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^2\rightarrow 0 $ and $\sigma_p^2 \rightarrow \infty$ 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.

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.