I Does ΔPΔX = ΔEΔT? Solving the Uncertainty Principle

[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 ?