# Canonical commutation relations for a particle

1. Dec 11, 2006

### stunner5000pt

1. The problem statement, all variables and given/known data
The canonical commutation relations for a particl moving in 3D are
$$[\hat{x},\hat{p_{x}}]= i\hbar$$
$$[\hat{y},\hat{p_{y}}]= i\hbar$$
$$[\hat{z},\hat{p_{z}}]= i\hbar$$

and all other commutators involving x, px, y ,py, z , pz (they should all have a hat on eahc of them signifying that htey are operators) are zero. These relations can be used to show that the operators for the orbital angular mometum obey the following commutation relations

$$[\hat{L_{x}},\hat{L_{y}}]= i\hbar \hat{L_{z}}$$
$$[\hat{L_{y}},\hat{L_{z}}]= i\hbar \hat{L_{x}}$$
$$[\hat{L_{z}},\hat{L_{x}}]= i\hbar \hat{L_{y}}$$

Using
$$\hat{L_{x}} = \hat{y}\hat{p_{z}} - \hat{z}\hat{p_{y}}$$
$$\hat{L_{y}} = \hat{z}\hat{p_{x}} - \hat{x}\hat{p_{z}}$$

Verify that
$$[\hat{L_{x}},\hat{L_{y}}] = [\hat{y}\hat{p_{z}},\hat{z}\hat{p_{x}}]+[\hat{z}\hat{p_{y}},\hat{x}\hat{p_{z}}]$$

3. The attempt at a solution
I tried opening up the commutators and it really did get me anywhere

here is what i did

$$[\hat{y}\hat{p_{z}},\hat{z}\hat{p_{x}}]+[\hat{z}\hat{p_{y}},\hat{x}\hat{p_{z}}] = yp_{y}zp_{x} - zp_{x}yp_{z} + zp_{y}xp_{z} - xp_{z}zp_{y}$$

and the left hand side yields

$$yp_{z}zp_{x} - yp_{z}xp_{z} - zp_{y}zp_{x} + zp_{y}xp_{z} + zp_{x} yp_{z} - zp_{x} zp_{y} - xp_{z}yp_{z} + xp_{z} z p_{y}$$

nothing seems to simplify... or is there something im missing...?

o and i did not put hats on eahc of them because it would just too much typing...

Last edited: Dec 11, 2006
2. Dec 11, 2006

### StatusX

It's easier to start from the other side, ie, expand [Lx,Ly]. The commutator is linear, in that [a+b,c]=[a,c]+[b,c], and after expanding like this several of the terms will be zero.

3. Dec 11, 2006

### Physics Monkey

You will find the identity listed by StatusX very useful. The other identity you will find very useful is $$[AB,C] = A[B,C] + [A,C]B$$. Use these two identities to reduce every angular momentum commutator to commutators of position with momentum (or position with position which is zero, etc).

4. Dec 11, 2006

### stunner5000pt

currently working on it ... ill post what i got if i got it right... when i complet eit

thanks for hte help so far...

Last edited: Dec 11, 2006