Question for large quantum system

[JasonWuzHear]

I'm trying to understand the concept of uncertainty in relation to derivatives for a large quantum system, i.e. one with many degrees of freedom.

When is it true that σE/σt ~ dE/dt? ---- 'σ' is the uncertainty

First, I know there is no time operator in quantum mechanics. I'm not sure how to work around that, perhaps we could try the question with momentum and position. The point is, for a large quantum system, can I approximate the rate of change of energy with the uncertainty of energy divided by the uncertainty of its age?I've tried unconventional methods to answer the question. First using an assumption that:

i/ħ d²/dt² ψ = dE/dt ψ

then trying to compare it to the uncertainty calculation assumed as follows:

(σE)² = (∫ ψ*Eψ dt)² - ∫ ψ* E² ψ dt

Which gives equal results for a Gaussian wave function. But I'm not sure how to go about this question in a more conventional way.

 

[PeterDonis]

When is it true that σE/σt ~ dE/dt? ---- 'σ' is the uncertainty

Why do you think this should be true at all?

 

[JasonWuzHear]

Why do you think this should be true at all?

I was thinking of a couple reasons why it might be true.

For one, there is the Griffith's proof that Exact decoherence implies conservation of quantities that commute with the Hamiltonian. Exact decoherence in energy states is not different than saying the quantum uncertainty in energy is zero (there can still be some classical probabilities). This leads to the idea that when

dE/dt = 0 ⇒ σE = 0

which means that

σE/σt = 0 for σt ≠ 0

Additionally, if we take the momentum operator, P

if σP = 0, then
(P) (P) Ψ Ψ = (PΨ)²
and
σP/σx = 0 for σx ≠ 0

the particle is not influenced by a potential in this scenario, and momentum is conserved through space translations

dP/dx = 0

So it at least seems true to me for when the system conserves energy.

You can define the uncertainty as the average deviation from the mean value when the system collapses in that basis. If you take a system, periodically measure its energy with a period, T, and allow it to grow in energy uncertainty, σE, in between measurements, then the change in energy over time on average will be:

<dE/dt> = σE/T

I feel like there ought to be some relationship between T and t relating to the success of the measurement. If σt is much larger than σT, then the measurement is less likely to succeed. The larger your T, the larger σE can grow. This makes me want to multiply the thing above by (T σT)/(σt) if the probability of measurement success, P_s, is proportional to σT/σt which gives

σE = ("σE")T the energy uncertainty in quotes is some value not dependent on T
P_s ∝ σT/σt
<dE/dt> ∝ "σE"σT/σtThat's all I've got so far

 

[PeterDonis]

Griffith's proof that Exact decoherence in energy states is not different than saying the quantum uncertainty in energy is zero

Are you referring to the proof discussed in section 2 of this paper?

https://arxiv.org/pdf/gr-qc/9410006.pdf

 

[JasonWuzHear]

Are you referring to the proof discussed in section 2 of this paper?

https://arxiv.org/pdf/gr-qc/9410006.pdf

yes, that is where I got it.