Canonical commutation relations for a particle

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Homework Help Overview

The discussion revolves around the canonical commutation relations for a particle moving in three dimensions and their implications for the commutation relations of orbital angular momentum operators. Participants are exploring how to verify specific commutation relations using the definitions of angular momentum operators.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to expand the commutators directly but finds the process complex and unclear. Some participants suggest starting from the other side of the equation to simplify the problem. Others mention useful identities that could aid in reducing the complexity of the commutators.

Discussion Status

The discussion is ongoing, with participants providing hints and suggestions for approaches without reaching a consensus. There is a recognition of the need to simplify the problem using established identities, and some participants are actively working on their solutions.

Contextual Notes

There is mention of the difficulty in typing the hat notation for operators, which may affect clarity in the discussion. The original poster expresses uncertainty about their approach and seeks further assistance.

stunner5000pt
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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...
 
Last edited:

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