Eigenfunction of Angular Momentum Squared Operator

[Silversonic]

Homework Statement

The square of the angular momentum operator is (in SPC);

http://img6.imageshack.us/img6/67/54712598.png

Show that Y([itex]\theta[/itex],[itex]\phi[/itex]) = C[itex]sin^{2}[/itex][itex]\theta[/itex][itex]e^{2i\phi}[/itex]

Is an eigenfunction. C is a constant.

Homework Equations

The Attempt at a Solution

I am not able to get it in the form of (A number) times Y([itex]\theta[/itex],[itex]\phi[/itex]) = C[itex]sin^{2}[/itex][itex]\theta[/itex][itex]e^{2i\phi}[/itex]. No matter what, I always end up with;

-2C[itex]sin^{2}[/itex][itex]\theta[/itex][itex]e^{2i\phi}[/itex] - 4C[itex]e^{2i\phi}[/itex] - and I don't know what to do with that.

This is clearly not what I want, there is an extra term that I just can't seem to eradicate. What am I doing wrong?

 

[fzero]

You might want to show some of the earlier steps. You've made a mistake combining the first 2 terms.

 

[Silversonic]

You might want to show some of the earlier steps. You've made a mistake combining the first 2 terms.

Hmm, well with

[itex]\frac{\delta^{2}}{\delta\theta^{2}}[/itex] of Y([itex]\theta[/itex],[itex]\phi[/itex]) I get

(having taken the constant C out)

[itex]-2e^{2i\phi}[/itex]([itex]sin^{2} - cos^{2}[/itex])With [itex]cot\theta \frac{\delta}{\delta\theta}[/itex]

I get

[itex]-2e^{2i\phi}[/itex][itex]\frac{cos\theta}{sin\theta} sin\theta cos\theta[/itex] = [itex]-2e^{2i\phi}[/itex][itex]cos^{2}\theta[/itex]

and lastly with

[itex]\frac{1}{sin^{2}\theta}\frac{\delta^{2}}{\delta {\phi}^{2}}[/itex]

I get [itex]-4e^{2i\phi}[/itex]

Adding all the terms gives me;

[itex]-2e^{2i\phi}[/itex][itex]sin^{2} \theta[/itex] [itex]-4e^{2i\phi}[/itex]

Have I done something wrong?

 

[fzero]

With [itex]cot\theta \frac{\delta}{\delta\theta}[/itex]

I get

[itex]-2e^{2i\phi}[/itex][itex]\frac{cos\theta}{sin\theta} sin\theta cos\theta[/itex] = [itex]-2e^{2i\phi}[/itex][itex]cos^{2}\theta[/itex]

You've factored out the [itex]-\hbar^2[/itex] so there's no minus sign in front of this term.

 

[Silversonic]

You've factored out the [itex]-\hbar^2[/itex] so there's no minus sign in front of this term.

[itex]\frac{\delta}{\delta\theta}[/itex] of

[itex]sin^{2}[/itex][itex]\theta[/itex][itex]e^{2i\phi}[/itex]

is

[itex]2sin\theta(-cos\theta)e^{2i\phi}[/itex] = [itex]-2sin\theta cos\theta e^{2i\phi}[/itex]

surely?

So times by [itex] cot\theta [/itex]

is

[itex]-2 cos^{2} \theta e^{2i\phi}[/itex]

?

 

[fzero]

[itex]\frac{\delta}{\delta\theta}[/itex] of

[itex]sin^{2}[/itex][itex]\theta[/itex][itex]e^{2i\phi}[/itex]

is

[itex]2sin\theta(-cos\theta)e^{2i\phi}[/itex] = [itex]-2sin\theta cos\theta e^{2i\phi}[/itex]

surely?

No,

[itex] \frac{d}{d\theta}\sin\theta = \cos\theta,~~~\frac{d}{d\theta}\cos\theta = -\sin\theta.[/itex]

You'll want to doublecheck the first term as well then, but you had the overall sign correct there.

 

[Silversonic]

No,

[itex] \frac{d}{d\theta}\sin\theta = \cos\theta,~~~\frac{d}{d\theta}\cos\theta = -\sin\theta.[/itex]

You'll want to doublecheck the first term as well then, but you had the overall sign correct there.

Oh my good lord.

This is just embarrassing. It's not like I'm even tired or anything, I just continuously made an elementary mistake.

Thanks. I'll be handing in my University resignation form tomorrow ._.