- #1

- 511

- 1

## 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?