Quantum Physics, find the commutator

[Antti]

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:

-i \hbar \bold r \times \nabla

I rewrote this as:

i \hbar \nabla \times \bold r

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

\nabla \times \bold r = 0

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

 

[alphysicist]

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 f, then:

<br /> \hat L f = -i\hbar \vec r\times \vec\nabla f = i \hbar \left(\vec \nabla f\right) \times \vec r<br />

The important things is that the del operator is not operating on \vec r, but is operating on whatever the angular momentum operator is operating on.

 

[Antti]

Thanks for your reply!

When i calculate the components i get

<br /> \hat L f = -i\hbar \vec r\times \vec\nabla f = i \hbar \left(\vec \nabla f\right) \times \vec r<br />

<br /> \left(\vec \nabla f\right) \times \vec r = \hat x ( f_y&#039; z - f_z&#039; y ) - \hat y ( f_x&#039; z - f_z&#039; x ) + \hat z ( f_x&#039; y - f_y&#039; x )<br />

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

<br /> \hat L_x f = i \hbar ( f_y&#039; z - f_z&#039; y )
\hat L_y f = - i \hbar ( f_x&#039; z - f_z&#039; x ) = i \hbar ( f_z&#039; x - f_x&#039; z )<br />

Am I right so far?

 

[alphysicist]

Those components look right to me.

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

<br /> \hat L = -i\hbar \vec r\times \vec \nabla <br />

and operating on something:

<br /> \hat L f= -i\hbar \vec r\times \vec \nabla f<br />

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

<br /> \hat L = \hat r\times \hat p<br />

The specific form you are using (with the del operator for the momentum) is in a position space representation.

 

[Antti]

Ok, the answer I'm expecting is the following. It's from a solution to an exam:

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

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

<br /> \hat L_x f = i \hbar ( f_y&#039; z - f_z&#039; y )
\hat L_y f = - i \hbar ( f_x&#039; z - f_z&#039; x ) = i \hbar ( f_z&#039; x - f_x&#039; z )<br />

So my commutator is:

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

\hat L_x \hat L_y f = \hat L_x i \hbar ( f_z&#039; x - f_x&#039; z ) = i \hbar \ ( \ \ z (i \hbar ( f_z&#039; x - f_x&#039; z ))_y&#039; - y(i \hbar ( f_z&#039; x - f_x&#039; z ))_z&#039; \ \ )
= (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)

\hat L_y \hat L_x f = \hat L_y i \hbar ( f_y&#039; z - f_z&#039; y ) = i \hbar \ ( \ \ x (i \hbar ( f_y&#039; z - f_z&#039; y ))_z&#039; - z(i \hbar ( f_y&#039; z - f_z&#039; y ))_x&#039; \ \ )
= (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)

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.

 

[alphysicist]

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: \hat L_x acting on some function g is:

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

and the \hat L_y operating on some function f is:

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

Now if you want \hat L_x \hat L_y f, you just plug the entire form of \hat L_y f (the last equation) into the spot for g in the L_x equation above. What's important is that the derivative \frac{\partial}{\partial z} that occurs in the x component acts on the entire quantity (\hat L_y f), not just f. So you'll get five terms for \hat L_x \hat L_y f. What do you get?

 

[Antti]

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. I am posting the correct solution too so this thread is complete.

\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)

\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)

[\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 ) \ \ )

Recall from my earlier post that

<br /> \left(\vec \nabla f\right) \times \vec r = \hat x ( f_y&#039; z - f_z&#039; y ) - \hat y ( f_x&#039; z - f_z&#039; x ) + \hat z ( f_x&#039; y - f_y&#039; x )<br />

Thus

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

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

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

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