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Given a classical observable, $a(x,p),$ Weyl quantization gives the correspondent QM observable as: $$\langle x | \hat{A} | \phi \rangle=\hbar^{-3}\int \int a \left(\frac{x+y}{2},p\right)\phi(y) e^{2\pi ip(x-y)/\hbar} dy dp.$$ The correspondent inverse transformation (Wigner transform; which if I understand correctly would be the symbol of the pseudo-differential operator defined above, but maybe this is not very important) is: $$a(x,p)= \int \langle x-\frac{y}{2} \vert \hat{A} \vert x+\frac{y}{2} \rangle e^{-2\pi i yp/\hbar}dy.$$

I wanted to test this in a few simple examples. It is always easy, except for when $\langle x \vert \hat{A} \vert x^\prime \rangle $ is actually a distribution and cannot be understand as a function.

A basic case which is confusing me (even ignoring the mathematical subtleties of the matter) is the case in which $\hat{A}$ is $\hat{T},$ the quantum mechanical kinetic operator. If you start with $a(x,p)=\dfrac{p^2}{2},$ it is very easy to see that the first formula above gives the correct $\hat{T},$ i.e, that $\langle x \vert \hat{T} \vert \phi \rangle$ is just the laplacian. But what if want to check the second formula, the inversion of the quantization? Well, in position representation, we have $\langle x \vert \hat{T} \vert x^\prime \rangle=-\dfrac{1}{2}\Delta_{x} \delta(x-x^\prime),$ where $\Delta$ is the laplacian. Therefore, one would expect: $$\dfrac{p^2}{2}=\int -\dfrac{1}{2}\Delta_{x-\frac{y}{2}} \delta(x-\frac{y}{2}-x-\frac{y}{2}) \exp \left(-2\pi i yp/\hbar \right) dy,$$ but I don't what to do with this; I don't even undertand what is the exact meaning of this. This can be simplified: $$-\dfrac{1}{2}\int\Delta_{x-\frac{y}{2}} \ \delta(y) \exp \left(-2\pi i yp/\hbar \right) dy,$$ but I don't know how to continue from this.

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    $\begingroup$ by translation invariance, is $\Delta_{x-y/2}\delta(y)=\Delta_{-y/2}\delta(y)$? ($x$ is a dummy variable). By the chain rule, this equals $=4\Delta_y\delta(y)$, from which your identity follows, right? $\endgroup$ – AccidentalFourierTransform Oct 6 '17 at 10:20
  • $\begingroup$ You are totally right, it is easy from translational invariance of the derivative... I still have some issues with the precise value of the premultiplicative constants, but that is of no much relevance $\endgroup$ – Qwertuy Oct 6 '17 at 12:59
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I'll tweak your notations to correct the normalizations and conform with the sound Wikipedia conventions, instead:

The Weyl map is $$\langle x | \hat{A} | y \rangle=\frac{1}{2\pi\hbar}\int dp ~ a \left(\frac{x+y}{2},p\right) e^{ ip(x-y)/\hbar},$$ with the inverse, Wigner map, $$a(x,p)=2 \int dy ~\langle x+y \vert \hat{A} \vert x-y \rangle ~e^{-2 i yp/\hbar}.$$

For $a=p^2$, the Weyl map yields, $$ \frac{1}{2\pi\hbar}\int dp ~ p^2 e^{ ip(x-y)/\hbar}\\ =\frac{-\hbar^2}{2\pi\hbar}\int dp ~ \partial_x^2 ~ e^{ ip(x-y)/\hbar}\\ =-\hbar^2 \partial_x^2 \delta(x-y)= -\hbar^2\partial_x^2 \langle x| y\rangle =\langle x| \hat{p}^2|y\rangle ~. $$

Conversely, in the Wigner map, $$2 \int dy ~\langle x+y \vert \hat{p}^2 \vert x-y \rangle ~e^{-2 i yp/\hbar}\\ =2 \int dy dp' ~\langle x+y \vert \hat{p}^2|p'\rangle\langle p' \vert x-y \rangle ~e^{-2 i yp/\hbar}\\ =2 \int dy dp' ~p'^2 ~\langle x+y \vert p'\rangle\langle p' \vert x-y \rangle ~e^{-2 i yp/\hbar}\\ =\frac{2}{2\pi\hbar} \int dy dp' ~p'^2 ~ ~e^{i(-2 yp +2p'y)/\hbar}\\ =\int dp' ~p'^2 ~ \delta (p-p')=p^2.$$


But for the wrong normalization and exponential factors, etc... your first equation follows from this first equation here for the Weyl map, upon operating on the latter with $\int dy \langle y|\phi\rangle$.

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  • $\begingroup$ A very nice answer indeed! I must have the premultiplicative constants wrong, because I managed to prove the result up to some silly factor $4\pi^2.$ Thanks for your clarity $\endgroup$ – Qwertuy Oct 7 '17 at 18:05
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Hint: Use Integration by parts. You have $\Delta = \nabla^2$ and you can use Substitution $z = - \frac{y}{2}$ and then for any functions $X,Y$

$\int (\nabla_z X) Ydz = \int (\nabla_z(XY) - X \nabla_z Y)dz$.

The first term on the right Hand side gives a boundary term and These will always give you Zero.

You do the Integration by part twice and finally you can evaluate the Delta function easily (at $z = 0$).

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