Intermediate-field case of the Zeeman effect

The intermediate-field case of the Zeeman effect is the regime in which an atom or molecule in an external static magnetic field exhibits splitting of its spectral lines into multiple components, with the field strength comparable to the internal spin–orbit interaction.

To analyse the intermediate-field regime, we consider a hydrogen atom in an external magnetic field whose strength is comparable to that of the internal spin-orbit coupling ( $\boldsymbol{\mathit{B}}_{ext}\approx\boldsymbol{\mathit{B}}_{so}$ ). In this case, neither the coupled basis $\vert n,l,j,m_j\rangle$ nor the uncoupled basis $\vert n,l,m_l,m_s\rangle$ is strictly a good quantum state. Nevertheless, the Hamiltonian $\hat{H}$ remains invariant under rotations about the direction of the external magnetic field (taken as the lab $z$ -axis).

With reference to eq330 and eq331, where the spin-orbit Hamiltonian is $\hat{H}_{so}=A\boldsymbol{\mathit{L}}\cdot\boldsymbol{\mathit{S}}$ , the perturbed part of the Hamiltonian becomes

$\hat{H}'=A\boldsymbol{\mathit{L}}\cdot\boldsymbol{\mathit{S}}+\frac{eB}{2m}(\hat{L}_z+2\hat{S}_z)$

if we take $\boldsymbol{\mathit{B}}_{ext}=B\hat{\boldsymbol{\mathit{k}}}$ , where $\hat{\boldsymbol{\mathit{k}}}$ is a unit vector.

The spin-orbit term $A\boldsymbol{\mathit{L}}\cdot\boldsymbol{\mathit{S}}$ commutes with $\hat{J}^2$ and $\hat{J}_z$ but not with $\hat{L}_z$ or $\hat{S}_z$ , while the Zeeman term $\frac{eB}{2m}(\hat{L}_z+2\hat{S}_z)$ commutes with $\hat{L}_z$ and $\hat{S}_z$ , and hence with $\hat{J}_z$ , but not with $\hat{J}^2$ . Consequently, $[\hat{H},\hat{J}_z]=0$ and $m_j=m_l+m_s$ may still be regarded as a good quantum number.

Question

Show that $\hat{L}_x\hat{S}_x+\hat{L}_y\hat{S}_y=\frac{1}{2}(\hat{L}_{+}\hat{S}_{-}+\hat{L}_{-}\hat{S}_{+})$ .

Answer

The ladder operators of quantum orbital angular momentum are defined as $\hat{L}_+=\hat{L}_{x}+i\hat{L}_{y}$ and $\hat{L}_-=\hat{L}_{x}-i\hat{L}_{y}$ . Therefore, $\hat{L}_x=\frac{1}{2}(\hat{L}_{+}+\hat{L}_{-})$ and $i\hat{L}_y=\frac{1}{2}(\hat{L}_{+}-\hat{L}_{-})$ . Similarly, the ladder operators of spin angular momentum are $\hat{S}_+=\hat{S}_{x}+i\hat{S}_{y}$ and $\hat{S}_-=\hat{S}_{x}-i\hat{S}_{y}$ , with $\hat{S}_x=\frac{1}{2}(\hat{S}_{+}+\hat{S}_{-})$ and $i\hat{S}_y=\frac{1}{2}(\hat{S}_{+}-\hat{S}_{-})$ . It follows that $\hat{L}_x\hat{S}_x=\frac{1}{4}(\hat{L}_{+}\hat{S}_{+}+\hat{L}_{+}\hat{S}_{-}+\hat{L}_{-}\hat{S}_{+}+\hat{L}_{-}\hat{S}_{-})$ and $\hat{L}_y\hat{S}_y=-\frac{1}{4}(\hat{L}_{+}\hat{S}_{+}-\hat{L}_{+}\hat{S}_{-}-\hat{L}_{-}\hat{S}_{+}+\hat{L}_{-}\hat{S}_{-})$ . Therefore, $\hat{L}_x\hat{S}_x+\hat{L}_y\hat{S}_y=\frac{1}{2}(\hat{L}_{+}\hat{S}_{-}+\hat{L}_{-}\hat{S}_{+})$ .

Substituting $\boldsymbol{\mathit{L}}\cdot\boldsymbol{\mathit{S}}=\hat{L}_x\hat{S}_x+\hat{L}_y\hat{S}_y+\hat{L}_z\hat{S}_z=\frac{1}{2}(\hat{L}_{+}\hat{S}_{-}+\hat{L}_{-}\hat{S}_{+})+\hat{L}_z\hat{S}_z$ into $\hat{H}'$ gives:

$\hat{H}'=\frac{A}{2}(\hat{L}_{+}\hat{S}_{-}+\hat{L}_{-}\hat{S}_{+})+A\hat{L}_z\hat{S}_z+\frac{eB}{2m}(\hat{L}_z+2\hat{S}_z)\;\;\;\;\;\;\;\;340$

To determine the energy levels, we can no longer rely on the simple first-order perturbation formulas derived in eq334 or eq336. Instead, the spin-orbit Hamiltonian $\hat{H}_{so}$ and the Zeeman Hamiltonian $\hat{H}_Z$ must be treated on equal footing. One qualitative approach is to interpolate between the energy sublevels obtained in the weak-field and strong-field limits. However, a more rigorous method involves the following steps:

- choosing the basis state,
- identifying the eigenstates, and
- using the degenerate perturbation theory to determine the energy levels.

To illustrate this method, we consider the 2p¹ configuration of the hydrogen atom, with $l=1$ and $s=1/2$ . In the absence of an external magnetic field, spin-orbit coupling (using LS coupling) splits the configuration into two energy levels, $^2P_{3/2}$ and $^2P_{1/2}$ , with degeneracies $2J+1=4$ and $2J+1=2$ respectively. Even though only $m_j$ remains a good quantum number, we can choose $\vert m_l,m_s\rangle$ as the uncoupled basis state and form linear combinations as needed to generate the eigenstates.

Question

Explain further why $\vert m_l,m_s\rangle$ can be chosen as the basis state even though $m_l$ and $m_s$ are not good quantum numbers.

Answer

In Hilbert space, the set of states $\vert m_l,m_s\rangle$ forms a complete orthonormal basis and therefore spans the entire space, regardless of whether $m_l$ and $m_s$ are good quantum numbers. Quantum numbers only need to be good if we want the Hamiltonian to be diagonal in the chosen basis.

As mentioned earlier, the spin-orbit Hamiltonian and the Zeeman Hamiltonian must be treated on equal footing. The basis $\vert j,m_j\rangle$ diagonalises the spin-orbit Hamiltonian but not the Zeeman Hamiltonian, while the basis $\vert m_l,m_s\rangle$ diagonalises the Zeeman Hamiltonian but not the spin-orbit Hamiltonian. Since neither term dominates, no basis diagonalises the full Hamiltonian. Nevertheless, $\vert m_l,m_s\rangle$ is a convenient choice because the Zeeman term is diagonal in this basis and the matrix elements of the spin–orbit operator $\boldsymbol{\mathit{L}}\cdot\boldsymbol{\mathit{S}}$ can be calculated relatively easily.

Since $m_j$ is the only conserved quantity, we group the eigenstates according to their $m_j$ values. For $l=1$ and $s=1/2$ , the possible $m_j$ values are $\pm 3/2$ and $\pm 1/2$ .

The basis states $\vert m_l,m_s\rangle=\vert 1,1/2\rangle$ and $\vert -1,-1/2\rangle$ corresponding to $m_j=3/2$ and $m_j=-3/2$ respectively are regarded as unmixed states because each value of $m_j$ corresponds to a unique pair of $m_l$ and $m_s$ . Consequently, these states, $\vert\psi_{3/2}\rangle=c_1\vert 1,1/2\rangle$ and $\vert\psi_{-3/2}\rangle=c_6\vert -1,-1/2\rangle$ , are eigenstates of the perturbed part of the Hamiltonian.

For $m_j=1/2$ , there are two possible basis states, $\vert 1,-1/2\rangle$ and $\vert 0,1/2\rangle$ , and the corresponding eigenstates are linear combinations of these states: $\vert\psi_{1/2}\rangle=c_2\vert 1,-1/2\rangle+c_3\vert 0,1/2\rangle$ . Similarly, for $m_j=-1/2$ , the eigenstates are linear combinations of $\vert -1,1/2\rangle$ and $\vert 0,-1/2\rangle$ , i.e. $\vert\psi_{-1/2}\rangle=c_4\vert -1,1/2\rangle+c_5\vert 0,-1/2\rangle$ . We refer to these linear combinations as mixed states.

Having identified the eigenstates, the remaining task is to determine the energy levels. The expectation value for $\hat{H}_{3/2}$ is $\langle\hat{H}_{3/2}\rangle=\langle 1,1/2\vert\hat{H}'\vert 1,1/2\rangle$ , in which $\hat{H}'$ is given by eq340. From eq144, and noting that $\vert m_l,m_s\rangle$ can be expressed as the Krönecker product $\vert m_l,m_s\rangle=\vert l,m_l\rangle\otimes\vert s,m_s\rangle$ ,

$\hat{L}_+\hat{S}_-\vert m_l,m_s\rangle=\sqrt{l(l+1)-m_l(m_l+1)}\vert l,m_l+1\rangle\otimes\hat{S}_-\vert s,m_s\rangle$

For $\vert m_l,m_s\rangle=\vert 1,1/2\rangle$ , we have $\hat{L}_+\hat{S}_-\vert m_l,m_s\rangle=0$ . Similarly, using the spin analogue of eq144,

$\hat{L}_-\hat{S}_+\vert m_l,m_s\rangle=\hat{L}_-\vert l,m_l\rangle\otimes\sqrt{l(l+1)-m_l(m_l+1)}\vert s,m_s+1\rangle=0$

Furthermore, $\hat{L}_z\hat{S}_z\vert m_l,m_s\rangle=m_lm_s\hbar^2\vert m_l,m_s\rangle$ and $(\hat{L}_z+2\hat{S}_z)\vert m_l,m_s\rangle=\hbar(m_l+2m_s)\vert m_l,m_s\rangle$ . Therefore,

$\langle\hat{H}_{3/2}\rangle=\frac{A\hbar^2}{2}+2\mu_BB\;\;\;\;\;\;\;\;341$

where $\mu_B=\frac{e\hbar}{2m}$ .

Using eq147 and its spin analogue, and repeating the above logic,

$\langle\hat{H}_{-3/2}\rangle=\frac{A\hbar^2}{2}-2\mu_BB\;\;\;\;\;\;\;\;342$

For the mixed state $\vert \psi_{1/2}\rangle=\vert \phi_1\rangle+\vert \phi_2\rangle$ , where $\vert \phi_{1}\rangle=c_2\vert 1,-1/2\rangle$ and $\vert \phi_{2}\rangle=c_3\vert 0,1/2\rangle$ , the Hamiltonian is:

$\hat{H}'=\begin{pmatrix} \langle\phi_1 \vert\hat{H}'\vert \phi_{1}\rangle & \langle\phi_1 \vert\hat{H}'\vert \phi_{2}\rangle\\\langle\phi_2 \vert\hat{H}'\vert \phi_{1}\rangle &\langle\phi_2 \vert\hat{H}'\vert \phi_{2}\rangle\end{pmatrix}$

Repeating the steps used for the unmixed states to compute the matrix elements gives:

$\hat{H}'=\begin{pmatrix} -\frac{A\hbar^2}{2} & \frac{A\hbar^2\sqrt{2}}{2}\\\frac{A\hbar^2\sqrt{2}}{2} &\mu_BB\end{pmatrix}$

To find the eigenvalues, we solve the secular equation $\vert\hat{H}'-EI\vert=0$ or equivalently,

$\begin{vmatrix} -\frac{A\hbar^2}{2}-E & \frac{A\hbar^2\sqrt{2}}{2}\\\frac{A\hbar^2\sqrt{2}}{2} &\mu_BB-E\end{vmatrix}=0$

Expanding the determinant gives the characteristic equation:

$E^2 -\biggr$\mu_BB-\frac{A\hbar^2}{2}\biggr$E-\frac{A\hbar^2\mu_BB}{2}-\frac{A^2\hbar^4}{2}=0\;\;\;\;\;\;\;\;343$

Solving the quadratic equation yields two energy levels that depend on $B$ :

$E=\frac{\biggr$\mu_BB-\frac{A\hbar^2}{2}\biggr$\pm\sqrt{(\mu_BB)^2+A\hbar^2\mu_BB+\frac{9A^2\hbar^4}{4}}}{2}\;\;\;\;\;\;\;\;344$

A similar analysis for $\vert \psi_{-1/2}\rangle=\vert \phi_3\rangle+\vert \phi_4\rangle$ , where $\vert \phi_{3}\rangle=c_4\vert -1,1/2\rangle$ and $\vert \phi_{4}\rangle=c_5\vert 0,-1/2\rangle$ leads to:

$\hat{H}'=\begin{pmatrix} -\frac{A\hbar^2}{2} & \frac{A\hbar^2\sqrt{2}}{2}\\\frac{A\hbar^2\sqrt{2}}{2} &-\mu_BB\end{pmatrix}$

with

$E^2 +\biggr$\mu_BB+\frac{A\hbar^2}{2}\biggr$E+\frac{A\hbar^2\mu_BB}{2}-\frac{A^2\hbar^4}{2}=0\;\;\;\;\;\;\;\;345$

and

$E=\frac{-\biggr$\mu_BB+\frac{A\hbar^2}{2}\biggr$\pm\sqrt{(\mu_BB)^2-A\hbar^2\mu_BB+\frac{9A^2\hbar^4}{4}}}{2}\;\;\;\;\;\;\;\;346$

Therefore, the intermediate-field case results in six energy levels for the 2p¹ configuration of the hydrogen atom, consistent with the weak-field and strong-field cases, with the perturbed portion of the Hamiltonian given by a $6\times 6$ matrix:

$\hat{H}'=\begin{pmatrix} \frac{A\hbar^2}{2}+2\mu_BB &0&0&0&0&0\\0& -\frac{A\hbar^2}{2}&\frac{A\hbar^2\sqrt{2}}{2}&0&0&0\\0&\frac{A\hbar^2\sqrt{2}}{2} &\mu_BB&0&0&0\\0&0&0&-\frac{A\hbar^2}{2}&\frac{A\hbar^2\sqrt{2}}{2}&0\\0&0&0&\frac{A\hbar^2\sqrt{2}}{2}&- \mu_BB&0\\0&0&0&0&0&\frac{A\hbar^2}{2}-2\mu_BB\end{pmatrix}$

To show that the matrix elements of $\hat{H}'$ are consistent with the eigenvalues at the extreme limits of $B$ , we consider the cases $B\rightarrow 0$ and $B\rightarrow\infty$ .

When $B\rightarrow 0$ eq343 and eq345 reduce to the same equation: $E^2+\frac{A\hbar^2}{2}E-\frac{A^2\hbar^4}{2}=0$ , with solutions $E_1=\frac{A\hbar^2}{2}$ and $E_2=-A\hbar^2$ . These correspond to the spin-orbit energy levels $^2P_{3/2}$ and $^2P_{1/2}$ in the absence of an external magnetic field. Similarly, the eigenvalues of the unmixed states collapse onto the same energy as the $^2P_{3/2}$ level, namely $\langle\hat{H}_{\pm3/2}\rangle=\frac{A\hbar^2}{2}$ . In other words, the degeneracies are restored, with $^2P_{3/2}$ being four-fold degenerate and $^2P_{1/2}$ being two-fold degenerate.

In the opposite limit $B\rightarrow\infty$ , the unmixed-state eigenvalues become $\langle\hat{H}_{\pm3/2}\rangle=\frac{A\hbar^2}{2}\pm2\mu_BB$ , which agrees with the strong-field regime for the states $(m_l,m_s)=(\pm 1,\pm 1/2)$ . For the mixed states, we refer to eq346, where the $(\mu_BB)^2$ and $A\hbar^2\mu_BB$ terms dominate the discriminant, allowing it to be approximated by the perfect square $\biggr$\mu_BB+\frac{A\hbar^2}{2}\biggr$^2$ . So,

$E\approx\frac{\biggr$\mu_BB-\frac{A\hbar^2}{2}\biggr$\pm\biggr$\mu_BB+\frac{A\hbar^2}{2}\biggr$}{2}=\biggr\{\begin{matrix}\mu_BB\\-\frac{A\hbar^2}{2}\end{matrix}\;\;\;\;\;\;\;\;347$

Similarly, eq346 yields

$E\approx\frac{-\biggr$\mu_BB+\frac{A\hbar^2}{2}\biggr$\pm\biggr$\mu_BB-\frac{A\hbar^2}{2}\biggr$}{2}=\biggr\{\begin{matrix}-\frac{A\hbar^2}{2}\\-\mu_BB\end{matrix}\;\;\;\;\;\;\;\;348$

The energies given by eq347 and eq348 are again consistent with those obtained in the strong-field case.

Question

Why is the intermediate-field case of the Zeeman effect analysed using degenerate perturbation theory, while the weak-field and strong-field cases are not?

Answer

Degenerate perturbation theory can be applied to all three cases. However, in the weak-field and strong-field regimes the eigenvalues are usually obtained more simply by computing expectation values, since the relevant Hamiltonians are already diagonal in the corresponding unmixed eigenstate bases.

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