# Momentum probability density from Wigner distribution

I want to prove that $$|\hat{\psi}(p)|^2= \frac{1}{2\pi} \int W_\psi \mathrm{d}x$$ where $$W_\psi$$ is the Wigner function. Starting with the definition I get ($$z=-y$$ and $$u=x+z/2$$): $$\frac{1}{2\pi}\iint \psi^*\left(x-\frac{y}{2}\right)\mathrm{e}^{\mathrm{i}py} \psi\left(x+\frac{y}{2}\right) \mathrm{d}y \mathrm{d}x= \frac{1}{2\pi}\iint \psi^*\left(u\right)\psi\left(u-z\right) \mathrm{d}u\: \mathrm{e}^{-\mathrm{i}pz}\mathrm{d}z$$ Next I want to use the Convolution Theorem to get the product of the Fourier Transforms: $$\frac{1}{2\pi}\iint \psi^*\left(u\right)\phi\left(z-u\right) \mathrm{d}u\: \mathrm{e}^{-\mathrm{i}pz}\mathrm{d}z= \frac{1}{\sqrt{2\pi}}\mathcal{F}(\psi^{*}*\phi)(p)=\hat\psi^*(-p)\hat{\psi}(-p)$$ But with the definition $$\phi(x):=\psi(-x)$$ I get $$|\hat{\psi}(-p)|^2$$. Does somebody know where the problem is?

• Your exponent on the left hand side has a very wrong sign ! This amounts to a p flipped around.... May 18, 2022 at 15:29
• @CosmasZachos in the first line? May 18, 2022 at 15:35
• On the left hand side of the first line. You are mis-defining p. May 18, 2022 at 15:41
• @CosmasZachos but is $W_\psi(x,p)=\int\psi^*\left(x-\frac{y}{2}\right)\mathrm{e}^{\mathrm{i}py} \psi\left(x+\frac{y}{2}\right) \mathrm{d}y$ not the definition of the Wigner function? May 18, 2022 at 15:45
• @CosmasZachos ahh yes thanks. I didn‘t see that the WP definition has the plus sign in the complex conjugated wavefunction. May 18, 2022 at 16:50

Use $$\psi\left(x\pm\frac{y}{2}\right)=\int dx e^{ip\left(x\pm\frac{y}{2}\right)}\bar{\psi}(p)$$ and $$\int dxe^{i(p-p')}=2\pi\delta(p-p')$$