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## Question: Correlation function ([1] pr. 2.16)

A correlation function can be defined as

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

C(t) = \expectation{ x(t) x(0) }.

\end{equation}

Using a Heisenberg picture \( x(t) \) calculate this correlation for the one dimensional SHO ground state.

## Answer

The time dependent Heisenberg picture position operator was found to be

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

x(t) = x(0) \cos(\omega t) + \frac{p(0)}{m \omega} \sin(\omega t),

\end{equation}

so the correlation function is

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

\begin{aligned}

C(t)

&=

\bra{0} \lr{ x(0) \cos(\omega t) + \frac{p(0)}{m \omega} \sin(\omega t)} x(0) \ket{0} \\

&=

\cos(\omega t) \bra{0} x(0)^2 \ket{0} + \frac{\sin(\omega t)}{m \omega} \bra{0} p(0) x(0) \ket{0} \\

&=

\frac{\Hbar \cos(\omega t) }{2 m \omega} \bra{0} \lr{ a + a^\dagger}^2 \ket{0} – \frac{i \Hbar}{m \omega} \sin(\omega t),

\end{aligned}

\end{equation}

But

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

\begin{aligned}

\lr{ a + a^\dagger} \ket{0}

&=

a^\dagger \ket{0} \\

&=

\sqrt{1} \ket{1} \\

&=

\ket{1},

\end{aligned}

\end{equation}

so

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

C(t) = x_0^2 \lr{ \inv{2} \cos(\omega t) – i \sin(\omega t) },

\end{equation}

where \( x_0^2 = \Hbar/(m \omega) \), not to be confused with \( x(0)^2 \).

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## Question: Partition function and ground state energy ([1] pr. 2.32)

Define the partition function as

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

Z = \int d^3 x’ \evalbar{ K( \Bx’, t ; \Bx’, 0 ) }{\beta = i t/\Hbar},

\end{equation}

Show that the ground state energy is given by

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

-\inv{Z} \PD{\beta}{Z}, \qquad \beta \rightarrow \infty.

\end{equation}

## Answer

The propagator evaluated at the same point is

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

\begin{aligned}

K( \Bx’, t ; \Bx’, 0 )

&=

\sum_{a’} \braket{\Bx’}{a’} \ket{a’}{\Bx’} \exp\lr{ -\frac{i E_{a’} t}{\Hbar}} \\

&=

\sum_{a’} \Abs{\braket{\Bx’}{a’}}^2 \exp\lr{ -\frac{i E_{a’} t}{\Hbar}} \\

&=

\sum_{a’} \Abs{\braket{\Bx’}{a’}}^2 \exp\lr{ -E_{a’} \beta}.

\end{aligned}

\end{equation}

The derivative is

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

\PD{\beta}{Z}

=

-\int d^3 x’ \sum_{a’} E_{a’} \Abs{\braket{\Bx’}{a’}}^2 \exp\lr{ -E_{a’} \beta}.

\end{equation}

In the \( \beta \rightarrow \infty \) this sum will be dominated by the term with the lowest value of \( E_{a’} \). Suppose that state is \( a’ = 0 \), then

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

\lim_{ \beta \rightarrow \infty }

-\inv{Z} \PD{\beta}{Z}

= \frac{

\int d^3 x’ E_{0} \Abs{\braket{\Bx’}{0}}^2 \exp\lr{ -E_{0} \beta}

}

{

\int d^3 x’ \Abs{\braket{\Bx’}{0}}^2 \exp\lr{ -E_{0} \beta}

}

= E_0.

\end{equation}

# References

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

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