Undergrad Calculating the eigenvalue of orbital angular momentum

[TheBlueDot]

Hello,

I'm trying to calculate the measurement of the orbital angular momentum of the state l=1 and m = -1. The operator for the angular momentum squared is
$L^2 = -\hbar (\frac{1}{sin\theta}(\frac{\partial}{\partial \theta}(sin\theta\frac{\partial}{\partial \theta})) +\frac{1}{sin\theta^2}\frac{\partial^2}{\partial\phi^2})$,

which expands to
$L^2 = -\hbar (\frac{1}{sin\theta}(\frac{\partial sin\theta}{\partial \theta}\frac{\partial}{\partial \theta}+sin\theta \frac{\partial^2}{\partial \theta^2}) +\frac{1}{sin\theta^2}\frac{\partial^2}{\partial\phi^2})$

When operate this on the wave function $Y_{1,-1} = csin\theta e^{-i\phi}$, I got

$Y_{1,-1}*(\frac{cos^2\theta}{sin\theta} -\frac{sin^2\theta}{sin\theta}-\frac{1}{sin\theta})$,

which is zero. The answer should be $2\hbar^2$.

If the cosine term is zero, then I'll get the right result.

Please help!

Thanks

 

[stevendaryl]

Hello,

I'm trying to calculate the measurement of the orbital angular momentum of the state l=1 and m = -1. The operator for the angular momentum squared is
$L^2 = -\hbar (\frac{1}{sin\theta}(\frac{\partial}{\partial \theta}(sin\theta\frac{\partial}{\partial \theta})) +\frac{1}{sin\theta^2}\frac{\partial^2}{\partial\phi^2})$,

which expands to
$L^2 = -\hbar (\frac{1}{sin\theta}(\frac{\partial sin\theta}{\partial \theta}\frac{\partial}{\partial \theta}+sin\theta \frac{\partial^2}{\partial \theta^2}) +\frac{1}{sin\theta^2}\frac{\partial^2}{\partial\phi^2})$

When operate this on the wave function $Y_{1,-1} = csin\theta e^{-i\phi}$, I got

$Y_{1,-1}*(\frac{cos^2\theta}{sin\theta} -\frac{sin^2\theta}{sin\theta}-\frac{1}{sin\theta})$,

which is zero. The answer should be $2\hbar^2$.

If the cosine term is zero, then I'll get the right result.

Please help!

Thanks

$\frac{cos^2(\theta)}{sin(\theta)} - \frac{sin^2(\theta)}{sin(\theta)} - \frac{1}{sin(\theta)} = \frac{1-sin^2(\theta) - sin^2(\theta) - 1}{sin(\theta)} = \frac{-2 sin^2(\theta)}{sin(\theta)} = -2 sin(\theta)$

 

[TheBlueDot]

$\frac{cos^2(\theta)}{sin(\theta)} - \frac{sin^2(\theta)}{sin(\theta)} - \frac{1}{sin(\theta)} = \frac{1-sin^2(\theta) - sin^2(\theta) - 1}{sin(\theta)} = \frac{-2 sin^2(\theta)}{sin(\theta)} = -2 sin(\theta)$

@stevendaryl,
Thanks for your response. I feel silly now. I looked at the $sin^2(\theta) -1$ and thought it was $cos^2(\theta)$.

 

[BvU]

It doesn't say $\ \ \sin^2(\theta) -1\$, but $\ \ - \sin^2(\theta) -1\ \$