Is the overlap of coherent states circular symmetric?

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
1 reply · 2K views
Danny Boy
Messages
48
Reaction score
3
What I am interested in doing, is considering the angular momentum eigenstate for a spin ##1## system: ##|J=1, M=1\rangle = \begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}##, forming the coherent state ##|CS \rangle = \begin{bmatrix}
0.5 \\
-\frac{i}{\sqrt{2}} \\
-0.5
\end{bmatrix}## by taking the rotation ##\text{exp}(-\frac{i \pi}{2}\hat{J}_x)\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix} = \begin{bmatrix}
0.5 \\
-\frac{i}{\sqrt{2}} \\
-0.5
\end{bmatrix}##.

I then want to work out the modulus squared of the overlap of the coherent state with a rotation of the coherent state about the ##x##-axis and the ##z##-axis for various values of ##\theta## and ##\phi##: $$|\langle CS| \text{exp}(-i \phi \hat{J}_z) \text{exp}(-i \theta \hat{J}_x)|CS\rangle|^2.$$ The result I get is http://farm5.staticflickr.com/4211/34692246053_e0fd2d7cd8_b.jpg. Is this the type of plot you would have predicted? I would have expected something more circular symmetric?
Thanks for any assistance.
 
Physics news on Phys.org
Looks quite circular symmetric to me, given that the origin is not in the middle of the figure.