1. Sep 1, 2010

fluidistic

$$\Delta E \Delta t \geq \frac{\hbar}{2}$$.
If I understand well, if I measure the energy of a particle (or system of particles) with a great precision, I cannot know well at all when the system had this energy... right?

My doubt is: The system had (or will have?!) the energy I measured, but when? Well, it could be long ago, a few seconds ago, or... in the future?
I don't really know how to form my question.
Say I measure with a 100% accuracy the energy of a particle. I will have a 0% accuracy in the time the system had this energy. However I know it can't be in future (right?), so there's a restriction in time. It can only be present or past, but not future... unless I'm wrong.
My common sense tells me I can't measure an energy the system never had if I measured with a perfect accuracy (or almost perfect). However from Heisenberg principle, all seems to indicate that I can measure very accurately an energy that the system will have within say $$10 ^9$$ years, which makes no sense to me.

Can someone explain clear my doubts?

In a sketch, say I have the "time axis" on the real numbers. Delta t would be an interval. On another real line I could put the value I measured for the energy. The interval being very small or even vanishing if I measured perfectly. So I know "the" value of the energy of the system. In this case, the Delta t interval would be the whole real numbers axis. However if the positive t's means future, I know I can't have measured the energy the system will be in the future! So I can reduce the interval from $$-\infty$$ to $$0$$. And so writing $$\Delta E \Delta t \geq \frac{\hbar}{2}$$ is wrong although $$\Delta x \Delta p \geq \frac{\hbar}{2}$$ is correct.

I hope you can understand what I mean. In case not, I'll try to clarify but please let me know.

2. Sep 1, 2010

maxverywell

If $$\Delta E=0$$, which means that the state of the system is some energy eigenstate, then $$\Delta t=\infty$$ and these means that the system will remain stationary (no changes in the state of the system).

But if $$\Delta E\neq 0$$, for example the system is in superposition of two energy eigenstates: $$\psi=c_1 \psi_1+\psi=c_2 \psi_2$$, then $$\Delta t$$ is finite which means that the state of the system is not stationary, it will evolve in time: $$\psi(t)=c_1 \psi_1 exp(-iE_1 t/\hbar)+c_2 \psi_2 exp(-iE_2 t/\hbar)$$.

So $$\Delta t$$ is actually the time we need to wait for some reasonable change in the state of the system.

Last edited: Sep 1, 2010
3. Sep 1, 2010

Iforgot

Ummmm,

The problem with quantum mechanics is that the classes are not taught with an emphasis on experimental methods. To answer your question

1) I'm not sure you can directly measure the energy of a particle.

2) You can probe the momentum of a free particle and calculate the energy..... using a magnetic field and measuring deflection, or using scintillators?

2) Or you can probe the energy absorbed or emitted during some transition. The is a measurement of DE

3) I think your question would be readily answered if you could come up with some experiment, that describes how the quantities are measured.

4. Sep 2, 2010

fluidistic

I'm all confused on the meaning of $$\Delta t$$. Is it what you say, i.e. it's the time needed for the system to change its state (I don't even know what a state is yet)?
Or is it a time interval like $$[t_1, t_2]$$ where the present time is in the middle of the interval?
Or both meanings?

I'll try to rephrase my original doubt: If I measure with a good accuracy $$\Delta E$$, I will get a large value for $$\Delta t$$. Since \Delta t include both past and future (and even present), can I simply discard half of the interval $$\Delta t$$, namely the future?

You might be right. My understanding is that in practice is that $$\Delta E \Delta t > \frac{\hbar}{2}$$ while in theory it could be $$\Delta E \Delta t = \frac{\hbar}{2}$$ if I can get the best measure ever.
But since I've been thrown the formulas without any explanation (not even how to derive them, yet), I certainly has almost no understanding of it.

5. Sep 2, 2010

fluidistic

I just found in Pfeffer's book "Modern Physics, an introductory text"

which seems to confirm that I can really discard any t in the future!
Therefore $$\Delta t$$ can't be the whole real axis. In the worst case I can measure very, very accurately the energy of a particle, but I can't know well at all when the particle had this energy. However, I know that it couldn't have it "before the big bang" and I also know it can't be in the future, therefore the worst $$\Delta t$$ would be around [0, today] where 0 means the big bang's time.
So it's impossible that $$\Delta t \to +\infty$$!

6. Sep 2, 2010

alxm

The time-uncertainty uncertainty principle is not a true uncertainty relation; time is not an operator. The equation posted holds, but only in some specific circumstances. (there's been quite a few threads about this)

For more information, check out http://arxiv.org/abs/quant-ph/0609163" [Broken] (from a PF contributor no less)

Last edited by a moderator: May 4, 2017
7. Sep 2, 2010

fluidistic

Well, thanks a lot. Although I can't understand everything on the topic yet, I get the main idea and that there's a big difference between saying $$\Delta x \Delta p \geq \frac{\hbar}{2}$$ and $$\Delta E \Delta t \geq \frac{\hbar}{2}$$.
By the way do you know if the author of the paper is a Ph.D.? Or at least that the paper is not crackpotry? I'll be studying QM in the next semester so I can't really judge the paper and I don't want to learn false facts; although I'd love to fully understand the whole paper as it can clear many doubts I believe.

Last edited by a moderator: May 4, 2017
8. Sep 2, 2010

woodyallen1

Is it helpful to imafine ΔΕ as energy fluctuation and Δt the time a process lasts?

9. Sep 2, 2010

fluidistic

I don't think so... why would it be?!

10. Sep 2, 2010

woodyallen1

Imagine ΔΕ as an energy loan. The shortest the process is the bigger the loan is.

11. Sep 4, 2010

WarPhalange

That's one way to think of it.

An example of this working is with lifetimes of electronic excitation states. If a state has a very short lifetime (Δt), the energy it emits will be distributed quite widely (i.e. if you try to measure the energy of the state, your sigma will be quite large), whereas a state with a really long lifetime will give you a distribution that is very peaked, meaning small ΔE.

12. Sep 4, 2010

jcsd

The author of the article is a theoretical physicist and the article itself appeared in the journal Foundations of Physics in 2007. On this board he is one of the most respected contributors to the quantum physics forum on this board. He has his own opinions on the subject, but these opinions come from a deep knowledge of the subject.

I have to say I'm a big fan of that article, I felt that it was written in such an easy to digest way and I also felt it genuienly improved my understanding of some ideas I had not previously been able to wrap my head around fully. The only very tiny criticism I have is that the pedant in me says you should always use 'spatial' instead of 'spacial', but that really is just pedantry!