Time-energy uncertainty relation

[quasar987]

How should $\Delta E \Delta t \geq \hbar/2$ be interpreted? When does it apply? Delta t refers to the time of what? etc.

Thx.

 

[Galileo]

For a given time independent observable A that doesn't commute with the Hamiltonian and a state |psi>, interpret $\Delta E = \Delta H$ (H is the hamiltonian and delta means standard deviation) and define:

$$\Delta t := \frac{\Delta A}{|d\langle A \rangle/dt|}$$

So $$\Delta t$$ is the characteristic time it takes for the observable to change by one standard deviation.
I think this is the best interpretation of the inequality. No mystic mojo is involved.

Then by Heisenberg's inequality:

$$\Delta H \Delta A \geq \frac{1}{2}|\langle [H,A] \rangle|$$

And using:
$$\frac{d\langle A\rangle}{dt}=\frac{i}{\hbar}\langle [H,A]\rangle$$

you can rewrite it as $\Delta E \Delta t \geq \frac{\hbar}{2}$

 

[FunkyDwarf]

It sets a lower limit to the products of the uncertainty in those two quantities, ie for heisenburgs uncertainty principle its displacement and velocity

 

[dextercioby]

See the explanation in Sakurai's book.

Daniel.

 

[pmb_phy]

How should $\Delta E \Delta t \geq \hbar/2$ be interpreted? When does it apply? Delta t refers to the time of what? etc.

Thx.

To be precise there is no such thing as a time-energy uncertainty principle. Uncertainty principles are the relationship between two operators and while there is an energy operator there is no such thing as a time operator. Best to call it the time-energy uncertainty relation. The meaning of this relation is dt is the amount of time it takes for a system to evolve and dE represents the average change in the amount of energy during this time of evolution.

Pete

 

[exponent137]

But, I heard somewhere that in one case dW dt>=hbar/2 is correct. What is this example? Maybe for photons?

 

[Pythagorean]

I have two questions:

1) what's the motivations for this expression?

And using:
$$\frac{d\langle A\rangle}{dt}=\frac{i}{\hbar}\langle [H,A]\rangle$$

2) why is 1 standarad deviation chosen? Wouldn't the inequality change for any other choice?

 

[nucl34rgg]

The time-energy uncertainty principle is a little bit different from the position-momentum one. This is perhaps best illustrated by Landau's quote, "To violate the time-energy uncertainty relation all I have to do is measure the energy very precisely and then look at my watch!"

 

[Ken G]

I'd say the physical meaning behind the time-energy uncertainty relation has to do with the fact that a state of definite energy is characterized (physically) by having a definite frequency of change of its phase. To decide what that frequency is, you need to watch many cycles of time, and the more cycles you follow, the more precisely you know that frequency. But the more cycles you watch, the less you can say about the actual time at which you "looked", for you looked over a range of times. Conversely, if you look at "your watch" at a very specific time, then you cannot say what is the frequency at which the phase of the state is changing. Note that if you divide through by h, the expression becomes uncertainty in frequency times uncertainty in time exceeds 1 cycle.

 

[facenian]

All I know about it is that you have to be carefull with the interpretation. It seems that there is not a single version of it and not even experts agree on it.
There is however a simple version that can be derived clearly from first principles an is the one Galileo expleined, but that is not the only interpretation.

 

[Demystifier]

For a given time independent observable A that doesn't commute with the Hamiltonian and a state |psi>, interpret $\Delta E = \Delta H$ (H is the hamiltonian and delta means standard deviation) and define:

$$\Delta t := \frac{\Delta A}{|d\langle A \rangle/dt|}$$

So $$\Delta t$$ is the characteristic time it takes for the observable to change by one standard deviation.
I think this is the best interpretation of the inequality. No mystic mojo is involved.

Then by Heisenberg's inequality:

$$\Delta H \Delta A \geq \frac{1}{2}|\langle [H,A] \rangle|$$

And using:
$$\frac{d\langle A\rangle}{dt}=\frac{i}{\hbar}\langle [H,A]\rangle$$

you can rewrite it as $\Delta E \Delta t \geq \frac{\hbar}{2}$

In this approach, Delta t is NOT UNCERTAINTY of time, but a time DURATION of a physical process. There is however a similar way to introduce UNCERTAINTY of time measured by a clock, as explained, e.g., in
http://xxx.lanl.gov/abs/1203.1139 (v3)
Eqs. (19)-(23)