I Does ΔPΔX = ΔEΔT? Solving the Uncertainty Principle

[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.

 

[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.

 

[Karolus]

Does ΔPΔX = ΔEΔT?

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!

 

[mike1000]

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

Why not?

 

[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.

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?

 

[Demystifier]

Why not?

Why not what? I had two negations.

 

[mike1000]

Why not what? I had two negations.

Please see my immediately previous post.

 

[Demystifier]

Why can't you take the limits because it is an inequality?

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

 

[mike1000]

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

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.

 

[Demystifier]

While I absolutely love your logic, the same logic is true for ΔE and ΔT.

This only reinforces my argument.

 

[mike1000]

This only reinforces my argument.

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

 

[PeroK]

Does ΔPΔX = ΔEΔT?

From the Uncertainty Principle we know that ΔPΔX ≥ hbar/2 and ΔEΔT ≥ hbar/2.

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

 

[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.

 

[PeroK]

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.

What's a standard deviation of time?

 

[mikeyork]

What's a standard deviation of time?

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

 

[PeroK]

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

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

 

[mike1000]

What's a standard deviation of time?

https://lh3.googleusercontent.com/HrcUajVIBZL6SbnxV7ZXSh_yckMegKNZYk9_Z_ut4QrR-UIIW8OR96Bjqcr0bwt3e7_3lMdKWGbPsvDS_Kj26_Xa5TDipG_dOrl2CV3223CD02hJZHDzDTIv-CldsoblcYrD2xGTr9Yblag_GqGR_0G9O48HAHKkmUTDALIKZs3qyI8SKxG6J3dhXaNUHqeQO2PPbtn1DmKLO_7n7Y1-J5fI6ZolbBzIkO0iXMRQOj3vIsS485bSAR-muLhExqiuiRGu_a5q3snjHwOK01ZaqGKKCffHHj__0GNmH08MdoMB5Tfj2lDuKa8_OO3uP55aTEq-okRoo7Oac5s5GSbmtaYoorWSPOsRGEfpfZDk4ZjJsp3ZorzHBNn9twqzQBh157tsiOd7A8xOoqE5w86kyRKkYRGlGcnHn3bGVE9brISS2IewLokU7D-y2ZcI9s6V5RoMDfd-lW9KJ3Y3EuGsbUP1tTbkM_TGvITtJ1wrho1H9IiQKiiWIU_MmyGIrc3FbJZSNTVakHo1gafMML25opeiiRs3Fwdz9hzhmzBZ5GjOPp8uTLSZme8WM7KrIeJu06IY1ZJel1hDVC3kBk6h4HOfrxUdGmpdal3mH_Be0Xbk1vLY0v3Y=w133-h114-no

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

 

[mikeyork]

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

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.

 

[mike1000]

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.

I think you guys are getting off topic.

 

[mike1000]

https://lh3.googleusercontent.com/HrcUajVIBZL6SbnxV7ZXSh_yckMegKNZYk9_Z_ut4QrR-UIIW8OR96Bjqcr0bwt3e7_3lMdKWGbPsvDS_Kj26_Xa5TDipG_dOrl2CV3223CD02hJZHDzDTIv-CldsoblcYrD2xGTr9Yblag_GqGR_0G9O48HAHKkmUTDALIKZs3qyI8SKxG6J3dhXaNUHqeQO2PPbtn1DmKLO_7n7Y1-J5fI6ZolbBzIkO0iXMRQOj3vIsS485bSAR-muLhExqiuiRGu_a5q3snjHwOK01ZaqGKKCffHHj__0GNmH08MdoMB5Tfj2lDuKa8_OO3uP55aTEq-okRoo7Oac5s5GSbmtaYoorWSPOsRGEfpfZDk4ZjJsp3ZorzHBNn9twqzQBh157tsiOd7A8xOoqE5w86kyRKkYRGlGcnHn3bGVE9brISS2IewLokU7D-y2ZcI9s6V5RoMDfd-lW9KJ3Y3EuGsbUP1tTbkM_TGvITtJ1wrho1H9IiQKiiWIU_MmyGIrc3FbJZSNTVakHo1gafMML25opeiiRs3Fwdz9hzhmzBZ5GjOPp8uTLSZme8WM7KrIeJu06IY1ZJel1hDVC3kBk6h4HOfrxUdGmpdal3mH_Be0Xbk1vLY0v3Y=w133-h114-no

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

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 )

 

[Demystifier]

Nothing you have stated says they cannot be equal to each other.

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.

 

[Demystifier]

They only differ by the presence of a minus sign

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

 

[DrClaude]

From the Uncertainty Principle we know that ΔPΔX ≥ hbar/2 and ΔEΔT ≥ hbar/2.

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.

 

[hadrian_athena]

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.

 

[mike1000]

https://lh3.googleusercontent.com/HrcUajVIBZL6SbnxV7ZXSh_yckMegKNZYk9_Z_ut4QrR-UIIW8OR96Bjqcr0bwt3e7_3lMdKWGbPsvDS_Kj26_Xa5TDipG_dOrl2CV3223CD02hJZHDzDTIv-CldsoblcYrD2xGTr9Yblag_GqGR_0G9O48HAHKkmUTDALIKZs3qyI8SKxG6J3dhXaNUHqeQO2PPbtn1DmKLO_7n7Y1-J5fI6ZolbBzIkO0iXMRQOj3vIsS485bSAR-muLhExqiuiRGu_a5q3snjHwOK01ZaqGKKCffHHj__0GNmH08MdoMB5Tfj2lDuKa8_OO3uP55aTEq-okRoo7Oac5s5GSbmtaYoorWSPOsRGEfpfZDk4ZjJsp3ZorzHBNn9twqzQBh157tsiOd7A8xOoqE5w86kyRKkYRGlGcnHn3bGVE9brISS2IewLokU7D-y2ZcI9s6V5RoMDfd-lW9KJ3Y3EuGsbUP1tTbkM_TGvITtJ1wrho1H9IiQKiiWIU_MmyGIrc3FbJZSNTVakHo1gafMML25opeiiRs3Fwdz9hzhmzBZ5GjOPp8uTLSZme8WM7KrIeJu06IY1ZJel1hDVC3kBk6h4HOfrxUdGmpdal3mH_Be0Xbk1vLY0v3Y=w133-h114-no

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

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.

 

[PeroK]

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.

You should read post #23. The terms here are not consistently defined. In fact, you can derive your equation simply. Using your style and defining nothing precisely we have:

$E = \frac{p^2}{2m} + V$

$\frac{dE}{dp} = \frac{p}{m} = \frac{dx}{dt}$

$dE dt = dp dx$

$\Delta E \Delta t = \Delta p \Delta x \ge \frac{\hbar}{2}$

Exercise: How many flaws can you find in that argument?

 

[mike1000]

You should read post #23. The terms here are not consistently defined. In fact, you can derive your equation simply. Using your style and defining nothing precisely we have:

$E = \frac{p^2}{2m} + V$

$\frac{dE}{dp} = \frac{p}{m} = \frac{dx}{dt}$

$dE dt = dp dx$

$\Delta E \Delta t = \Delta p \Delta x \ge \frac{\hbar}{2}$

Exercise: How many flaws can you find in that argument?

I was not trying to derive the equation, I was trying to figure out if the equality I stated in the first post is true. The fact that the logic I used lead to Hamiltons equations lead me to believe that it may be true.

As I worked with this, and realized that the Uncertainty Principle is using statistical standard deviations and not deltas as in differential calculus, it actually did become clear to me that as we take the limit of the standard deviations as they go to zero, the equation describes behavior in the world of Classical Physics, where there is no uncertainty in the measurements of the parameters. The fact that the Uncertainty Principle can lead to the kinematic equations that describe Classical motion I find wonderfully consistent.

 

[Dadface]

If the quantities on opposite sides of an equation have the same units and can be the same numerically it doesn't follow that they must represent the same thing. As an example an energy of 100Nm is not the same as a torque of 100Nm.

 

[mike1000]

If the quantities on opposite sides of an equation have the same units and can be the same numerically it doesn't follow that they must represent the same thing. As an example an energy of 100Nm is not the same as a torque of 100Nm.

I never said that nor did I assume it. That was not the motivation to ask the question ... Are they equal?

 

[PeroK]

I wasn't trying to derive the equation, I am trying to figure out if the equality that I suspect to be trueI was not trying to derive the equation, I was trying to figure out if the equality I stated in the first post is true. The fact that the logic I used lead to Hamiltons equations lead me to believe that it may be true.

First, you cannot prove or justify something by showing that it leads to a known equation. Second, surely it's much better to derive the equation (as I have done), rather than assume it is true?

Third, the point that you are missing entirely, is that your equation mixes and matches quantities and terminology. You are only able to persist with your analysis because you have failed to define or recognise what $\Delta$ even means. Your analysis, as is my derivation of your equation, is physically meaningless symbolic manipulation, I'm sorry to say.

You even admit yourself that one of Hamilton's equations emerges but the other has the wrong sign.

 

[mike1000]

First, you cannot prove or justify something by showing that it leads to a known equation. Second, surely it's much better to derive the equation (as I have done), rather than assume it is true?

Third, the point that you are missing entirely, is that your equation mixes and matches quantities and terminology. You are only able to persist with your analysis because you have failed to define or recognise what $\Delta$ even means. Your analysis, as is my derivation of your equation, is physically meaningless symbolic manipulation, I'm sorry to say.

You even admit yourself that one of Hamilton's equations emerges but the other has the wrong sign.

No, I have not failed to recognize what delta means. It means the standard deviation of a set of observations. When the standard deviation = 0, what does that mean?

Also, if delta is the standard deviation then what does the Uncertainty Principle tell us? It tells us that the standard deviations when multiplied together are greater than or equal to planks constant. If that is true, doesn't that imply that there could exist a set of observations within the data set of observations where the momentum and position were measured precisely and do not obey the Uncertainty Principle?

 

[PeroK]

No, I have not failed to recognize what delta means. It means the standard deviation of a set of observations.

And then we're back to the question of what on Earth the standard deviation of a set of time measurements means?

Moreover, as the energy of a system can be well-defined, the standard deviation of a set of energy measurements can be 0.

This is what post #23 was all about.

 

[mike1000]

And then we're back to the question of what on Earth the standard deviation of a set of time measurements means?

Moreover, as the energy of a system can be well-defined, the standard deviation of a set of energy measurements can be 0.

This is what post #23 was all about.

So obviously, delta is not standard deviation, right?

 

[PeroK]

So obviously, delta is not standard deviation, right?

Yes, the deltas in the time-energy relation are not standard deviations and refer to something very different.

 

[mike1000]

Yes, the deltas in the time-energy relation are not standard deviations and refer to something very different.

You seem to be implying that delta is standard deviation for momentum and position. If that is so, can you please comment on the following...

If delta is the standard deviation then what does the Uncertainty Principle tell us? It tells us that the standard deviations when multiplied together are greater than or equal to planks constant. If that is true, doesn't that imply that there could exist a set of observations within the data set of observations where the momentum and position were measured precisely and do not obey the Uncertainty Principle?

 

[PeroK]

You seem to be implying that delta is standard deviation for momentum and position. If that is so, can you please comment on the following...

If delta is the standard deviation then what does the Uncertainty Principle tell us? It tells us that the standard deviations when multiplied together are greater than or equal to planks constant. If that is true, doesn't that imply that there could exist a set of observations within the data set of observations where the momentum and position were measured precisely and do not obey the Uncertainty Principle?

The UP is a statistical law; and, as with all statistical laws, says nothing definite about any specific measurement or set of measurements. In the same way you could throw a die and get 0 variance (for a small number of throws), likewise (for an ensemble of identically prepared systems) you could get 0 variance for position measurements (for a small number of experiments).

 

[vanhees71]

Maybe it's helpful to think about the meaning of the "energy-time uncertainty" relation first. One should be very clear that time in QT is not an observable but a parameter. This must be so, because otherwise energy and time would just be like momentum and position, and then the energy of any system could take just all real values, and this would mean that nothing could be stable, because the energy is not bounded from below. This is fortunately not what we observe in nature, because if this would be true, there'd be no stable matter and consequently no human beings there to observe anything. Thus, the time-energy uncertainty relation has a very different meaning than the usual Heisenberg-Robertson uncertainty relation, which reads
$$\Delta A \Delta B \geq \frac{1}{2} |\langle [\hat{A},\hat{B}] \rangle|,$$
where $A$ and $B$ are arbitrary observable and $\hat{A}$ and $\hat{B}$ their representing self-adjoint operators on Hilbert space. The averaging is taken over any arbitrary (pure or mixed) state, and $\Delta A$ and $\Delta B$ are the standard deviations of the observables for the system being prepared in that state.

For a derivation of what's usually meant when talking about energy-time uncertainty relations, have a look here:

http://math.ucr.edu/home/baez/uncertainty.html

 

[mike1000]

The UP is a statistical law; and, as with all statistical laws, says nothing definite about any specific measurement or set of measurements. In the same way you could throw a die and get 0 variance (for a small number of throws), likewise (for an ensemble of identically prepared systems) you could get 0 variance for position measurements (for a small number of experiments).

I am going to paraphrase, I hope I do it justice.

Individual measurements may not obey the Uncertainty Principle.

 

[vanhees71]

I have no clue what you mean by this sentence. From an individual observation (i.e., one meausrement) you can't even get the standard deviations of the quantities. It's all about statistics, i.e., many independent individual measurements on an ensemble of independently prepared systems in a given state!

 

[mike1000]

I have no clue what you mean by this sentence. From an individual observation (i.e., one meausrement) you can't even get the standard deviations of the quantities. It's all about statistics, i.e., many independent individual measurements on an ensemble of independently prepared systems in a given state!

The individual measurements do not have to obey the Uncertainty Principle? You can perform an experiment where the momentum and position that is measured does not obey the Uncertainty Principle?

A data set having only 1 data point does have a standard deviation and a mean and a varaince.

Does the Uncertainty Principle hold for datasets that have only one element?

 

[vanhees71]

I don't know, which experiment you have in mind. So far no deviation from QT has been found, and thus also no violation of the uncertainty principle. A single measurement doesn't tell you anything about standard deviations and thus cannot test the uncertainty principle to begin with.

 

[mike1000]

I don't know, which experiment you have in mind. So far no deviation from QT has been found, and thus also no violation of the uncertainty principle. A single measurement doesn't tell you anything about standard deviations and thus cannot test the uncertainty principle to begin with.

A data set having only 1 data point does have a standard deviation and a mean and a varaince.

Does the Uncertainty Principle hold for datasets that have only one element?

 

[vanhees71]

A data point with standard deviation is obtained by many measurements of the observable on a sufficiently large ensemble. That's all I'm saying.

 

[mike1000]

A data point with standard deviation is obtained by many measurements of the observable on a sufficiently large ensemble. That's all I'm saying.

When you perform an experiment, what does the distribution of momentum measurements and position measurements look like? Can you point me somewhere where I can see a graph of them? Are they always symetrical? Are they the probability distributions?. QM tries to predict those distributions?

 

[Mentz114]

When you perform an experiment, what does the distribution of momentum measurements and position measurements look like? Can you point me somewhere where I can see a graph of them? Are they always symetrical? Are they the probability distributions?. QM tries to predict those distributions?

Have a look at this https://en.wikipedia.org/wiki/Stern–Gerlach_experiment
[PLAIN]https://en.wikipedia.org/wiki/Stern–Gerlach_experiment[/PLAIN]
If a beam with only 1 atom is used what informatin does it give us ( assuming we can detect a sigle atom ) about the various amplitudes in the model ?