) \psi{n}^{0}$ to the superposition (9.121). Measurement of the energy then changes $\Psi$ to one of the energy eigenfunctions $\exp \left(-i E{m}^{0} t / \hbar\right) \psi{m}^{0}$ (reduction of the wave function, Section 7.9). The net result is a transition from stationary state $n$ to stationary state $m$, the probability of such a transition being $\left|b_{m}\left(t^{\prime}\right)\right|^{2}$.