# Translation operators and positive-semidefinite condition

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.

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.

• This conclusion is what I got from the Fourier analises of the integral. We can't say anything of the positive semidefinitness of the integral in the end. However, that problem is confusing me a lot since its a part of a bigger one (I edited my question, maybe you can give a look). Any comments would be good since I really stuck.
– Kim
Commented Apr 17 at 8:55
• We can say things about the positive semi-definiteness of the integral: it is not, in general positive semi-definite. Looking at your comment, I am not familiar with the formalism you are using, but I would note that the product of positive semi-definite operators is not necessarily positive semi-definite. On the other hand, unitary transforms $\rho \rightarrow U^\dagger \rho U$ do preserve positive semi-definiteness, and looking at the ingredients you have there that is the route I would try to go Commented Apr 17 at 9:10
• Thats a good point, then positive semidefinitness is not the right idea to look at. I just see that the integral in general must be nonnegative for all $|\chi\rangle$, so $\phi(1|\mu,\nu)$ the characteristic function of a distribution can't influence the sign in the case. I know its an even function, bounded by one. I thought the if I can say something about the sign of $\langle \chi|e^{-i(\mu q+nu p)}|\chi\rangle$ then it will show me why the integral is nonnegative for all states. It feels like $\langle \chi|e^{-i(\mu q+nu p)}|\chi\rangle$ must be even too, for example.
– Kim
Commented Apr 17 at 9:15