Consider a hydrogen atom in an external electric field: \[\hat{H}_0=\frac{\hat{p}^2}{2\mu}-\frac{e^2}{\hat{r}},
\ \text{and}\
\hat{V}_{FS}=-\frac{1}{8}\frac{\hat{p}^4}{\mu^3c^2}+\frac{1}{2}\left(\frac{\hbar}{mc}\right)^2\hat{\mathbf{L}}\cdot\hat{\mathbf{S}}\frac{e^2}{\hat{r}^3}\]
and if we choose the electric field to be in the \(z\)-direction, we have that \[\hat{V}_{Stark}=e\mathbf{\varepsilon}\cdot\hat{\mathbf{r}}=e\varepsilon
z=e\varepsilon r\cos{\theta}.\] What are good quantum numbers?
For \(\hat{H_0}:\) \[\hat{L}^2,\hat{L}_z,\hat{S}^2,\hat{S}_z,\hat{J}^2,\hat{J}_z\]
For \(\hat{H}_0+\hat{H}_{FS}:\) \[\hat{L}^2,\hat{S}^2,\hat{J}^2,\hat{J}_z\]
and for \(\hat{H}_0+\hat{H}_{FS}+\hat{H}_{Stark}:\)
\[\hat{S}^2,\hat{J}_z.\] \(\hat{V}_{Stark}\) has odd parity so it
connects states with different parity (\(l\) odd with \(l\) even).
Since the ground state is an \(s\)-wave, there is no linear Stark shift.
But, for \(n\ge 2\) different \(l\) are degenerate for \(\hat{H}_0\) so a first order shift is
possible.
The ground state has \(n=1, l=0,
s=\frac{1}{2},\) and \(j=\frac{1}{2}\) with degeneracy coming from
the fact that \(m_j=\pm\frac{1}{2}\).
But \(m_j\) is a good quantum number so
there is no first order shift. So, \[E_{gs}=E_{FS}+E_{Stark}^{(2)}\] since
\(E_{Stark}^{(1)}=0\). Then, \[E_{Stark}^{(2)}=\sum_{njl,
n\ne1}\frac{|\left\langle
njm_sl\middle|\hat{V}_{Stark}\middle|n{=}1,j{=}\frac{1}{2},m_j,l{=}0\right\rangle|^2}{E_1^{(0)}-E_{njl}^{(0)}}\]
The smallest value of the denominator is for \(n=2\). The denominator goes like \(\sim\frac{e^2}{a_0}\) and the numerator
goes like \(e^2\varepsilon^2a_0^2\) so
the shift is on the order of \(\varepsilon^2a_0^3\). \[|E^{(2)}|\le\frac{e^2\varepsilon^2}{E_1^{(0)}-E_2^{(0)}}\sum_{njlm'}\left\langle\frac{1}{2}
\ 1 \ m_j \ 0\middle|z\middle|njm'l\right\rangle\left\langle
njm'l\middle|z\middle|\frac{1}{2} \ 1 \ m_j \
0\right\rangle\] But, by completeness \[\sum_{njlm'}\left|njm'l\right\rangle\left\langle
njm'l\right|=\mathbb{I}\] so, \[|E^{(2)}|\le\frac{e^2\varepsilon^2}{E_1^{(0)}-E_2^{(0)}}\left\langle
1 \ \frac{1}{2} \ m \ 0\middle|z^2\middle|1 \ \frac{1}{2} \ m \
0\right\rangle\] Therefore, \[|E^{(2)}|\le\frac{8}{3}\varepsilon^2a_0^3.\]
This method is similar to the method of Dalgarno and Lewis (Shiff pg
266) which you may want to look at.
Here, we can neglect fine structure. The first nontrivial case is \(n=2\). This gives us \(j=\frac{3}{2},\frac{1}{2}\), \(m_j=\pm\frac{3}{2}\) (which have no linear shift) and \(m_j=\pm\frac{1}{2}\) (which can mix states). It turns out that \(m_j=\pm\frac{1}{2}\) are degenerate and called Kramer’s doublets. This gives us three states \(\left|njml\right\rangle\): \[\left|2 \ \frac{3}{2} \ \frac{1}{2} \ 1\right\rangle,\left|2 \ \frac{1}{2} \ \frac{1}{2} \ 1\right\rangle,\left|2 \ \frac{1}{2} \ \frac{1}{2} \ 0\right\rangle\] In the degenerate subspace we have \[\begin{pmatrix}E_2^{(0)} & 0 & a \\ 0 & E_2^{(0)} & b \\ a^* & b^* & E_2^{(0)}\end{pmatrix}\] which is similar to a HW problem. Note that \[\begin{aligned} a=e\varepsilon\left\langle 2 \ \frac{3}{2} \ \frac{1}{2} \ 1\middle|z\middle|2 \ \frac{1}{2} \ \frac{1}{2} \ 0\right\rangle=a^* \\ b=e\varepsilon\left\langle 2 \ \frac{1}{2} \ \frac{1}{2} \ 1\middle|z\middle|2 \ \frac{1}{2} \ \frac{1}{2} \ 0\right\rangle=b^* \end{aligned}\] After converting to wavefunctions and integrating, one can find \[\begin{aligned} a=-\sqrt{6}a_0e\varepsilon \\ b=-\sqrt{3}a_0e\varepsilon \end{aligned}\] Now, we find the eigenvalues, \[\det\begin{pmatrix}E_2^0-E & 0 & a \\ 0 & E_2^0-E & \frac{1}{\sqrt{2}}a \\ a & \frac{1}{\sqrt{2}}a & E_2^0-E\end{pmatrix}(E_2^0-E)^3-(E_2^0-E)a^2\frac{3}{2}=0\] \[\implies E=E_2^0,E=E_2^0\pm3a_0e\varepsilon\]
We normally take relativistic effects and spin orbit coupling first then add the Stark effect but we can do it all at once. The \(2p_{3/2}\) level has energy \(E_{3/2}\) and the \(2p_{1/2}2s_{1/2}\) level has energy \(E_{1/2}\). \[\det\begin{pmatrix}E_{3/2} & 0 & a \\ 0 & E_{1/2}-E & \frac{1}{\sqrt{2}}a \\ a & \frac{1}{\sqrt{2}}a & E_{1/2}-E\end{pmatrix}=(E_{3/2}-E)(E_{1/2}-E)^2-(E_{1/2}-E)|a|^2-(E_{3/2}-E)\frac{|a|^2}{2}=0\] which we solve to get the roots. The algebra is straightforward but not too illuminating.
Consider 3 spins on a triangle: \[\hat{H}=A(\hat{\mathbf{S}_1}\cdot\hat{\mathbf{S}_2}+\hat{\mathbf{S}_2}\cdot\hat{\mathbf{S}_3}+\hat{\mathbf{S}_3}\cdot\hat{\mathbf{S}_1})+BS_1^z\] for a small \(B\)-field. Here \(J_z\) is a good quantum number. Here there are \(2^3=8\) states labeled with \(J,m_j\). \(j=\frac{3}{2}\) has 4 states, \(j=\frac{1}{2}\) has 2 states, and \(j=\frac{1}{2}\) has 2 states. Use \(S_{tot}^-=S_1^-+S_2^-+S_3^-\) to get all of the states. \[\begin{aligned} \left|j=\frac{3}{2},m_j=\frac{3}{2}\right\rangle=& \ \left|\uparrow\uparrow\uparrow\right\rangle \\ \left|j=\frac{3}{2},m_j=\frac{1}{2}\right\rangle=& \ \frac{1}{\sqrt{3}}(\left|\downarrow\uparrow\uparrow\right\rangle+\left|\uparrow\downarrow\uparrow\right\rangle+\left|\uparrow\uparrow\downarrow\right\rangle) \\ \left|j=\frac{3}{2},m_j=-\frac{1}{2}\right\rangle=& \ \frac{1}{\sqrt{3}}(\left|\downarrow\downarrow\uparrow\right\rangle+\left|\downarrow\uparrow\downarrow\right\rangle+\left|\uparrow\downarrow\downarrow\right\rangle) \\ \left|j=\frac{3}{2},m_j=-\frac{3}{2}\right\rangle=& \ \left|\downarrow\downarrow\downarrow\right\rangle \\ \end{aligned}\] Then, \[\begin{aligned} \left|j=\frac{1}{2},m_j=\frac{1}{2}\right\rangle=& \ \frac{1}{\sqrt{2}}(\left|\downarrow\uparrow\uparrow\right\rangle-\left|\uparrow\downarrow\uparrow\right\rangle)\equiv\left|1\right\rangle \\ \left|j=\frac{1}{2},m_j=-\frac{1}{2}\right\rangle=& \ \frac{1}{\sqrt{2}}(\left|\downarrow\uparrow\downarrow\right\rangle-\left|\uparrow\downarrow\downarrow\right\rangle) \end{aligned}\] and \[\begin{aligned} \left|j=\frac{1}{2},m_j=\frac{1}{2}\right\rangle'=& \ \frac{1}{\sqrt{6}}(\left|\downarrow\uparrow\uparrow\right\rangle+\left|\uparrow\downarrow\uparrow\right\rangle-2\left|\uparrow\uparrow\downarrow\right\rangle)\equiv\left|2\right\rangle \\ \left|j=\frac{1}{2},m_j=-\frac{1}{2}\right\rangle'=& \ \frac{1}{\sqrt{6}}(\left|\downarrow\uparrow\downarrow\right\rangle+\left|\uparrow\downarrow\downarrow\right\rangle-2\left|\downarrow\downarrow\uparrow\right\rangle) \end{aligned}\] Note \(S_1\cdot S_2+S_2\cdot S_3+S_3\cdot S_1=\frac{1}{2}((S_1+S_2+S_3)^2-S_1^2-S_2^2-S_3^2)\). Now, consider the \(j=\frac{1}{2},m_j=\frac{1}{2}\) state. \[E_0^0=-\frac{3}{4}A\] which is two-fold degenerate. Then, \[\begin{aligned} \left\langle 1\middle|S_1^z\middle|1\right\rangle=& \ 0 \\ \left\langle 1\middle|S_1^z\middle|2\right\rangle=& \ \frac{1}{\sqrt{2}}(\left\langle\downarrow\uparrow\uparrow\right|-\left\langle\uparrow\downarrow\uparrow\right|)\frac{1}{\sqrt{6}}\cdot\frac{1}{2}(\left|\downarrow\uparrow\uparrow\right\rangle+\left|\uparrow\downarrow\uparrow\right\rangle-2\left|\uparrow\uparrow\downarrow\right\rangle) \\ =& \ -\frac{1}{2\sqrt{3}} \\ \left\langle 2\middle|S_1^z\middle|1\right\rangle=& \ -\frac{1}{2\sqrt{3}} \\ \left\langle 2\middle|S_1^z\middle|2\right\rangle=& \ \frac{1}{6}\cdot\frac{1}{2}(\left\langle\downarrow\uparrow\uparrow\right|+\left\langle\uparrow\downarrow\uparrow\right|-2\left\langle\uparrow\uparrow\downarrow\right|)(\left|\downarrow\uparrow\uparrow\right\rangle+\left|\uparrow\downarrow\uparrow\right\rangle-2\left|\uparrow\uparrow\downarrow\right\rangle) \\ =& \frac{1}{3} \end{aligned}\] Then, we find the eigenvalues \[\det\begin{pmatrix}-\frac{3}{4}A-E & -\frac{B}{2\sqrt{3}} \\ -\frac{B}{2\sqrt{3}} & -\frac{3}{4}A+\frac{B}{3}-E\end{pmatrix}=0\] \[E^2-E\left(-\frac{3}{2}A+\frac{B}{3}\right)+\frac{9}{16}A^2-\frac{AB}{4}-\frac{B^2}{12}=0\] Hence, \[E=-\frac{3}{4}A+\frac{B}{6}\pm\frac{1}{3}B\] Compare with the exact solution which comes from a cubic equation. \[\det\begin{pmatrix}-\frac{3}{4}A-E & -\frac{B}{2\sqrt{3}} & -\frac{B}{\sqrt{6}} \\ -\frac{B}{2\sqrt{3}} & -\frac{3}{4}A+\frac{1}{3}B-E & -\frac{B}{3\sqrt{2}} \\ -\frac{B}{\sqrt{6}} & -\frac{B}{3\sqrt{2}} & \frac{3}{4}A+\frac{1}{6}B-E\end{pmatrix}=0\] Using Mathematica gives \[\begin{aligned} E=& \ -\frac{3}{4}A+\frac{1}{2}B \checkmark \\ E=& \ -\frac{1}{4}\sqrt{9A^2+4AB+4B^2}=-\frac{3}{4}A-\frac{1}{6}B +\cdots\checkmark \\ E=& \ \frac{1}{4}\sqrt{9A^2+4AB+4B^2}=\frac{3}{4}A+\frac{1}{6}B+\cdots \end{aligned}\]