Physical meaning of the operator $\exp(-a {\hat{p}}^2)$

Question

I am curious about the physical meaning of the operator $\exp(-a {\hat{p}}^2)$ with $a$ being a positive constant. With respect to the coordinate basis, I find that $\langle x |\exp(-a {\hat{p}}^2)|x' \rangle = \exp(a \hbar^2 \frac{\partial^2}{\partial x^2})\langle x|x' \rangle $. However, I cannot deduce any physically meaningful implications both from the operator itself and the matrix element in coordinate basis. Can anyone give me some insights? Thanks in advance.

Answer 1

You are asking about the "physical meaning" of the celebrated Weierstrass transform, which is used routinely in physics, of course: $$ \bbox[yellow]{ e^{\partial_x^2}f(x) =\frac{1}{\sqrt{4\pi}} \int_{-\infty}^\infty f(x-y)~ e^{-y^2/4}\;dy}~. $$ In your case, $$\langle x |\exp(-a {\hat{p}}^2)| \psi \rangle = \exp(a \hbar^2 \partial_x^2)~\langle x|\psi \rangle\\ =e^{a\hbar^2 \partial_x^2} ~\psi(x) =\frac{\sqrt{a} \hbar}{\sqrt{4\pi}} \int_{-\infty}^\infty \!\!dy ~~\psi(x-y)~ e^{-a\hbar^2 y^2/4} ~, $$ a Gaussian smoothing, (low pass filtering) of the wavefunction.

• For the formal δ(x-x')-wavefunction you chose, you naturally smooth that to a Gaussian, $\sqrt{a/4\pi} \hbar \exp (-a\hbar^2 (x-x')^2/4)$, your matrix element. In a sense, you undo the limit of the Gaussian you took to get the δ.

There is a bevy of applications of this transform in phase-space quantization.

NB. In point of fact, through Fourier transformation, you may find the corresponding integral kernels for any power n of the momentum operator in the exponential. They are generalized hypergeometrics $_0 F_{n-2} (y)$, so, extending the Gaussian exemplified above.

Answer 2

The kinetic energy of a free particle is described by Hamiltonian \begin{equation} \hat{H} = \frac{\hat{p}^2}{2m}, \end{equation} which means that the corresponding partition function is \begin{equation} Z(\beta) = \mathbf{Tr} [e^{-\beta H}] = \mathbf{Tr} [e^{-a \hat{p}^2}], \end{equation} where $a = \frac{\beta}{2m}$ is a positive constant. Naturally, the trace can be calculated in any basis, including the coordinate basis - as it is often done in quantum statistical physics texts, e.g., when developing the Green's function formulation.

This is just one example - anything with combination $\hat{p}^2$ asks for an interpretation in terms of kinetic energy.