The energy-time uncertainty principle

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I'm watching a freshman-level lecture trying to take students through the energy-time uncertainty principle. (They've covered the position-momentum uncertainty principle).

In the lecture, the professor starts by saying that we have a particle with some momentum, but that we can't know the momentum of this particle with full certainty. So we have a particle with uncertainty ## \Delta p ##.

Now, he takes the energy of this particle, ## E = \frac{p^2}{2m} ## and asks how we might find the change in energy (or the uncertainty in energy). We take the differential of ## E ## to get ## \Delta E = \frac{p}{m}\Delta p ##.

My question is this: in the energy uncertainty relationship, we have a ## \Delta p ##. But we also have a ## p ##. We have this momentum term. But what does that refer to? Because we started this conversation saying we can't know the momentum exactly. And we called ## \Delta E ## the uncertainty in energy.

So what does it mean to have the ## p ## term directly in our ## \Delta E ## expression?

Thanks for any insight.
 

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  • #2
Chandra Prayaga
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Since the momentum is uncertain, that means there is a spread in the values of the momentum. The p in your equation is the average of this spread.
 
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...we have a particle with some momentum, but that we can't know the momentum of this particle with full certainty
Please clarify. We can't know the momentum with full certainty because it is unknowable, or because our measuring apparatus is imprecise?
 
  • #4
Chandra Prayaga
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In the lecture, the professor starts by saying that we have a particle with some momentum, but that we can't know the momentum of this particle with full certainty. So we have a particle with uncertainty ΔpΔp \Delta p .
That the particle has a momentum uncertainty is evidently the starting point for establishing the energy-time uncertainty. The uncertainty in the momentum itself is a consequence of the position momentum uncertainty priciple. Once you have a particle located in a certain region of space, there is a corresponding uncerainty in momentum. Unless you are able to give up completely the knowledge of position, in other words, you are saying that you don't know where the particle is in the universe, the momentum is unknowable. This has nothing to do with lack of precision in any measuring apparatus.
 
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If we have a momentum measuring apparatus consisting of a metal plate with a hole, and the diameter of the hole is small in comparison to the size of an electron, then when the electron passes through the hole, we know exactly where it is. And later when it impacts the screen, we also know exactly where it is. By measuring the time it takes to travel from the hole to the screen, we should be able calculate the momentum of the electron to any arbitrary level of precision.
 

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  • #6
Chandra Prayaga
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1. It cannot be done. The electrons in the metal plate are in extended states and will promptly fill the hole.
2. By trying to constrain the electron to such a small region in the transverse direction, you are only increasing the uncertainty in the transverse momentum
 
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Correct, but I think increasing the uncertainty of momentum just means it is impossible to predict where the electron will land.
 
  • #8
Chandra Prayaga
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Quite right. It also means you don't know what the momentum is, and there is a probability that it turns out to be very high (or low), which means that there is a large uncertainty in energy.
 
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I believe the energy of the electron depends on its speed, whereas the momentum depends on its velocity.
 
  • #10
Chandra Prayaga
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I believe the energy of the electron depends on its speed, whereas the momentum depends on its velocity.
Both correct. The energy depends on the magnitude of the momentum, which is the mass times the speed
 
  • #11
jfizzix
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I'm watching a freshman-level lecture trying to take students through the energy-time uncertainty principle. (They've covered the position-momentum uncertainty principle).

In the lecture, the professor starts by saying that we have a particle with some momentum, but that we can't know the momentum of this particle with full certainty. So we have a particle with uncertainty ## \Delta p ##.

Now, he takes the energy of this particle, ## E = \frac{p^2}{2m} ## and asks how we might find the change in energy (or the uncertainty in energy). We take the differential of ## E ## to get ## \Delta E = \frac{p}{m}\Delta p ##.

My question is this: in the energy uncertainty relationship, we have a ## \Delta p ##. But we also have a ## p ##. We have this momentum term. But what does that refer to? Because we started this conversation saying we can't know the momentum exactly. And we called ## \Delta E ## the uncertainty in energy.

So what does it mean to have the ## p ## term directly in our ## \Delta E ## expression?

Thanks for any insight.
Starting with [itex]E^2 = \frac{p^4}{4m^2}[/itex]
The formula for the standard deviation [itex]\Delta E[/itex] is given by:
[itex]\Delta E=\sqrt{\langle (E - \langle E\rangle) ^2\rangle}[/itex]
where [itex]E[/itex] is the mean value of [itex]E[/itex] when measured.

I think in this case, if you work out the math, we are intended to see that [itex]p[/itex] refers to the root-mean-square value of [itex]p[/itex], or equivalently, [itex]\sqrt{\langle p^2 \rangle}[/itex].
 
  • #12
PeterDonis
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the energy-time uncertainty principle
It's important to understand that this "uncertainty principle" is not like the others (although many introductory presentations, including textbooks, gloss over this important fact), because time is not an operator, and the standard uncertainty principle is a statement about the limitations on states in the Hilbert space on which a pair of non-commuting operators operates. So any derivation of an energy-time uncertainty principle is going to be heuristic and hand-waving, unlike derivations of standard uncertainty principles like the position-momentum one, which can be rigorously formulated.

So what does it mean to have the ##p## term directly in our ##\Delta E## expression?
It means your professor is waving his hands and making a heuristic argument, not a rigorous one. See above.
 
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  • #14
PeterDonis
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If we have a momentum measuring apparatus consisting of a metal plate with a hole, and the diameter of the hole is small in comparison to the size of an electron, then when the electron passes through the hole, we know exactly where it is.
Not exactly, only to the precision of the size of the hole. And the electron's momentum will become correspondingly uncertain--in other words, knowledge of what the electron's momentum was before it went through the hole will not let you predict where it will land on the screen; and the smaller the hole, the greater the uncertainty in where the electron will land on the screen, for a given momentum before it went through the hole.

And later when it impacts the screen, we also know exactly where it is.
Not exactly, only to the precision of the size of the spot that appears on the screen on impact.

By measuring the time it takes to travel from the hole to the screen
How?

we should be able calculate the momentum of the electron to any arbitrary level of precision.
You can calculate the momentum the electron had from the hole to the screen, yes. But that doesn't help you predict the electron's momentum after it hits the screen from knowledge of its momentum before it hits the screen; that latter prediction is what the position-momentum uncertainty principle puts restrictions on.
 
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You can calculate the momentum the electron had from the hole to the screen, yes. But that doesn't help you predict the electron's momentum after it hits the screen from knowledge of its momentum before it hits the screen;
Please elaborate. After it hits the screen, its momentum is zero. Before it hits the screen, the momentum is uncertain. What am I missing?
 
  • #17
PeterDonis
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After it hits the screen, its momentum is zero.
No, it isn't. You aren't keeping track of the electron after it hits the screen, so you don't know what its momentum is.
 
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You can calculate the momentum the electron had from the hole to the screen, yes.
And if I understand it correctly you can also calculate the position of that electron. Precision limited only by your measuring technology.
 
  • #19
PeterDonis
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if I understand it correctly you can also calculate the position of that electron
From what? You can measure it directly, but if you didn't do that, how would you calculate it?
 
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Right. "Calculate" isn't the appropriate term. You can pinpoint the position of the electron the moment it passes through the hole, and then when it hits the screen.
 
  • #21
PeterDonis
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You can pinpoint the position of the electron the moment it passes through the hole
To the accuracy of the size of the hole, yes, as I said before.

and then when it hits the screen
To the accuracy of the size of the spot on the screen, yes.

But you aren't measuring the momentum in either one of these measurements. You can calculate a momentum for the electron between those two measurements, and the accuracy of that calculation is not limited by the uncertainty principle. But the momentum you get from that calculation tells you nothing about what the momentum of the electron will be after the second measurement--which is what the uncertainty principle limits the accuracy of predicting.
 

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