1
$\begingroup$

Background

The top voted answer to this question seems to make some assumptions I'm uncertain about (to say the least): What is $\Delta t$ in the time-energy uncertainty principle?

Assumptions

  • The uncertainty principle is usually a statement about the measurement (while it is true they are applicable during the unitary process as well. They usually are not used in that context).
  • Since they usually the measurement is considered a discontinuous process is he considering an interpretation where there is no measurement such as many worlds?

Question

Is there a version of this derivation (compatible with the discontinuous measurement interpretations of quantum mechanics) where one is allowed to use the calculus in the way he does (both integration and differentiation)?

P.S: I have already commented on this.

$\endgroup$
  • 3
    $\begingroup$ I don't understand the question. Is there a version of this derivation where one is allowed to use the calculus in the way he does (both integration and differentiation)? Yes there is, it is the one you are referring to. $\endgroup$ – Aaron Stevens Sep 10 at 16:26
  • $\begingroup$ @AaronStevens I also mention the measurement is discontinuous (at least in some interpretations). In that case I want a version of this derivation where one can do those manipulation there too. $\endgroup$ – More Anonymous Sep 10 at 16:28
  • $\begingroup$ Since, I'm getting a down vote despite explaining my objection (in the comment). Can someone clarify why the objection is wrong? $\endgroup$ – More Anonymous Sep 10 at 16:31
  • $\begingroup$ Baez's summary is based on the resolution of an optimal clock, Δt, the approximate amount of time it takes for the expectation value of an observable to change by a standard deviation provided the system is in a pure state. What is your question?? $\endgroup$ – Cosmas Zachos Sep 10 at 16:48
  • $\begingroup$ @CosmasZachos I'm discussing this in the chat at the moment ... (I can't do both simultaneously) $\endgroup$ – More Anonymous Sep 10 at 16:57
2
$\begingroup$

The WP summary of the Mandeshtam-Tamm relation is, for an observable $\hat B$, $$ \sigma_E ~~~\frac{\sigma_B}{\left| \frac{\mathrm{d}\langle \hat B \rangle}{\mathrm{d}t}\right |} \ge \frac{\hbar}{2} ~~, $$ where the second factor on the l.h.s., with dimensions of time, is a lifetime of the state ψ with respect to the hermitean observable $\hat B$. Roughly, the time interval (Δt) after which the expectation value ⟨$\hat B$⟩ changes appreciably. For a stationary state, the drift rate of ⟨$\hat B$⟩ goes to zero, and the variance of energy goes to 0 as well, as it should.

This is all in standard QM, unitarily evolving, with or without measurements. You may do any and all measurements discontinuous, delirious, expialidocious, whatever, and plot your results, but you must be talking about the same state ψ all the time. The distribution in B will have a variance, which is what is under discussion.

(Heuristically, a state ψ that only exists for a short time cannot have a definite energy. To have a definite energy, the frequency of the state must be defined accurately, and this requires the state to persist for many cycles, the reciprocal of the required accuracy. In spectroscopy, excited states have a finite lifetime. By above, they do not have a definite energy, and, each time they decay, the energy they release is slightly different. The average energy of the outgoing photon has a peak at the theoretical energy of the state, but the distribution has a finite width called the natural linewidth. Fast-decaying states have a broad linewidth, while slow-decaying states have a narrow linewidth.)

I have not fully appreciated your misgivings, but they seem to me to also apply to the standard Δx Δp uncertainty principle: A pure state will have corresponding distributions for x and p with nontrivial variances, computable through standard continuous QM, which your measurements will probe.

$\endgroup$
  • $\begingroup$ I feel my I came to the conclusion that my question was based on legimate doubts but if you have the patience to understand what they were here is the link: chat.stackexchange.com/transcript/message/51655733#51655733 Also mentioning $\Delta x \Delta p$ uncertainty makes me think you suspect the derivation in question enables you to make the claim that $2$ measurements cannot be arbitrarily close together (in time) ... But in the chat we have users comparable to your points (and one even more) who agree this derivation cannot warrant that. $\endgroup$ – More Anonymous Sep 10 at 21:30
  • $\begingroup$ I'm sorry, I must then not understand your question. The relation, and my answer have little to do with measurement. They provide probabilities of outcomes when/if you make a measurement. Once you have made a measurement, you have altered the state, and you are not talking about that state anymore. Probabilities are estimated experimentally by repeatedly measuring the "same" state, no? $\endgroup$ – Cosmas Zachos Sep 10 at 21:37
  • $\begingroup$ Yes, see assumption $1$. And I asked the chatroom what was the experimental context of this derivation and the (accepted) answer seemed to be of very limited scope (which I suspect most upvoters of the original answer probably do not know). $\endgroup$ – More Anonymous Sep 10 at 21:40
  • 1
    $\begingroup$ Oh... If you are thinking about measurements, I have nothing to say. All derivations of this ilk use standard QM and leave it to measurements to experimentally probe the answers. They really, really, deal with the same state ψ... $\endgroup$ – Cosmas Zachos Sep 10 at 21:43
  • 1
    $\begingroup$ Of course. We might agree there. This is (highly abstractly) what spectroscopists do to extract lifetimes out of linewidths. $\endgroup$ – Cosmas Zachos Sep 10 at 21:48
-7
$\begingroup$

Time is dissipation of energy. Delta t in that context is the ratio of original to current energy density. Delta E is the remaining original energy. As energy dissipates, Delta t increases, and Delta E goes to zero.

For an analogy consider meausring a wave in water. Early in the wave the amplitude is high and similar to when the wave was created. Delta T is low and Delta E is high. Over time, the wave peters out. Delta T is high because little amplitude is left and Delta E is low.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.