We now consider the interaction of an atom or molecule with electromagnetic radiation. A proper quantum-mechanical approach would treat both the atom and the radiation quantum mechanically, but we shall simplify things by using the classical picture of the light as an electromagnetic wave of oscillating electric and magnetic fields.
A detailed investigation, which we omit, shows that usually the interaction between the radiation's magnetic field and the atom's charges is much weaker than the interaction between the radiation's electric field and the charges, so we shall consider only the latter interaction. (In NMR spectroscopy the important interaction is between the magnetic dipole moments of the nuclei and the radiation's magnetic field. We shall not consider this case.)
Let the electric field $\mathscr{E}$ of the electromagnetic wave point in the $x$ direction only. (This is plane-polarized radiation.) The electric field is defined as the force per unit charge, so the force on charge $Q{i}$ is $F=Q{i} \mathscr{E}{x}=-d V / d x$, where (4.24) was used. Integration gives the potential energy of interaction between the radiation's electric field and the charge as $V=-Q{i} \mathscr{C}{x} x$, where the arbitrary integration constant was taken as zero. For a system of several charges, $V=-\sum{i} Q{i} x{i} \mathscr{E}{x}$. This is the time-dependent perturbation $\hat{H}^{\prime}(t)$. The space and time dependence of the electric field of an electromagnetic wave traveling in the $z$ direction with wavelength $\lambda$ and frequency $\nu$ is given by (see a first-year physics text) $\mathscr{E}{x}=$ $\mathscr{E}{0} \sin (2 \pi \nu t-2 \pi z / \lambda)$, where $\mathscr{E}{0}$ is the maximum value of $\mathscr{E}_{x}$ (the amplitude). Therefore,
\(
\hat{H}^{\prime}(t)=-\mathscr{E}{0} \sum{i} Q{i} x{i} \sin \left(2 \pi \nu t-2 \pi z_{i} / \lambda\right)
\)
where the sum goes over all the electrons and nuclei of the atom or molecule.
Defining $\omega$ and $\omega_{m n}$ as
\(
\begin{equation}
\omega \equiv 2 \pi \nu, \quad \omega{m n} \equiv\left(E{m}^{0}-E_{n}^{0}\right) / \hbar \tag{9.122}
\end{equation}
\)
and substituting $\hat{H}^{\prime}(t)$ into (9.120), we get the coefficients in the expansion (9.116) of the state function $\Psi$ as
\(
b{m} \approx \delta{m n}+\frac{i \mathscr{E}{0}}{\hbar} \int{0}^{t^{\prime}} \exp \left(i \omega{m n} t\right)\left\langle\psi{m}^{0}\right| \sum{i} Q{i} x{i} \sin \left(\omega t-2 \pi z{i} / \lambda\right)\left|\psi_{n}^{0}\right\rangle d t
\)
The integral $\left\langle\psi{m}^{0}\right| \sum{i} \cdots\left|\psi{n}^{0}\right\rangle$ in this equation is over all space, but significant contributions to its magnitude come only from regions where $\psi{m}^{0}$ and $\psi{n}^{0}$ are of significant magnitude. In regions well outside the atom or molecule, $\psi{m}^{0}$ and $\psi{n}^{0}$ are vanishingly small, and such regions can be ignored. Let the coordinate origin be chosen within the atom or molecule. Since regions well outside the atom can be ignored, the coordinate $z{i}$ can be considered to have a maximum magnitude of the order of one nm. For ultraviolet light, the wavelength $\lambda$ is on the order of $10^{2} \mathrm{~nm}$. For visible, infrared, microwave, and radiofrequency radiation, $\lambda$ is even larger. Hence $2 \pi z{i} / \lambda$ is very small and can be neglected, and this leaves $\sum{i} Q{i} x{i} \sin \omega t$ in the integral.
Use of the identity (Prob. 1.29) $\sin \omega t=\left(e^{i \omega t}-e^{-i \omega t}\right) / 2 i$ gives
\(
b{m}\left(t^{\prime}\right) \approx \delta{m n}+\frac{\mathscr{E}{0}}{2 \hbar}\left\langle\psi{m}^{0}\right| \sum{i} Q{i} x{i}\left|\psi{n}^{0}\right\rangle \int{0}^{t^{\prime}}\left[e^{i\left(\omega{m n}+\omega\right) t}-e^{i\left(\omega_{m n}-\omega\right) t}\right] d t
\)
Using $\int_{0}^{t^{\prime}} e^{a t} d t=a^{-1}\left(e^{a t^{\prime}}-1\right)$, we get
\(
\begin{equation}
b{m}\left(t^{\prime}\right) \approx \delta{m n}+\frac{\mathscr{E}{0}}{2 \hbar i}\left\langle\psi{m}^{0}\right| \sum{i} Q{i} x{i}\left|\psi{n}^{0}\right\rangle\left[\frac{e^{i\left(\omega{m n}+\omega\right) t^{\prime}}-1}{\omega{m n}+\omega}-\frac{e^{i\left(\omega{m n}-\omega\right) t^{\prime}}-1}{\omega{m n}-\omega}\right] \tag{9.123}
\end{equation}
\)
For $m \neq n$, the $\delta{m n}$ term equals zero.
As noted at the end of Section 9.8, $\left|b{m}\left(t^{\prime}\right)\right|^{2}$ gives the probability of a transition to state $m$ from state $n$. There are two cases where this probability becomes of significant magnitude. If $\omega{m n}=\omega$, the denominator of the second fraction in brackets is zero and this fraction's absolute value is large (but not infinite; see Prob. 9.27). If $\omega{m n}=-\omega$, the first fraction has a zero denominator and a large absolute value.
For $\omega{m n}=\omega$, Eq. (9.122) gives $E{m}^{0}-E{n}^{0}=h \nu$. Exposure of the atom to radiation of frequency $\nu$ has produced a transition from stationary state $n$ to stationary state $m$, where (since $\nu$ is positive) $E{m}^{0}>E_{n}^{0}$. We might suppose that the energy for this transition came from the system's absorption of a photon of energy $h \nu$. This supposition is confirmed by a fully quantum-mechanical treatment (called quantum field theory) in which the radiation is treated quantum mechanically rather than classically. We have absorption of radiation with a consequent increase in the system's energy.
For $\omega{m n}=-\omega$, we get $E{n}^{0}-E{m}^{0}=h \nu$. Exposure to radiation of frequency $\nu$ has induced a transition from stationary state $n$ to stationary state $m$, where (since $\nu$ is positive) $E{n}^{0}>E_{m}^{0}$. The system has gone to a lower energy level, and a quantum-field-theory treatment shows that a photon of energy $h \nu$ is emitted in this process. This is stimulated emission of radiation. Stimulated emission occurs in lasers.
A defect of our treatment is that it does not predict spontaneous emission, the emission of a photon by a system not exposed to radiation, the system falling to a lower energy level in the process. Quantum field theory does predict spontaneous emission.
Note from (9.123) that the probability of absorption is proportional to $\left.\left|\left\langle\psi{m}^{0}\right| \sum{i} Q{i} x{i}\right| \psi{n}^{0}\right\rangle\left.\right|^{2}$. The quantity $\sum{i} Q{i} x{i}$ is the $x$ component of the system's electric-dipole-moment operator $\hat{\boldsymbol{\mu}}$ (see Section 14.2 for details), which is [Eqs. (14.14) and (14.15)]
$\hat{\boldsymbol{\mu}}=\mathbf{i} \sum{i} Q{i} x{i}+\mathbf{j} \sum{i} Q{i} y{i}+\mathbf{k} \sum{i} Q{i} z{i}=\mathbf{i} \hat{\mu}{x}+\mathbf{j} \hat{\mu}{y}+\mathbf{k} \hat{\mu}{z}$, where $\mathbf{i}, \mathbf{j}, \mathbf{k}$ are unit vectors along the axes and $\hat{\mu}{x}, \hat{\mu}{y}, \hat{\mu}_{z}$ are the components of $\hat{\boldsymbol{\mu}}$. We assumed polarized radiation with an electric field in the $x$ direction only. If the radiation has electric-field components in the $y$ and $z$ directions also, then the probability of absorption will be proportional to
\(
\left.\left.\left.\left.\left|\left\langle\psi{m}^{0}\right| \hat{\mu}{x}\right| \psi{n}^{0}\right\rangle\left.\right|^{2}+\left|\left\langle\psi{m}^{0}\right| \hat{\mu}{y}\right| \psi{n}^{0}\right\rangle\left.\right|^{2}+\left|\left\langle\psi{m}^{0}\right| \hat{\mu}{z}\right| \psi{n}^{0}\right\rangle\left.\right|^{2}=\left|\left\langle\psi{m}^{0}\right| \hat{\boldsymbol{\mu}}\right| \psi_{n}^{0}\right\rangle\left.\right|^{2}
\)
where Eq. (5.25) was used. The integral $\left\langle\psi{m}^{0}\right| \hat{\boldsymbol{\mu}}\left|\psi{n}^{0}\right\rangle=\boldsymbol{\mu}_{m n}$ is the transition (dipole) moment.
When $\boldsymbol{\mu}{m n}=0$, the transition between states $m$ and $n$ with absorption or emission of radiation is said to be forbidden. Allowed transitions have $\boldsymbol{\mu}{m n} \neq 0$. Because of approximations made in the derivation of (9.123), forbidden transitions may have some small probability of occurring.
Consider, for example, the particle in a one-dimensional box (Section 2.2). The transition dipole moment is $\left\langle\psi{m}^{0}\right| Q x\left|\psi{n}^{0}\right\rangle$, where $Q$ is the particle's charge and $x$ is its coordinate and where $\psi{m}^{0}=(2 / l)^{1 / 2} \sin (m \pi x / l)$ and $\psi{n}^{0}=(2 / l)^{1 / 2} \sin (n \pi x / l)$. Evaluation of this integral (Prob. 9.28) shows it is nonzero only when $m-n= \pm 1, \pm 3, \pm 5, \ldots$. and is zero when $m-n=0, \pm 2, \ldots$ The selection rule for a charged particle in a one-dimensional box is that the quantum number must change by an odd integer when radiation is absorbed or emitted.
Evaluation of the transition moment for the harmonic oscillator and for the twoparticle rigid rotor gives the selection rules $\Delta v= \pm 1$ and $\Delta J= \pm 1$ stated in Sections 4.3 and 6.4 .
The quantity $\left|b{m}\right|^{2}$ in (9.123) is sharply peaked at $\omega=\omega{m n}$ and $\omega=-\omega{m n}$, but there is a nonzero probability that a transition will occur when $\omega$ is not precisely equal to $\left|\omega{m n}\right|$, that is, when $h \nu$ is not precisely equal to $\left|E{m}^{0}-E{n}^{0}\right|$. This fact is related to the energy-time uncertainty relation (5.15). States with a finite lifetime have an uncertainty in their energy.
Radiation is not the only time-dependent perturbation that produces transitions between states. When an atom or molecule comes close to another atom or molecule, it suffers a time-dependent perturbation that can change its state. Selection rules derived for radiative transitions need not apply to collision processes, since $\hat{H}^{\prime}(t)$ differs for the two processes.