# I Does ΔPΔX = ΔEΔT?

1. Feb 16, 2017

### mike1000

Does ΔPΔX = ΔEΔT?

From the Uncertainty Principle we know that ΔPΔX ≥ hbar/2 and ΔEΔT ≥ hbar/2. I know it is an inequality and not an identity. However, when I do the math it appears to me that ΔPΔX = ΔEΔT. Taking the limits it appears to me that it leads to the differential equation dP/dt = dE/dx, or in other words, the time derivative of momentum equals the space derivative of energy, which I think is correct.

2. Feb 16, 2017

### Demystifier

Due to the inequality you have written down, it does not make sense to consider the infinitesimal limit. ΔPΔX is not equal to ΔEΔT.

3. Feb 16, 2017

### Karolus

The relation of identity can not be written as well. These are two distinct equations that must be distinguished not there is an equality
On the derivation of one from the other we say that the way how you obtained is not very strict, although not foolish. Physicists use mathematics with a certain ease (and do well) but be careful not to overdo it!

4. Feb 16, 2017

### mike1000

Why not?

5. Feb 16, 2017

### mike1000

Let me try it a different way.

Lets ASSUME that it is true and you can take the limits. If we do, we get the differential equation ∂P/∂t = ∂E/∂x. In words, it is saying that the time derivative of momentum equals the space derivative of energy. Is that false?

6. Feb 16, 2017

### Demystifier

Why not what? I had two negations.

7. Feb 16, 2017

### mike1000

Please see my immediately previous post.

8. Feb 16, 2017

### Demystifier

Because in the infinitesimal limit $\Delta p\Delta x=0$, and $0$ is not larger or equal than $\hbar/2$.

9. Feb 16, 2017

### mike1000

While I absolutely love your logic, the same logic is true for ΔE and ΔT. ħ is not really involved in the equation I have written down.

10. Feb 16, 2017

### Demystifier

This only reinforces my argument.

11. Feb 16, 2017

### mike1000

I don't think so. Nothing you have stated says they cannot be equal to each other.

Last edited: Feb 16, 2017
12. Feb 16, 2017

### PeroK

How are $\Delta P, \Delta X, \Delta E, \Delta T$ defined here?

13. Feb 16, 2017

### mikeyork

Think of $\Delta$ as meaning something like a standard deviation. The uncertainty principle tells us that if $\Delta X \rightarrow 0$ then $\Delta P$ blows up. So you cannot make them both infinitesimal simultaneously. This does not mean that the standard deviations cannot obey the relation $\Delta P/\Delta T = \Delta E/\Delta X$. But there is no corresponding relationship for the derivatives.

14. Feb 16, 2017

### PeroK

What's a standard deviation of time?

15. Feb 16, 2017

### mikeyork

Depends on context. Standard deviation of decay times would be one well-known example.

16. Feb 16, 2017

### PeroK

How would decay times be related to the position and momentum of a particle?

17. Feb 16, 2017

### mike1000

The image shows Hamiltons equations. I am only concerned with the top equation. The top equation states that the time derivative of the momentum is equal to the negative of the space derivative of the total energy.

If we ASSUME that the equality I stated in my first post is true and we ASSUME we can take limits, we get a differential equation, ∂P/∂t = ∂E/∂x, that is almost identical to Hamiltons first equation. They only differ by the presence of a minus sign, which could be accounted for by an i2

18. Feb 16, 2017

### mikeyork

A decaying particle can be described in either energy-momentum space or co-ordinate space-time. As you well know, the uncertainty principle is a way of describing the limitations of doing both at once. Your question is about whether we can measure standard deviations in both descriptions for i.i.d. systems and I see no reason why, in principle, we cannot.

Consider a resonance in a scattering experiment. We know that the "lifetime" of the resonance is inversely related to the width in the energy of the resonance. Likewise the uncertainty in position at which the collision occurs can be related to the uncertainty in net momentum of the system. The fact that we idealize scattering experiments as viewed in the CM frame does not in any way contradict the reality that in any real repeatable scattering experiment there are standard deviations to both the location and net momentum of the collision, regardless of whether or not they are ever measured.

But we should not take any of this as particularly useful. I was merely trying to help the OP understand the mathematical limitations of what he was proposing.

19. Feb 16, 2017

### mike1000

I think you guys are getting off topic.

20. Feb 16, 2017

### mike1000

What does it mean when we take the limit as it approaches 0 of a variance? It means there is no variance. It means we know the momentum exactly and the position exactly and the energy exactly and the time exactly. Looked at it that way, all that we are doing when we take the differentials is converting the equation into a classical equation. And when we do that we find that the resulting differential equation is equal to the top Hamiltonian equation except for a factor of -1 (i2 )

Last edited: Feb 16, 2017
21. Feb 17, 2017

### Demystifier

But that's my second negation. You said that you ask about my first negation.

Concerning the second negation, nobody said that they cannot be equal to each other. They can. But they don't must to. Your argument that they must is wrong, because in this argument you use an illegitimate limit.

22. Feb 17, 2017

### Demystifier

And why do you think that the sign doesn't matter?

23. Feb 17, 2017

### Staff: Mentor

These two equations have a very different footing. Only the first one comes from uncertainty principle, i.e., from the non-commutation of two observables. That is not the case for the second one, since there is no observable corresponding to time in QM. Time is a parameter.

24. Feb 17, 2017

It is true but... rate of change of momentum is equal to the applied force (Newton2). Path integral of scalar product of force and distance equals work done and energy is the capacity to do work. Hey presto classical mechanics. The Heisenberg uncertainty products tell you that zeros in the equations are not permitted. The classical mechanics says that they are permitted. Getting rid of the ≥½hbar just takes you out of the realm of quantum mechanics.

25. Feb 17, 2017

### mike1000

In my previous post I focused on the top Hamiltonian equation and showed that the ASSUMPTION leads to a differential equation that confirms, except for the minus sign, the top Hamiltonian equation.

I now show that the ASSUMPTION leads to a second differential equation that confirms, exactly, the second Hamiltonian equation...

If we assume ΔPΔX = ΔEΔT and take limits we can rearrange the resulting differential equation into the following alternative, (but equivalent) form...

dx/dt = dE/dp,

which is equivalent to the second Hamiltonian equation shown in the image.