# Prove angular momentum operator identity

1. Apr 4, 2013

### Tom_12

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

Using the operator identity:
$$\hat{L}^2=\hat{L}_-\hat{L}_+ +\hat{L}_z^2 + \hbar\hat{L}_z$$ show explicitly:
$$\hat{L}^2 = -\hbar^2 \left[ \frac{1}{\sin^2\theta} \frac{\partial^2}{\partial\phi^2} + \frac{1}{\sin\theta} \frac{\partial}{\partial\theta} \left(\sin\theta\frac{\partial}{\partial\theta}\right) \right]$$(Note: all L are operators, i.e. L(hat))

2. Relevant equations
$$\hat{L}_\pm = \hbar e^{\pm i\phi}\left(\pm \frac{\partial}{\partial\theta} + i\cot\theta\frac{\partial}{\partial\phi}\right) \\ \hat{L}_z = -i\hbar \frac{\partial}{\partial\phi}$$

3. The attempt at a solution

\begin{align*}
\hat{L}^2 &= \hat{L}_-\hat{L}_+ + \hat{L}_z^2 + \hbar \hat{L}_z \\
&= \hbar e^{-i\phi}\left(- \frac{\partial}{\partial\theta} + i\cot\theta\frac{\partial}{\partial\phi}\right) \times \hbar e^{+i\phi}\left(+ \frac{\partial}{\partial\theta} + i\cot\theta\frac{\partial}{\partial\phi}\right) + \left(-i\hbar \frac{\partial}{\partial\phi}\right)^2 + \hbar \left(-i\hbar \frac{\partial}{\partial\phi}\right) \\
&= \hbar^2\left[\left(- \frac{\partial}{\partial\theta} + i\cot\theta\frac{\partial}{\partial\phi}\right)\left( \frac{\partial}{\partial\theta} + i\cot\theta\frac{\partial}{\partial\phi}\right)\right] -\hbar^2\frac{\partial^2}{\partial\phi^2} - i\hbar^2\frac{\partial}{\partial\phi} \\
&= -\hbar^2\left[\left( \frac{\partial}{\partial\theta}\right)^2 + \left(\cot\theta\right)^2 + \left(\frac{\partial}{\partial\phi}\right)^2 + i\frac{\partial}{\partial\phi}\right] \\
&= -\hbar^2\left[\left( \frac{\partial}{\partial\theta}\right)^2 + \left(\frac{\partial}{\partial\phi}\right)^2\left(\cot^2\theta+1\right) + i\frac{\partial}{\partial\phi}\right] \\
&= -\hbar^2\left[\left( \frac{\partial}{\partial\theta}\right)^2 + \left(\frac{\partial}{\partial\phi}\right)^2\left(\frac{1}{\sin^2\theta}\right) + i\frac{\partial}{\partial\phi}\right]
\end{align*}

not sure how to procced from here, it's close to the required form but I do not know how to deal with the $i\frac{\partial}{\partial\phi}$ term or I might have made mistakes....

Hope someone can help, thanks

Last edited by a moderator: Apr 4, 2013
2. Apr 4, 2013

### Staff: Mentor

Hi Tom, welcome to PF!

In this step, you seem to have forgotten that you are dealing with operators. For instance, you can't just cancel out $e^{-i\phi}$ and $e^{+i\phi}$ because there is a $∂/∂\phi$ inbetween them.

3. Apr 4, 2013

### Tom_12

Ok, I have no idea if what I'm doing is right, but would really appreciate some guidance here:
\begin{align*}
&= \hbar e^{-i\phi} \left(-\frac{\partial}{\partial\theta} + i\cot\theta\frac{\partial}{\partial\phi}\right)
\hbar e^{+i\phi} \left(+\frac{\partial}{\partial\theta} + i\cot\theta\frac{\partial}{\partial\phi}\right) +
\left(-i\hbar\frac{\partial}{\partial\phi}\right)^2 -
i\hbar \frac{\partial}{\partial\phi} \\
&= \hbar \left(e^{-i\phi}\left(-\frac{\partial}{\partial\theta}\right) + ie^{-i\phi}\cot\theta\frac{\partial}{\partial\phi}\right) \hbar \left(e^{+i\phi}\left(\frac{\partial}{\partial\theta}\right) + ie^{+i\phi}\cot\theta\frac{\partial}{\partial\phi}\right) +\left(-i\hbar \frac{\partial}{\partial\phi}\right)^2 -i\hbar \frac{\partial}{\partial\phi} \\
&= -\hbar^2 \left[e^{-2i\phi} \left(\frac{\partial}{\partial\theta}\right)^2 - e^{-2i\phi}\cot\theta\frac{\partial}{\partial\theta} - \cot\theta\frac{\partial}{\partial\theta} + \cot^2\theta + \left(\frac{\partial}{\partial\phi}\right)^2 + i\frac{\partial}{\partial\phi}\right] \\
&= -\hbar^2 \left[e^{-2i\phi} \left(\frac{\partial}{\partial\theta}\right)^2 + e^{-2i\phi}\csc^2\theta + \csc^2\theta + \cot^2\theta +\left(\frac{\partial}{\partial\phi}\right)^2 + i\frac{\partial}{\partial\phi}\right]
\end{align*}
I think I'm doing something where as this doesn't seem to be going anyway.....

Last edited by a moderator: Apr 4, 2013
4. Apr 4, 2013

### Staff: Mentor

Start by considering only $\hat{L}_- \hat{L}_+$:
$$\hat{L}_- \hat{L}_+ = \hbar^2 e^{-i \phi} \left( - \frac{\partial}{\partial \theta} + i \cot \theta \frac{\partial}{\partial \phi} \right) e^{i \phi} \left( \frac{\partial}{\partial \theta} + i \cot \theta \frac{\partial}{\partial \phi} \right)$$
If you distribute $\hat{L}_-$ on $\hat{L}_+$,
$$\frac{\hat{L}_- \hat{L}_+}{\hbar^2} = - e^{-i \phi} \frac{\partial}{\partial \theta} e^{i \phi} \left( \frac{\partial}{\partial \theta} + i \cot \theta \frac{\partial}{\partial \phi}\right) + e^{-i \phi} i \cot \theta \frac{\partial}{\partial \phi} \left[ e^{i \phi} \left( \frac{\partial}{\partial \theta} + i \cot \theta \frac{\partial}{\partial \phi} \right) \right]$$
Then
$$- e^{-i \phi} \frac{\partial}{\partial \theta} \left( e^{i \phi} \frac{\partial}{\partial \theta} \right) = - \frac{\partial^2}{\partial \theta^2}$$
and
$$- e^{-i \phi} \frac{\partial}{\partial \theta} \left( e^{i \phi} i \cot \theta \frac{\partial}{\partial \phi}\right) = i \csc^2 \theta \frac{\partial}{\partial \phi} - i \cot \theta \frac{\partial}{\partial \theta} \frac{\partial}{\partial \phi}$$
and so on.

It is a bit tedious, but just remember the "rule"
$$\frac{\partial}{\partial x} f(x) g(y) = \frac{df}{dx} g(y) + f(x) g(y) \frac{\partial}{\partial x}$$

5. Apr 4, 2013

### Tom_12

I see, thank you very much