Angular momentum commutation relation, extra terms?

[rwooduk]

Homework Statement

What is the commutation relation between the x and y components 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:$$L_{x} = (-i \hbar) (y \frac{d}{dz} - z \frac{d}{dy})$$
$$L_{y} = (-i \hbar) (z \frac{d}{dx} - x \frac{d}{dz})$$
$$L_{z} = (-i \hbar) (x \frac{d}{dy} - y \frac{d}{dx})$$

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

My question:

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

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

 

[Orodruin]

Homework Statement

What is the commutation relation between the x and y components 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:$$L_{x} = (-i \hbar) (y \frac{d}{dz} - z \frac{d}{dy})$$
$$L_{y} = (-i \hbar) (z \frac{d}{dx} - x \frac{d}{dz})$$
$$L_{z} = (-i \hbar) (x \frac{d}{dy} - y \frac{d}{dx})$$

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

My question:

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

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.

 

[rwooduk]

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!

 

[rwooduk]

Just to complete this thread, this derivation is MUCH easier using commutators! Here's an excellent web page detailing the process:

http://physicspages.com/2011/07/19/angular-momentum-commutators/