P4 Interaction with Radiation

These problems explore electric-dipole transitions, selection rules, Einstein rate equations, and spectral linewidths.

Electric-Dipole Transitions

Problem P4.1: The \(2p\rightarrow1s\) Transition in Hydrogen

A hydrogen atom is in the excited \(2p\) state with wavefunction

\[ \langle\mathbf r|2,1,0\rangle=\psi_{210}(\mathbf r) =\frac{1}{4\sqrt{2\pi a_0^3}}\frac{r}{a_0}e^{-r/(2a_0)}\cos\vartheta. \]

The ground-state wavefunction is

\[ \langle\mathbf r|1,0,0\rangle=\psi_{100}(\mathbf r) =\frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0}. \]

Tasks

  1. Before evaluating any integrals, use symmetry to determine which Cartesian components of the electric-dipole transition matrix element

    \[ \langle1,0,0|\mathbf r|2,1,0\rangle \]

    can be nonzero.

  2. Calculate the complete transition matrix element as a three-dimensional vector.

  3. The Einstein coefficient for spontaneous decay is

    \[ A_{21}=\frac43\alpha c\left(\frac{\omega}{c}\right)^3 \left|\langle1,0,0|\mathbf r|2,1,0\rangle\right|^2. \]

    Calculate the lifetime \(\tau=1/A_{21}\) of the excited \(2p\) state in seconds.

You may use

\[ E_n=-\frac{13.6\,\mathrm{eV}}{n^2},\qquad \hbar\omega=E_2-E_1, \]

and

\[ \alpha\simeq\frac1{137},\quad a_0=5.292\times10^{-11}\,\mathrm m,\quad \hbar=6.582\times10^{-16}\,\mathrm{eV\,s}. \]

The \(1s\) state is spherically symmetric, while the \(2p,m=0\) wavefunction is proportional to \(\cos\vartheta=z/r\). Examine the parity of the integrands for the \(x\), \(y\), and \(z\) components.

For the nonzero component, use \(z=r\cos\vartheta\), \(d^3r=r^2\sin\vartheta\,dr\,d\vartheta\,d\varphi\), and

\[ \int_0^\infty r^n e^{-\beta r}\,dr=\frac{n!}{\beta^{n+1}}. \]

1. Symmetry

The product \(\psi_{100}^*\psi_{210}\) is proportional to \(\cos\vartheta=z/r\). Multiplication by \(x\) or \(y\) produces an integrand that is odd under \(x\to-x\) or \(y\to-y\), respectively. Both integrals vanish. Multiplication by \(z\) gives an even integrand, so only the \(z\) component can be nonzero.

2. Transition matrix element

The nonzero component is

\[ \begin{aligned} \langle100|z|210\rangle &=\int\psi_{100}^*(\mathbf r)\,r\cos\vartheta\,\psi_{210}(\mathbf r)\,d^3r\\ &=\frac{1}{4\sqrt2\pi a_0^4} \left(\int_0^\infty r^4e^{-3r/(2a_0)}\,dr\right) \left(\int_0^{2\pi}d\varphi\int_0^\pi\cos^2\vartheta\sin\vartheta\,d\vartheta\right). \end{aligned} \]

The angular integral is \(4\pi/3\), and the radial integral is

\[ \int_0^\infty r^4e^{-3r/(2a_0)}\,dr =4!\left(\frac{2a_0}{3}\right)^5. \]

Thus,

\[ \boxed{\langle100|\mathbf r|210\rangle =\frac{256}{243\sqrt2}a_0\,\hat{\mathbf z}.} \]

Numerically, its magnitude is approximately \(3.94\times10^{-11}\,\mathrm m\).

3. Spontaneous decay and lifetime

The transition energy is

\[ \hbar\omega=E_2-E_1=-3.4\,\mathrm{eV}-(-13.6\,\mathrm{eV}) =10.2\,\mathrm{eV}, \]

so

\[ \omega=\frac{10.2\,\mathrm{eV}}{6.582\times10^{-16}\,\mathrm{eV\,s}} \approx1.55\times10^{16}\,\mathrm{s^{-1}}. \]

Substituting into the Einstein coefficient,

\[ \begin{aligned} A_{21} &=\frac43\alpha\frac{\omega^3}{c^2} \left(\frac{256}{243\sqrt2}a_0\right)^2\\ &\approx6.27\times10^8\,\mathrm{s^{-1}}. \end{aligned} \]

Therefore,

\[ \boxed{\tau=\frac1{A_{21}}\approx1.60\times10^{-9}\,\mathrm s=1.60\,\mathrm{ns}.} \]

Problem P4.2: Electric-Dipole Selection Rules

For electric-dipole (E1) transitions in LS coupling, the relevant selection rules are

\[ \Delta L=\pm1,\qquad \Delta S=0,\qquad \Delta J=0,\pm1, \]

with transitions between two states with \(J=0\) forbidden.

Tasks

Determine which of the following transitions are electric-dipole allowed. If a transition is forbidden, identify the violated selection rule or rules.

  1. \(4\,{}^2P_{3/2}\rightarrow1\,{}^2S_{1/2}\)
  2. \(3\,{}^2D_{3/2}\rightarrow2\,{}^2S_{1/2}\)
  3. \(4\,{}^2F_{5/2}\rightarrow3\,{}^2D_{3/2}\)
  4. \(5\,{}^4S_{1/2}\rightarrow3\,{}^4P_{5/2}\)
  5. \(2\,{}^2P_{1/2}\rightarrow1\,{}^2S_{1/2}\)
  6. \(2\,{}^3P_0\rightarrow1\,{}^1S_0\)

One of the term symbols listed above is itself not physically possible. Which one is it? Explain why its values of \(L\), \(S\), and \(J\) are inconsistent.

The letter \(S,P,D,F\) specifies \(L=0,1,2,3\), respectively. The superscript \(2S+1\) gives the spin multiplicity. For a given \(L\) and \(S\), the allowed values of \(J\) run from \(|L-S|\) to \(L+S\) in integer steps.

Transition E1 allowed? Reason
\(4\,{}^2P_{3/2}\to1\,{}^2S_{1/2}\) Yes \(\Delta L=-1\), \(\Delta S=0\), \(\Delta J=-1\)
\(3\,{}^2D_{3/2}\to2\,{}^2S_{1/2}\) No \(\Delta L=-2\)
\(4\,{}^2F_{5/2}\to3\,{}^2D_{3/2}\) Yes \(\Delta L=-1\), \(\Delta S=0\), \(\Delta J=-1\)
\(5\,{}^4S_{1/2}\to3\,{}^4P_{5/2}\) Invalid term symbol \({}^4S_{1/2}\) is impossible; the displayed \(\Delta J=2\) also violates the E1 rule
\(2\,{}^2P_{1/2}\to1\,{}^2S_{1/2}\) Yes \(\Delta L=-1\), \(\Delta S=0\), \(\Delta J=0\)
\(2\,{}^3P_0\to1\,{}^1S_0\) No \(\Delta S\ne0\) and \(J=0\to J=0\) is forbidden

The impossible term symbol is

\[ \boxed{{}^4S_{1/2}.} \]

Its multiplicity \(4\) implies \(S=3/2\), while the letter \(S\) implies \(L=0\). Angular-momentum addition then requires

\[ J=|L-S|=\frac32, \]

not \(J=1/2\). The only possible \({}^4S\) term has \(J=3/2\).

Einstein Rate Equations

Problem P4.3: Rate Equations for Hydrogen Populations

Consider a sample of \(N\) hydrogen atoms. Let \(N_1\) denote the number in the \(1^2S_{1/2}\) ground state and \(N_2\) the number in the \(2^2P_{3/2}\) excited state.

The populations evolve according to

\[ \frac{dN_1}{dt}=-N_1B_{12}w(\nu) +N_2\left[B_{21}w(\nu)+A_{21}\right], \]

\[ \frac{dN_2}{dt}=-\frac{dN_1}{dt}. \]

Here \(w(\nu)\) is the spectral energy density of the radiation field, while \(B_{12}\), \(B_{21}\), and \(A_{21}\) are the Einstein coefficients for absorption, stimulated emission, and spontaneous emission. The total number of atoms is conserved:

\[ N_1(t)+N_2(t)=N. \]

Tasks

  1. Suppose all atoms are initially excited, \(N_2(0)=N\), and the radiation field is absent, \(w(\nu)=0\). Derive \(N_2(t)\).

  2. Suppose again that \(N_2(0)=N\), but the atoms are exposed to a radiation field with spectral energy density

    \[ w(\nu)=\frac{4\pi h\nu^3}{5c^3}. \]

    Determine the steady-state excited population \(N_2(\infty)\).

  3. Show that the time-dependent solution in the presence of this field is

    \[ N_2(t)=\frac N{13}\left(11e^{-\frac{13}{10}A_{21}t}+2\right). \]

    Check the initial population at \(t=0\) and the steady-state limit as \(t\to\infty\).

The Einstein coefficients obey

\[ g_1B_{12}=g_2B_{21},\qquad \frac{A_{21}}{B_{21}}=\frac{8\pi h\nu^3}{c^3}. \]

For \(1^2S_{1/2}\), \(g_1=2\); for \(2^2P_{3/2}\), \(g_2=4\).

Eliminate \(N_1=N-N_2\) to obtain a first-order differential equation of the form

\[ \frac{dN_2}{dt}=R-\Gamma N_2. \]

1. Spontaneous decay without radiation

With \(w(\nu)=0\),

\[ \frac{dN_2}{dt}=-A_{21}N_2. \]

Integrating and applying \(N_2(0)=N\) gives

\[ \boxed{N_2(t)=Ne^{-A_{21}t}.} \]

2. Steady-state population in the radiation field

From the Einstein relations and the degeneracies,

\[ B_{12}=\frac{g_2}{g_1}B_{21}=2B_{21}. \]

The specified radiation field gives

\[ B_{21}w(\nu) =B_{21}\frac{4\pi h\nu^3}{5c^3} =\frac{A_{21}}{10}, \]

and hence

\[ B_{12}w(\nu)=\frac{A_{21}}5. \]

Using \(N_1=N-N_2\), the excited-state rate equation becomes

\[ \begin{aligned} \frac{dN_2}{dt} &=(N-N_2)\frac{A_{21}}5 -N_2\left(\frac{A_{21}}{10}+A_{21}\right)\\ &=\frac{A_{21}N}{5}-\frac{13A_{21}}{10}N_2. \end{aligned} \]

At steady state, \(dN_2/dt=0\), so

\[ \boxed{N_2(\infty)=\frac{2N}{13}.} \]

3. Time-dependent solution

Rearrange the rate equation as

\[ \frac{dN_2}{dt} =-\frac{13A_{21}}{10}\left(N_2-\frac{2N}{13}\right). \]

Its solution is

\[ N_2(t)=\frac{2N}{13}+C e^{-13A_{21}t/10}. \]

The initial condition \(N_2(0)=N\) gives \(C=11N/13\). Therefore,

\[ \boxed{ N_2(t)=\frac N{13}\left(11e^{-\frac{13}{10}A_{21}t}+2\right). } \]

At \(t=0\), the exponential is \(1\) and \(N_2(0)=N(11+2)/13=N\). As \(t\to\infty\), the exponential vanishes and \(N_2\to2N/13\), as required.

Physical interpretation: Rate equations describe atomic populations. Coherent atom–light dynamics, including Rabi oscillations, require a treatment of coherences, such as the optical Bloch equations.

Spectral Linewidths

Problem P4.4: Lithium Spectroscopy

You perform fluorescence spectroscopy on the lithium transition

\[ 2\,{}^2P_{3/2}\rightarrow2\,{}^2S_{1/2}. \]

The excited state has a lifetime of \(\tau=27.1\,\mathrm{ns}\).

Tasks

  1. What is the natural linewidth \(\delta\omega_{\mathrm{nat}}\)? Use

    \[ \delta\omega_{\mathrm{nat}}=\frac1\tau. \]

  2. The lithium atoms are evaporated from an oven at \(400^\circ\mathrm C\). What is the Doppler broadening \(\delta\omega_D\) of the observed line? Use the Doppler full width at half maximum (FWHM),

    \[ \delta\omega_D=\omega_0\sqrt{\frac{8k_BT\ln2}{mc^2}}. \]

  3. Compare the Doppler linewidth with the natural linewidth at \(400^\circ\mathrm C\). Which mechanism dominates, and by approximately what factor?

  4. At which temperature are Doppler broadening and natural linewidth of the same magnitude?

Use

\[ \omega_0=2.81\times10^{15}\,\mathrm{s^{-1}},\quad m\simeq7u,\quad u=\text{(not needed)},\quad u_0=\frac{\omega_0}{2\pi}, \]

where \(u=1.6605\times10^{-27}\,\mathrm{kg}\). You may also use \(k_B=1.38065\times10^{-23}\,\mathrm{J/K}\) and \(c=2.99792\times10^8\,\mathrm{m/s}\).

Convert the oven temperature to kelvin. Both linewidths are expressed as angular frequencies, in \(\mathrm{rad/s}\) (or equivalently \(\mathrm{s^{-1}}\)). To find the crossover temperature, set \(\delta\omega_D=\delta\omega_{\mathrm{nat}}\) and solve for \(T\).

1. Natural linewidth

\[ \delta\omega_{\mathrm{nat}} =\frac1{27.1\times10^{-9}\,\mathrm s} \approx\boxed{3.69\times10^7\,\mathrm{s^{-1}}}. \]

In ordinary frequency units, this corresponds to \(\delta\nu_{\mathrm{nat}}=\delta\omega_{\mathrm{nat}}/(2\pi)\approx5.87\,\mathrm{MHz}\).

2. Doppler linewidth

The oven temperature and atomic mass are

\[ T=400+273.15=673.15\,\mathrm K, \]

\[ m=7(1.6605\times10^{-27})\approx1.1624\times10^{-26}\,\mathrm{kg}. \]

Substitution gives

\[ \begin{aligned} \delta\omega_D &=2.81\times10^{15} \sqrt{\frac{8(1.38065\times10^{-23})(673.15)\ln2} {(1.1624\times10^{-26})(2.99792\times10^8)^2}}\\ &\approx\boxed{1.97\times10^{10}\,\mathrm{s^{-1}}}. \end{aligned} \]

This is \(\delta\nu_D\approx3.14\,\mathrm{GHz}\) in ordinary frequency units.

3. Comparison

\[ \frac{\delta\omega_D}{\delta\omega_{\mathrm{nat}}} \approx\frac{1.97\times10^{10}}{3.69\times10^7} \approx\boxed{535}. \]

Doppler broadening dominates the observed line at \(400^\circ\mathrm C\) by more than two orders of magnitude.

4. Temperature at equal widths

Set the two linewidths equal:

\[ \frac1\tau=\omega_0\sqrt{\frac{8k_BT\ln2}{mc^2}}. \]

Squaring and solving for \(T\) gives

\[ \boxed{T=\frac{mc^2}{8k_B\ln2}\frac1{(\omega_0\tau)^2}.} \]

Numerically,

\[ \boxed{T\approx2.35\times10^{-3}\,\mathrm K=2.35\,\mathrm{mK}.} \]

This estimate shows why cooling atoms strongly suppresses Doppler broadening. The calculation assumes the stated thermal Doppler formula remains applicable and neglects other broadening mechanisms.

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