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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.