Angular momentum commutation relation, extra terms?

  • Thread starter rwooduk
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Homework Statement


What is the commutation relation between the x and y componants of angular momentum L = r X P

Homework Equations


None.

The Attempt at a Solution


I do r X p and get the angular momentum componants:


[tex]L_{x} = (-i \hbar) (y \frac{d}{dz} - z \frac{d}{dy})[/tex]
[tex]L_{y} = (-i \hbar) (z \frac{d}{dx} - x \frac{d}{dz})[/tex]
[tex]L_{z} = (-i \hbar) (x \frac{d}{dy} - y \frac{d}{dx})[/tex]

then when I attempt to put into the commutation relation [Lx,Ly] it comes out very complicated.

My question:

I found a derivation BUT it has extra terms in it (circled in red)

9EVMLQo.jpg


Why are there 5 terms when there should only be 4 when multiplying out 2 brackets? It's important because those 2 extra terms enable the correct solution. Or have I missed something?

Thankyou for any help in advance.

Source of derivation: http://www2.ph.ed.ac.uk/~ldeldebb/docs/QM/lect8.pdf
 

Answers and Replies

  • #2
Orodruin
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Homework Statement


What is the commutation relation between the x and y componants of angular momentum L = r X P

Homework Equations


None.

The Attempt at a Solution


I do r X p and get the angular momentum componants:


[tex]L_{x} = (-i \hbar) (y \frac{d}{dz} - z \frac{d}{dy})[/tex]
[tex]L_{y} = (-i \hbar) (z \frac{d}{dx} - x \frac{d}{dz})[/tex]
[tex]L_{z} = (-i \hbar) (x \frac{d}{dy} - y \frac{d}{dx})[/tex]

then when I attempt to put into the commutation relation [Lx,Ly] it comes out very complicated.

My question:

I found a derivation BUT it has extra terms in it (circled in red)

9EVMLQo.jpg


Why are there 5 terms when there should only be 4 when multiplying out 2 brackets? It's important because those 2 extra terms enable the correct solution. Or have I missed something?

Thankyou for any help in advance.

Source of derivation: http://www2.ph.ed.ac.uk/~ldeldebb/docs/QM/lect8.pdf
There are four terms when you multiply the brackets. However, one of them can be rewritten with two terms, namely
$$
y\partial_z z \partial_x = y \partial_x + yz \partial_z \partial_x.
$$
This is just the product rule for differentiation.
 
  • #3
747
55
There are four terms when you multiply the brackets. However, one of them can be rewritten with two terms, namely
$$
y\partial_z z \partial_x = y \partial_x + yz \partial_z \partial_x.
$$
This is just the product rule for differentiation.
thats great, many thanks for your help!!
 

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