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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.