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

1. Nov 7, 2015

### Icaro Lorran

1. The problem statement, all variables and given/known data
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.

2. Nov 8, 2015

### 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.

3. Nov 8, 2015

### 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?

4. Nov 8, 2015

### 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 $z \rightarrow -z$.

5. Nov 8, 2015

### Icaro Lorran

I'll try that, thank you

6. Nov 8, 2015

### Icaro Lorran

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

7. Nov 8, 2015

### blue_leaf77

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

8. Nov 8, 2015

### Icaro Lorran

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

9. Nov 8, 2015

### 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 $Y_{l,-m}(\theta,\phi)$.

10. Nov 8, 2015

### Icaro Lorran

But that is what I did

11. Nov 8, 2015

### blue_leaf77

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

Last edited: Nov 8, 2015
12. Nov 8, 2015

### Icaro Lorran

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

13. Nov 8, 2015

### blue_leaf77

Yes, it should be that.

14. Nov 8, 2015

### Icaro Lorran

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

15. Nov 8, 2015

### 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.

16. Nov 8, 2015

Got it