# Angular momentum operator acting on |j,m>

1. Aug 9, 2012

### PhyPsy

1. The problem statement, all variables and given/known data
Prove that $e^{-i \pi J_x} \mid j,m \rangle =e^{-i \pi j} \mid j,-m \rangle$

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

3. 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?

2. Aug 9, 2012

### Oxvillian

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

3. Aug 10, 2012

### 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.

4. Aug 11, 2012

### dextercioby

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