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### Q: [1] pr. 5.18

Work out the quadratic Zeeman effect for the ground state hydrogen atom due to the usually neglected \( e^2 \BA^2/2 m_e c^2 \) term in the Hamiltonian.

### A:

The first order energy shift is

For a z-oriented magnetic field we can use

\begin{equation}\label{eqn:quadraticZeeman:20}

\BA = \frac{B}{2} \setlr{ -y, x, 0 },

\end{equation}

so the perturbation potential is

\begin{equation}\label{eqn:quadraticZeeman:40}

\begin{aligned}

V

&= \frac{e^2 \BA^2}{2 m_e c^2} \\

&= \frac{e^2 \BB^2 (x^2 + y^2)}{8 m_e c^2} \\

&= \frac{ e^2 \BB^2 r^2 \sin^2\theta }{8 m_e c^2}

\end{aligned}

\end{equation}

The ground state wave function is

\begin{equation}\label{eqn:quadraticZeeman:60}

\begin{aligned}

\psi_0

&= \braket{\Bx}{0} \\

&= \inv{\sqrt{\pi a_0^3}} e^{-r/a_0},

\end{aligned}

\end{equation}

so the energy shift is

\begin{equation}\label{eqn:quadraticZeeman:80}

\begin{aligned}

\Delta

&= \bra{0} V \ket{0} \\

&= \inv{ \pi a_0^3 } 2 \pi \frac{ e^2 \BB^2 }{8 m_e c^2} \int_0^\infty r^2 \sin\theta e^{-2r/a_0} r^2 \sin^2\theta dr d\theta \\

&=

\frac{ e^2 \BB^2 }{4 a_0^3 m_e c^2}

\int_0^\infty r^4 e^{-2r/a_0} dr \int_0^\pi \sin^3\theta d\theta \\

&= –

\frac{ e^2 \BB^2 }{4 a_0^3 m_e c^2}

\frac{4!}{(2/a_0)^{4+1} } \evalrange{\lr{u – \frac{u^3}{3}}}{1}{-1} \\

&=

\frac{ e^2 a_0^2 \BB^2 }{4 m_e c^2}.

\end{aligned}

\end{equation}

If this energy shift is written in terms of a diamagnetic susceptibility \( \chi \) defined by

\begin{equation}\label{eqn:quadraticZeeman:100}

\Delta = -\inv{2} \chi \BB^2,

\end{equation}

the diamagnetic susceptibility is

\begin{equation}\label{eqn:quadraticZeeman:120}

\chi = -\frac{ e^2 a_0^2 }{2 m_e c^2}.

\end{equation}

# References

[1] Jun John Sakurai and Jim J Napolitano. *Modern quantum mechanics*. Pearson Higher Ed, 2014.