# For Angular Momentum Operator L, prove [Lx,Ly] = ihLz

1. Oct 17, 2011

### Diomarte

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

For an angular momentum operator ~L =ˆiLx +ˆjˆLy + ˆkˆLz = ˆr × ˆp, prove that
[ˆLx, ˆLy] = i$\hbar$ˆLz,
[ˆLx, ˆLz] = −i$\hbar$ˆLy,
[ˆL2, ˆLx] = 0,
[$p^{2}$, ˆLx] = 0,
[$r^{2}$, ˆLx] = 0,
[ˆLx, ˆy] = i$\hbar$ˆz,
[ˆLx, ˆpy] = i$\hbar$ˆPz.

**Note: I'm really only looking for help for the first part here, but any suggestions on how to approach the others would be appreciated. I figure I can't approach those, until I understand fully how to work through the first one.**

2. Relevant equations

Px = −i$\hbar$ d/dx
Py = −i$\hbar$ d/dy
Pz = −i$\hbar$ d/dz

Lx = yPz - zPy
Ly = zPx - xPz
Lz = xPy - yPx

3. The attempt at a solution

I'm simply looking for help on [Lx,Ly] = −i$\hbar$ Lz right now:

Take [Lx,Ly] = [yPz - zPy, zPx - xPz] = [yPz, zPx]-[yPz, xPz]-[zPy, zPx]+[zPy, xPz]

Looking at the first term:
[yPz,zPx] = (yPz zPx - zPx yPz)
==> first term here: yPz zPx = y(-i$\hbar$)d/dz z(-i$\hbar$)d/dx
==> second term here: zPx yPz = z(-i$\hbar$)d/dx y(-i$\hbar$)d/dz

==> first term simplified: -$\hbar^{2}$yz d/dz d/dx ????
==> second term simplified: -$\hbar^{2}$zy d/dx d/dz ????

Subtracting the second term from the first term, I get:
[ -$\hbar^{2}$yz d/dz d/dx + $\hbar^{2}$zy d/dx d/dz ] ????

The "????" is to represent where I'm having trouble with the algebra and differentiation.
I'm not sure that these operations actually work out like this. Also, the next step I have in my notes here says:

[yPz,zPx] = -$\hbar^{2}$yz $\partial^{2}$/dzdx - $\hbar^{2}$y $\partial$/dx + $\hbar^{2}$yz $\partial^{2}$/dxdz = -$\hbar^{2}$y $\partial$/dx

It could be a simple chain rule issue that I'm just not seeing, but would somebody please explain to me how you obtain each of these three terms? The first and third terms I seem to have no problem coming up with, but the second term is throwing me, especially since there isn't a matching term to follow before the $\hbar^{2}$yz $\partial^{2}$/dxdz term. I think the second term is actually more common sense than the first and third terms since d/dz z = 1.... As I said, anyone who can explain how to arrive at each of these terms, your assistance would be greatly appreciated.

Last edited: Oct 17, 2011
2. Oct 17, 2011

### mathfeel

Don't bother with actual representation in space (that momentum become a partial derivative operator).

Since $L_z = x p_y - y p_x$ etc, just write out the commutator in Cartesian coordinates:
$[L_x, L_y] = [y p_z - z p_y, z p_x - x p_z]$ and use properties of the commutator to get rid/factor out anything that commutes.

You should end up with $[L_x, L_y] = y p_x [p_z, z] + x p_y [z, p_z]$. Then apply the canonical commutation relation.

Last edited: Oct 17, 2011
3. Oct 17, 2011

### vela

Staff Emeritus
You're not evaluating the products correctly. You need to apply the product rule, not the chain rule.
$$y\frac{\partial}{\partial z} z\frac{\partial}{\partial x} = y \left(\frac{\partial}{\partial z} z\right) \frac{\partial}{\partial x} + y z \left(\frac{\partial}{\partial z}\frac{\partial}{\partial x}\right) = y\frac{\partial}{\partial x} + \frac{\partial^2}{\partial z\partial x}$$
But as mathfeel implied, this is the long, tedious way of doing it. I suggest you follow his suggestion.

4. Oct 18, 2011

### Diomarte

Thank you both. I was able to solve that problem, I have moved forward in this problem and am stuck on the [r^2, Lx] portion. I know that the x[x,Lx] + [x,Lx]x = 0 and that I'm left with the y and z terms, but I'm having trouble with those. Do you have any suggestions towards that approach?

I've got [y,Lx] = [y, xPz-zPy] ==> [y,yPz] - [y, zPy]
and following [A,BC] = [A,B]C + B[A,C] this expands out to be

[y,y]Pz + y[y,Pz] - [y, z]Py + z[y,Py]

If I understand this correctly, the [y,y] term goes to zero, as it must, and so must [y,z]
so I'm left with two terms y[y,Pz] - z[y,Py]

Is this correct?

5. Oct 18, 2011

### vela

Staff Emeritus
That's correct, but [y, pz] also vanishes.

6. Oct 18, 2011

### Diomarte

so I'm left only with -z[y,Py] huh?

I've worked it out a few times, and if I remember correctly [y,Pz] clearly vanishes when operating on some form of dummy function, but without actually working it out, can you explain briefly why exactly it vanishes?

7. Oct 18, 2011

### Diomarte

"The only operator in Lz that does not commute with x, is Px..." but why?

8. Oct 18, 2011

### vela

Staff Emeritus
Because when you take the partial with respect to z, you hold y constant.

9. Oct 18, 2011

### Diomarte

Most definitely, but forgive my lack of understanding at 1:00am , but is there a physical explanation, or is this just one of those things that is only mathematical?

10. Oct 18, 2011

### Diomarte

Never mind, Vela, you're amazing and awesome, and I really appreciate all the help you've given me. You answer just about every question I ask, and for that I thank you. I totally just woke up, and now things are making sense again. I think I need a short break.

11. Oct 18, 2011

### Diomarte

I do have one last question for you, does [Px, Py] commute? (Do each part of [Pi,Pj] commute?

12. Oct 18, 2011

### dextercioby

P_x and P_y commute, of course. This is postulated in a tensorial notation [p_i,p_j] = 0, $\forall$ $i,j=\bar{1,3}$. When i=1. j=2, it means exactly [p_x,p_y]=0.

13. Oct 18, 2011

### mathfeel

If you are in doubt, think intuitively back to differential operator. Can you change the order of taking partial derivative in the x and in the y direction?