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

sunrah
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Homework Statement


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

Homework Equations



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

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
 
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sunrah said:

Homework Statement


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

Homework Equations



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

L_{z}Y^{m}_{l}= l Y^{m}_{l}

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
 
vela said:
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.

Hi, should I be using

Y^{m}_{l} \propto P^{m}_{l}(cos\theta)e^{i m \varphi} ?
 
Yes.
 
vela said:
Yes.

just to confirm is

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

thanks!
 
No, that's not correct. Don't you have a list of spherical harmonics in your textbook? If not, just google it.
 
thank you no I wasn't aware of spherical harmonics, but i am now!
 
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