Fine Structure of the Hydrogen Atom
Where fine structure sits in the hierarchy of corrections to the Bohr energy, measured in powers of \(\alpha\).
The relativistic correction \(-p^{4}/8m^{3}c^{2}\) and its expectation value.
Spin–orbit coupling: the magnetic field in the electron's frame, the electron magnetic moment, and the Thomas factor of \(\tfrac12\).
The Darwin term, and why it matters only for \(s\) states.
The combined fine structure formula — which depends on \(n\) and \(j\) but not on \(\ell\).
Selection rules, the resulting line pattern of H\(_\alpha\), and the Lamb shift.
Summary: the Quantum Mechanical Hydrogen Atom
Everything in this chapter is a perturbation applied to the solutions of the non-relativistic Schrödinger equation for a one-electron atom. Those solutions are assumed known from quantum mechanics; this section collects the results that will be used, in the form in which they will be used.
The eigenvalue problem and its separation
For a single electron bound to a point nucleus by the Coulomb potential, \begin{equation} H\,\psi_{n\ell m}(r,\theta,\phi) = E_{n}\,\psi_{n\ell m}(r,\theta,\phi), \qquad \psi_{n\ell m}(r,\theta,\phi) = R_{n\ell}(r)\,Y_{\ell}^{m}(\theta,\phi). \label{eq:h-eigen} \end{equation} The potential is central, so the angular dependence separates off completely into the spherical harmonics \(Y_{\ell}^{m}\) and all the physics of the binding sits in the radial factor. Writing \(u(r) = r\,R_{n\ell}(r)\) turns the radial problem into a one-dimensional Schrödinger equation, \begin{equation} -\frac{\hbar^{2}}{2m}\frac{\dd^{2}u}{\dd r^{2}} + \left[\, -\frac{e^{2}}{4\pi\varepsilon_{0}}\frac{1}{r} + \frac{\hbar^{2}}{2m}\frac{\ell(\ell+1)}{r^{2}} \right] u = E\,u , \label{eq:radial-eq} \end{equation} in which the second bracketed term is the centrifugal barrier — the only place \(\ell\) enters. Demanding that \(u\) be normalisable quantizes the energy: \begin{equation} \boxed{\; E_{n} = -\left[\frac{m}{2\hbar^{2}} \left(\frac{e^{2}}{4\pi\varepsilon_{0}}\right)^{2}\right]\frac{1}{n^{2}} = \frac{E_{1}}{n^{2}}, \qquad n = 1,2,3,\dots \;} \label{eq:E-n-summary} \end{equation} with \(E_{1} = -13.6\,\mathrm{eV}\). The three quantum numbers run over \begin{equation} n = 1,2,3,\dots, \qquad \ell = 0,1,2,\dots,n-1, \qquad -\ell \le m \le \ell . \label{eq:qnum-ranges} \end{equation}
Note what \(\eqref{eq:E-n-summary}\) does not contain: \(\ell\) and \(m\). The Schrödinger energy depends on \(n\) alone, so each level carries an \(n^{2}\)-fold orbital degeneracy (\(2n^{2}\) with spin). Fine structure is precisely the story of how that accidental degeneracy is broken — and the whole of this chapter is about which quantum numbers it survives in.
The radial functions
Solving \(\eqref{eq:radial-eq}\) by the standard series method, with \(\rho = r/n\abohr\), gives \begin{equation} R_{n\ell}(r) = \frac{1}{r}\,\rho^{\,\ell+1}\,e^{-\rho}\,v(\rho), \label{eq:Rnl-form} \end{equation} where \(v(\rho)\) is a polynomial (an associated Laguerre polynomial) of degree \(n-\ell-1\). The structure of \(\eqref{eq:Rnl-form}\) is worth reading off directly: the factor \(\rho^{\ell+1}/r \sim r^{\ell}\) controls the behaviour at the origin, the exponential \(e^{-r/n\abohr}\) controls the tail, and the polynomial supplies exactly \(n-\ell-1\) radial nodes. The three lowest functions are
| State | Radial function |
|---|---|
| \(1s\) | \(R_{10} = 2\,a^{-3/2}\,\exp(-r/a)\) |
| [4pt] \(2s\) | \(R_{20} = \dfrac{1}{\sqrt{2}}\,a^{-3/2} \left(1 - \dfrac{1}{2}\dfrac{r}{a}\right)\exp(-r/2a)\) |
| [8pt] \(2p\) | \(R_{21} = \dfrac{1}{\sqrt{24}}\,a^{-3/2}\,\dfrac{r}{a}\,\exp(-r/2a)\) |
Radial expectation values
Every perturbation in this chapter reduces, after the angular integrals are done, to a radial expectation value \begin{equation} \big\langle r^{k}\big\rangle_{n\ell m} = \int_{0}^{\infty} \lvert R_{n\ell}(r)\rvert^{2}\, r^{k}\, r^{2}\,\dd r . \label{eq:rk-def} \end{equation} For a hydrogenic atom of nuclear charge \(Z\), writing \(a_{\mu}\) for the Bohr radius corrected for reduced mass, the standard results are \begin{align} \big\langle r \big\rangle_{n\ell m} &= a_{\mu}\,\frac{n^{2}}{Z} \left\{1 + \frac{1}{2}\left[1 - \frac{\ell(\ell+1)}{n^{2}}\right]\right\}, \label{eq:exp-r}\\[4pt] \big\langle r^{2} \big\rangle_{n\ell m} &= a_{\mu}^{2}\,\frac{n^{4}}{Z^{2}} \left\{1 + \frac{3}{2} \left[1 - \frac{\ell(\ell+1) - \tfrac13}{n^{2}}\right]\right\}, \label{eq:exp-r2} \end{align} and, for the negative powers that the perturbations actually need, \begin{equation} \boxed{\; \left\langle \frac{1}{r}\right\rangle_{n\ell m} = \frac{Z}{a_{\mu}n^{2}}, \qquad \left\langle \frac{1}{r^{2}}\right\rangle_{n\ell m} = \frac{Z^{2}}{a_{\mu}^{2}n^{3}\left(\ell+\tfrac12\right)}, \qquad \left\langle \frac{1}{r^{3}}\right\rangle_{n\ell m} = \frac{Z^{3}}{a_{\mu}^{3}n^{3}\,\ell\left(\ell+\tfrac12\right)(\ell+1)} . \;} \label{eq:exp-inverse} \end{equation}
The positive powers, \(\eqref{eq:exp-r}\) and \(\eqref{eq:exp-r2}\), are governed mainly by \(n\) — \(\langle r\rangle \sim n^{2}a_{\mu}/Z\) is the “size” of the atom, inversely proportional to \(Z\) as expected. The negative powers in \(\eqref{eq:exp-inverse}\) depend strongly on \(\ell\) as well, and it is these that control the fine structure. Observe that \(\langle 1/r^{3}\rangle\) carries a factor \(\ell\) in the denominator and therefore diverges as \(\ell \to 0\). That divergence is not a disaster: it is exactly cancelled by the vanishing of the \(\vec S\cdot\vec L\) matrix element for \(s\) states, and what is left over is supplied by the Darwin term. See 2.4.
Of the six formulas above, \(\eqref{eq:exp-inverse}\) is the set that is asked directly and used constantly. The mnemonic is the power of \(n\): \(1/r\) goes as \(n^{-2}\), while both \(1/r^{2}\) and \(1/r^{3}\) go as \(n^{-3}\), picking up one and two extra \(\ell\)-factors respectively.
What fine structure looks like: the \(2 \to 1\) doublet
Before the calculation, the observation. Examined at ordinary resolution, the hydrogen lines are single. Examined at very high resolution, they are not: each is a closely spaced multiplet. This splitting is what the term fine structure names, and it was among the first pieces of experimental evidence for electron spin.
Take the \(n=2 \to n=1\) transition, Lyman-\(\alpha\). The gross energy difference is \(10.2\,\mathrm{eV}\), corresponding to \(121.6\,\mathrm{nm}\) in the ultraviolet. The upper level, however, is not one level. The \(2p\) state has \(\ell = 1\), which combines with the electron spin \(s = \tfrac12\) to give two possible values of the total angular momentum, \(j = \tfrac32\) and \(j = \tfrac12\). These two have slightly different energies, and the single line is therefore a doublet, split by only \(4.5\times 10^{-5}\,\mathrm{eV}\) — about one part in \(2\times10^{5}\) of the line energy.
The splitting is attributed to an interaction between the electron spin \(\vec S\) and the orbital angular momentum \(\vec L\) — the spin–orbit interaction of 2.3. In the electron's rest frame the proton circulates around it and produces a magnetic field \(\vec B\); the electron's spin magnetic moment then has energy \(-\vec\mu\cdot\vec B\), which is positive or negative according to the relative orientation. That is the whole mechanism, and the rest of this chapter is the business of getting the size right.
2.1 The Hierarchy of Corrections
The term fine structure describes the splitting of the spectral lines of hydrogen arising from two effects:
a relativistic correction to the non-relativistic Schrödinger equation, and
spin–orbit coupling.
Both are tiny perturbations, smaller than the gross Bohr structure by a factor \(\alpha^{2} \approx (1/137)^{2} \approx 5.3\times 10^{-5}\), where \begin{equation} \alpha = \frac{1}{4\pi\varepsilon_{0}}\frac{e^{2}}{\hbar c} \approx \frac{1}{137} \label{eq:alpha2} \end{equation} is the fine structure constant introduced in Chapter 1.
Measured against the electron rest mass energy \(mc^{2} = 511\,\mathrm{keV}\), the successive corrections to the hydrogen levels form a clean hierarchy in powers of \(\alpha\):
| Contribution | Order | Magnitude (\(n=2\)) | Origin |
|---|---|---|---|
| Bohr energy | \(\alpha^{2}mc^{2}\) | \(\sim10\,\mathrm{eV}\) | Coulomb binding |
| Fine structure | \(\alpha^{4}mc^{2}\) | \(\sim-4\,\mathrm{eV}\) | relativity \(+\) spin–orbit |
| Lamb shift | \(\alpha^{5}mc^{2}\) | \(\sim-6\,\mathrm{eV}\) | QED radiative correction |
| Hyperfine structure | \(\dfrac{m}{M}\alpha^{4}mc^{2}\) | \(\sim-7\,\mathrm{eV}\) | nuclear spin coupling |
The powers of \(\alpha\) in Table 2.2 are asked directly, and they are also the fastest way to sanity-check any numerical answer in this chapter. Fine structure is \(\alpha^{2}\) times the gross structure; the Lamb shift is another factor of \(\alpha\) down; hyperfine is suppressed by the mass ratio \(m/M \sim 1/1836\) instead.
Express the Bohr energy of hydrogen in terms of the fine structure constant and the electron rest mass energy.
Solution. From Chapter 1, \[ E_{n} = -\frac{m}{2\hbar^{2}}\left(\frac{e^{2}}{4\pi\varepsilon_{0}}\right)^{2} \frac{1}{n^{2}} . \] Insert \(e^{2}/4\pi\varepsilon_{0} = \alpha\hbar c\) from \(\eqref{eq:alpha2}\): \[ E_{n} = -\frac{m}{2\hbar^{2}}\,\alpha^{2}\hbar^{2}c^{2}\,\frac{1}{n^{2}} , \] so \begin{equation} \boxed{\;E_{n} = -\frac{1}{2}\,\frac{\alpha^{2}mc^{2}}{n^{2}}\;} \label{eq:bohr-alpha} \end{equation} Numerically, \(\tfrac12\alpha^{2}mc^{2} = \tfrac12 \times 5.33\times 10^{-5} \times 511\,\mathrm{keV} = 13.6\,\mathrm{eV}\), as it must be. This form makes the \(\alpha^{2}mc^{2}\) entry of Table 2.2 explicit.
2.2 Relativistic Correction to the Kinetic Energy
The non-relativistic Hamiltonian of the hydrogen atom is \begin{equation} H_{0} = \frac{p^{2}}{2m} - \frac{1}{4\pi\varepsilon_{0}}\frac{e^{2}}{r} , \label{eq:H0} \end{equation} in which the first term is the classical kinetic energy.
But the electron's speed is not small: from Chapter 1, \(v_{1} = \alpha c \approx c/137\), and for a hydrogenic ion \(v \sim Z\alpha c\). So the correct kinetic energy is the relativistic one, \begin{equation} T = \sqrt{p^{2}c^{2} + m^{2}c^{4}} - mc^{2} . \label{eq:T-rel} \end{equation} Expanding for \(pc \ll mc^{2}\), \begin{equation} T = mc^{2}\left[\left(1+\frac{p^{2}}{m^{2}c^{2}}\right)^{1/2} - 1\right] = \frac{p^{2}}{2m} - \frac{p^{4}}{8m^{3}c^{2}} + \frac{p^{6}}{16m^{5}c^{4}} - \cdots \label{eq:T-expansion} \end{equation} The leading term is the classical kinetic energy; the first correction is the perturbation we want. Writing \(H = H_{0} + H'_{r}\), \begin{equation} \boxed{\;H'_{r} = -\frac{p^{4}}{8m^{3}c^{2}}\;} \label{eq:H-rel} \end{equation}
The correction is negative — relativity lowers the kinetic energy relative to the classical estimate, hence binds the electron slightly more tightly. Since \(v \propto Z\), this term grows as \(Z^{4}\) and becomes very important for heavy hydrogenic ions, where the electron's kinetic energy can approach its rest mass energy.
2.2.1 First-order energy shift
By first-order perturbation theory, \begin{equation} E_{r}^{(1)} = \big\langle \psi_{n\ell m} \big| H'_{r} \big| \psi_{n\ell m}\big\rangle = -\frac{1}{8m^{3}c^{2}}\big\langle p^{4}\big\rangle . \label{eq:Er-def} \end{equation} Evaluating \(\langle p^{4}\rangle\) directly is painful. The trick is to use the unperturbed equation: since \(\dfrac{p^{2}}{2m} = H_{0} - V\), \begin{equation} \big\langle p^{4}\big\rangle = 4m^{2}\big\langle (H_{0}-V)^{2}\big\rangle = 4m^{2}\!\left[E_{n}^{2} - 2E_{n}\langle V\rangle + \langle V^{2}\rangle\right]. \label{eq:p4-trick} \end{equation} With \(V = -\dfrac{1}{4\pi\varepsilon_{0}}\dfrac{e^{2}}{r}\) this needs only two standard expectation values: \begin{equation} \left\langle \frac{1}{r}\right\rangle = \frac{1}{n^{2}\abohr}, \qquad \left\langle \frac{1}{r^{2}}\right\rangle = \frac{1}{\left(\ell+\tfrac12\right)n^{3}\abohr^{2}} . \label{eq:expectation-r} \end{equation} Substituting and simplifying gives the standard result \begin{equation} \boxed{\; E_{r}^{(1)} = -\frac{E_{n}^{2}}{2mc^{2}} \left[\frac{4n}{\ell+\tfrac12} - 3\right]\;} \label{eq:Er-result} \end{equation}
Memorize \(\eqref{eq:expectation-r}\) along with \(\left\langle 1/r^{3}\right\rangle\) from \(\eqref{eq:expectation-r3}\) below. These three expectation values are all you need to reproduce the entire fine structure calculation, and they are themselves asked as standalone questions.
Find the lowest-order relativistic correction to the energy levels of a one-dimensional harmonic oscillator.
Solution. The perturbation is again \(H'_{r} = -p^{4}/8m^{3}c^{2}\), so \[ \Delta E_{n} = -\frac{\langle p^{4}\rangle}{8m^{3}c^{2}} . \] For the oscillator eigenstates, \[ \big\langle p^{2}\big\rangle = m\hbar\omega\left(n+\tfrac12\right), \qquad \big\langle p^{4}\big\rangle = \frac{3}{4}\left(m\hbar\omega\right)^{2}\!\left(2n^{2}+2n+1\right). \] Hence \begin{equation} \Delta E_{n} = -\frac{3\hbar^{2}\omega^{2}}{32\,mc^{2}}\left(2n^{2}+2n+1\right). \label{eq:sho-rel} \end{equation} For the ground state \(n=0\) this reduces to \[ \Delta E_{0} = -\frac{3\hbar^{2}\omega^{2}}{32\,mc^{2}} = -\frac{3}{8}\,\frac{\left(\tfrac12\hbar\omega\right)^{2}}{mc^{2}} . \]
Note the structure: the correction is of order \((\text{unperturbed energy})^{2}/mc^{2}\) — exactly as in \(\eqref{eq:Er-result}\) for hydrogen. That is the generic signature of the \(p^{4}\) term and a good way to check any answer of this type.
2.3 Spin–Orbit Coupling
From the electron's point of view it is the proton that circulates around it. That moving positive charge sets up a magnetic field at the electron's location, which exerts a torque on the spinning electron, tending to align its magnetic moment with the field. The interaction energy is \begin{equation} H_{\text{so}} = -\,\vec\mu_{s}\cdot\vec B . \label{eq:H-so-def} \end{equation}
2.3.1 The magnetic field of the proton in the electron's frame
Treating the circulating proton as a current loop of radius \(r\) and current \(i = e/T\), the Biot–Savart law gives a field at the centre of magnitude \begin{equation} B = \frac{\mu_{0} i}{2r} = \frac{\mu_{0} e}{2rT} = \frac{\mu_{0} e\, v}{4\pi r^{2}} . \label{eq:B-loop} \end{equation} The orbital angular momentum of the relative motion is \(L = m v r\), so \(v = L/mr\), and using \(\mu_{0} = 1/\varepsilon_{0}c^{2}\), \begin{equation} \boxed{\;\vec B = \frac{1}{4\pi\varepsilon_{0}}\frac{e}{mc^{2}r^{3}}\,\vec L\;} \label{eq:B-field} \end{equation} The field is parallel to \(\vec L\), as Figure 2.3 shows.
2.3.2 The magnetic moment of the electron
The magnetic dipole moment of a spinning charge is proportional to its angular momentum; the constant of proportionality is the gyromagnetic ratio. Consider charge \(q\) smeared uniformly around a ring of radius \(r\) rotating with period \(T\). Its dipole moment is current times area: \begin{equation} \mu = i\,A = \frac{q}{T}\,\pi r^{2} = \frac{q\,\omega r^{2}}{2} , \qquad \omega = \frac{2\pi}{T} . \label{eq:ring-moment} \end{equation} If the ring carries mass \(m\), its angular momentum is \(S = I\omega = mr^{2}\omega\), so \begin{equation} \frac{\mu}{S} = \frac{q}{2m} \quad\Longrightarrow\quad \vec\mu = \frac{q}{2m}\,\vec S . \label{eq:gyromagnetic} \end{equation}
For the electron (\(q = -e\)) the true result carries an extra factor \(g_{s}\), the Landé \(g\) factor, which for electron spin is \(g_{s} \approx 2\): \begin{equation} \vec\mu_{s} = -\,g_{s}\frac{e}{2m}\,\vec S = -\frac{e}{m}\,\vec S . \label{eq:mu-spin} \end{equation}
\(g_{s}=2\) is a prediction of the Dirac equation, not something derivable from the classical ring picture — the classical model gives \(g=1\). QED corrects it to \(g_{s} = 2.00232\). The factor of \(2\) is essential to getting the fine structure right.
2.3.3 The Thomas factor
Substituting \(\eqref{eq:B-field}\) and \(\eqref{eq:mu-spin}\) into \(\eqref{eq:H-so-def}\) gives \[ H_{\text{so}} = \frac{1}{4\pi\varepsilon_{0}}\frac{e^{2}}{m^{2}c^{2}r^{3}}\, \vec S\cdot\vec L . \] But this is too large by a factor of two. The calculation above treats the electron's frame as inertial, whereas it is in fact accelerating; the correct relativistic kinematics introduces the Thomas precession factor \(\tfrac12\). The corrected perturbation is \begin{equation} \boxed{\; H'_{\text{so}} = \frac{1}{8\pi\varepsilon_{0}} \frac{e^{2}}{m^{2}c^{2}}\,\frac{\vec S\cdot\vec L}{r^{3}}\;} \label{eq:H-so} \end{equation}
2.3.4 The coupling term \(\vec S\cdot\vec L\)
Because \(H'_{\text{so}}\) contains \(\vec S\cdot\vec L\), neither \(\vec L\) nor \(\vec S\) is separately conserved — they exert torques on one another. What is conserved is the total angular momentum \begin{equation} \vec J = \vec L + \vec S . \label{eq:J-def} \end{equation} Squaring, \begin{equation} J^{2} = L^{2} + S^{2} + 2\,\vec L\cdot\vec S \quad\Longrightarrow\quad \vec S\cdot\vec L = \tfrac12\left(J^{2} - L^{2} - S^{2}\right), \label{eq:SL-identity} \end{equation} so the eigenvalues are \begin{equation} \big\langle \vec S\cdot\vec L\big\rangle = \frac{\hbar^{2}}{2}\left[\,j(j+1) - \ell(\ell+1) - \tfrac34\,\right], \label{eq:SL-eigen} \end{equation} using \(s=\tfrac12\) so that \(S^{2} \to \tfrac34\hbar^{2}\).
Equation \(\eqref{eq:SL-identity}\) is the single most reused identity in Chapters 2–4. Any question of the form “the splitting between \(^{2}P_{3/2}\) and \(^{2}P_{1/2}\) for \(H = a\,\vec L\cdot\vec S\)” is one line of \(\eqref{eq:SL-eigen}\).
2.3.5 First-order energy shift
The remaining radial expectation value is \begin{equation} \left\langle \frac{1}{r^{3}}\right\rangle = \frac{1}{\ell\left(\ell+\tfrac12\right)(\ell+1)\,n^{3}\abohr^{3}} . \label{eq:expectation-r3} \end{equation} Combining \(\eqref{eq:H-so}\), \(\eqref{eq:SL-eigen}\) and \(\eqref{eq:expectation-r3}\): \begin{equation} \boxed{\; E_{\text{so}}^{(1)} = \frac{E_{n}^{2}}{mc^{2}}\, \frac{n\left[\,j(j+1)-\ell(\ell+1)-\tfrac34\,\right]} {\ell\left(\ell+\tfrac12\right)(\ell+1)}\;} \label{eq:Eso-result} \end{equation}
The good quantum numbers for the perturbed problem are therefore \(n\), \(\ell\), \(s\), \(j\) and \(m_{j}\) — not \(m_{\ell}\) and \(m_{s}\) separately.
The spin–orbit effect splits the \(2P \to 2S\) transition of lithium (\(\lambda = 6521\,\mathrm{Å}\)) into two lines separated by \(\Delta\lambda = 0.14\,\mathrm{Å}\). Find the corresponding energy difference between the two lines, in eV.
Solution. From \(E = hc/\lambda\), \[ \left\lvert\Delta E\right\rvert = \frac{hc}{\lambda^{2}}\,\Delta\lambda . \] With \(hc = 1.24\times 10^{-6}\,\mathrm{eV}\,\mathrm{m}\): \[ \left\lvert\Delta E\right\rvert = \frac{1.24\times10^{-6}}{\left(6521\times10^{-10}\right)^{2}} \times 0.14\times10^{-10} = 4.08\times 10^{-5}\,\mathrm{eV} . \] So \(\Delta E \approx 4.08\times 10^{-5}\,\eV\), i.e.\ about \(0.33\,\mathrm{cm}^{-1}\).
The relation \(\lvert\Delta E\rvert = (hc/\lambda^{2})\,\Delta\lambda\) converts a measured wavelength splitting into an energy splitting, and appears in this chapter, Chapter 4 and Chapter 9. Note the \(\lambda^{2}\): the same \(\Delta\lambda\) corresponds to a much smaller \(\Delta E\) in the infrared than in the ultraviolet.
2.4 The Darwin Term
There is a third contribution at the same order in \(\alpha\), whose full explanation lies in QED — vacuum fluctuations and the quantization of the electromagnetic field.
At relativistic velocities the electron does not move smoothly but undergoes extremely rapid small-scale fluctuations in position, a phenomenon called zitterbewegung (“trembling motion”). The amplitude of this trembling is of the order of the Compton wavelength, \begin{equation} \lambda_{C} = \frac{\hbar}{mc} = 3.86\times 10^{-13}\,\mathrm{m} . \label{eq:compton} \end{equation} The consequence is that the electron does not see the sharp Coulomb potential \(V(r)\) of the nucleus, but a version smeared over a distance \(\sim\lambda_{C}\). Expanding \(\langle V\rangle\) about \(r\) and keeping the leading term gives the Darwin correction \begin{equation} E_{D} = \frac{\hbar^{2}}{8m^{2}c^{2}}\,\nabla^{2}V . \label{eq:H-darwin} \end{equation}
2.4.1 Evaluating the Darwin shift
The evaluation is short, and every step of it is worth following because the same \(\delta\)-function manoeuvre recurs in hyperfine structure.
2.4.2 Step 1: the Laplacian of the Coulomb potential.
With \[ V = -\frac{e^{2}}{4\pi\varepsilon_{0}}\,\frac{1}{r} \qquad\Longrightarrow\qquad \nabla^{2}V = -\frac{e^{2}}{4\pi\varepsilon_{0}}\,\nabla^{2}\!\left(\frac1r\right). \] The standard identity of electrostatics is \begin{equation} \nabla^{2}\!\left(\frac{1}{r}\right) = -4\pi\,\delta^{3}(\vec r), \label{eq:laplacian-identity} \end{equation} so that \begin{equation} \nabla^{2}V = +\frac{e^{2}}{4\pi\varepsilon_{0}}\cdot 4\pi\,\delta^{3}(\vec r). \label{eq:laplacian-V} \end{equation}
2.4.3 Step 2: substitute into \(\eqref{eq:H-darwin}\).
\[ E_{D} = \frac{\hbar^{2}}{8m^{2}c^{2}}\cdot \frac{e^{2}}{4\pi\varepsilon_{0}}\cdot 4\pi\,\delta^{3}(\vec r) \qquad\Longrightarrow\qquad E_{D} = \frac{\pi\hbar^{2}}{2m^{2}c^{2}}\, \frac{e^{2}}{4\pi\varepsilon_{0}}\, \big\langle \delta^{3}(\vec r)\big\rangle , \] using \(4\pi/8 = \pi/2\).
2.4.4 Step 3: the \(\delta\)-function expectation value.
A \(\delta\)-function simply samples the wavefunction at the origin: \[ \big\langle \delta^{3}(\vec r)\big\rangle = \int \lvert\psi_{n\ell m}\rvert^{2}\,\delta^{3}(\vec r)\,\dd^{3}r = \big\lvert\psi_{n\ell m}(0)\big\rvert^{2}, \] so \begin{equation} E_{D} = \frac{\pi\hbar^{2}}{2m^{2}c^{2}}\, \frac{e^{2}}{4\pi\varepsilon_{0}}\, \big\lvert\psi_{n\ell m}(0)\big\rvert^{2}. \label{eq:E-darwin} \end{equation}
2.4.5 Step 4: only \(s\) states survive.
From \(\eqref{eq:Rnl-form}\), \(R_{n\ell}(r) \propto r^{\ell}\) near the origin, so \[ R_{n\ell}(0) = 0 \quad\text{if}\quad \ell \ne 0 . \] Only the spherically symmetric \(\ell = 0\) states have non-vanishing amplitude at the nucleus, and for those \[ \big\lvert\psi_{n00}(0)\big\rvert^{2} = \frac{1}{\pi n^{3}\abohr^{3}} . \] Therefore \begin{equation} E_{D} = \frac{\pi\hbar^{2}}{2m^{2}c^{2}} \left(\frac{e^{2}}{4\pi\varepsilon_{0}}\right) \frac{1}{\pi n^{3}\abohr^{3}} = \frac{\hbar^{2}}{2m^{2}c^{2}} \left(\frac{e^{2}}{4\pi\varepsilon_{0}}\right) \frac{1}{n^{3}\abohr^{3}} . \label{eq:E-darwin-s} \end{equation}
2.4.6 Step 5: eliminate \(\abohr\).
Insert \(\abohr = 4\pi\varepsilon_{0}\hbar^{2}/me^{2}\), i.e.\ \(\abohr^{-3} = \left(me^{2}/4\pi\varepsilon_{0}\hbar^{2}\right)^{3}\): \[ E_{D} = \frac{\hbar^{2}}{2m^{2}c^{2}} \left(\frac{e^{2}}{4\pi\varepsilon_{0}}\right) \frac{1}{n^{3}} \left(\frac{m}{\hbar^{2}}\right)^{3} \left(\frac{e^{2}}{4\pi\varepsilon_{0}}\right)^{3} = \frac{m}{2c^{2}\hbar^{4}}\,\frac{1}{n^{3}} \left(\frac{e^{2}}{4\pi\varepsilon_{0}}\right)^{4} . \] Now recognise the Bohr energy \(\eqref{eq:E-n-summary}\): squaring it gives \[ E_{n}^{2} = \frac{m^{2}}{4\hbar^{4}} \left(\frac{e^{2}}{4\pi\varepsilon_{0}}\right)^{4}\frac{1}{n^{4}} \qquad\Longrightarrow\qquad \left(\frac{e^{2}}{4\pi\varepsilon_{0}}\right)^{4} = \frac{4\hbar^{4}n^{4}E_{n}^{2}}{m^{2}} . \] Substituting, \begin{equation} \boxed{\; E_{D}^{(1)} = \frac{4n}{2mc^{2}}\,E_{n}^{2} = \frac{2n\,E_{n}^{2}}{mc^{2}} \qquad (\ell = 0 \text{ only}) \;} \label{eq:E-darwin-final} \end{equation}
Since \(E_{n}^{2} > 0\), the Darwin term is positive: it pushes \(s\) states slightly upward. In terms of \(\alpha\), using \(E_{n} = -\tfrac12\alpha^{2}mc^{2}/n^{2}\), \[ E_{D}^{(1)} = \frac{\alpha^{4}mc^{2}}{2n^{3}}, \] which is of order \(\alpha^{4}mc^{2}\) — the same order as the relativistic and spin–orbit terms, as promised by Table 2.2.
Two things are asked repeatedly. First, why only \(s\) states: because \(R_{n\ell}\propto r^{\ell}\) makes \(\psi(0)\) vanish for every \(\ell\ne0\). Second, the sign: the Darwin shift is upward. If a question gives you \(E_{n}\) and asks for the Darwin correction, \(\eqref{eq:E-darwin-final}\) in the form \(2nE_{n}^{2}/mc^{2}\) is the fastest route — compare it with the relativistic result \(\eqref{eq:Er-result}\), which carries \(-E_{n}^{2}/2mc^{2}\) outside.
Look at the spin–orbit result \(\eqref{eq:Eso-result}\): for \(\ell=0\) the denominator vanishes while the numerator also vanishes (\(j=\tfrac12\) forces \(j(j+1)-\ell(\ell+1)-\tfrac34 = 0\)), so the expression is \(0/0\) — undefined. Physically there is no orbital motion for an \(s\) state, hence no internal magnetic field and no spin–orbit coupling. The Darwin term steps in at exactly those states, and it does so with precisely the magnitude needed to make the combined formula \(\eqref{eq:E-fs}\) valid for all \(\ell\) including \(\ell=0\). This is why the total fine structure need not treat \(s\) states as a special case.
2.5 Total Fine Structure of Hydrogen
Adding the relativistic correction \(\eqref{eq:Er-result}\) to the spin–orbit correction \(\eqref{eq:Eso-result}\) produces a remarkable simplification. There are two cases, and both give the same answer.
Case 1: \(j = \ell + \tfrac12\) (so \(\ell = j - \tfrac12\))
The bracket in \(\eqref{eq:SL-eigen}\) becomes \[ j(j+1) - \ell(\ell+1) - \tfrac34 = \ell , \] so \[ E_{\text{so}}^{(1)} = \frac{E_{n}^{2}}{mc^{2}}\cdot\frac{n\,\ell}{\ell(\ell+\tfrac12)(\ell+1)} = \frac{E_{n}^{2}}{mc^{2}}\cdot\frac{n}{(\ell+\tfrac12)(\ell+1)} = \frac{E_{n}^{2}}{mc^{2}}\cdot\frac{n}{j\left(j+\tfrac12\right)} . \] Meanwhile \(\eqref{eq:Er-result}\) with \(\ell + \tfrac12 = j\) gives \[ E_{r}^{(1)} = -\frac{E_{n}^{2}}{2mc^{2}}\left[\frac{4n}{j}-3\right]. \] Adding: \[ E_{\text{fs}} = \frac{E_{n}^{2}}{2mc^{2}} \left[3 - \frac{4n}{j} + \frac{2n}{j\left(j+\tfrac12\right)}\right] = \frac{E_{n}^{2}}{2mc^{2}}\left[3 - \frac{4n}{j+\tfrac12}\right]. \]
Case 2: \(j = \ell - \tfrac12\) (so \(\ell = j + \tfrac12\))
Now the bracket is \[ j(j+1) - \ell(\ell+1) - \tfrac34 = -(\ell+1) , \] giving \[ E_{\text{so}}^{(1)} = -\frac{E_{n}^{2}}{mc^{2}}\cdot\frac{n}{\ell\left(\ell+\tfrac12\right)} , \qquad E_{r}^{(1)} = -\frac{E_{n}^{2}}{2mc^{2}}\left[\frac{4n}{\ell+\tfrac12}-3\right], \] and after substituting \(\ell = j+\tfrac12\) the sum collapses to the same expression.
2.5.1 The fine structure formula
In both cases \begin{equation} \boxed{\; E_{\text{fs}}^{(1)} = \frac{E_{n}^{2}}{2mc^{2}} \left(3 - \frac{4n}{\,j+\tfrac12\,}\right)\;} \label{eq:E-fs} \end{equation} and therefore the total energy including fine structure is \begin{equation} \boxed{\; E_{n j} = -\frac{13.6\,\mathrm{eV}}{n^{2}} \left[\,1 + \frac{\alpha^{2}}{n^{2}} \left(\frac{n}{\,j+\tfrac12\,} - \frac{3}{4}\right)\right]\;} \label{eq:E-nj} \end{equation} For a hydrogenic ion of charge \(Z\), replace \(\alpha \to Z\alpha\) and multiply the leading term by \(Z^{2}\), so the splitting scales as \(Z^{4}\).
Compare \(\eqref{eq:E-nj}\) with Sommerfeld's result from Chapter 1: they are identical in form, with Sommerfeld's \(n_{\phi}\) replaced by \(j+\tfrac12\). Sommerfeld got the right formula from the wrong physics. This is a favourite discussion question.
The energy depends on \(n\) and \(j\) only — not on \(\ell\). Hence states with the same \(n\) and \(j\) but different \(\ell\) remain degenerate. For example \(2s_{1/2}\) and \(2p_{1/2}\) have exactly the same energy in Dirac theory, as do \(3p_{3/2}\) and \(3d_{3/2}\). This degeneracy is only lifted by the Lamb shift (2.7).
Note also that no separate Darwin term appears in \(\eqref{eq:E-fs}\): it is already accounted for, and \(\eqref{eq:E-fs}\) is correct including \(\ell=0\).
2.6 Selection Rules for One-Electron Atoms
The quantum numbers describing a one-electron state are
principal quantum number \(n = 1,2,3,\dots\)
orbital quantum number \(\ell = 0,1,2,\dots,n-1\)
magnetic quantum number \(m_{\ell} = -\ell,\dots,+\ell\)
spin quantum number \(s=\tfrac12\), with \(m_{s} = \pm\tfrac12\)
total angular momentum \(j = \ell \pm \tfrac12\) (and \(j = \tfrac12\) only, when \(\ell=0\)), with \(m_{j} = -j,\dots,+j\)
For electric dipole (E1) transitions the allowed changes are \begin{equation} \Delta \ell = \pm 1, \qquad \Delta j = 0, \pm 1, \qquad \Delta m_{j} = 0, \pm 1, \qquad \Delta s = 0, \label{eq:selection} \end{equation} with \(\Delta n\) unrestricted.
Two traps. First, \(\Delta\ell = 0\) is forbidden, so \(3s_{1/2}\to 1s_{1/2}\) cannot occur even though \(\Delta j = 0\) is allowed. Second, \(\Delta j = 0\) is allowed (except \(j=0 \to j=0\)), which surprises people who remember only \(\Delta j = \pm1\).
2.6.1 Degeneracy counting
\begin{equation} \text{For a given } n:\ 2n^{2}, \qquad \text{for a given } \ell:\ 2(2\ell+1), \qquad \text{for a given } j:\ (2j+1). \label{eq:degeneracy} \end{equation}
The most prominent feature of the hydrogen spectrum in the visible region is the red Balmer line, from the transition \(n=3 \to n=2\).
Determine its wavelength and frequency according to Bohr theory.
Fine structure splits it into several closely spaced lines. What is the spacing between the sublevels?
Solution. (a) \[ \Delta E = 13.6\left(\frac{1}{4}-\frac{1}{9}\right) = 13.6\times\frac{5}{36} = 1.89\,\mathrm{eV}, \] \[ \lambda = \frac{12400\,\mathrm{eV}\,\mathrm{Å}}{1.89\,\mathrm{eV}} = 6563\,\mathrm{Å}, \qquad \nu = \frac{c}{\lambda} = 4.57\times 10^{14}\,\mathrm{Hz}. \]
(b) Use \(\eqref{eq:E-fs}\). With \(E_{n} = -13.6/n^{2}\) eV and \(mc^{2} = 511\,\mathrm{keV}\), the prefactor is \[ \frac{E_{n}^{2}}{2mc^{2}} = \frac{(13.6)^{2}}{2\times 511\times10^{3}\,n^{4}} = \frac{1.81\times 10^{-4}}{n^{4}}\ \eV . \]
For \(n=2\): \(\ell = 0\) or \(1\), so \(j = \tfrac12\) or \(\tfrac32\). The level splits into two: \[ E_{\text{fs}}\!\left(j=\tfrac12\right) = 1.13\times 10^{-5}\left(3 - \frac{8}{1}\right) = -5.66\times 10^{-5}\,\mathrm{eV}, \] \[ E_{\text{fs}}\!\left(j=\tfrac32\right) = 1.13\times 10^{-5}\left(3 - \frac{8}{2}\right) = -1.13\times 10^{-5}\,\mathrm{eV}, \] a splitting of \(4.53\times 10^{-5}\,\mathrm{eV}\) between \(2p_{1/2}\) and \(2p_{3/2}\).
For \(n=3\): \(\ell = 0,1,2\), so \(j = \tfrac12, \tfrac32, \tfrac52\) — three levels. With prefactor \(1.81\times 10^{-4}/81 = 2.24\times 10^{-6}\) eV: \[ j=\tfrac12:\ 2.24\times 10^{-6}\left(3-\tfrac{12}{1}\right) = -2.01\times 10^{-5}\,\mathrm{eV}, \] \[ j=\tfrac32:\ 2.24\times 10^{-6}\left(3-\tfrac{12}{2}\right) = -6.71\times 10^{-6}\,\mathrm{eV}, \] \[ j=\tfrac52:\ 2.24\times 10^{-6}\left(3-\tfrac{12}{3}\right) = -2.24\times 10^{-6}\,\mathrm{eV}. \]
So the sublevel spacings are of order \(10^{-5}\)–\(10^{-6}\) eV, i.e.\ about \(10^{-5}\) of the \(1.89\,\mathrm{eV}\) line energy — consistent with the \(\alpha^{2}\) estimate of Table 2.2.
Consider the fine structure of the \(n=3 \to n=2\) transition.
Theoretically, how many transitions are allowed by the selection rules?
How many distinct lines are actually observed, and why does this differ from (a)?
Solution. (a) The available levels are \[ n=3:\ 3s_{1/2},\ 3p_{1/2},\ 3p_{3/2},\ 3d_{3/2},\ 3d_{5/2}; \qquad n=2:\ 2s_{1/2},\ 2p_{1/2},\ 2p_{3/2}. \] Applying \(\Delta\ell = \pm1\) and \(\Delta j = 0,\pm1\):
\(3s_{1/2} \to 2p_{1/2}\)
\(3s_{1/2} \to 2p_{3/2}\)
\(3p_{1/2} \to 2s_{1/2}\)
\(3p_{3/2} \to 2s_{1/2}\)
\(3d_{3/2} \to 2p_{1/2}\)
\(3d_{3/2} \to 2p_{3/2}\)
\(3d_{5/2} \to 2p_{3/2}\)
So 7 transitions are allowed.
(b) Dirac theory predicts 5 distinct lines. Because the energy depends only on \(n\) and \(j\), two pairs among the seven transitions have identical photon energies:
\(3s_{1/2}\to2p_{1/2}\) and \(3p_{1/2}\to2s_{1/2}\) are both \(j=\tfrac12 \to j=\tfrac12\), hence coincide.
\(3p_{3/2}\to2s_{1/2}\) and \(3d_{3/2}\to2p_{1/2}\) are both \(j=\tfrac32 \to j=\tfrac12\), hence coincide.
Two coincidences reduce \(7\) transitions to \(7-2 = 5\) theoretical lines. These are the five positions marked in the synthetic spectrum of Figure 2.5, spread over about \(0.36\,\mathrm{cm}^{-1}\) in total.
The experimental count is 7, not 5. The two coincidences above rest on the Dirac degeneracy of \(2s_{1/2}\) with \(2p_{1/2}\) and of \(3s_{1/2}\) with \(3p_{1/2}\). That degeneracy is not exact. The Lamb shift (2.7) lifts it, and once it is lifted both coincidences break: each of the two merged positions separates into two, and the pattern becomes \(5 + 2 = 7\) resolvable components. So the accounting for H\(_\alpha\) runs \[ \underbrace{7}_{\substack{\text{allowed}\\\text{transitions}}} \ \longrightarrow\ \underbrace{5}_{\substack{\text{Dirac}\\\text{lines}}} \ \longrightarrow\ \underbrace{7}_{\substack{\text{observed at highest}\\\text{resolution (QED)}}} . \]
The counting logic — “list the levels, apply \(\Delta\ell\) and \(\Delta j\), then merge transitions with equal \((j_i \to j_f)\)” — is the whole method, and it recurs for every H\(_\alpha\)-type question. Read the question wording carefully: transitions is 7, lines predicted by Dirac theory is 5, and components observable at the highest resolution is 7 again. An examiner who says “according to fine-structure theory” wants 5.
2.7 The Lamb Shift
According to Dirac theory, levels of a one-electron atom with the same \(n\) and \(j\) but different \(\ell\) have exactly the same energy. The Lamb–Retherford experiment (1947) showed this is not quite true: \(2s_{1/2}\) and \(2p_{1/2}\) do not coincide. The \(2s_{1/2}\) level lies above \(2p_{1/2}\) by \begin{equation} \Delta E_{\text{Lamb}} = 4.37\times 10^{-6}\,\mathrm{eV} \qquad\text{(about 1057\,\mathrm{MHz})}. \label{eq:lamb} \end{equation}
The explanation requires QED. Radiative corrections to Dirac theory are obtained by taking into account the interaction of the electron with the quantized electromagnetic field. A quantized radiation field in its lowest energy state is not identical with zero field — there remain zero-point oscillations. So even in vacuum there are fluctuations in this zero-point field, and they act on the electron, causing it to execute rapid oscillatory motion. Its charge is effectively smeared out, and the point electron behaves as a small sphere.
An electron bound in a non-uniform electric field therefore experiences a potential slightly different from the point-particle case. Electrons in \(s\) states, which are most sensitive to short-distance modifications of the potential, are raised in energy relative to states of higher \(\ell\), for which the modification is much smaller. Hence only \(s\) states shift appreciably upward — structurally similar to the Darwin term of 2.4.
Order of magnitude \(\alpha^{5}mc^{2}\), one power of \(\alpha\) below fine structure. Lamb and Retherford's measurement was the experimental trigger for the development of modern QED, and won Lamb the 1955 Nobel Prize.
2.7.1 Seeing the Lamb shift clearly: high-\(Z\) hydrogenic ions
In hydrogen itself the Lamb shift is a \(4.4\times 10^{-6}\,\mathrm{eV}\) effect on levels separated by electron-volts — exquisitely hard to measure. The radiative corrections grow much faster with nuclear charge than the Dirac energies do, so the cleanest tests use hydrogen-like ions of heavy elements, where a single electron orbits a bare high-\(Z\) nucleus. Figure 2.6 shows the case of hydrogen-like uranium, \(\mathrm{U}^{91+}\).
Formula Summary
Hierarchy Bohr \(\alpha^{2}mc^{2}\) \ \(>\) \ fine structure \(\alpha^{4}mc^{2}\) \ \(>\) \ Lamb \(\alpha^{5}mc^{2}\) \ \(>\) \ hyperfine \(\tfrac{m}{M}\alpha^{4}mc^{2}\) \[ E_{n} = -\frac{1}{2}\frac{\alpha^{2}mc^{2}}{n^{2}}, \qquad mc^{2} = 511\,\mathrm{keV}, \qquad \alpha = \tfrac{1}{137} \] Relativistic correction \[ H'_{r} = -\frac{p^{4}}{8m^{3}c^{2}}, \qquad E_{r}^{(1)} = -\frac{E_{n}^{2}}{2mc^{2}}\left[\frac{4n}{\ell+\tfrac12}-3\right] \] Spin–orbit coupling \[ H'_{\text{so}} = \frac{1}{8\pi\varepsilon_{0}}\frac{e^{2}}{m^{2}c^{2}} \frac{\vec S\cdot\vec L}{r^{3}}, \qquad \vec S\cdot\vec L = \frac{\hbar^{2}}{2}\!\left[j(j{+}1)-\ell(\ell{+}1)-\tfrac34\right] \] \[ E_{\text{so}}^{(1)} = \frac{E_{n}^{2}}{mc^{2}} \frac{n\!\left[j(j{+}1)-\ell(\ell{+}1)-\tfrac34\right]} {\ell\!\left(\ell+\tfrac12\right)(\ell+1)} \] Expectation values \[ \left\langle\frac1r\right\rangle = \frac{1}{n^{2}\abohr}, \qquad \left\langle\frac{1}{r^{2}}\right\rangle = \frac{1}{\left(\ell+\frac12\right)n^{3}\abohr^{2}}, \qquad \left\langle\frac{1}{r^{3}}\right\rangle = \frac{1}{\ell\left(\ell+\frac12\right)(\ell+1)n^{3}\abohr^{3}} \] Darwin term \(H'_{D} = \dfrac{\hbar^{2}}{8m^{2}c^{2}}\nabla^{2}V\) \ —\ acts only on \(\ell=0\), shifts \(s\) states up \[ \lambda_{C} = \frac{\hbar}{mc} = 3.86\times 10^{-13}\,\mathrm{m} \] Total fine structure \[ E_{\text{fs}}^{(1)} = \frac{E_{n}^{2}}{2mc^{2}}\left(3-\frac{4n}{j+\frac12}\right), \qquad E_{nj} = -\frac{13.6\,\mathrm{eV}}{n^{2}} \left[1+\frac{\alpha^{2}}{n^{2}}\left(\frac{n}{j+\frac12}-\frac34\right)\right] \] Depends on \(n\) and \(j\) only. Scales as \(Z^{4}\). Good quantum numbers: \(n,\ell,s,j,m_{j}\).
Selection rules (E1) \(\Delta\ell = \pm1\), \ \(\Delta j = 0,\pm1\), \ \(\Delta m_{j} = 0,\pm1\), \ \(\Delta s = 0\), \ \(\Delta n\) free
Degeneracy given \(n\): \(2n^{2}\); \ given \(\ell\): \(2(2\ell+1)\); \ given \(j\): \((2j+1)\)
Lamb shift \(2s_{1/2}\) above \(2p_{1/2}\) by \(4.37\times 10^{-6}\,\mathrm{eV} \approx 1057\,\mathrm{MHz}\)
Useful conversions \[ \lvert\Delta E\rvert = \frac{hc}{\lambda^{2}}\,\Delta\lambda, \qquad hc = 12400\,\mathrm{eV}\,\mathrm{Å} = 1.24\times 10^{-6}\,\mathrm{eV}\,\mathrm{m} \]
Previous Year Questions
The spin–orbit interaction term of an electron moving in a central field is written as \(f(r)\,\vec\ell\cdot\vec s\), where \(r\) is the radial distance of the electron from the origin. If the electron moves inside a uniformly charged sphere, then
- (a)
\(f(r) = \text{constant}\)
- (b)
\(f(r) \propto r^{-1}\)
- (c)
\(f(r) \propto r^{-2}\)
- (d)
\(f(r) \propto r^{-3}\)
\(4\,\mathrm{MeV}\) \(\gamma\)-rays emitted by the de-excitation of \(^{19}\mathrm{F}\) are attributed, assuming spherical symmetry, to the transition of protons from the \(1d_{3/2}\) state to the \(1d_{5/2}\) state. If the contribution of the spin–orbit term to the total energy is written as \(C\big\langle \vec\ell\cdot\vec s\big\rangle\), the magnitude of \(C\) is \(\mathrm{MeV}\) (up to one decimal place).
The spin–orbit effect splits the \(^{2}P \to {}^{2}S\) transition
(wavelength \(\lambda = 6521\,\mathrm{Å}\)) in lithium into two lines with
separation \(\Delta\lambda = 0.14\,\mathrm{Å}\). The corresponding positive
value of the energy difference between the two lines, in eV, is
\(m\times10^{-5}\). The value of \(m\) (rounded off to the nearest integer) is
.
[2pt]
(Given: \(h = 4.125\times 10^{-15}\,\mathrm{eV}\,\mathrm{s}\), \(c = 3\times 10^{8}\,\mathrm{m}\,\mathrm{s}^{-1}\).)
The spin–orbit interaction in a hydrogen-like atom is given by the Hamiltonian \[ H' = -k\,\vec{L}\cdot\vec{S}, \] where \(k\) is a real constant. The splitting between the levels \(^{2}P_{3/2}\) and \(^{2}P_{1/2}\) due to this interaction is
- (a)
\(\tfrac12 k\hbar^{2}\)
- (b)
\(\tfrac32 k\hbar^{2}\)
- (c)
\(\tfrac34 k\hbar^{2}\)
- (d)
\(2k\hbar^{2}\)
Which of the following option(s) is/are correct for the ground state of a hydrogen atom?
- (a)
The linear Stark effect is zero
- (b)
It has definite parity
- (c)
Spin–orbit coupling is zero
- (d)
Hyperfine splitting is zero
The spin–orbit interaction in an atom is given by \(H = a\,\vec L\cdot\vec S\), where \(\vec L\) and \(\vec S\) denote the orbital and spin angular momenta of the electron. The splitting between the levels \(^{2}P_{3/2}\) and \(^{2}P_{1/2}\) is
- (a)
\(\tfrac32 a\hbar^{2}\)
- (b)
\(\tfrac12 a\hbar^{2}\)
- (c)
\(3a\hbar^{2}\)
- (d)
\(\tfrac52 a\hbar^{2}\)
If the fine structure splitting between the \(2\,^{2}P_{3/2}\) and \(2\,^{2}P_{1/2}\) levels in the hydrogen atom is \(0.4\,\mathrm{cm}^{-1}\), the corresponding splitting in \(\mathrm{Li}^{2+}\) will approximately be
- (a)
\(1.2\,\mathrm{cm}^{-1}\)
- (b)
\(10.8\,\mathrm{cm}^{-1}\)
- (c)
\(32.4\,\mathrm{cm}^{-1}\)
- (d)
\(36.8\,\mathrm{cm}^{-1}\)
Two Stern–Gerlach apparatuses \(S_{1}\) and \(S_{2}\) are kept in a line along the \(x\)-axis. The directions of their magnetic fields are along the positive \(z\)- and \(y\)-axes respectively. Each apparatus transmits only particles with spins aligned along the direction of its magnetic field. If an initially unpolarized beam of spin-\(\tfrac12\) particles passes through this configuration, the ratio of intensities \(I_{0}:I_{f}\) of the initial and final beams is
- (a)
\(16:1\)
- (b)
\(2:1\)
- (c)
\(4:1\)
- (d)
\(1:0\)
A hydrogen atom, excited to the electronic configuration \(3S_{1/2}\) (\(nL_{j}\) notation), relaxes to the ground state via electric dipole transitions. Considering only fine structure and ignoring hyperfine structure, the maximum number of emitted spectral lines is
- (a)
\(3\)
- (b)
\(6\)
- (c)
\(1\)
- (d)
\(4\)
For a system of two electrons, define an operator \[ \hat A = \frac{3}{a^{2}}\left(\hat{\vec S}_{1}\cdot\vec a\right) \left(\hat{\vec S}_{2}\cdot\vec a\right) - \hat{\vec S}_{1}\cdot\hat{\vec S}_{2} , \] where \(\vec a\) is an arbitrary vector and \(\hat{\vec S}_{1}\), \(\hat{\vec S}_{2}\) are spin operators. The eigenvalues of \(\hat A\) (in units of \(\hbar^{2}\)) are
- (a)
\(-1,\ 1,\ \tfrac32,\ \tfrac32\)
- (b)
\(-1,\ -\tfrac12,\ -\tfrac12,\ 0\)
- (c)
\(\tfrac12,\ 1,\ \tfrac32,\ \tfrac32\)
- (d)
\(0,\ \tfrac12,\ \tfrac12,\ -1\)
The energy difference between the \(3p\) and \(3s\) levels in Na is \(2.1\,\mathrm{eV}\). Spin–orbit coupling splits the \(3p\) level, resulting in two emission lines differing by \(6\,\mathrm{Å}\). The splitting of the \(3p\) level is approximately
- (a)
\(2\,\mathrm{eV}\)
- (b)
\(0.2\,\mathrm{eV}\)
- (c)
\(0.02\,\mathrm{eV}\)
- (d)
\(2\,\mathrm{meV}\)
What is the difference between the maximum and the minimum eigenvalues of a system of two electrons whose Hamiltonian is \(H = J\,\vec S_{1}\cdot\vec S_{2}\), where \(\vec S_{1}\) and \(\vec S_{2}\) are the spin angular momentum operators of the two electrons?
- (a)
\(\dfrac{J}{4}\)
- (b)
\(\dfrac{J}{2}\)
- (c)
\(\dfrac{3J}{4}\)
- (d)
\(J\)
Solutions to Previous Year Questions
Solution. Inside a uniformly charged sphere of radius \(R\) and charge \(Q\), the
electric potential at \(r
Solution. Using \(\vec\ell\cdot\vec s = \tfrac12\left(j^{2}-\ell^{2}-s^{2}\right)\), so that \[ \big\langle \vec\ell\cdot\vec s\big\rangle = \frac{\hbar^{2}}{2}\left[j(j+1)-\ell(\ell+1)-s(s+1)\right] . \] For the two \(1d\) levels (\(\ell=2\)) the \(\ell\) and \(s\) terms are common and cancel in the difference: \[ \Delta E = C\left[\big\langle\vec\ell\cdot\vec s\big\rangle_{5/2} - \big\langle\vec\ell\cdot\vec s\big\rangle_{3/2}\right] = C\,\frac{\hbar^{2}}{2}\left[\frac52\cdot\frac72 - \frac32\cdot\frac52\right] = C\,\frac{\hbar^{2}}{2}\cdot\frac{20}{4} = \frac{20}{8}\,C\hbar^{2} . \] Setting this equal to \(4\,\mathrm{MeV}\), \[ \frac{20}{8}C = 4 \quad\Longrightarrow\quad C = \frac{32}{20} = 1.6\,\mathrm{MeV} . \]
Solution. \[ \lvert\Delta E\rvert = \frac{hc}{\lambda^{2}}\,\Delta\lambda = \frac{1.24\times 10^{-6}\,\mathrm{eV}\,\mathrm{m}}{\left(6521\times10^{-10}\,\mathrm m\right)^{2}} \times 0.14\times10^{-10}\,\mathrm m = 4.08\times 10^{-5}\,\mathrm{eV} , \] so \(m \approx 4\).
Solution. Use the standard trick \(\vec{J}^{2} = \vec{L}^{2}+\vec{S}^{2}+2\vec{L}\cdot\vec{S}\), so that in a state of definite \(j\), \(l\), \(s\) \[ \langle \vec{L}\cdot\vec{S}\rangle = \tfrac12\left[j(j+1)-l(l+1)-s(s+1)\right]\hbar^{2}. \] For a \(P\) state, \(l=1\) and \(s=\tfrac12\): \[ j=\tfrac32:\quad \tfrac12\left[\tfrac{15}{4}-2-\tfrac34\right]\hbar^{2} = +\tfrac12\hbar^{2}, \qquad j=\tfrac12:\quad \tfrac12\left[\tfrac34-2-\tfrac34\right]\hbar^{2} = -\hbar^{2}. \] With \(H' = -k\,\vec{L}\cdot\vec{S}\) the energies are \(E_{3/2} = -\tfrac12 k\hbar^{2}\) and \(E_{1/2} = +k\hbar^{2}\), so the splitting is \[ \left|E_{1/2}-E_{3/2}\right| = \left|k\hbar^{2} + \tfrac12 k\hbar^{2}\right| = \boxed{\tfrac32 k\hbar^{2}} . \]
The Landé interval rule gives it in one line. For \(H' = \xi\,\vec{L}\cdot\vec{S}\) the gap between adjacent fine-structure levels is \(\xi\hbar^{2}J_{\text{upper}}\), here \(\xi\hbar^{2}\times\tfrac32\). The answer is the same because only the magnitude is asked.
The minus sign inverts the multiplet. Physical hydrogen has \(H' = +\xi\vec{L}\cdot\vec{S}\) with \(\xi>0\), so \(^{2}P_{1/2}\) lies below \(^{2}P_{3/2}\) (a normal multiplet). The Hamiltonian as written here, with \(k>0\), reverses that order. The question asks only for the separation, but a student who has understood the physics should notice.
Solution. The ground state is \(1s\): \(n=1\), \(l=0\), and it is the only state with that \(n\).
(a) True. A linear Stark shift is \(\langle 100|{-}eEz|100\rangle\), which vanishes because \(z\) is odd under parity and \(|100\rangle\) has definite parity. A first-order shift needs degenerate states of opposite parity to mix, which is exactly what \(n=2\) has (\(2s\) and \(2p\)) and \(n=1\) does not.
(b) True. \(\psi_{100}\propto e^{-r/a_{0}}\) is even; the parity of a hydrogen state is \((-1)^{l}\), here \(+1\).
(c) True. \(H_{\text{SO}} \propto \vec{L}\cdot\vec{S}\) and \(\vec{L}=0\) for an \(s\) state.
(d) False. The hyperfine splitting of the hydrogen ground state is the \(21\,\mathrm{cm}\) line — the single most observed spectral feature in radio astronomy. It is emphatically not zero (Chapter 4).
Options (c) and (d) look like the same statement and are not. The spin–orbit term needs \(l\neq0\), so it dies for \(s\) states. The dominant hyperfine term is the Fermi contact interaction \(\propto|\psi(0)|^{2}\,\vec{I}\cdot\vec{S}\), which is non-zero only for \(s\) states, since only they have finite amplitude at the nucleus. The two couplings are complementary, not parallel.
Solution. From \(\eqref{eq:SL-eigen}\) with \(\ell = 1\), \(s=\tfrac12\): \[ j=\tfrac32:\ \big\langle\vec L\cdot\vec S\big\rangle = \frac{\hbar^{2}}{2}\left[\tfrac{15}{4}-2-\tfrac34\right] = +\frac{\hbar^{2}}{2}, \qquad j=\tfrac12:\ \frac{\hbar^{2}}{2}\left[\tfrac34-2-\tfrac34\right] = -\hbar^{2}. \] Hence \[ \Delta E = a\left[\frac{\hbar^{2}}{2}+\hbar^{2}\right] = \frac{3}{2}a\hbar^{2} . \]
The general result is \(\Delta E = a\hbar^{2}\left(\ell+\tfrac12\right)\): for \(\ell=1\) this gives \(\tfrac32 a\hbar^{2}\), for \(\ell=2\), \(\tfrac52 a\hbar^{2}\), and so on. One line answers every question of this form.
Solution. Fine structure splitting scales as \(Z^{4}\) (see \(\eqref{eq:E-nj}\) with \(\alpha \to Z\alpha\)). For \(\mathrm{Li}^{2+}\), \(Z = 3\): \[ \frac{(\Delta E)_{\mathrm{Li}}}{(\Delta E)_{\mathrm H}} = \frac{3^{4}}{1^{4}} = 81 \quad\Longrightarrow\quad (\Delta E)_{\mathrm{Li}} = 81 \times 0.4 = 32.4\,\mathrm{cm}^{-1} . \]
Solution. The unpolarized beam entering \(S_{1}\) (field along \(\hat z\)) is transmitted with probability \(\tfrac12\), giving \(I_{0}/2\) of pure \(\lvert{\uparrow_{z}}\rangle\). Expanding in the \(\hat y\) basis, \[ \lvert{\uparrow_{z}}\rangle = \frac{1}{\sqrt2}\left(\lvert{\uparrow_{y}}\rangle + \ii\lvert{\downarrow_{y}}\rangle\right), \] so \(S_{2}\) transmits a further \(\tfrac12\): \[ I_{f} = \frac{I_{0}}{4} \quad\Longrightarrow\quad \frac{I_{0}}{I_{f}} = 4 . \]
Solution. E1 transitions require \(\Delta\ell = \pm1\) and \(\Delta j = 0,\pm1\). From \(3s_{1/2}\) the only allowed steps are to \(2p_{1/2}\) and \(2p_{3/2}\); the direct \(3s_{1/2}\to1s_{1/2}\) is forbidden (\(\Delta\ell = 0\)). Each \(2p\) level then decays to \(1s_{1/2}\): \[ 3s_{1/2}\!\to\!2p_{1/2},\quad 3s_{1/2}\!\to\!2p_{3/2},\quad 2p_{1/2}\!\to\!1s_{1/2},\quad 2p_{3/2}\!\to\!1s_{1/2}, \] giving \(\boxed{4}\) lines.
Solution. Choose \(\vec a\) along \(\hat z\), so \(\hat A = 3S_{1z}S_{2z} - \vec S_{1}\cdot\vec S_{2}\). Writing \(\vec S_{1}\cdot\vec S_{2} = S_{1z}S_{2z} + \tfrac12\left(S_{1+}S_{2-}+S_{1-}S_{2+}\right)\), \[ \hat A = 2S_{1z}S_{2z} - \tfrac12\left(S_{1+}S_{2-}+S_{1-}S_{2+}\right). \] In units of \(\hbar^{2}\):
\(\lvert{\uparrow\uparrow}\rangle\) and \(\lvert{\downarrow\downarrow}\rangle\): the flip term gives zero and \(2S_{1z}S_{2z} \to +\tfrac12\). Eigenvalue \(+\tfrac12\) each.
On \(\{\lvert{\uparrow\downarrow}\rangle, \lvert{\downarrow\uparrow}\rangle\}\): \[ \hat A \to \begin{pmatrix} -\tfrac12 & -\tfrac12\\[2pt] -\tfrac12 & -\tfrac12\end{pmatrix}, \] with eigenvalues \(0\) and \(-1\).
The spectrum is \(\left\{0,\ \tfrac12,\ \tfrac12,\ -1\right\}\).
\(\hat A\) is the tensor (dipole–dipole) operator of hyperfine structure. Its trace vanishes — a fast check: \(0+\tfrac12+\tfrac12-1 = 0\).
Solution. The two emission lines are \(3p_{3/2}\to3s_{1/2}\) and \(3p_{1/2}\to3s_{1/2}\), of wavelengths \(\lambda_{2}\) and \(\lambda_{1}\). Their energy difference is the \(3p\) splitting: \[ \Delta E = \frac{12400}{\lambda_{2}} - \frac{12400}{\lambda_{1}} = 12400\left(\frac{\lambda_{1}-\lambda_{2}}{\lambda_{1}\lambda_{2}}\right) \ \eV \quad (\lambda \text{ in \mathrm{Å}}) . \] Since the splitting is tiny, \(\lambda_{1}\lambda_{2} \approx \lambda^{2}\) where \(\lambda\) is the \(3p \to 3s\) wavelength: \[ \lambda = \frac{12400}{2.1} = 5905\,\mathrm{Å} . \] With \(\Delta\lambda = 6\,\mathrm{Å}\), \[ \Delta E = \frac{12400\times 6}{(5905)^{2}} = \frac{2.1\times2.1\times6}{12400} \approx 2\times 10^{-3}\,\mathrm{eV} = 2\,\mathrm{meV} . \]
Solution. For two spin-\(\tfrac12\) particles, \[ \vec S_{1}\cdot\vec S_{2} = \frac{\hbar^{2}}{2}\left[s(s+1) - \tfrac34 - \tfrac34\right] = \begin{cases} +\dfrac{\hbar^{2}}{4}, & s=1 \ \text{(triplet)},\\[6pt] -\dfrac{3\hbar^{2}}{4}, & s=0 \ \text{(singlet)} . \end{cases} \] Hence \[ E_{\max} - E_{\min} = J\left[\frac{\hbar^{2}}{4} + \frac{3\hbar^{2}}{4}\right] = J\hbar^{2} . \]