Canonical commutation relations for a particle

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The discussion focuses on the canonical commutation relations for a particle in three dimensions, specifically highlighting the relationships between position and momentum operators. It emphasizes that these relations can be applied to demonstrate the commutation relations for orbital angular momentum operators. A user shares their attempt to verify the commutation relation [Lx, Ly] but struggles with simplification. They receive advice to utilize the linearity of commutators and specific identities to simplify the calculations effectively. The conversation indicates ongoing collaboration and problem-solving among participants.
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Homework Statement


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}}]

The Attempt at a Solution


I tried opening up the commutators and it really did get me anywherehere 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 I am missing...?

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

thanks for your help!
 
Last edited:
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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.
 
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).
 
currently working on it ... ill post what i got if i got it right... when i complet eit

thanks for hte help so far...
 
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