# Uncertainty principle, commutator.

1. Feb 8, 2012

### LogicX

1. The problem statement, all variables and given/known data

Calculate [x,px] = (xpx - pxx)

Do this for a function f(x).

Now calculate [x,py] for f(x,y)

2. Relevant equations

px is actually px hat, I'm just not familiar with latex code.

px= -i (d/dx)

3. The attempt at a solution

I believe I got the first part, for f(x). This is just the canonical commutator relation with f(x) instead of a wavefunction like my book had.

[x,px]= i f(x)

I'm a little stuck on the second part because my derivative skills are shaky right now. I believe that the momentum operator changes to py= -i (d/dy)

So, if I do the question I get (using the product rule):

$$x p_y f(x,y) ~=~ x \times -i \frac{df(x,y)}{dy}$$

$$p_y x f(x,y) ~=~ -i x \frac{df(x,y)}{dy} + i f(x,y) \frac{dx}{dy}$$

Then two things cancel out when you do the subtraction and I am left with:

$$x p_y f(x,y) - p_y x f(x,y) ~=~ i f(x,y) \frac{dx}{dy}$$

I am not sure where to go from here. This is where I say that I am shaky with derivatives because I don't know what f(x,y) dx/dy means. In the first part, instead of dx/dy I just got dx/dx which I assume cancels to one. So that is why I could do the first part but not this second part.

2. Feb 8, 2012

### Dick

Review partial derivatives because that's what these derivatives are. $\frac{\partial f(x,y)}{\partial y}$ means take the derivative of f(x,y) with respect to y with x held constant. If f(x,y) is just x, then that is 0. It's the derivative of a constant.

3. Feb 8, 2012

### LogicX

So how did you know that it was a partial derivative? Just because it couldn't be anything else with that specific derivative that I presented? I just didn't recognize it because my HW used d instead of $\partial$.

And I think I see what you are saying. I understand that the partial derivative of x with respect to y is 0. So you are tell me that the f(x,y) in the numerator of the derivative is referring to the x as being the function f(x,y)?

Here is where I am a little confused. I thought that f(x,y) represented an actual function, it just happens to be the general case for any function with independent variables x and y. So, if you wanted to do it for a specific function, say f(x,y)=x2y2, you could write that in place of f(x,y).

Do you see what I'm getting at? In what you are doing, f(x,y) seems to just be part of the derivative notation in order to take the partial derivative of x. In my specific example, f(x,y) is an actual function and you must include that in the what you are taking a derivative of. I am confused that it just kind of disappears in what you are telling me to do.

I had thought that having f(x,y) meant you couldn't simplify it anymore because you don't know what f(x,y) is, it is just a general function.

(sorry for being so longwinded, I'm just trying to be as specific as possible so that you can get a sense for what my trouble is, since it seems to be some sort of fundamental misunderstanding of derivatives.)

4. Feb 8, 2012

### Dick

I'm trying to say that you are doing everything right. The HW is sloppy if it is just using d instead of $\partial$. When you are dealing with multiple variables like x and y you should use $\partial$. And your answer is correct. It's $i f(x,y) \frac{\partial x}{\partial y}$. But $\frac{\partial x}{\partial y}=0$. So you can simplify the whole answer to 0.

5. Feb 8, 2012

### LogicX

So $\frac{\partial x}{\partial y}=0$ only because it is a partial derivative, right? $\frac{\partial x}{\partial y}$ means take the derivative of x with respect to y with x as a constant. Thus it is zero.

I guess I'm just a little confused because I haven't seen partial derivatives in a while. Normally in the types of problems I do, derivatives look like dy/dx, and that just means take the derivative of whatever is in front of it, not that it equals zero.

Okay one last thing and again I apologize that my problem seems to be with the fundamentals. When I did this part of the problem:

$$p_y x f(x,y) ~=~ -i \left[xf(x,y)\frac{\partial}{\partial y}\right] ~=~ -i x \frac{\partial f(x,y)}{\partial y} + i f(x,y) \frac{\partial x}{\partial y}$$

I had to use the product rule. What is different about $x\frac{\partial f(x,y)}{\partial y}$ that it is not considerd a product rule? f(x,y) and x are being multiplied together in both the first equation where they are being acted on by an operator, and in the latter example as well. Or is something different about the latter example that you cannot factor out both x and f(x,y) to have:

$xf(x,y)\frac{\partial}{\partial y}$

Last edited: Feb 8, 2012
6. Feb 8, 2012

### Dick

It IS a product rule and you did it almost correctly. $p_y x f(x,y) ~=~ -i \frac{\partial }{\partial y} (x f(x,y)) ~=~ -i x \frac{ \partial f(x,y)}{\partial y} - i f(x,y) \frac{\partial x}{\partial y}$.

7. Feb 8, 2012

### Dick

This part I don't get. How do you get the partial derivative hanging at the end? It used to be at the beginning.

8. Feb 8, 2012

### LogicX

Ok, ignore that right now, I'm not sure what I was going for. When you do the subtraction:

$-i x \frac{\partial f(x,y)}{\partial y} - \left[-i x \frac{ \partial f(x,y)}{\partial y} - i f(x,y) \frac{\partial x}{\partial y}\right] ~=~ i f(x,y) \frac{\partial x}{\partial y}$

And this is where my previous' post was in error, since:

$i f(x,y) \frac{\partial x}{\partial y}\neq i x \frac{\partial f(x,y)}{\partial y}$

The derivative only operates on what is to its right, and you cannot just factor anything in and out of the derivative as you please (?).

Last edited: Feb 8, 2012
9. Feb 8, 2012

### Dick

Sure, you can't do that. You can't just move d/dy around where ever you please. It doesn't commute with everything. But you had the right answer to begin with. And I hope you'll agree that it's zero.

10. Feb 8, 2012

### LogicX

Yes I do, and I think I've sorted my issues out now. Thanks very much.