# 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?

## Answers and Replies

Homework Helper
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.

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?

Homework Helper
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.

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.

Homework Helper
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?

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

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

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!