(adsbygoogle = window.adsbygoogle || []).push({}); 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[itex]\hbar[/itex]ˆLz,

[ˆLx, ˆLz] = −i[itex]\hbar[/itex]ˆLy,

[ˆL2, ˆLx] = 0,

[[itex]p^{2}[/itex], ˆLx] = 0,

[[itex]r^{2}[/itex], ˆLx] = 0,

[ˆLx, ˆy] = i[itex]\hbar[/itex]ˆz,

[ˆLx, ˆpy] = i[itex]\hbar[/itex]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[itex]\hbar[/itex] d/dx

Py = −i[itex]\hbar[/itex] d/dy

Pz = −i[itex]\hbar[/itex] 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[itex]\hbar[/itex] 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[itex]\hbar[/itex])d/dz z(-i[itex]\hbar[/itex])d/dx

==> second term here: zPx yPz = z(-i[itex]\hbar[/itex])d/dx y(-i[itex]\hbar[/itex])d/dz

==> first term simplified: -[itex]\hbar^{2}[/itex]yz d/dz d/dx ????

==> second term simplified: -[itex]\hbar^{2}[/itex]zy d/dx d/dz ????

Subtracting the second term from the first term, I get:

[ -[itex]\hbar^{2}[/itex]yz d/dz d/dx + [itex]\hbar^{2}[/itex]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] = -[itex]\hbar^{2}[/itex]yz [itex]\partial^{2}[/itex]/dzdx - [itex]\hbar^{2}[/itex]y [itex]\partial[/itex]/dx + [itex]\hbar^{2}[/itex]yz [itex]\partial^{2}[/itex]/dxdz = -[itex]\hbar^{2}[/itex]y [itex]\partial[/itex]/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 [itex]\hbar^{2}[/itex]yz [itex]\partial^{2}[/itex]/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.

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

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