correlation function

Second update of aggregate notes for phy1520, Graduate Quantum Mechanics

October 20, 2015 phy1520 No comments , , , , , , , , , , , ,

I’ve posted a second update of my aggregate notes for PHY1520H Graduate Quantum Mechanics, taught by Prof. Arun Paramekanti. In addition to what was noted previously, this contains lecture notes up to lecture 9, my ungraded solutions for the second problem set, and some additional worked practise problems.

Most of the content was posted individually in the following locations, but those original documents will not be maintained individually any further.

Correlation function. Partition function and ground state energy.

September 5, 2015 phy1520 No comments , , , , ,

[Click here for a PDF of this problem with nicer formatting]

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 \( x(0)^2 \).

[Click here for a PDF of this problem with nicer formatting]

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.