Angular momentum operator acting on |j,m>

[PhyPsy]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

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

[itex]\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[/itex]
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 [itex]\mid j,-m \rangle[/itex]. This means that the coefficient for [itex]\mid j,m \rangle[/itex] should be 0 for all cases except when [itex]m=0[/itex].

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 [itex]\mid j,m \rangle[/itex]. Then, I can use the Wigner D-matrix to simplify the expression:
[itex]R(\alpha , \beta , \gamma) \mid j,m \rangle =\sum _{m'=-j} ^j D _{m'm} ^{(j)} (\alpha , \beta , \gamma) \mid j,m' \rangle[/itex]

[itex]e^{-i \pi J_x}[/itex] can be written as [itex]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})[/itex].

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

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

 

[dextercioby]

[tex] (-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} [/tex]

 

[paranormal]

As a scientist, it is important to understand that mathematical proofs can be complex and require a thorough understanding of the underlying concepts. In this case, it may be helpful to approach the problem from a different perspective.

One approach could be to use the identity e^{i\theta} = \cos\theta + i\sin\theta and express the exponential operator in terms of trigonometric functions. From there, you can use the trigonometric identities to simplify the expression and show that it is equal to the right side of the equation.

Another approach could be to use the commutation relation [J_x, J_y] = i\hbar J_z and the fact that e^{-i\pi J_x} = e^{-i\pi/2}e^{-i\pi J_z}e^{-i\pi/2} to simplify the expression. This approach may be more straightforward and require less manipulation of the series.

In any case, it is important to carefully consider the properties and relationships of the operators involved and use them to simplify the expression. With a thorough understanding and careful manipulation, you should be able to prove the given equation.