# Spherical Polar Proof for Non-Commutativity of Indivdual Quantum Angular Momentum Operators

How can the following commutation relation be solved through spherical polar coordinates

$[\hat{L}_{z}$,$\hat{L}_{x}]$ = $\imath\hbar\hat{L}_{y}$

I understand the derivation through partial derivatives, and the canonical method, and have successfully derived the values for the three $\hat{L}_{x,y,z}$ operators in spherical polars, but I can not get my head around how the proof works with spherical polars?

• Can you be more precise about what confuses you? – TLDR Feb 21 '16 at 5:09
• Can't you just plug in what $\hat{L_z}$ and $\hat{L_x}$ are in spherical polar coordinates, act it on a wavefunction and then you're good? – Spaderdabomb Feb 21 '16 at 5:27

So I'm not going to work this all out as it's just algebra, but I believe the following is what you'll want to do.

$$L_z = \frac{\hbar}{i}\frac{\partial}{\partial \phi}$$ $$L_x = i\hbar \left(\sin \phi \frac{\partial}{\partial \theta} + \cot \theta \cos \phi \frac{\partial}{\partial \phi}\right)$$

Now since derivatives are involved, you can't just compute a simple commutator and get the result you want. You have to "act it" on a test wavefunction.

$$[L_z, L_x]\Psi = i\hbar L_y \Psi$$

If you can show that is true, then you have proved it.

My Solution

Here is the solution fully done out. The algebra really isn't all that complicated. I recommend getting a big whiteboard and writing everything out explicitly and just go one step at a time and don't get overwhelmed. There are no complex tricks here, just follow the simple rules of derivatives (there are a few chain rules involved however).

$$L_z = \frac{\hbar}{i}\frac{\partial}{\partial \phi}$$ $$L_x = i\hbar \left(\sin \phi \frac{\partial}{\partial \theta} + \cot \theta \cos \phi \frac{\partial}{\partial \phi}\right)$$ $$L_y = i\hbar\left(-\cos \phi \frac{\partial}{\partial \theta} + \cot \theta \sin \phi \frac{\partial}{\partial \phi}\right)$$

We now act on a test wavefunction (make a habit of this when computing two sides of commutators, and ALWAYS do this if derivatives are involved)

$$[L_z, L_x]\Psi = i\hbar L_y \Psi$$

Expanding the commutator on the left side, we get $$[L_z, L_x]\Psi = L_zL_x\Psi - L_xL_z\Psi$$ $$(1) \to L_zL_x\Psi$$ $$(2) \to L_xL_z\Psi$$

For Equation (1) we have

$$(1) \to \left(\frac{\hbar}{i}\frac{\partial}{\partial \phi}\right) \left(i\hbar \left(\sin \phi \frac{\partial}{\partial \theta} + \cot \theta \cos \phi \frac{\partial}{\partial \phi}\right)\right)\Psi$$ $$\to \hbar^2 \left(\left(\frac{\partial}{\partial \phi}\right)\left(\sin \phi \frac{\partial}{\partial \theta} + \cot \theta \cos \phi \frac{\partial}{\partial \phi}\right)\right)\Psi$$ $$\to \hbar^2 \left(\cos \phi \frac{\partial}{\partial \theta} - \cot \theta \sin \phi \frac{\partial}{\partial \phi} \right) \Psi + \hbar^2\left(\sin \phi \frac{\partial}{\partial \theta} + \cot \theta \cos \phi \frac{\partial}{\partial \phi}\right) \frac{\partial \Psi}{\partial \phi}$$ $$(1) \to \hbar^2 \left(\cos \phi \frac{\partial \Psi}{\partial \theta} - \cot \theta \sin \phi \frac{\partial \Psi}{\partial \phi} \right) + \hbar^2\left(\sin \phi \frac{\partial^2 \Psi}{\partial \theta \partial \phi} + \cot \theta \cos \phi \frac{\partial^2 \Psi}{\partial \phi^2}\right)$$

For Equation (2) we have

$$(2) \to i\hbar\left(\sin \phi \frac{\partial}{\partial \theta} \cot \theta \cos \phi \frac{\partial}{\partial \phi}\right)\left(\frac{\hbar}{i}\frac{\partial}{\partial \phi}\right)\Psi$$ $$\to \hbar^2\left(\sin \phi \frac{\partial}{\partial \theta} + \cot\theta \cos \phi \frac{\partial}{\partial \phi}\right)\frac{\partial \Psi}{\partial \phi}$$ $$(2) \to \hbar^2\left(\sin\phi \frac{\partial^2 \phi}{\partial \theta \partial \phi} + \cot \theta \cos \phi \frac{\partial^2 \Psi}{\partial \phi^2}\right)$$

Taking (1)-(2), we get see that the second term from (1) is identically (2), so we get

$$(1)-(2) \to \hbar^2\left(\cos \phi \frac{\partial \Psi}{\partial \theta} - \cot \theta \sin \phi \frac{\partial \Psi}{\partial \phi}\right)$$

Which is now our left hand side. Computing the right side, we plug in $$L_y$$ to get

$$i\hbar L_y \Psi = (i\hbar)(i\hbar)\left(-\cos \phi \frac{\partial}{\partial \theta} + \cot \theta \sin \phi \frac{\partial}{\partial \phi}\right) \Psi$$ $$\to \hbar^2\left(\cos \phi \frac{\partial \Psi}{\partial \theta} - \cot \theta \sin \phi \frac{\partial \Psi}{\partial \phi}\right)$$

Thus we have equated sides and proved the theorem.

• Thanks, it's the algebra I just can't get to work properly. It's a big ask but if you get the time please could you demonstrate this because I've spent hours trying and got no where! – James Jeans Feb 21 '16 at 14:36
• Alright I gave a pretty thorough derivation, Cheers. – Spaderdabomb Feb 21 '16 at 21:34