# Time-energy uncertainty principle

1. Nov 13, 2009

### Phyisab****

Why is there a time energy uncertainty principle? I thought that uncertainty was a characteristic of two operators not commuting. There is no operator for time, which, as I am constantly reminded, is a parameter of evolution, not an observable.

2. Nov 13, 2009

### Count Iblis

3. Nov 13, 2009

### Phyisab****

Wow, i guess that my concern was justified. That guy knows a hell of a lot about quantum mechanics.

4. Nov 13, 2009

### Phyisab****

Upon further reading, the author makes some claims which seem quite radical. His interpretation of quantum mechanics is very different from what I have been taught.

5. Nov 13, 2009

### Count Iblis

The author is a fan of the Bohm interpretation. He actually posts here on PhysicsForums under the name "Demystifier".

6. Nov 13, 2009

### George Jones

Staff Emeritus
The standard interpretation is given in

https://www.physicsforums.com/showpost.php?p=1102314&postcount=2

and starting at the bottom of page 319 of the Messiah's classic text

7. Nov 14, 2009

### MaxwellsDemon

I just thought I'd say that in a relativistic quantum theory, I would expect the energy time uncertainty relation to just be a consequence of the 4-position 4-momentum uncertainty. I would imagine that your evolution parameter would be the proper time and that energy and time would be observables/operators.

8. Nov 14, 2009

### dx

The E/t uncertainty relation is correct. It has a subtler interpretation, and the derivation is different from the P/x analgoue, but it is true nevertheless.

9. Nov 15, 2009

### haushofer

Funny to see how different people can say such different things about something which is explained in a second years QM course. I've also had problems with this uncertainty.

10. Nov 15, 2009

### mikeph

We always had it told to us as if we knew it already, it was never explained in any detail.

11. Nov 19, 2009

### mpkannan

I am a chemistry student. We use this energy-time uncertainty quite a lot especially in spectroscopy and it works and explains. This relation explains tunneling also. So I do believe that this uncertainty relation is quite genuine.

According to the operator postulate of Quantum Mechanics, the operator form of px is (h/2$$i)d/dx whereas, that of t is itself and of x is x itself. You can construct any other operator using these basic operators. The operator form of Energy is -(h/2[tex]i)d/dt (You easily get it from the time-dependent Schrodinger equation) . We can show that this operator does not commute with t operator and the commutator is -(h/2[tex]i). Thus, the fact that uncertainty exits only in the case of observables that do do not commute is not violated and I do not see any problem! I expect your response. 12. Nov 19, 2009 ### Einstein Mcfly You can't use "t" the way you're saying. It's not an operator in QM, it's a parameter. It's not like "x" at all in that regard. There has been some work done trying to come up with something like a "time operator" in QM, but as far as I know (and can follow) there's nothing terribly useful that's been accepted by the wider community. The time/energy relation does "mean" something, but its interpretation is very different from those derived from proper operators. I suspect many others on here will be able to explain this far better than I. 13. Nov 19, 2009 ### jambaugh Right. Let me add that the fact that the momentum position uncertainty relation is derived one way doesn't mean an energy-time uncertainty relation with a different interpretation and a different derivation cannot also hold. Recall also that when we move into the relativistic realm with unified space-time then spatial position must be treated as a parameter on par with time. We loose even the traditional position operators and thus we must also re-interpret the traditional position-momentum uncertainty relations in a similar vein as the energy-time relations. Delta x becomes a characteristic distance and no longer an error bar of an observable. The alternative is to indeed invoke a time coordinate operator and make time an observable on par with spatial coordinate observables. That is of course a possibility. However in so doing we may need to move away from thinking we are observing properties of a quantum system and rather think in terms of observing properties of a quantum event. Maybe that is more appropriate. I don't know. For all of its success there are still a few cobwebs in quantum theory which need to be cleared away. 14. Nov 20, 2009 ### Neo_Anderson Time/energy uncertainty is used in only very restricted cases. The uncertainty is relegated to particle/high-energy physics, mostly. Let's not confuse it with Heisenberg's Uncertainty relation. 15. Nov 20, 2009 ### Fredrik Staff Emeritus Your observation is correct so far. The same way we get [x,p]=i from the definition p=-id/dx, we can get [t,H]=-i from the definition H=id/dt. But you need to think about what sort of functions these operators are supposed to act on. To get [x,p]=i, you have to assume that x and p act on functions of x, and to get get [t,H]=-i, you have to assume that t and H act on functions of t. To get both, you have to assume that all these operators are acting on functions of both t and x. At first glance, this doesn't look like a problem since something called $\psi(\vec x,t)$ appears in the Schrödinger equation. To see why it is a problem, we must look at the uncertainty relation: [tex]\Delta A\Delta B\geq\frac 1 2|\langle[A,B]\rangle|$$

Note that there's an expectation value on the right. The expectation value is defined using an inner product. This means that if we use an inner product on the set of functions of the form $\psi:\mathbb R^4\rightarrow\mathbb R$, we can derive both [x,p]=i and [t,H]=-i. But what inner product do we actually use in QM? It looks like this:

$$\langle f,g\rangle=\int_{\mathbb R^3} f(\vec x)^*g(\vec x) d^3x$$

As you can see, it's defined for functions of 3 variables, not 4. This means that the right-hand side of the uncertainty relation makes sense when A=x, B=p, but not when A=t, B=H.

This is why "time is not an observable in QM".

However, there is an energy-time uncertainty relation that makes sense in QM. Use the first link in George's post. In that one $\Delta t$ is the time it takes the expectation value to change by one standard deviation.

Last edited: Nov 20, 2009
16. Nov 20, 2009

### dextercioby

I think the reasons to reject a time operator are not really indisputable

http://atkinson.fmns.rug.nl/public_html/Time_in_QM.pdf [Broken]

Last edited by a moderator: May 4, 2017
17. Nov 20, 2009

### Phyisab****

Uncertainty in the measurement of an observable is an idea that makes sense. I can't think of any way to measure time, and thus it doesn't make much sense to have an uncertainty in time. When you make a measurement, you know exactly what time it is, thats just how making a measurement works. But, I know there is truth to the time-energy uncertainty in some sense. In what sense is it true? Someone above mentioned spreading of a wavepacket, that makes sense to me. To chemistry student who posted above, in what way have you seen it used?

18. Nov 20, 2009

### Phyisab****

To clarify what I just said. I just said I don't know of any way to measure time. What I mean is there is no absolute time. Furthermore, time is not a property of an object. You can't say that this particle has 30 seconds, but you can make a statement like that for every other observable. With time, all you can measure is a change in time. In what sense can time be on the same footing as space. But upon further thought, there is no absolute space or momentum either. But when you make a measurement, you can measure the momentum or the position. It does not make any sense to measure the time of a particle. But furthermore, yes it does. Just as when you measure the position of a particle, you have to set a zero point. To measure the time of a particle, you just need to set a zero of time. Now I feel like I am getting into relativity which I do not have much experience with at all.

19. Nov 20, 2009

### Einstein Mcfly

To the first question: The way that the time energy principal is used the most in my experience is when discussing linewidths. For an electron in an excited state, the longer it stays in that state before re-emitting a photon (the larger delta t) the more narrow the linewidth (smaller delta E). It's not really an issue that the constant of inequality is what it is, mainly just that delta E and delta t are inversely proportional.

The times I've heard delta t described as "the time it takes for a quantum state (some wave packet) to change appreciably", which to me is ruined because "change appreciably" is way too vague. This only really makes much sense to me in the limit where you have a stationary state that (with nothing else acting on it) lasts forever (delta t is infinite) and precise energy (delta E =0).

20. Nov 22, 2009

### mpkannan

It is possible to deduce the time-energy uncert. relation from the "true" position-momentum uncert. relation as many books show:

For a free particle, E=p^2/2m. So, the error in the measurement of the energy is
dE = p/m dp (1)
where dp is the error in the measurement of momentum, which can be written as
delta p = delta E/(p/m) (2)

p/m, which is the velocity, can be measured by measuring the time dt required for the particle to get displaced by dx:
p/m = dx/dt or p/m = delta x/ delta t. (3)

But, delta x is the uncertainty in the position of the particle during the measurement of p and from (3) it is
delta x = (p/m) delta t (4)

From (2) and (4) we get
delta x. delta p = (p/m) delta t. [delta E/(p/m)] >= h/4pi (5)
so that,
delta E. delta t >= h/4pi (6)
Here delta t is the duration of the measurement (observable?).