# Homework Help: Quantum Physics, find the commutator

1. Jun 9, 2008

### 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?

2. Jun 9, 2008

### 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:

$$\hat L f = -i\hbar \vec r\times \vec\nabla f = i \hbar \left(\vec \nabla f\right) \times \vec r$$

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.

3. Jun 10, 2008

### Antti

When i calculate the components i get

$$\hat L f = -i\hbar \vec r\times \vec\nabla f = i \hbar \left(\vec \nabla f\right) \times \vec r$$

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

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

$$\hat L_x f = i \hbar ( f_y' z - f_z' y )$$
$$\hat L_y f = - i \hbar ( f_x' z - f_z' x ) = i \hbar ( f_z' x - f_x' z )$$

Am I right so far?

4. Jun 10, 2008

### alphysicist

Those components look right to me.

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

$$\hat L = -i\hbar \vec r\times \vec \nabla$$

and operating on something:

$$\hat L f= -i\hbar \vec r\times \vec \nabla f$$

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:

$$\hat L = \hat r\times \hat p$$

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

5. Jun 10, 2008

### 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 :)

$$\hat L_x f = i \hbar ( f_y' z - f_z' y )$$
$$\hat L_y f = - i \hbar ( f_x' z - f_z' x ) = i \hbar ( f_z' x - f_x' z )$$

So my commutator is:

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

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

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

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.

6. Jun 10, 2008

### 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:

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

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

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

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?

7. Jun 10, 2008

### 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. Im 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

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

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!