Angular momentum operator acting on |j,m>

[PhyPsy]

Homework Statement

Prove that e^{-i \pi J_x} \mid j,m \rangle =e^{-i \pi j} \mid j,-m \rangle

Homework Equations

J_x \mid j,m \rangle =\frac{\hbar}{2} [\sqrt{(j-m)(j+m+1)} \mid j,m+1 \rangle +
\sqrt{(j+m)(j-m+1)} \mid j,m-1 \rangle]

The Attempt at a Solution

Expanding e^{-i \pi J_x} to a power series and applying the equation in #2, I come up with an expression with coefficients for \mid j,m \rangle, \mid j,m+1 \rangle, \mid j,m-1 \rangle, \mid j,m+2 \rangle, \mid j,m-2 \rangle, and so on as far as the quantum number j allows. I will focus on the \mid j,m \rangle term.

\left\{ 1+ \sum _{n=1} ^\infty \frac{(i \pi \hbar)^{2n} \left[\sqrt{(j-m)^{2n} (j+m+1)^{2n}} +\sqrt{(j+m)^{2n} (j-m+1)^{2n}}\right]}{2^{n+1} (2n)!} +\sum _{n=1} ^\infty \frac{(i \pi \hbar)^{2n+2} \left[\sqrt{(j-m)^{2n} (j+m+1)^{2n} (j-m-1)^{2n} (j+m+2)^{2n}} +\sqrt{(j+m)^{2n} (j-m+1)^{2n} (j+m-1)^{2n} (j-m+2)^{2n}}\right]}{2^{n+3} (2n+2)!} +... \right\} \mid j,m \rangle
This coefficient goes on for as long as quantum number j allows.

The right side of the equation I am trying to prove has no operators, so the only non-zero coefficient I should have is the one for \mid j,-m \rangle. This means that the coefficient for \mid j,m \rangle should be 0 for all cases except when m=0.

The above series does converge, but to what it converges seems to depend on j and m. I have manipulated the series a couple different ways, but I have been unable to show that it equals 0 for all values of m except 0. I wonder if I am making this more difficult that it has to be. Any ideas?

 

[Oxvillian]

Well you could always cheat and notice that all you're doing is turning a spherical harmonic upside-down

 

[PhyPsy]

Your tip led me to realize that it would be simpler to think of it as a rotation applied to \mid j,m \rangle. Then, I can use the Wigner D-matrix to simplify the expression:
R(\alpha , \beta , \gamma) \mid j,m \rangle =\sum _{m'=-j} ^j D _{m'm} ^{(j)} (\alpha , \beta , \gamma) \mid j,m' \rangle

e^{-i \pi J_x} can be written as e^{-i (-\frac{\pi}{2}) J_z} e^{-i \pi J_y} e^{-i \frac{\pi}{2} J_z}=R(-\frac{\pi}{2}, \pi , \frac{\pi}{2}).

Now I can use the Wigner D-matrix equation:
R(-\frac{\pi}{2}, \pi , \frac{\pi}{2}) \mid j,m \rangle =\sum _{m'=-j} ^j D _{m'm} ^{(j)} (-\frac{\pi}{2} , \pi , \frac{\pi}{2}) \mid j,m' \rangle
=\sum _{m'=-j} ^j e^{-i(-m' \pi /2 +m \pi /2)} d _{m'm} ^{(j)} \mid j,m' \rangle
=(-1)^{j-m} \sum _{m'=-j} ^j e^{-i(-m' \pi /2 +m \pi /2)} \delta _{m',-m} \mid j,m' \rangle
=(-1)^{j-m} e^{-i(m \pi /2 + m \pi /2)} \mid j,-m \rangle

So I'm close now, but it seems like they set m=j to get to e^{-i \pi j} \mid j,-m \rangle, and I don't know why that step would be made.

 

[dextercioby]

(-1)^{j-m} e^{-i(m\pi/2+m\pi/2)} = (-1)^{j-m} e^{-im\pi} = (-1)^{j-m} (e^{-i\pi})^m = (-1)^{j-m} (-1)^m = (-1)^{j-m+m} = (-1)^j = (e^{-i\pi})^j = e^{-i\pi j}