Can the operator Exp[-I*Pi*L_x/h] be faced as parity?

[Icaro Lorran]

Homework Statement

The problem originally asks to evaluate ##exp(\frac{-i\pi L_x}{h})## in a ket |l,m>. So I am wondering if I can treat the operator as a parity operator or if I really have to expand that exponential, maybe in function of ##L_+## and ##L_-##.

2. The attempt at a solution
If ##exp(\frac{-i\pi L_x}{h})## operates in |y,z> it returns |-y,-z>. Then, if it operates in |l,m> it would just change the ket's signal.

 

[blue_leaf77]

Well, proving it for general ##l## might be a tedious task. But if you are only interested in the final answer, think of the fact that ##\exp \left( -\frac{i\pi L_x}{\hbar} \right)## is a rotation operator around ##x## axis through an angle of ##\pi## radian. Now ##|l,m \rangle ## is a state which points in the ##z## direction because it is an eigenstate of ##L_z##. Applying ##\exp \left(- \frac{i\pi L_x}{\hbar} \right)## to this state is the same as rotating this state around ##x## axis through an angle of ##\pi## radian.

 

[Icaro Lorran]

By expanding ##\exp \left( - \frac{i \pi L_x}{\hbar}\right) |l,m\rangle## into a Cartesian basis like ##\int \exp \left( - \frac{i \pi L_x}{\hbar}\right)| x,y,x \rangle \langle x,y,z |l,m\rangle \mathrm{dV}## it would be possible to use the operator in the xyz ket so that it could be simplified into this: ##\int | x,y,x \rangle \langle x,-y,-z |l,m\rangle \mathrm{dV}##. With that said, is there a way to write this answer with the ##|l,m \rangle## ket alone?

 

[blue_leaf77]

If you insist that way, you can start from the fact that ##\langle x,y,z | l,m\rangle = Y_{lm}(\theta,\phi)## with ##Y_{lm}## being spherical harmonics defined as
$$
Y_{lm}(\theta,\phi) = \sqrt{\frac{(l-m)!}{(l+m)!}} P_l^m(\cos \theta) e^{i m\phi}
$$
Now first find out how it will affect ##Y_{lm}(\theta,\phi)## when you make the changes ##y \rightarrow -y## and ##z \rightarrow -z##.

 

[Icaro Lorran]

I'll try that, thank you

 

[Icaro Lorran]

Just to verify the answer, should it be ##(-1)^{(l+m)} |l,m \rangle##?

 

[blue_leaf77]

Nope, it's not. It seems that to arrive at that equation you changed ##\theta## only, keeping ##\phi## unchanged.

 

[Icaro Lorran]

How about this? If it isn't, I'll need help.

##(-1)^l \frac{(l-m)!}{(l+m)!}|l,-m \rangle##

 

[blue_leaf77]

Almost there.
So, following the aforementioned changes, you should get (I hope)
$$
\cos \theta \rightarrow -\cos \theta\\
\phi \rightarrow 2\pi -\phi
$$
Applying these changes to the ##Y_{lm}(\theta,\phi)## yields
$$
Y_{lm}(\theta,\phi) = \sqrt{\frac{(l-m)!}{(l+m)!}} P_l^m(\cos \theta) e^{i m\phi} \rightarrow \sqrt{\frac{(l-m)!}{(l+m)!}} P_l^m(-\cos \theta) e^{-i m\phi}
$$
Using the parity property of the associated Legendre polynomial ##P_l^m(-x) = (-1)^{l+m} P_l^m(x)## and ##P_l^{-m}(x) = (-1)^m \frac{(l-m)!}{(l+m)!} P_l^m(x)##, the right hand side of the above arrow sign can be manipulated to give
$$
(-1)^l \frac{(l+m)!}{(l-m)!} \sqrt{\frac{(l-m)!}{(l+m)!}} P_l^{-m}(\cos \theta) e^{-i m\phi}
$$
Now try to manipulate this further such that it becomes proportional to ##Y_{l,-m}(\theta,\phi)##.

 

[Icaro Lorran]

But that is what I did

 

[blue_leaf77]

No, it's different from yours. Look what happen when you simplify those fractions with factorial signs.

 

[Icaro Lorran]

So just ##(-1)^l |l,-m \rangle ## then

 

[blue_leaf77]

Yes, it should be that.

 

[Icaro Lorran]

Wow, that was a hell of a evaluation. Thank you for helping me out!

 

[blue_leaf77]

You can further check if this is true by applying the second ##\exp \left( \frac{i \pi L_x}{\hbar} \right)## to the last result. You should easily see that it will end up with ##(-1)^{2l} |l,m \rangle##, due to the fact that a double ##\exp \left( \frac{i \pi L_x}{\hbar} \right)##'s is equivalent to a full rotation about x axis. However, for ##l## equal to an half-odd integer, a full rotation results in a negative sign, but this is indeed a characteristic of half-odd integer angular momenta.

 

[Icaro Lorran]

Got it