energy shift

Quadratic Zeeman effect

December 13, 2015 phy1520 No comments , , , ,

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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.

Harmonic oscillator with energy shift

December 5, 2015 phy1520 No comments , ,

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Q: [1] pr 5.1

Given a perturbed 1D SHO Hamiltonian

\begin{equation}\label{eqn:harmonicOscillatorEnergyShiftPertubation:20}
H = \inv{2m} p^2 + \inv{2} m \omega^2 x^2 + \lambda b x,
\end{equation}

calculate the first non-zero perturbation to the ground state energy. Then solve for that energy directly and compare.

A:

The first order energy shift is seen to be zero

\begin{equation}\label{eqn:harmonicOscillatorEnergyShiftPertubation:40}
\begin{aligned}
\Delta_0^{(0)}
&= V_{00} \\
&= \bra{0} b x \ket{0} \\
&= \frac{x_0}{\sqrt{2}} \bra{0} a + a^\dagger \ket{0} \\
&= \frac{x_0}{\sqrt{2}} \braket{0}{1} \\
&= 0.
\end{aligned}
\end{equation}

The first order perturbation to the ground state is

\begin{equation}\label{eqn:harmonicOscillatorEnergyShiftPertubation:60}
\begin{aligned}
\ket{0^{(1)}}
&= \sum_{m \ne 0} \frac{ \ket{m} \bra{m} b x \ket{0} }{ \Hbar \omega/2 – \Hbar
\omega (m – 1/2) } \\
&= -b \frac{x_0}{\sqrt{2} \Hbar \omega} \sum_{m \ne 0} \frac{ \ket{m}
\braket{m}{1} }{ m } \\
&= -b \frac{x_0}{\sqrt{2} \Hbar \omega} \ket{1}.
\end{aligned}
\end{equation}

The second order ground state energy perturbation is

\begin{equation}\label{eqn:harmonicOscillatorEnergyShiftPertubation:80}
\begin{aligned}
\Delta_0^{(2)}
&=
\bra{0} b x \ket{0^{(1)}} \\
&=
\frac{b x_0}{\sqrt{2}} \bra{0} a + a^\dagger \lr{ -b \frac{x_0}{\sqrt{2} \Hbar \omega} \ket{1} } \\
&=
\frac{b x_0}{\sqrt{2}} \lr{ -b \frac{x_0}{\sqrt{2} \Hbar \omega} } \\
&=
-\frac{b^2 x_0^2}{ 2 \Hbar \omega } \\
&=
-\frac{b^2 }{ 2 \Hbar \omega } \frac{\Hbar}{m \omega} \\
&=
-\frac{b^2 }{ 2 m \omega^2 },
\end{aligned}
\end{equation}

so the total energy perturbation up to second order is

\begin{equation}\label{eqn:harmonicOscillatorEnergyShiftPertubation:100}
\Delta_0 = -\lambda^2 \frac{b^2 }{ 2 m \omega^2 }.
\end{equation}

To compare to the exact result, rewrite the Hamiltonian as

\begin{equation}\label{eqn:harmonicOscillatorEnergyShiftPertubation:120}
\begin{aligned}
H
&= \inv{2m} p^2 + \inv{2} m \omega^2 \lr{ x^2 + \frac{2 \lambda b x}{m \omega^2} } \\
&= \inv{2m} p^2 + \inv{2} m \omega^2 \lr{ x + \frac{\lambda b }{m \omega^2} }^2 – \inv{2} m \omega^2 \lr{ \frac{\lambda b }{m \omega^2} }^2.
\end{aligned}
\end{equation}

The Hamiltonian is subject to a constant energy shift

\begin{equation}\label{eqn:harmonicOscillatorEnergyShiftPertubation:140}
\begin{aligned}
\Delta E
&=
– \inv{2} m \omega^2 \frac{\lambda^2 b^2 }{m^2 \omega^4} \\
&=
– \frac{\lambda^2 b^2 }{2 m \omega^2}.
\end{aligned}
\end{equation}

This is an exact match with the second order perturbation result of \ref{eqn:harmonicOscillatorEnergyShiftPertubation:100}.

References

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