Interesting question!  For an eigenstate, the variance in energy is zero, as you already mentioned (del(E) = 0).  This would imply that the variance in time is infinite, in order to satisfy the uncertainty relation.  This is similar to when the particle is completely localized, its variance in position is zero (del(x) = 0), so its variance in momentum must be infinite.

But what does it mean that the variance in time is infinite?  Perhaps it is helpful to think in terms of measurement.  Determining the position of a particle to infinite precision means that its momentum will be completely unknown.  Similarly, determining the energy of a particle to infinite precision means the "time" will be completely unknown.  That is, your experiment to determine the energy of the particle perfectly must require an infinite amount of time.  You can determine the energy to less precision in a more reasonable amount of time.  At the other extreme, if you make a measurement over a very small period of time, your value for the energy will have a large error/variance.

There is an analogy with sound:  to hear a note at a given pitch, that note has to last a certain amount of time.  If the note lasts only for a very short amount of time, you will be more uncertain about what the pitch actually was.  But if the note lasts for a long time, you will be able to measure its pitch very accurately.

Interesting question!  For an eigenstate, the variance in energy is zero, as you already mentioned (del(E) = 0).  This would imply that the variance in time is infinite, in order to satisfy the uncertainty relation.  This is similar to when the particle is completely localized, its variance in position is zero (del(x) = 0), so its variance in momentum must be infinite.

But what does it mean that the variance in time is infinite?  Perhaps it is helpful to think in terms of measurement.  Determining the position of a particle to infinite precision means that its momentum will be completely unknown.  Similarly, determining the energy of a particle to infinite precision means the "time" will be completely unknown.  That is, your experiment to determine the energy of the particle perfectly must require an infinite amount of time.  You can determine the energy to less precision in a more reasonable amount of time.  At the other extreme, if you make a measurement over a very small period of time, your value for the energy will have a large error/variance.

There is an analogy with sound:  to hear a note at a given pitch, that note has to last a certain amount of time.  If the note lasts only for a very short amount of time, you will be more uncertain about what the pitch actually was.  But if the note lasts for a long time, you will be able to measure its pitch very accurately.

That would be true of any system with discrete eigenstates, not only the particle in a box. Eigenstates of the hamiltonian are stationary states, i.e. they are of the form F(x,t) = exp(i a t) f(x), where f(x) are the time-independent states. Therefore, the uncertainty in time is - ^2, = Integrate[t*F(t,x)*conjg(F(t,x)),t between -inf and inf], which is 0, and a similar integral, which is infinite, therefore Variance(t) = infinite. Thus making dE go to 0, makes dt go to infinite, in accordance with the uncertainty principle.

In reality, those eigenstates will have some linewidth associated with them.  This gives del(E).  The linewidth corresponds to a finite lifetime before the state radiatively decays back down to the ground state.  This gives del(t).  

The "Particle in a Box" model is typically used as a first introduction to quantum mechanical ideas and typically neglects effects of spontaneous emission, which is a more advanced topic. 

Matthew, I don't agree. For the model, dE is 0. The fact that the model cannot be experimentally implemented is a different issue. Stationary states of bounded isolated systems do not have "linewidths". 

Julian, fair point. I was thinking a little too "realistically".  I suppose the fact that dE = 0 and dt = infinite is ok within the model.  But isn't it a bit more subtle than that, since the product still must be greater than hbar/2?  This would mean that, starting from a finite linewidth (dE), and taking the limit as dE goes to zero, dt would have to change in such a way that the dE*dt >= hbar/2.

Thanks all for the valuable comments.

I would be grateful if someone would answer to my following question, too:

https://www.researchgate.net/post/Is_Uncertainty_Principle_satisfied_according_to_Bohms_model_of_Quantum_Mechanics

I recommend that you go through a rigorous proof of the time-energy uncertainty relation. It is somewhat different from the coordinate-momentum one. See, for example Messiah, pp. 319-320.

There is no operator for time in quantum mechanics. So, there is no as such energy-time uncertainty relationship. The time you refer here is the life time of the state. Then if del E = 0 that means del T = infinite implies you can never measure/calculate the life time of that state.

Time-energy uncertainty relation has a different status in comparison with standard uncertainty relations for two observables, because time is not an observable operator. The two most frequent interpretations are

• time uncertainty is life time of the state and energy uncertainty is width of energy level,
• time uncertainty is the time of observation and energy uncertainty is energy uncertainty after the observation.

In your case, you need infinite time to find exact value of energy.

In my opinion, whether there is a time operator is beside the point. The Uncertainty Principle is a physical relationship, and if your formalism cannot deal with it, your formalism is inadequate. Back to the question. The Schrödinger equation is fully deterministic, but in this variable ψ, which is expressed in terms of how it varies with changes of energy. Therefore, if you have a stationary state, where the energy terms are not functions of time, you have a precise energy. The Uncertainty Principle is not violated because the uncertainty in time is effectively infinite for a stationary state. In my ebook "Guidance Waves" I derive the Uncertainty Principle from the premise that action is quantised. If so, ΔE = ΔS/Δt, and since ΔS is h, and Δt can be a long as you wish for a stationary state, ΔE can be precise.

Energy defined precisely = infinite lifetime. That's what a stationary state is. (Indeed, a state's linewidth is the inverse of its lifetime.)

Infinite lifetime (zero linewidth) = energy defined precisely, Vincenzo.

Yes of course, Eugene ! :D

Vincenzo,

the problem is that in reality there is no zero linewidth except of ground state.

I think you give a good example to explain the uncertainty principle.

You may have a look @ this article,

Origin of Heisenberg's Uncertainty Principle

American Journal of Modern Physics

Volume 4, Issue 4, July 2015, Pages: 203-211

Author Bhushan Bhoja Poojary

Here : Article Origin of Heisenberg's Uncertainty Principle

A very good Review article

Uncertainty in quantum mechanics: faith or fantasy?

Philosophical Transactions of the Royal Society A (2011) 369, 4864–4890

doi:10.1098/rsta.2011.0179

Full Text : http://rsta.royalsocietypublishing.org/content/roypta/369/1956/4864.full.pdf

I agree with Mikhail Ivanov to a large extent. Actually, time-energy uncertainty principle applies when two systems with well measured energy E1 and E2, respectively, start to interact with each other. If the time interval over which they interact is del t, then the amount of energy exchanged between them is uncertain by at least hbar/(2 del t).

i.h = H.dt extremely

xtreme state

if the uncertainty principle is valid

is obtained   x. dp> - h / 2. dt