How can I prove that (x p)^2 is not equal to (x)^2 (p)^2 in quantum mechanics?

[neelakash]

Homework Statement

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.

Homework Equations

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?

 

[malawi_glenn]

(xp)^2 = xpxp

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

 

[kdv]

Homework Statement

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.

Homework Equations

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?

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).

 

[kdv]

kdv: isn't easier to see that if xpxp = xxpp, then px = xp ?

But xpxp is NOT equal to xxpp!

 

[kdv]

Iknow, but you missed my point.

IF xpxp = xxpp, then px = xp since

xpxp = x(px)p and xxpp = x(xp)p, so the things inside the paranthesis must be the same, i.e:

px = xp, and if that holds, what can one say about the commutator: [x,p] = xp - px ?

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

 

[malawi_glenn]

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

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 ;)

 

[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.

 

[malawi_glenn]

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.

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.