# Angular momentum Commutator

1. Oct 1, 2012

### Xyius

This isn't really a homework question, it came along in my studying of the chapter, but it is a homework "type" question so I assumed this would be the best place to post this.

I am trying to show that
$$[L_x,L_y]=y[p_z,z]p_x+x[z,p_z]p_y=i \hbar L_z$$
This is all the work the book showed. So I tried to come to this conclusion myself and it is proving to be more difficult than I had originally thought. So here is what I did..

Here are the commutator properties I used..
http://imageshack.us/a/img823/9923/proby1l.png [Broken]

And here is my work using those properties..
http://imageshack.us/a/img441/8048/proby2.png [Broken]

My question is, is there an easier way to do this? Also, how in the hell does that big block of terms simplify to those two terms? It looks like the first and last term in that whole mess are the only ones that survive somehow.

Can anyone help?

Last edited by a moderator: May 6, 2017
2. Oct 1, 2012

### gabbagabbahey

Notice, for example, $[p_z,p_x]=0$ and $[y,z]=0$ In index notation, you have $[x_i,x_j]=[p_i,p_j]=0$, so half your terms in that big block with 16 terms are zero right off the bat. Then, use the fact that $[x_j,p_k]=i\hbar\delta_{jk}$, and all your other terms are zero except for the first and last.

A better way might have been to do the entire problem using index notation (with the Einstein summation convention), where you have $L_i=\epsilon_{ijk}x_jp_k$, so $[L_j,L_k]=[\epsilon_{jlm}x_lp_m, \epsilon_{kmn}x_mp_n]$. Then you would have been able to calculate all the permutations of $[L_j,L_k]$ in one shot, and with only 4 terms instead of 16.

Last edited by a moderator: May 6, 2017
3. Oct 1, 2012

### Xyius

I am still new to Quantum mechanics, while this is a senior level course, its the first course which really delves into quantum mechanics. That being said, I am still not used to dealing with operators and commutators. I was getting overly paranoid about commutation of operators that I didn't even realize that yz-zy will equal zero if we are in position space. (If I am not mistaken, these only take on values which involve a derivative if we are in momentum space.)

My only problem now is showing that (as you said)
$$[x_j,p_k]=i \hbar \delta_{j,k}$$

If I do $[y,p_x]$ I get..

$$-i \hbar [y\frac{\partial}{\partial x}-\frac{\partial y}{\partial x}]$$
The second term is obviously zero, but what to do with the first one??

4. Oct 1, 2012

### gabbagabbahey

Commutation relations are more abstract than that. For example, $[\hat{y}, \hat{z}]=0$ (I'll use ^ to distinguish between operator and scalar) regardless of whether you act on a wavefunction in the position basis or momentum basis, or any other basis. To see why this is true in the momentum basis, just consider that $\frac{\partial^2}{\partial y \partial z} = \frac{\partial^2}{\partial z \partial y}$.

This is usually derived early on in any good QM textbook, so it will probably be in yours.

However, you've expanded the commutator wrong. To see how, try acting on a specific wavefunction $\psi ( \mathbf{r} )$ in the position basis:

$[y,p_x]\psi ( \mathbf{r} )=y\left( -i\hbar \frac{\partial \psi}{\partial x}\right) + i\hbar \frac{\partial }{\partial x} \left( y\psi \right)$

the derivative acts on the product $y\psi$, not just on $y$, so you will need to use the product rule, from which the desired result follows. In more abstract terms, when you act on a state with a product of operators, $\hat{A}\hat{B}$, you first act on the state with $\hat{B}$ and then act on the result with $\hat{A}$.

Last edited: Oct 1, 2012
5. Oct 1, 2012

### Xyius

Thank you for your reply! (Oh man you got rid of the bit where you showed that [x,y]=0 in any basis. I was going to go through that later! Oh well)

Thanks again for clearing things up :]

6. Oct 1, 2012

### gabbagabbahey

You're welcome. The reason I deleted the other bit was that there were some errors that I didn't have time to fix (it takes a while to write it out in $\LaTeX$), but the general idea is to insert the identity operator $\hat{I}=\int_{\text{all space}} d^3x |\mathbf{x}\rangle \langle \mathbf{x}|$ (in $\langle \mathbf{n} | \psi \rangle = \langle \mathbf{n} | \hat{I} |\psi \rangle$), and then pull everything inside the integral, so that the commutator acts on $\psi(\mathbf{x}) \equiv \langle \mathbf{x} | \psi \rangle$ (since you know that the result of that is zero) and the result gets multiplied by some complex scalar ($\langle \mathbf{n} | \mathbf{x} \rangle$ ) and then integrated over all space, which obviously gives zero.