# Uncertainty principle on P and E exchange

1. Feb 16, 2010

### luxiaolei

Hi,all, I am wondering, momentum can expressed as energy, so:

dxdP can expresses as dxdE ???

then position, momentum, energy,time are all related, and they all form uncertainty relations??

in the view of commute the operators, all these four operators can NOT commute??

where am I wrong?

2. Feb 16, 2010

### LostConjugate

The total energy depends on position as well, thus does not commute. You could say for a free particle everything commutes however in a universe with only a single particle I am not sure such a term as position would even exist as it is a relative term.

3. Feb 16, 2010

### luxiaolei

Thanks for replay,LostConjugate. I am very interested in what you said free particle's all operators commute, would you please explain it more? Thanks

4. Feb 16, 2010

### LostConjugate

Actually it still does not commute with position. A free particle's energy only depends on the momentum and not the position because it is in a universe where there are no other forces. So the total energy would commute with the momentum.

5. Feb 16, 2010

### LostConjugate

This is true.

6. Feb 16, 2010

### SpectraCat

It is true that in classical mechanics, kinetic energy can be expressed in terms of the momentum as p2/2m. However, in QM, momentum is an operator, so you need to be a little more careful. You can construct the analogous kinetic energy operator as:

$$\hat{K}=\frac{\hat{p}^{2}}{2m}=-\frac{\hbar^{2}}{2m}\nabla^{2}$$

Also, you seem to be neglecting potential energy in your question, V(x), which is in turn a function of position. The total energy is of course always the sum of the kinetic and potential energies.

1) momentum and kinetic energy
2) position and potential energy

The following pairs of operators never commute:
1) momentum and position
2) kinetic energy and position

The following pairs of operators do not commute in general, but can in special cases (like the free particle, where V=0):
1) momentum and potential energy
2) kinetic energy and potential energy

HTH

7. Feb 16, 2010

### luxiaolei

LostConjugate:
right, start to make sense now, thanks alot. so it comes an conclusion:
if precisely measure time then enlarger E and P, then result in precise the position, am I right? position and time, if measure one of them precisely, the other one will locate precisely automatically?

8. Feb 16, 2010

### LostConjugate

Better answer Cat :) The way I always remember it is that you cant have an exact position if you have a momentum. Simply because momentum is dx/dt and an exact position would require dx = 0. Therefore... uncertainty existed far before QM.

Since E = p = x/t everything is uncertain.

9. Feb 16, 2010

### LostConjugate

If you measures position to an exact measurement you will no longer be able to measures momentum, because dx = 0 and p = dx/dt (or in QM d^2/dx^2) so now how to you calculate that?

Same goes for measuring momentum, you will no longer get an exact position, it is either one or the other.

10. Feb 16, 2010

### Frame Dragger

@luxiaoloei: This is the whole reason that QM is probablilstic and not deterministic... we can't ignore that. If not, why $$\Psi$$ ? right?

11. Feb 16, 2010

### luxiaolei

Thanks all! :)