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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.
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