# Wigner function of position eigenket?

I am trying to derive the Wigner function of position eigenket $\rho = |x'\rangle \langle x'|$. One method is to use the formal expression for the Wigner function and then solve: $$W(q,p) = \frac{1}{2 \pi \hbar}\int_{-\infty}^{\infty}\langle q+\frac{1}{2}x| \rho |q-\frac{1}{2}\rangle e^{ipx/\hbar} dx$$

$$= \frac{1}{2 \pi \hbar}\int_{-\infty}^{\infty}\langle q+\frac{1}{2}x| x'\rangle \langle x' |q-\frac{1}{2}\rangle e^{ipx/\hbar} dx$$

$$= \frac{1}{2 \pi \hbar}\int_{-\infty}^{\infty}\delta(q+\frac{1}{2}x-x')\delta(x'-q+\frac{1}{2}x) e^{ipx/\hbar} dx$$

I am stuck here. Question related to integral involving two dirac delta functions has been previously asked here.

I have another idea using the Wigner function of coherent state:

$$W_{|\alpha\rangle }(q,p)= \frac{1}{\sigma_q \sigma_p}exp\left[-\frac{1}{2\hbar}\frac{(q-\overline{q})^2}{\sigma_q^2}-\frac{1}{2\hbar}\frac{(p-\overline{p})^2}{\sigma_p^2}\right]$$

$\sigma_q^2$ and $\sigma_p^2$ are variances of $q$ and $p$ quadrature (equal to 1/2 if $\lambda =1$. (Notation in the book)).

Now $\rho = |x'\rangle \langle x'|$ is an infinitely squeezed state in $q$ quadrature, we can take the limit $\sigma_q\rightarrow 0$ and $\sigma_p \rightarrow \infty$ (such that the product of $\sigma_q \sigma_p$ remains constant and equal to that of for the coherent state) and apply these limits on the above Wigner function for coherent state to get the desired answer. Here also I am stuck.

Any help would be appreciated.

$$W(q,p) = \frac{1}{2 \pi \hbar}\int_{-\infty}^{\infty}\delta\left(q+\frac{1}{2}x-x'\right)\delta\left(x'-q+\frac{1}{2}x\right) ~ e^{ipx/\hbar} dx\\ = \frac{2}{ \pi \hbar}\int_{-\infty}^{\infty}dx ~\delta(x-2x'+2q)~~\delta(x+2x'-2q) e^{ipx/\hbar} \\ = \frac{2}{ \pi \hbar} \delta(2x'-2q+2x'-2q+) e^{ip2(x'-q)/\hbar}\\ = \frac{1}{2 \pi \hbar} ~ \delta(q-x') ~.$$ This is what you would expect: Your density matrix is but a spike at x', and a p-indifferent razor ridge in p.
• In your squeezed-state representation, the momentum Gaussian disappears in the limit; and all you have to do is recall the limit of the sharp Gaussian, $\exp (-q^2/\epsilon) /\sqrt{\pi \epsilon}$ going to $\delta(q)$ as $\epsilon =2\hbar \sigma_q^2$ goes to 0.
• There is a flip side to this W, extreme squeezing the other way. This is to say the Wigner function of Dirac's translationally invariant standard ket, $\rho=|\varpi\rangle \langle \varpi|\equiv \lim_{p\to 0} |p\rangle \langle p|$, where $\langle x| \varpi\rangle=1/\sqrt{2\pi \hbar}$. It then follows that $W=\delta (p)/ 2\pi \hbar$, also as $\epsilon \to \infty$.
• I am ecstatic to see your response! I was able to reach to your second last expression, however, with unsurety. But still I am not clear about how you got the last expression from the previous one that is dropped the term $e^{i p2(x'-q)/ \hbar}$? – Mark Robinson Nov 14 '17 at 16:03