1
$\begingroup$

A classical way to define the Wigner function ($\hbar=2$) of a density operator $\rho$ is as follows for $x=(x_{1}, x_{2})^{T}$: $$W(x) = \frac{1}{4\pi} \int^{\infty}_{-\infty} d\xi \exp(\frac{-i}{2}x_{2}\xi)\langle x_{1}+\frac{\xi}{2}| \rho |x_{1} - \frac{\xi}{2} \rangle \tag{1}$$ which seems to be the treatment I've seen in most textbooks.

Many modern papers on quantum information (for example: Weedbrook et al. "Gaussian Quantum Information", Rev. Mod. Phys. 84, 621 (2012) [arXiv:1110.3234]) define the Wigner function as the inverse symplectic Fourier transform of the Wigner characteristic function that is (one mode) $$W(x) = \int_{\mathbb{R}^{2}} \frac{d^{2}\xi}{(2\pi)^{2}}\exp(-ix^{T}\omega\xi)\chi(\xi) \tag{2} $$ where $$\chi (\xi) = \mathrm{Tr}[\rho \hat{D}_{\xi}]$$ is the characteristic function and $$D_{\xi} =\mathrm{exp}(i \hat{X}^{\mathrm{T}} \! \omega \, \xi)$$ is the displacement operator is the phase-space displacement operator, $$ \omega = \left( \begin{array}{cc} 0& 1\\-1&0 \end{array}\right),$$ and $\hat{X} = [\hat{q}, \hat{p}]$is the vector of position $\hat{q}$ and momentum operators $\hat{p}$.

I'm wondering how to reconcile these two definitions (1) and (2) of the Wigner function, i.e., how can one prove these two definitions are equivalent?

$\endgroup$
1

1 Answer 1

1
$\begingroup$

You have jammed up your shift variables, making them do double duty; and, importantly, you have used different conventions for $\hbar$: (1) requires that it be equal to 1, while (2) to 2.

In any case, going from (2) to (1) is straightforward, and literally covered in the first week of most deformation quantization mini-courses.

First, denote your symplectic vectors $\vec{x}\equiv (q,p)^T$, and $\vec{\xi}\equiv (\xi,\eta)^T$, so that $$W({\mathbf x}) = \int {d \xi ~d\eta\over (2\pi)^{2}} e^{-i(q\eta-p\xi)} \mathrm{Tr}(\hat \rho e^{i(\hat q \eta-\hat p \xi)} ). \tag{2} $$

Now, recall the cyclicity of the trace, the routine degenerate CBH identity, and the representation of the trace, $$ \mathrm{Tr} \hat O = \int\!\! dx \langle x|\hat O|x\rangle ~~~\leadsto \\ \mathrm{Tr}(\hat \rho e^{i(\hat q \eta-\hat p \xi)} )=\int\!\! dx~ \langle x| e^{-i\eta \xi } e^{i\hat q \eta} e^{-i\hat p \xi }\hat \rho|x\rangle \\ = \int\!\! dx~ e^{i\eta(x-\xi)} \langle x| \hat e^{-i\hat p \xi }\rho |x\rangle = \int\!\! dx ~ e^{i\eta(x-\xi)} \langle x-2\xi| \hat \rho |x\rangle , $$ whence $$ (2)= \int \!\!\frac{d\xi}{2\pi}dx ~\delta(x-q-\xi) ~e^{ip\xi} \langle x-2\xi| \hat \rho |x \rangle \\ =\int\!\! \frac{d\xi}{2\pi} e^{-ip\xi} \langle q+\xi| \hat \rho |q-\xi\rangle, $$ upon changing of the dummy variable by reflection.

But this amounts to just (1), recalling the peculiar unit of action you used, to choose $\hbar=2$, implying nonstandard commutation relations. If your source wrote (1) the way you have, it would be quite disorienting...

$\endgroup$
7
  • $\begingroup$ One step, I find a little hard to follow is what I read as $\langle x |e^{-i\hat{q}\eta} = e^{i\eta{x}}\langle x |$. Is this a property of displacement operators? $\endgroup$
    – user135520
    Dec 20, 2021 at 3:49
  • $\begingroup$ ? There is no such step. The sign of i preceding the operator is + in your symplectic setup ! $\endgroup$ Dec 20, 2021 at 3:54
  • $\begingroup$ My apologies, I’m asking about the $e^{\eta{i}(x-\xi)}$ that comes after you apply BCH in the expansion of the trace. $\endgroup$
    – user135520
    Dec 20, 2021 at 4:20
  • 1
    $\begingroup$ $\exp (A-B)=\exp([A,B]/2)~ \exp(A) ~ \exp(-B)$ for central commutator [A,B]. $\endgroup$ Dec 20, 2021 at 6:52
  • 1
    $\begingroup$ $\hat q |x\rangle= x |x\rangle$, and the hermitian transpose likewise, so $\langle x| \hat q= \langle x| x$. $\endgroup$ Dec 20, 2021 at 16:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.