### Q: [1] pr 5.1

Given a perturbed 1D SHO Hamiltonian

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

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

\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}

The first order perturbation to the ground state is

\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}

The second order ground state energy perturbation is

\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}

so the total energy perturbation up to second order is

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

To compare to the exact result, rewrite the Hamiltonian as

\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}

The Hamiltonian is subject to a constant energy shift

\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}

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.