Translation operators and positive-semidefinite condition

Question

Good day. I have an operator $\exp{(-i\mu\hat{q})}\exp{(-i\nu\hat{p})}$, where $\hat{q}$, $\hat{p}$ are the position and momentum operators, respectively. The parameters $\mu,\nu$ are some real numbers. I am interested if $\exp{(-i\mu\hat{q})}\exp{(-i\nu\hat{p})}$ is positive semi-definite operator?

Then I need to check $\langle \chi|\exp{(-i\mu\hat{q})}\exp{(-i\nu\hat{p})}|\chi\rangle \geq 0$, where $|\chi\rangle$ is any state. My derivations are:

\begin{eqnarray} &&\langle \chi|\exp{(-i\mu\hat{q})}\exp{(-i\nu\hat{p})}|\chi\rangle\\\nonumber &=& \iint dx' dx'' \langle \chi|x''\rangle\langle x''|\exp{(-i\mu\hat{q})}\exp{(-i\nu\hat{p})}|x'\rangle\langle x'|\chi\rangle \\\nonumber &=&\iint dx' dx''\exp{(-i\mu x'')} \langle \chi|x''\rangle\langle x''|x'+\nu\rangle\langle x'|\chi\rangle \\\nonumber &=&\iint dx' dx''\exp{(-i\mu x'')}\chi^{\star}(x'')\langle x''|x'\rangle\chi(x'-\nu) \\\nonumber &=&\iint dx' dx''\exp{(-i\mu x'')}\chi^{\star}(x'')\delta( x''-x')\chi(x'-\nu) \\\nonumber &=&\int dx\exp{(-i\mu x)}\chi^{\star}(x)\chi(x-\nu). \end{eqnarray}

I wonder is it correct and can I simplify it more?

Edit: This operator arise in the density matrix representation by means of symplectic tomogram (that is a probability density function of the quadrature $X$). The inverse Radon transform is the following: \begin{eqnarray} \hat{\rho}=\frac{1}{2\pi}\int \mathcal{W}(X|\mu,\nu) \exp{(i(X\hat{1}-\mu\hat{q}-\nu\hat{p}))}dXd\mu d\nu. \end{eqnarray} It is know that the density matrix operator must be positive semidefinite. Then $\langle \chi|\hat{\rho}|\chi\rangle\geq 0$, $\forall |\chi\rangle$. This gives the condition on the integral on the right-hand side to be nonnegative: \begin{eqnarray} \frac{1}{2\pi}\int \phi(1;\mu,\nu) \langle \chi|\exp{(i(-\mu\hat{q}-\nu\hat{p}))}|\chi\rangle d\mu d\nu, \end{eqnarray} where the characteristic function of pdf is \begin{eqnarray} \phi(1;\mu,\nu)\equiv \int\limits_{-\infty}^{\infty}\mathcal{W}(X|\mu,\nu)e^{iX}dX \label{eqn pdf to charac_1} \end{eqnarray} and it known to be positive semidefinite function. Thus I thought that looking at $\langle \chi|\exp{(-i\mu\hat{q})}\exp{(-i\nu\hat{p})}|\chi\rangle$ would give me hint on the integral itself.

Answer 1

Your working looks correct, but there is a much simpler approach. It is easy to see that your operator is unitary. This implies that it has a complete set of eigenstates. Furthermore all its eigenvalues must have magnitude 1.

If we apply the positive semi-definite condition to an eigenstate of your operator we see that the eigenvalues of a positive semi-definite operator must be real and positive.

Combining this with requirement that unitary operators eigenvalues must have a magnitude of 1, we see that a positive semi-definite unitary operator must have 1 as its only eigenvalue and so is equal to the identity. Since your operator is clearly not the identity (unless $\mu = \nu = 0$), it is not semi-definite.