# Canonical commutation relations for a particle

#### 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:

#### StatusX

Homework Helper
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.

#### Physics Monkey

Homework Helper
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).

#### 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:

### The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving