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.
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2 Answers
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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.
answered Mar 23, 2020 at 20:25
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$\begingroup$ This is very informative! However your answer is more about the mathematical rather than physical meaning of this operator. $\endgroup$– Roger V.Mar 24, 2020 at 14:59
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1$\begingroup$ @Vadim "physical meaning" is hopelessly subjective. I desist from quoting Philip Morse's indictment. It is a little bit like asking for the "physical meaning" of the Fourier transform! I indicated that phase-space quantization bursts with examples, but it is not clear to me the OP wishes to delve into the Husimi distribution, in my own (strongly held) view the paradigmatic application... "Gaussian smoothing" and "low pass filtering" cover the generic bases adequately. $\endgroup$ Mar 24, 2020 at 17:53
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$\begingroup$ I disagree: "Physical" is objective, not subjective - it exists regardless of whether we know about its existence and how we describe it. Mathematical formalism, on the other hand, is a human construct. The whole point of doing physics is describing the real world, not making calculations for the sake of calculations. $\endgroup$– Roger V.Mar 24, 2020 at 18:04
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1$\begingroup$ @Vadim . Obviously. Only great physicists (like Feynman) winkle lessons out of idle calculations they do in private. But what different physicists hyperfocus on as the physical essence of a hyper-broad mathematical construct is not an object of plausible agreement. $\endgroup$ Mar 24, 2020 at 18:09
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1$\begingroup$ Statistics is physics as opposed to math? $\endgroup$ Mar 24, 2020 at 18:26
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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.
