Fourier transform of cross-spectral density space matrix elements

In order to derive phase space like equation of motion (e.g. the equation of motion for the Wigner function of a single particle in one-dimension), it is an advantage to work with the Fourier transform (Schleich, Sec. D.2:

$$t^{(\pm)}(\hat{A}) = \frac{1}{2\pi\hbar}\int_{-\infty}^\infty d\xi e^{-ip\xi/\hbar}\mathcal{M}^{\pm}(x,\xi) \, ,$$

where in this case I want to consider

$$\mathcal{M}^{+}(x,\xi) = \left(-\frac{\hbar^2}{2}\frac{\partial^2}{\partial x^2} - 2\hbar^2\frac{\partial^2}{\partial \xi^ 2} \right)\langle x+\xi/2|\hat{A}|x-\xi/2\rangle \, .$$

Now, in order to compute this $$t^+$$, the author says that we have to integrate by parts the partial derivatives with respect to $$\xi$$. For instance

$$\int_{-\infty}^\infty d\xi e^{-ip\xi/\hbar} \frac{\partial^2}{\partial \xi^ 2}\langle x+\xi/2|\hat{A}|x-\xi/2\rangle \, .$$

If I integrate by parts as suggested, why

$$e^{-ip\xi/\hbar} \frac{\partial}{\partial \xi}\langle x+\xi/2|\hat{A}|x-\xi/2\rangle$$

evaluates to zero when evaluated in the limits of the integral? It is not an odd function, and I can't figure out another argument why this element vanishes.

• It's probably subjective, but the language is needlessly stressful here. You might try, instead, Lemma 0.3, eqn (18-20) of this concise treatise, leaving the translation via Ch 0.18 aside. I will try to do the translation between the two books, but most of the work will go into translating to a triviality... – Cosmas Zachos Mar 12 '19 at 21:30
• Then... what's the rabbit at the end of the hole? Is the derivation in Schleich wrong? I didn't know your book until yesterday, a friend give me the reference. It's a pity that amazon does not have it on existence. – user2820579 Mar 12 '19 at 22:34
• PS the Fourier transform of the Wigner function is the cross-spectral density, $\tilde {f} (x,\xi)= \psi^* (x-\xi/2) \psi(x+\xi/2)$, and, as in our Lemma, it leads to the star-genvalue equations from Schroedinger's --or Heisenberg's--and, importantly, vice versa! – Cosmas Zachos Mar 12 '19 at 22:47
• Woops, it was only the hardbound, which is still odd indeed for amazon. – user2820579 Mar 12 '19 at 22:48
• Oh, but this is the non-boundary residue of the integration by parts you are talking about... Everything attenuates to 0 at the boundary, as the quantities are (mostly) local..... – Cosmas Zachos Mar 15 '19 at 18:41

I'll parallel your book's proof in my language of Lemma 3 in our booklet, but now for the off-diagonal Wigner function, setting $$\hbar=1$$ , easy to reinstate by dimensional analysis.

The time dependence is a canard, taken up at the very end: the crucial piece, including Wolfgang's integration by parts, visible further below, is for the stationary Wigner function, $$W_{E~E'}(x,p)=\frac{1}{2\pi} \int dy e^{-iyp} \tilde f(x,y),$$ where the all-central cross-spectral density amounts to $$\tilde f(x,y)=\langle x+y/2| (|E' \rangle\langle E|)|x-y/2 \rangle \\ =\psi^*_E (x-y/2)~~\psi_{E'} (x+y/2),$$ for TISE, $$\left (-\frac{1}{2m} \partial_x^2 + V(x)\right ) \psi_E (x) = E \psi_E (x).$$

Thus, by virtue of our eqn (12), the celebrated Bopp shift, $$\begin{eqnarray} && ~~~~H(x,p)\star W_{EE'}(x,p)= \nonumber \\ &=& {1\over 2\pi} \left( \frac{1}{2m}\left( p-{i\over 2} \overrightarrow{\partial }_x \right) ^2 +V(x) \right) \int\! dy~ e^{-iy(p+{i\over2} \overleftarrow {\partial }_x)} \tilde f(x,y) \nonumber \\ &=& {1\over 2\pi} \int\! dy~ e^{-iyp} ~ \left(\frac{1}{2m} \left( p-{i\over2} \overrightarrow{\partial }_x\right) ^2 +V(x+y/2)\right) \tilde f(x,y) \nonumber \\ &=& {1\over 2\pi} \int\! dy ~e^{-iyp} \left( \frac{1}{2m}\left( i\overrightarrow{\partial }_y +{i\over2} \overrightarrow{\partial }_x\right) ^2 + V(x+ y/2)\right) \psi^*_E(x- y/2) ~\psi_{E'}(x+ y/2) \nonumber \\ &=& {1\over 2\pi} \int\! dy ~e^{-iyp} \psi^*(x- y/2)~ E'~ \psi(x+ y/2) = E' ~W_{EE'}(x,p)~. \end{eqnarray}$$ The Fourier integration by parts yields the penultimate line, as the surface terms vanish: all wavefunctions, potentials, etc, are assumed to vanish at infinity. In the penultimate line, the operators only Schroedinger-act on the $$\psi$$, leaving the $$\psi^*$$ alone! (Think of left- versus right-movers.)

Mutatis mutandis, $$W_{EE'}(x,p) \star H (x,p)= E W_{EE'}(x,p).$$ So, then $$W_{EE'} \star H (x,p) + H\star W_{EE'}= (E+E') W_{EE'} ,$$ and $$W_{EE'} \star H (x,p) - H\star W_{EE'}= (E-E') W_{EE'} ,$$ for these stationary Wigner functions. Consequently, for the time-dependent ones, $${\cal W}_{EE'}(x,p;t) = e^{i(E-E')t} W_{EE'}(x,p),$$ one gets Moyal's evolution equation for the Wigner function, the Wigner transform of the von Neumann equation, $$H\star {\cal W}_{EE'}- {\cal W}_{EE'} \star H (x,p)= i\partial _t {\cal W}_{EE'}.$$

To compare with Wofgang's Summary D.5, first take into consideration our somewhat dyslexic adherence to Moyal's notation which reverses Schleich's E' E ordering; and further recall our stationary equations above amount to just $$W_{EE'} \star H (x,p) + H\star W_{EE'}\\ = \Biggl ( H \left (\left( x+{i\over 2} \overrightarrow{\partial }_p \right) ,\left( p-{i\over 2} \overrightarrow{\partial }_x \right) \right ) + H \left (\left( x-{i\over 2} \overrightarrow{\partial }_p \right) ,\left( p+{i\over 2} \overrightarrow{\partial }_x \right) \right ) \Biggr ) W_{EE'}(x,p)\\ =(E+E') W_{EE'},$$ and the corresponding one for the (-) equation that enters the Moyal evolution equation in phase space, $$\Biggl (- H \left (\left( x+{i\over 2} \overrightarrow{\partial }_p \right) ,\left( p-{i\over 2} \overrightarrow{\partial }_x \right) \right ) + H \left (\left( x-{i\over 2} \overrightarrow{\partial }_p \right) ,\left( p+{i\over 2} \overrightarrow{\partial }_x \right) \right ) \Biggr ) W_{EE'}(x,p)\\ =(E-E') W_{EE'}.$$

A ready check is looking at the fate of the kinetic term quadratic in the ps, etc... The fastidious practice of resolving the translation operators through infinite (Taylor) sums started with Wigner and has not abandoned some educators.

• The takeaway, emphasized by Fairlie in 1964, is that Moyal's equation does not suffice to determine the Wigner function completely, but relies on the above (+) equation as well. This has its analog in Hilbert space: solutions ρ of the von Neumann equation are not fully determined, unless their anticommutator with the hamiltonian is also fixed by an eigenvalue (+) equation.