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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.