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Quantum Physics, find the commutator

  • Thread starter Antti
  • Start date
  • #1
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This should be easy, since I'm sure I've misunderstood something here. The task is to find the commutator of the x- and y-components of the angular momentum operator. This operator is, according to physics handbook:

[tex]-i \hbar \bold r \times \nabla[/tex]

I rewrote this as:

[tex]i \hbar \nabla \times \bold r[/tex]

However there's a common rule for the del operator saying

[tex]\nabla \times \bold r = 0[/tex]

This doesn't make sense, what am I missing?
 

Answers and Replies

  • #2
alphysicist
Homework Helper
2,238
1
Hi Antti,

When you do the manipulations, you have to keep track of what the del operator is acting on. Let's say it is acting on a some general function [itex]f[/itex], then:

[tex]
\hat L f = -i\hbar \vec r\times \vec\nabla f = i \hbar \left(\vec \nabla f\right) \times \vec r
[/tex]

The important things is that the del operator is not operating on [itex]\vec r[/itex], but is operating on whatever the angular momentum operator is operating on.
 
  • #3
27
0
Thanks for your reply!

When i calculate the components i get

[tex]
\hat L f = -i\hbar \vec r\times \vec\nabla f = i \hbar \left(\vec \nabla f\right) \times \vec r
[/tex]

[tex]
\left(\vec \nabla f\right) \times \vec r = \hat x ( f_y' z - f_z' y ) - \hat y ( f_x' z - f_z' x ) + \hat z ( f_x' y - f_y' x )
[/tex]

And the x- and y-components of the angular mommentum operator are:

[tex]
\hat L_x f = i \hbar ( f_y' z - f_z' y )[/tex]
[tex]\hat L_y f = - i \hbar ( f_x' z - f_z' x ) = i \hbar ( f_z' x - f_x' z )
[/tex]

Am I right so far?
 
  • #4
alphysicist
Homework Helper
2,238
1
Those components look right to me.

I would suggest that you still consider the angular momentum operator to be your original form:

[tex]
\hat L = -i\hbar \vec r\times \vec \nabla
[/tex]

and operating on something:

[tex]
\hat L f= -i\hbar \vec r\times \vec \nabla f
[/tex]

In my last post I was just showing that if you do move it around you don't get [itex]\vec\nabla\times \vec r=0[/itex]. It makes no difference for the components of this problem; but that way it matches the definition of angular momentum:

[tex]
\hat L = \hat r\times \hat p
[/tex]

The specific form you are using (with the del operator for the momentum) is in a position space representation.
 
  • #5
27
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Ok, the answer I'm expecting is the following. It's from a solution to an exam:

[tex][\hat L_x, \hat L_y] = i \hbar \hat L_z[/tex]

You'll have to excuse me but I still write the components in this form out of pure LaTex-laziness :)

[tex]
\hat L_x f = i \hbar ( f_y' z - f_z' y )[/tex]
[tex]\hat L_y f = - i \hbar ( f_x' z - f_z' x ) = i \hbar ( f_z' x - f_x' z )
[/tex]

So my commutator is:

[tex][\hat L_x, \hat L_y] = \hat L_x \hat L_y f - \hat L_y \hat L_x f[/tex]

[tex]\hat L_x \hat L_y f = \hat L_x i \hbar ( f_z' x - f_x' z ) = i \hbar \ ( \ \ z (i \hbar ( f_z' x - f_x' z ))_y' - y(i \hbar ( f_z' x - f_x' z ))_z' \ \ )[/tex]
[tex]= (i \hbar)^{2} (xz \frac{d}{dy}\frac{d}{dz} f - z^{2} \frac{d}{dy} \frac{d}{dx} f - yx \frac{d^{2}}{dz^{2}} f + yz \frac{d}{dz} \frac{d}{dx} f)[/tex]

[tex]\hat L_y \hat L_x f = \hat L_y i \hbar ( f_y' z - f_z' y ) = i \hbar \ ( \ \ x (i \hbar ( f_y' z - f_z' y ))_z' - z(i \hbar ( f_y' z - f_z' y ))_x' \ \ )[/tex]
[tex]= (i \hbar)^{2} (xz \frac{d}{dz}\frac{d}{dy} f - yx \frac{d^{2}}{dz^{2}} f - z^{2} \frac{d}{dx} \frac{d}{dy} f + yz \frac{d}{dx} \frac{d}{dz} f)[/tex]

But these components are now equal so their difference is zero. This was not the answer I expected. I must still be doing something wrong.
 
  • #6
alphysicist
Homework Helper
2,238
1
Well, I'm all for laziness. However, here I think your notation is leading you down the wrong path. You've done just about everything right, but you are making one math error.

Let me write the operators: [itex]\hat L_x[/itex] acting on some function g is:

[tex]
\hat L_x g= -i\hbar\left[ y \frac{\partial}{\partial z} g - z \frac{\partial}{\partial y} g\right]
[/tex]

and the [itex]\hat L_y[/itex] operating on some function f is:

[tex]
\hat L_y f= -i\hbar\left[ z \frac{\partial}{\partial x} f - x \frac{\partial}{\partial z} f\right]
[/tex]

Now if you want [tex]\hat L_x \hat L_y f[/tex], you just plug the entire form of [itex]\hat L_y f[/itex] (the last equation) into the spot for g in the [itex]L_x[/itex] equation above. What's important is that the derivative [tex]\frac{\partial}{\partial z}[/tex] that occurs in the x component acts on the entire quantity [tex](\hat L_y f)[/tex], not just f. So you'll get five terms for [tex]\hat L_x \hat L_y f[/tex]. What do you get?
 
  • #7
27
0
I see. I had forgot to use the product rule when differentiating. As you say I got an extra term for both of the "chained" (?) operators. The other terms cancel out as before. Im posting the correct solution too so this thread is complete.

[tex]\hat L_x \hat L_y f = (i \hbar)^{2} (xz \frac{d}{dy}\frac{d}{dz} f - z^{2} \frac{d}{dy} \frac{d}{dx} f - yx \frac{d^{2}}{dz^{2}} f + yz \frac{d}{dz} \frac{d}{dx} f + y \frac{d}{dx} f)[/tex]

[tex]\hat L_y \hat L_x f = (i \hbar)^{2} (xz \frac{d}{dz}\frac{d}{dy} f - z^{2} \frac{d}{dx} \frac{d}{dy} f - yx \frac{d^{2}}{dz^{2}} f + yz \frac{d}{dx} \frac{d}{dz} f - x \frac{d}{dy} f)[/tex]

[tex][\hat L_x, \hat L_y] = \hat L_x \hat L_y f - \hat L_y \hat L_x f = (i \hbar)^{2} (y \frac{d}{dx} f) - (i \hbar)^{2} (x \frac{d}{dy} f) = i \hbar \ ( \ \ i \hbar ( y \frac{d}{dx} f - x \frac{d}{dy} f ) \ \ )[/tex]

Recall from my earlier post that

[tex]
\left(\vec \nabla f\right) \times \vec r = \hat x ( f_y' z - f_z' y ) - \hat y ( f_x' z - f_z' x ) + \hat z ( f_x' y - f_y' x )
[/tex]

Thus

[tex]\hat L_z f = i \hbar ( y \frac{d}{dx} f - x \frac{d}{dy} f )[/tex]

This is what I have in the end of the third line above. So

[tex][\hat L_x, \hat L_y] = i \hbar \hat L_z[/tex]

Which is what I wanted :)
Thanks for the help!
 

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