# Quantum operator problem

1. Jan 12, 2008

### neelakash

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

I got stuck to this problem:

To prove that (x p)^2 is not equal to (x)^2 (p)^2

where x and p are position and mometum operator in QM.

2. Relevant equations
3. The attempt at a solution

I approached this way:

Two operators A and B are equal iff Af=Bf for all f

So,here {(x p)^2 - (x)^2 (p)^2}f=0

Since this is valid for all f, we must have {(x p)^2 - (x)^2 (p)^2}=0

In other words, [(x p)^2 , (x)^2 (p)^2 ]=0

But I could not disprove this commutator.Calculations are big and after some steps I doubt whether this might at all be the correct way.

Can anyone please give some hint?

2. Jan 12, 2008

### malawi_glenn

(xp)^2 = xpxp

if xpxp = xxpp, then what can we say about [x,p] = xp - px ?

3. Jan 12, 2008

### kdv

You can indeed do it with commutators (the identity [AB,CD] = A[B,C]D + etc would be useful) or here it might be simpler to work with explicit differential operators. Just replace the operator p by its usual expression and apply the two expressions to a test function. Obviously, X^2 p^2 f is simply $$- \hbar^2 x^2 \frac{d^2f}{dx^2}$$. Calculating the other one won't be too hard and you will see if you get the same result (you won't).

4. Jan 12, 2008

### kdv

But xpxp is NOT equal to xxpp!

5. Jan 12, 2008

### kdv

Oh, I see your point. yes, that's a nice way to see right away that the two expressions can't be equal because they would imply that x and p commute. That's nice and quick indeed.

Regards

6. Jan 12, 2008

### malawi_glenn

Now we solved the problem for the OP, which is not so good. So I propose we delete our latest posts :)

EDIT: too late ;)

I should have taken the discussion with you kdv via PM ;)

7. Jan 12, 2008

### neelakash

I got glenn's point. Really, it is lot easier.

I also tried in the way kdv indicated.But that did not work.This is really much more simple.

8. Jan 13, 2008

### malawi_glenn

Doing it the other wat should aslo work, remember that an operator operates on EVERYTHING to the right of it, for example

$$px\psi (x)= p(x \psi (x) ) = \frac{d}{dx}(x \psi (x))$$

where the last thing is evaluated according to the leibniz product rule for derivatives. Here i have omitted hbar and the imaginary unit from the p-operator just to make things clearer.