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Simple check for operaters communting, think im missing something very obvious!

  • Thread starter hoch449
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  • #1
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Homework Statement



I am supposed to find if the following commutes: [Lx,Ly]



Homework Equations




Lx= -i[tex]\hbar[/tex][y([tex]\partial/[/tex][tex]\partial[/tex]z) - z([tex]\partial/[/tex][tex]\partial[/tex]y)]

Ly= -i[tex]\hbar[/tex][z([tex]\partial/[/tex][tex]\partial[/tex]x) - x([tex]\partial/[/tex][tex]\partial[/tex]z)]

where [Lx,Ly]=LxLy-LyLx

If it commutes then [Lx,Ly]=0

The Attempt at a Solution



[Lx,Ly]= (i[tex]\hbar[/tex])2{[y([tex]\partial/[/tex][tex]\partial[/tex]z) - z([tex]\partial/[/tex][tex]\partial[/tex]y)[z([tex]\partial/[/tex][tex]\partial[/tex]x) - x([tex]\partial/[/tex][tex]\partial[/tex]z)]}


After expanding this I got a result of 0. So my solution concluded that they commute.

The answer however is [Lx,Ly]= i[tex]\hbar[/tex]Lx

I clearly expanded it wrong. I was hoping if anyone could explain how they expanded the LxLy-LyLx part. In my calculations I cannot seem to figure out how the answer contains a few more parts in the expansion which results in a non-commutation..

Thanks!
 

Answers and Replies

  • #2
George Jones
Staff Emeritus
Science Advisor
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The answer however is [Lx,Ly]= i[tex]\hbar[/tex]Lx
I think you mean [Lx,Ly]= i[tex]\hbar[/tex]Lz.

Hint: Consider [itex]\left( L_x L_y - L_y L_x \right)f[/itex], where [itex]f[/itex] is an arbitrary function of the spatial coordinates. Take things one step at a time, don't forget about the product rule for differentiation, and remove [itex]f[/itex] at the end.
 
  • #3
525
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Keep in mind that you're working with operators, so it's not just a matter of simple expansion. To calculate such a commutator, let the whole thing act on a function.

[tex][L_x,L_y]f(x,y,z)[/tex]

Since you have all sort of derivatives you will find that the product comes into a play a number of times. For instance, one term gives:
[tex]y\frac{\partial}{\partial z} \left( z\frac{\partial}{\partial x}\right) f(x,y,z) =
y \frac{\partial}{\partial x} f(x,y,z) + yz \frac{\partial^2}{\partial x\partialz}f(x,y,z)[/tex]

One of these terms will be canceled by some other term, while the other one survives.

EDIT: OK, George beat me to it ;)
 
  • #4
Cyosis
Homework Helper
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Another way to do it is to use the two following commutator identities:

[tex]
\begin{align*}
&[A+B,C]=[A,C]+[B,C]
\\
&[A,BC]=[A,B]C+B[A,C]
\end{align*}
[/tex]
 
  • #5
13
0
Perfect it all makes sense now! You guys are great thanks!
 

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