A time-energy uncertainty relation

tzimie
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I am reading this: http://arxiv.org/pdf/quant-ph/0609163.pdf
And Demystifier claims that "The time-energy uncertainty relation is not fundamental"
However the proof is done in non-relativistic QM, where t and x are treated differently. My question is, what's about relativistic QM?
 
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The answer depends on what exactly one means by "time" (quantum observable or classical external parameter?), by "energy" (the Hamiltonian or merely the time derivative multiplied with ##i\hbar##?) and by "relativistic QM" (Bjorken Drell 1 or Bjorken Drell 2?).

For one (but not the only one) possible answer see http://lanl.arxiv.org/abs/0811.1905
 
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Demystifier said:

Thank you. I have a clearer picture now.
However, with every new article I read I have a feeling going into deeper and deeper circles of relativistic QM hell.
 
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Taking a step back, you can say that if any two observables are related by a Fourier transform, then there is a Heisenberg-style uncertainty relation connecting them.

Since position and momentum are related by a Fourier transform, you can prove the Heisenberg relation for position x and momentum p.
\sigma_{x}\sigma_{p}\geq\frac{\hbar}{2}
Since frequency \omega and time t, are also related by a Fourier transform, you can prove a Heisenberg relation between frequency and time.
\sigma_{t}\sigma_{\omega}\geq\frac{1}{2}
Though time is not an observable, you can still say that these are both "fundamental" in that they only rely on variables being related by Fourier transforms. Indeed, these previous two relations exist in other forms for classical waves.

For example, it's not possible for a pulse of sound to have a well-defined musical pitch, and for that sound to last an arbitrarily small time. If you were to play "concert A" for a second or two, it would be a well-defined note, but the smaller the duration of that note, the more the note just sounds like a chirp, or pop, without a well-defined pitch. (see for example http://newt.phys.unsw.edu.au/jw/uncertainty.html)

The other kind of energy-time uncertainty relation takes some extra derivation, and it relates the uncertainty of the energy E of a particle, to the uncertainty in the time evolution of an observable B of that particle.
\sigma_{E}\frac{\sigma_{B}}{|\frac{d\langle B\rangle}{dt}|}\geq\frac{\hbar}{2}
What this means is that if there is some aspect of the quantum state of a particle that is short lived or rapidly varies, then the uncertainty in the energy of that particle cannot also be arbitrarily small.
 
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jfizzix said:
What this means is that if there is some aspect of the quantum state of a particle that is short lived or rapidly varies, then the uncertainty in the energy of that particle cannot also be arbitrarily small.

I know about it, but however, this is interesting - quote from the article:

Although (5) is not a fundamental relation, in most practical situations it is still true that the uncertainty ∆E and the duration of the measurement process ∆t roughly satisfy the inequality (5). However, there exists also an explicit counterexample that demonstrates that it is possible in principle to measure energy with arbitrary accuracy during an arbitrarily short time-interval [9]
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
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