# Proving an eigenfunction (EF) from Lx as combination of EF of Lz

1. Apr 25, 2012

### sunrah

1. The problem statement, all variables and given/known data
Show that the eigenfunction of Lx can be written as a combination of eigenfunctions from Lz with the same l but different m. Using the eigenfunction
$Y_{x} = \frac{1}{\sqrt{2}}(Y^{-1}_{1} - Y^{1}_{1})$ as the eigenfunction of Lx

2. Relevant equations

$Y^{m}_{l} = \frac{1}{\sqrt{2 \pi}}e^{im\varphi}$ eigenfunction of Lz
$L_{z}Y^{m}_{l}= l Y^{m}_{l}$

3. The attempt at a solution
$Y_{x} = \frac{1}{\sqrt{2}}(Y^{-1}_{1} - Y^{1}_{1}) = \frac{1}{\sqrt{4\pi}}(e^{-i\varphi} - e^{i\varphi})$

$\widehat{L}_{x} = i \frac{h}{2\pi}(sin\theta \frac{d}{d\varphi} + cot\theta cos\varphi\frac{d}{d\varphi})$

$\widehat{L}_{x}Y_{x} = i \frac{h}{2\pi}cot\theta cos\varphi\frac{d}{d\varphi}\frac{1}{\sqrt{4\pi}}(e^{-i\varphi} - e^{i\varphi})$

$\widehat{L}_{x}Y_{x} = \frac{h}{2\pi}cot\theta cos\varphi\frac{1}{\sqrt{4\pi}}(e^{-i\varphi} + e^{i\varphi}) ≠ lY_{x}$

can't see where I'm going wrong

2. Apr 26, 2012

### vela

Staff Emeritus
The line above is your problem. $e^{i m \varphi}$ is an eigenfunction of Lz, but the spherical harmonic $Y^m_l$ is an eigenfunction of both Lz and L2.

3. Apr 26, 2012

### sunrah

Hi, should I be using

$Y^{m}_{l} \propto P^{m}_{l}(cos\theta)e^{i m \varphi}$ ?

4. Apr 26, 2012

### vela

Staff Emeritus
Yes.

5. Apr 26, 2012

### sunrah

just to confirm is

$P^{-1}_{1}(cos\theta) = cos\theta$

thanks!

6. Apr 26, 2012

### vela

Staff Emeritus
No, that's not correct. Don't you have a list of spherical harmonics in your textbook? If not, just google it.

7. Apr 26, 2012

### sunrah

thank you no I wasn't aware of spherical harmonics, but i am now!