P3 Hydrogen Atom: Shifts and Interactions
These problems explore electron spin, magnetic interactions, coupled spins, and the hyperfine structure of hydrogen.
Spin and Magnetic Interactions
Problem P3.1: Spin in a Magnetic Field
A sodium atom is placed in a homogeneous magnetic field \(\mathbf B=B_z\hat{\mathbf z}\). The interaction between the magnetic field and the magnetic moment of the valence electron is described by
\[ \hat H=-\boldsymbol{\mu}\cdot\mathbf B =2\frac{\mu_{\mathrm B}}{\hbar}B_z\hat S_z. \]
The spin operators in the Pauli spinor representation are
\[ \hat S_x=\frac{\hbar}{2} \begin{pmatrix}0&1\\1&0\end{pmatrix}, \qquad \hat S_y=\frac{\hbar}{2} \begin{pmatrix}0&-i\\i&0\end{pmatrix}, \qquad \hat S_z=\frac{\hbar}{2} \begin{pmatrix}1&0\\0&-1\end{pmatrix}. \]
Tasks
- Determine the two eigenvalues of the Hamiltonian and their corresponding normalized spinors.
- Determine the time dependence of each energy eigenstate.
Now suppose the atom is prepared at \(t=0\) in the state
\[ |\psi(0)\rangle=\frac{1}{\sqrt2} \begin{pmatrix}1\\1\end{pmatrix}. \]
- Express \(|\psi(0)\rangle\) as a superposition of the two energy eigenstates.
- Write down the time-dependent state \(|\psi(t)\rangle\).
- Calculate the time-dependent expectation values \(\langle\hat S_x\rangle\), \(\langle\hat S_y\rangle\), and \(\langle\hat S_z\rangle\).
- Describe physically what happens to the expectation value of the spin. About which axis does it precess?
An energy eigenstate with energy \(E\) evolves with the phase factor \(e^{-iEt/\hbar}\).
For the expectation values, use
\[ \langle\hat S_i\rangle =\langle\psi(t)|\hat S_i|\psi(t)\rangle. \]
1. Energy eigenvalues and eigenstates
Using the matrix representation of \(\hat S_z\),
\[ \hat H=\mu_{\mathrm B}B_z \begin{pmatrix}1&0\\0&-1\end{pmatrix}. \]
The eigenvalues are
\[ \boxed{E_+=+\mu_{\mathrm B}B_z,\qquad E_-=-\mu_{\mathrm B}B_z.} \]
The corresponding normalized spinors are
\[ |+\rangle=\begin{pmatrix}1\\0\end{pmatrix}, \qquad |-\rangle=\begin{pmatrix}0\\1\end{pmatrix}. \]
2. Time evolution of the eigenstates
Each energy eigenstate acquires a phase:
\[ |+(t)\rangle=e^{-i\mu_{\mathrm B}B_zt/\hbar}|+\rangle, \]
\[ |-(t)\rangle=e^{+i\mu_{\mathrm B}B_zt/\hbar}|-\rangle. \]
These phases do not change the spin expectation values of the individual eigenstates.
3. Initial superposition
The initial state is
\[ \boxed{ |\psi(0)\rangle=\frac{1}{\sqrt2} \left(|+\rangle+|-\rangle\right). } \]
4. Time-dependent state
Define
\[ \omega=\frac{2\mu_{\mathrm B}B_z}{\hbar}. \]
The state becomes
\[ \boxed{ |\psi(t)\rangle= \frac{1}{\sqrt2} \begin{pmatrix} e^{-i\omega t/2}\\ e^{+i\omega t/2} \end{pmatrix}. } \]
5. Spin expectation values
For the \(x\) component,
\[ \begin{aligned} \langle\hat S_x\rangle &=\frac{\hbar}{4} \begin{pmatrix} e^{i\omega t/2}&e^{-i\omega t/2} \end{pmatrix} \begin{pmatrix} 0&1\\1&0 \end{pmatrix} \begin{pmatrix} e^{-i\omega t/2}\\e^{i\omega t/2} \end{pmatrix}\\ &=\boxed{\frac{\hbar}{2}\cos(\omega t)}. \end{aligned} \]
Similarly,
\[ \boxed{ \langle\hat S_y\rangle =\frac{\hbar}{2}\sin(\omega t) } \]
and
\[ \boxed{\langle\hat S_z\rangle=0.} \]
6. Physical interpretation
The expectation value of the spin is
\[ \boxed{ \langle\hat{\mathbf S}\rangle =\frac{\hbar}{2} \begin{pmatrix} \cos(\omega t)\\ \sin(\omega t)\\ 0 \end{pmatrix}. } \]
Its magnitude remains constant, but its direction rotates in the \(xy\) plane.
The spin therefore precesses around the magnetic-field axis, which is the \(z\) axis, with angular frequency
\[ \boxed{ \omega=\frac{2\mu_{\mathrm B}B_z}{\hbar}. } \]
This is Larmor precession. For \(B_z>0\), the expectation value rotates from the positive \(x\) direction toward the positive \(y\) direction.
Problem P3.2: Two-Spin System
Consider two interacting particles, each with spin \(s_1=s_2=\frac12\). Their spins add to form the total spin
\[ \hat{\mathbf S}=\hat{\mathbf s}_1+\hat{\mathbf s}_2. \]
Tasks
What values of the total-spin quantum number \(S\) are possible? For each value of \(S\), list the allowed values of \(M_S\).
How many independent spin states are there altogether? Identify which states form the triplet and which form the singlet.
Starting from \(\hat{\mathbf S}=\hat{\mathbf s}_1+\hat{\mathbf s}_2\), show that
\[ \hat{\mathbf s}_1\cdot\hat{\mathbf s}_2 =\frac12\left( \hat{\mathbf S}^2-\hat{\mathbf s}_1^{\,2} -\hat{\mathbf s}_2^{\,2} \right). \]
Now consider the Hamiltonian
\[ \hat H= \frac{a}{\hbar^2}\hat{\mathbf s}_1\cdot\hat{\mathbf s}_2 +\frac{b}{\hbar}\hat S_z. \]
- Use the previous result to determine the energy eigenvalues in terms of \(S\) and \(M_S\).
- Determine the energies of all four spin states explicitly.
- Which states are degenerate when \(b=0\)? What happens to this degeneracy when \(b\neq0\)?
Use the angular-momentum addition rule
\[ S=|s_1-s_2|,\ldots,s_1+s_2. \]
The coupled states \(|S,M_S\rangle\) are simultaneous eigenstates of \(\hat{\mathbf S}^2\) and \(\hat S_z\).
For each individual spin,
\[ \hat{\mathbf s}_i^{\,2} =s_i(s_i+1)\hbar^2. \]
1. Total-spin quantum numbers
Adding two spins \(s_1=s_2=\frac12\) gives
\[ \boxed{S=1\quad\text{or}\quad S=0.} \]
For \(S=1\), the allowed projections are
\[ M_S=1,0,-1. \]
For \(S=0\), only \(M_S=0\) is possible.
2. Triplet and singlet states
The triplet consists of three states:
\[ |1,1\rangle=|\uparrow\uparrow\rangle, \]
\[ |1,0\rangle= \frac{|\uparrow\downarrow\rangle+ |\downarrow\uparrow\rangle}{\sqrt2}, \]
\[ |1,-1\rangle=|\downarrow\downarrow\rangle. \]
The singlet is
\[ |0,0\rangle= \frac{|\uparrow\downarrow\rangle- |\downarrow\uparrow\rangle}{\sqrt2}. \]
Thus, there are
\[ \boxed{3+1=4} \]
independent spin states, as expected from \(2\times2=4\).
3. Spin-coupling identity
Squaring the total-spin operator,
\[ \begin{aligned} \hat{\mathbf S}^2 &=(\hat{\mathbf s}_1+\hat{\mathbf s}_2)^2\\ &=\hat{\mathbf s}_1^{\,2} +\hat{\mathbf s}_2^{\,2} +2\hat{\mathbf s}_1\cdot\hat{\mathbf s}_2. \end{aligned} \]
Rearranging,
\[ \boxed{ \hat{\mathbf s}_1\cdot\hat{\mathbf s}_2 =\frac12\left( \hat{\mathbf S}^2-\hat{\mathbf s}_1^{\,2} -\hat{\mathbf s}_2^{\,2} \right). } \]
4. General energy eigenvalues
For a coupled state \(|S,M_S\rangle\),
\[ \hat{\mathbf S}^2|S,M_S\rangle =S(S+1)\hbar^2|S,M_S\rangle \]
and
\[ \hat S_z|S,M_S\rangle =M_S\hbar|S,M_S\rangle. \]
Since
\[ s_1(s_1+1)=s_2(s_2+1)=\frac34, \]
we obtain
\[ \boxed{ E(S,M_S)= \frac a2\left[S(S+1)-\frac32\right]+bM_S. } \]
5. Energies of the four states
For the triplet, \(S=1\) and \(S(S+1)=2\), giving
\[ E(1,M_S)=\frac a4+bM_S. \]
For the singlet, \(S=0\), giving
\[ E(0,0)=-\frac{3a}{4}. \]
The four energies are therefore:
| State | Energy |
|---|---|
| \(|1,1\rangle\) | \(a/4+b\) |
| \(|1,0\rangle\) | \(a/4\) |
| \(|1,-1\rangle\) | \(a/4-b\) |
| \(|0,0\rangle\) | \(-3a/4\) |
6. Degeneracies
When \(b=0\), all three triplet states have the same energy,
\[ \boxed{E_{\mathrm{triplet}}=\frac a4.} \]
The singlet has energy
\[ \boxed{E_{\mathrm{singlet}}=-\frac{3a}{4}.} \]
The singlet–triplet separation is \(a\).
When \(b\neq0\), the three triplet states split into three distinct energies separated by \(b\) (for \(b>0\)). The singlet remains unaffected because \(M_S=0\).
The magnetic field therefore lifts the triplet degeneracy. Additional crossings between singlet and triplet energies can occur for particular values of \(b\) relative to \(a\).
Hyperfine Structure
Problem P3.3: Hyperfine Splitting in Hydrogen
The ground state of hydrogen is split into two hyperfine levels with \(F=1\) and \(F=0\). This splitting arises from the magnetic interaction between the electron spin and the proton spin.
The magnetic moment of the proton is
\[ \boldsymbol{\mu}_p =g_p\mu_N\frac{\mathbf I}{\hbar}, \]
where \(I=\frac12\), \(g_p=5.5857\), and
\[ \mu_N=\frac{e\hbar}{2m_p} \]
is the nuclear magneton.
The magnetic field produced by an ideal magnetic dipole \(\boldsymbol{\mu}_e\) is
\[ \mathbf B(\mathbf r)= \frac{\mu_0}{4\pi r^3} \left[ 3(\boldsymbol{\mu}_e\cdot\hat{\mathbf r})\hat{\mathbf r} -\boldsymbol{\mu}_e \right] +\frac{2\mu_0}{3}\boldsymbol{\mu}_e\delta^3(\mathbf r). \]
The second term is the Fermi contact field,
\[ \mathbf B_{\mathrm{contact}} =\frac{2\mu_0}{3}\boldsymbol{\mu}_e\delta^3(\mathbf r), \]
where the electron magnetic moment is approximately
\[ \boldsymbol{\mu}_e =-g_e\mu_B\frac{\mathbf S}{\hbar}, \qquad g_e\simeq2. \]
For the hydrogen ground state,
\[ \psi_{100}(\mathbf r) =\frac{e^{-r/a_0}}{\sqrt{\pi a_0^3}}. \]
Tasks
Show that
\[ \left\langle\delta^3(\mathbf r)\right\rangle =|\psi_{100}(0)|^2 \]
and evaluate \(|\psi_{100}(0)|^2\).
Using
\[ H_{\mathrm{HF}} =-\boldsymbol{\mu}_p\cdot\mathbf B_{\mathrm{contact}}, \]
show that the effective hyperfine Hamiltonian can be written as
\[ H_{\mathrm{HF}}= \frac{A}{\hbar^2}\mathbf I\cdot\mathbf S \]
and determine an expression for the hyperfine constant \(A\).
Using
\[ \mathbf F=\mathbf I+\mathbf S, \]
express \(\mathbf I\cdot\mathbf S\) in terms of \(F\), \(I\), and \(S\). Calculate the hyperfine energies for \(F=1\) and \(F=0\), and determine their energy difference.
Evaluate the splitting numerically as a frequency
\[ \nu=\frac{\Delta E}{h} \]
in MHz. Then calculate the corresponding wavelength and confirm that the transition produces the well-known 21-cm line observed in radio astronomy.
Use the following constants:
\[ \begin{aligned} \mu_0&=4\pi\times10^{-7}\,\mathrm{N/A^2},\\ \mu_B&=9.2740\times10^{-24}\,\mathrm{J/T},\\ \mu_N&=5.0508\times10^{-27}\,\mathrm{J/T},\\ a_0&=5.2918\times10^{-11}\,\mathrm{m},\\ h&=6.6261\times10^{-34}\,\mathrm{J\,s},\\ c&=2.9979\times10^8\,\mathrm{m/s}. \end{aligned} \]
The delta function selects the value of the wavefunction at \(\mathbf r=0\). You do not need to perform a three-dimensional integral explicitly.
For the spin coupling, use
\[ \mathbf I\cdot\mathbf S =\frac12\left(\mathbf F^2-\mathbf I^2-\mathbf S^2\right). \]
Remember that both the electron and proton have spin \(\frac12\).
1. Electron probability density at the nucleus
By the defining property of the three-dimensional delta function,
\[ \begin{aligned} \left\langle\delta^3(\mathbf r)\right\rangle &=\int d^3r\,|\psi_{100}(\mathbf r)|^2 \delta^3(\mathbf r)\\ &=|\psi_{100}(0)|^2. \end{aligned} \]
Since
\[ \psi_{100}(0)=\frac1{\sqrt{\pi a_0^3}}, \]
we obtain
\[ \boxed{ \left\langle\delta^3(\mathbf r)\right\rangle =\frac1{\pi a_0^3}. } \]
2. Hyperfine Hamiltonian
Substituting the electron magnetic moment into the contact field gives
\[ \mathbf B_{\mathrm{contact}} =-\frac{2\mu_0}{3} g_e\mu_B\frac{\mathbf S}{\hbar} \delta^3(\mathbf r). \]
The interaction with the proton magnetic moment is
\[ H_{\mathrm{HF}} =-\boldsymbol{\mu}_p\cdot \mathbf B_{\mathrm{contact}}. \]
Therefore,
\[ H_{\mathrm{HF}} =\frac{2\mu_0}{3} g_eg_p\mu_B\mu_N \frac{\mathbf I\cdot\mathbf S}{\hbar^2} \delta^3(\mathbf r). \]
Taking the expectation value over the \(1s\) spatial wavefunction produces the effective spin Hamiltonian
\[ \boxed{ H_{\mathrm{HF}} =\frac{A}{\hbar^2}\mathbf I\cdot\mathbf S, } \]
with
\[ \boxed{ A=\frac{2\mu_0}{3} \frac{g_eg_p\mu_B\mu_N}{\pi a_0^3}. } \]
3. Hyperfine energies
Using \(\mathbf F=\mathbf I+\mathbf S\),
\[ \mathbf F^2=\mathbf I^2+\mathbf S^2 +2\mathbf I\cdot\mathbf S. \]
Thus,
\[ \boxed{ \mathbf I\cdot\mathbf S =\frac12(\mathbf F^2-\mathbf I^2-\mathbf S^2). } \]
The energy of a state with total angular momentum \(F\) is
\[ E_F=\frac A2 \left[F(F+1)-I(I+1)-S(S+1)\right]. \]
For \(I=S=\frac12\),
\[ I(I+1)=S(S+1)=\frac34. \]
For \(F=1\),
\[ E_{F=1} =\frac A2\left(2-\frac32\right) =\boxed{\frac A4}. \]
For \(F=0\),
\[ E_{F=0} =\frac A2\left(0-\frac32\right) =\boxed{-\frac{3A}{4}}. \]
The energy difference is
\[ \boxed{ \Delta E=E_{F=1}-E_{F=0}=A. } \]
The \(F=1\) triplet lies above the \(F=0\) singlet.
4. Numerical splitting and wavelength
Using \(g_e\simeq2\) and the supplied constants,
\[ A= \frac{2\mu_0}{3} \frac{g_eg_p\mu_B\mu_N}{\pi a_0^3} \approx9.42\times10^{-25}\,\mathrm J. \]
The corresponding frequency is
\[ \nu=\frac Ah \approx1.421\times10^9\,\mathrm{Hz}. \]
Thus,
\[ \boxed{\nu\approx1421\,\mathrm{MHz}.} \]
The wavelength is
\[ \lambda=\frac c\nu \approx0.211\,\mathrm m, \]
or
\[ \boxed{\lambda\approx21.1\,\mathrm{cm}.} \]
This agrees closely with the famous 21-cm hydrogen line. Its measured frequency is approximately \(1420.406\,\mathrm{MHz}\).
The small difference arises from approximations in this calculation, including the treatment of the electron magnetic moment and the neglect of higher-order corrections.
Physical interpretation
The contact interaction is nonzero because the hydrogen \(1s\) wavefunction has finite probability density at the nucleus.
For states with \(\ell>0\), the wavefunction vanishes at the origin, so the contact contribution is zero. Other magnetic interactions can nevertheless produce hyperfine structure in those states.