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?


2 Answers 2


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...

  • $\begingroup$ 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
    Commented Dec 20, 2021 at 3:49
  • $\begingroup$ ? There is no such step. The sign of i preceding the operator is + in your symplectic setup ! $\endgroup$ Commented 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
    Commented Dec 20, 2021 at 4:20
  • 1
    $\begingroup$ $\exp (A-B)=\exp([A,B]/2)~ \exp(A) ~ \exp(-B)$ for central commutator [A,B]. $\endgroup$ Commented Dec 20, 2021 at 6:52
  • 1
    $\begingroup$ $\hat q |x\rangle= x |x\rangle$, and the hermitian transpose likewise, so $\langle x| \hat q= \langle x| x$. $\endgroup$ Commented Dec 20, 2021 at 16:08

Let me slightly rephrase the question as follows: how to switch between the definition of Wigner function as $$W(x,p) = \int_{\mathbb R}\frac{d\xi}{2\pi} e^{-ip\xi} \langle x+\frac\xi2|\rho|x-\frac\xi2\rangle,\tag1$$ which is analogous to the one you find e.g. in the Wikipedia page, and the description in terms of coherent states, as $$W(\alpha) = \int\frac{d^2\eta}{\pi^2} \chi_W(\eta) e^{\alpha\bar\eta-\bar\alpha\eta}, \quad \chi_W(\eta)\equiv \operatorname{tr}[\rho\exp(\eta a^\dagger-\bar\eta a)],\tag2$$ where I used a notation analogous to the one found e.g. in Gerry&Knight's book.

Going from $W(\alpha)$ to $W(x,p)$

Start observing the equivalence between creation/annihilation operators and position/momentum ones. I'll use the common convention $$a=\frac{x+ip}{\sqrt2}, \quad x= \frac{a+a^\dagger}{\sqrt2}, \quad p =\frac{a-a^\dagger}{\sqrt2 i},$$ which ensures both $[a,a^\dagger]=1$ and $[x,p]=i$ hold. From these we observe that $$\eta a^\dagger - \bar\eta a=\sqrt2 i(\eta_2 x - \eta_1 p),$$ and thus $$\chi_W(\eta) \equiv \operatorname{tr}[\rho e^{\eta a^\dagger-\bar\eta a}] = \operatorname{tr}[\rho e^{\sqrt2 i(\eta_2 \hat x-\eta_1 \hat p)}].$$ Now, using the standard identity, holding for any pair of operators $A,B$ such that $[A,B]$ commutes with $A$ and $B$: $$e^{A+B} = e^A e^B e^{\frac12[B,A]} = e^B e^A e^{\frac12[B,A]},$$ together with the basic properties of position and momentum operators, we get $$\operatorname{tr}[e^{a \hat x+b \hat p}\rho] = e^{-\frac i2 ab}\int dx \, e^{a x} \langle x-i b|\rho|x\rangle.$$

Using a simple change of integration variables to get rid of stray $\sqrt2$ factors, and the identities above, we can then rewrite $W(\alpha)$ as the integral $$W(\alpha/\sqrt2) = \int \frac{d^2\eta}{(2\pi)^2} \operatorname{tr}[e^{i(\eta_2 \hat x-\eta_1 \hat p)}\rho] e^{i(\alpha_2\eta_1-\alpha_1\eta_2)} \\ = \int \frac{d^2\eta}{(2\pi)^2}\int dx e^{-\frac i2 \eta_1\eta_2} e^{i\eta_2 x} e^{i(\alpha_2\eta_1-\alpha_1\eta_2)} \langle x-\eta_1|\rho|x\rangle \\ = \int \frac{d\eta_1}{2\pi} e^{i\alpha_2\eta_1} \langle \alpha_1-\frac{\eta_1}{2}|\rho|\alpha_1+\frac{\eta_1}{2}\rangle.$$

Going from $W(x,p)$ to $W(\alpha)$

Let's now try going the other way around, i.e. starting from (1). This is essentially what we did before, but in reverse, but I think it's still worth it to see what the procedure looks like explicitly.

For starter we want to "desymmetrise" the expression $\langle x+\xi/2|\rho|x-\xi/2\rangle$. We can do that introducing a new variable $y$ to be integrated over, and ensuring $y=x+\xi/2$ with a delta, and then in turn converting the delta into an exponential via a Fourier transform: $$W(x,p) = \int \frac{d\xi}{(2\pi)^2} e^{-ip\xi} \int dy \delta(y-x-\xi/2) \langle y|\rho|y-\xi\rangle \\ = \int \frac{d\xi}{(2\pi)^2} \int dy \int d\eta e^{-ip\xi} e^{-i\eta(y-x-\xi/2)} \langle y|\rho|y-\xi\rangle .$$ Now let's bring some of the exponentials inside the expectation value, and convert everything into exponentials of operators acting on $|y\rangle$: $$= \int \frac{d\xi}{(2\pi)^2} \int d\eta e^{-ip\xi +ix\eta} \int dy \langle y|\rho \underbrace{e^{i\hat p\xi}e^{-i\hat x \eta}e^{i\xi\eta/2}|y\rangle}_{= \exp[i(\hat p\xi-\hat x\eta)] }\\ = \int \frac{d\xi d\eta}{2\pi^2} e^{\sqrt2 i(p\xi -x\eta)} \operatorname{tr}[ \rho \exp(\sqrt 2i(\hat x \eta - \hat p \xi) ] = \int \frac{d\xi d\eta}{2\pi^2} e^{\sqrt2 i(p\xi-x\eta)}\chi_W(\xi+i\eta).$$


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