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I am trying to solve the following problem on an old Quantum Mechanics exam as an exercise.

I know that the trace of an operator is the integral of its kernel.

\begin{equation}

Tr[K(x,y)] = \int K(x,x) dx

\end{equation}

I start out by finding the action of the operator in the trace:

\begin{equation}

(P_{\psi} V(-v) U(-u) \phi)(x) = \int \psi^{\star} e^{i v x} \phi(x+\hbar u) dx \psi(x)

\end{equation}

Which I think should, by sending $x\rightarrow y-\hbar u$, be equal to:

\begin{equation}

\int \psi^{\star}(y- \hbar u) e^{i v (y-\hbar u)} \phi(y) \psi(x) dy

\end{equation}

Which means that the integral kernel of this operator is:

\begin{equation}

K_{\psi}(x,y) = \psi^{\star}(y- \hbar u) e^{i v (y-\hbar u)} \psi(x)

\end{equation}

Which means that the trace is:

\begin{equation}

\int K_{\psi}(x,x) dx = \int \psi^{\star}(x- \hbar u) e^{i v (x-\hbar u)} \psi(x) dx

\end{equation}

So essentially I am stuck trying to solve this integral.

Any help would be greatly appreciated.

## Homework Statement

## Homework Equations

I know that the trace of an operator is the integral of its kernel.

\begin{equation}

Tr[K(x,y)] = \int K(x,x) dx

\end{equation}

## The Attempt at a Solution

I start out by finding the action of the operator in the trace:

\begin{equation}

(P_{\psi} V(-v) U(-u) \phi)(x) = \int \psi^{\star} e^{i v x} \phi(x+\hbar u) dx \psi(x)

\end{equation}

Which I think should, by sending $x\rightarrow y-\hbar u$, be equal to:

\begin{equation}

\int \psi^{\star}(y- \hbar u) e^{i v (y-\hbar u)} \phi(y) \psi(x) dy

\end{equation}

Which means that the integral kernel of this operator is:

\begin{equation}

K_{\psi}(x,y) = \psi^{\star}(y- \hbar u) e^{i v (y-\hbar u)} \psi(x)

\end{equation}

Which means that the trace is:

\begin{equation}

\int K_{\psi}(x,x) dx = \int \psi^{\star}(x- \hbar u) e^{i v (x-\hbar u)} \psi(x) dx

\end{equation}

So essentially I am stuck trying to solve this integral.

Any help would be greatly appreciated.