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## Question: Expectations for SHO Hamiltonian, and virial theorem. ([1] pr. 2.3)

## (a)

For a 1D SHO, compute \(

\bra{m} x \ket{n},

\bra{m} x^2 \ket{n},

\bra{m} p \ket{n},

\bra{m} p^2 \ket{n} \) and \( \bra{m} \symmetric{x}{p} \ket{n} \).

## (b)

Verify the virial theorem is satisfied for energy eigenstates.

## Answer

## (a)

Using

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

\begin{aligned}

x &= \frac{x_0}{\sqrt{2}} \lr{ a + a^\dagger } \\

p &= \frac{i\Hbar}{x_0 \sqrt{2}} \lr{ a^\dagger – a} \\

a(t) &= a(0) e^{-i \omega t} \\

a(0) \ket{n} &= \sqrt{n} \ket{n-1} \\

a^\dagger(0) \ket{n} &= \sqrt{n+1} \ket{n+1} \\

x_0^2 &= \frac{\Hbar}{\omega m},

\end{aligned}

\end{equation}

we have

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

\begin{aligned}

\bra{m} x \ket{n}

&=

\frac{x_0}{\sqrt{2}} \bra{m} \lr{ a + a^\dagger } \ket{n} \\

&=

\frac{x_0}{\sqrt{2}} \bra{m}

\lr{

e^{-i \omega t} \sqrt{n} \ket{n-1}

+

e^{i \omega t} \sqrt{n+1} \ket{n+1}

} \\

&=

\frac{x_0}{\sqrt{2}} \lr{

\delta_{m, n-1} e^{-i \omega t} \sqrt{n}

+

\delta_{m, n+1} e^{i \omega t} \sqrt{n+1}

},

\end{aligned}

\end{equation}

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

\begin{aligned}

\bra{m} x^2 \ket{n}

&=

\frac{x_0^2}{2} \bra{m} \lr{ a + a^\dagger }^2 \ket{n} \\

&=

\frac{x_0^2}{2}

\lr{

e^{i \omega t} \sqrt{m} \bra{m-1}

+

e^{-i \omega t} \sqrt{m+1} \bra{m+1}

}

\lr{

e^{-i \omega t} \sqrt{n} \ket{n-1}

+

e^{i \omega t} \sqrt{n+1} \ket{n+1}

} \\

&=

\frac{x_0^2}{2}

\lr{

\delta_{m+1,n+1} \sqrt{(m+1)(n+1)}

+\delta_{m+1,n-1} \sqrt{(m+1)n} e^{-2 i \omega t}

+\delta_{m-1,n+1} \sqrt{m(n+1)} e^{2 i \omega t}

+\delta_{m-1,n-1} \sqrt{m n}

},

\end{aligned}

\end{equation}

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

\begin{aligned}

\bra{m} p \ket{n}

&=

\frac{i\Hbar}{\sqrt{2} x_0} \bra{m} \lr{ a^\dagger – a} \ket{n} \\

&=

\frac{i\Hbar}{\sqrt{2} x_0} \bra{m} \lr{

e^{i \omega t} \sqrt{n+1} \ket{n+1}

–

e^{-i \omega t} \sqrt{n} \ket{n-1}

} \\

&=

\frac{i\Hbar}{\sqrt{2} x_0} \lr{

\delta_{m,n+1} e^{i \omega t} \sqrt{n+1}

–

\delta_{m,n-1} e^{-i \omega t} \sqrt{n}

},

\end{aligned}

\end{equation}

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

\begin{aligned}

\bra{m} p^2 \ket{n}

&=

\frac{\Hbar^2}{2 x_0^2} \ket{m} \lr{ a – a^\dagger } \lr{ a^\dagger – a}

\ket{n} \\

&=

\frac{\Hbar^2}{2 x_0^2}

\lr{

-e^{-i \omega t} \sqrt{m+1} \bra{m+1}

+

e^{i \omega t} \sqrt{m} \bra{m-1}

}

\lr{

e^{i \omega t} \sqrt{n+1} \ket{n+1}

–

e^{-i \omega t} \sqrt{n} \ket{n-1}

} \\

&=

\frac{\Hbar^2}{2 x_0^2}

\lr{

\delta_{m+1,n+1} \sqrt{(m+1)(n+1)}

+\delta_{m+1,n-1} \sqrt{(m+1)n} e^{-2 i \omega t}

+\delta_{m-1,n+1} \sqrt{m(n+1)} e^{2 i \omega t}

+\delta_{m-1,n-1} \sqrt{m n}

}.

\end{aligned}

\end{equation}

For the anticommutator \( \symmetric{x}{p} \), we have

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

\begin{aligned}

\symmetric{x}{p}

&=

\frac{i\Hbar}{2}

\lr{

\lr{ a e^{-i \omega t} + a^\dagger e^{i \omega t} } \lr{ a^\dagger e^{i \omega t} – a e^{-i \omega t} }

–

\lr{ a^\dagger e^{i \omega t} – a e^{-i \omega t} }

\lr{ a e^{-i \omega t} + a^\dagger e^{i \omega t} }

} \\

&=

\frac{i\Hbar}{2}

\lr{

– a^2 e^{- 2 i \omega t}

+ (a^\dagger)^2 e^{ 2 i \omega t}

+ a a^\dagger

– a^\dagger a

+ a^2 e^{- 2 i \omega t}

– (a^\dagger)^2 e^{ 2 i \omega t}

– a^\dagger a

+ a a^\dagger

} \\

&=

i\Hbar

\lr{

a a^\dagger – a^\dagger a

},

\end{aligned}

\end{equation}

so

\begin{equation}\label{eqn:shoExpectations:140}

\begin{aligned}

\bra{m} \symmetric{x}{p} \ket{n}

&=

i\Hbar

\bra{m}

\lr{

a a^\dagger – a^\dagger a

}

\ket{n} \\

&=

i\Hbar

\bra{m}

\lr{

\sqrt{(n+1)^2}\ket{n}

-\sqrt{n^2}\ket{n}

} \\

&=

i\Hbar

\bra{m}

\lr{

2 n + 1

}

\ket{n}.

\end{aligned}

\end{equation}

## (b)

For the SHO, the virial theorem requires \( \expectation{p^2/m} = \expectation{m \omega x^2} \). That momentum expectation with respect to the eigenstate \( \ket{n} \) is

\begin{equation}\label{eqn:shoExpectations:160}

\begin{aligned}

\expectation{p^2/m}

&=

\frac{\Hbar^2}{2 x_0^2 m}

\lr{

\sqrt{(n+1)(n+1)}

+

\sqrt{n n}

} \\

&=

\frac{\Hbar^2 m \omega}{2 \Hbar m} \lr{ 2 n + 1 } \\

&=

\Hbar \omega \lr{ n + \inv{2} }.

\end{aligned}

\end{equation}

For the position expectation we’ve got

\begin{equation}\label{eqn:shoExpectations:180}

\begin{aligned}

\expectation{m \omega x^2}

&=

\frac{m \omega^2 x_0^2}{2}

\lr{

\sqrt{(n+1)(n+1)}

+ \sqrt{n n}

} \\

&=

\frac{m \omega^2 \Hbar}{2 m \omega}

\lr{

\sqrt{(n+1)(n+1)}

+ \sqrt{n n}

} \\

&=

\frac{\omega \Hbar}{2 }

\lr{ 2 n + 1 } \\

&=

\omega \Hbar

\lr{ n + \inv{2} }.

\end{aligned}

\end{equation}

This shows that the virial theorem holds for the SHO Hamiltonian for eigenstates.

# References

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