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Oct 27, 2012 at 8:23 comment added twistor59 Sorry, in last comment "vacuum persistence ampltidue" should have just read "persistence amplitude". If an answer doesn't turn up here, I'll raise it as a separate question.
Oct 27, 2012 at 8:07 comment added twistor59 I must admit I get very confused about the energy/time-based explanations of HR. For a general state, I would start by looking at the time "width" of the vacuum persistence amplitude $\langle \Psi|U(t)|\Psi\rangle$, and the width of the energy spectrum $S(E)=|\langle E|\Psi \rangle|^2$, and from that derive the uncertainty relation. But with HR $|\Psi \rangle$ is the vacuum state, so has zero energy width. Moreover you have to be careful about which vacuum you're talking. The Bogoliubov tr. based explanations resolve this satisfactorily, but the E.T. based explanations leave me confused.
Oct 27, 2012 at 7:50 comment added anna v I think that it is not necessary to always think of the delta intervals as standard deviations from a gaussian/ statistical fit. It can be the one derived by the corresponding wave function expectation value, it can be just any interval, as in mathematics. The gaussian or wave function given the problem may be a better estimate, but a simple delta(variable) is enough for HUP, imo.
Oct 27, 2012 at 7:15 comment added anna v are you asking about the proportionality instead of the inequality? I would consider it a hand waving sloppiness unless hbar has been set to 1. One has to be talking at scales of hbar for quantum mechanics to make sense
Oct 27, 2012 at 7:11 comment added anna v lss.fnal.gov/archive/other/ift-p-036-93.pdf is the paper given in adsabs.harvard.edu/abs/1994PhRvA..50..933K . This is a reasonable solution to the problem which arises by the way one defines time.
Oct 27, 2012 at 7:05 comment added contrariwise @annv, I understand how particles are created. My question is about $\Delta E$ itself. I have similar-looking relations in the first and the second paragraph. But, the interpretation for $\Delta t$ in the first is very different from that in the second - at least at first glance. I am basically asking how they can be reconciled. Thanks.
Oct 27, 2012 at 6:47 comment added anna v continued: that these virtual particles are "real" comes from experimental verification of quantities computed by QFT.
Oct 27, 2012 at 6:46 comment added anna v you are not clear with what you mean by "the second relation": as you have written it is an algebraic manipulation keeping the inequalities. If you mean how could particles be presumed within the delta(E) it is another story and has to do with creation and annihilation operators in quantum field theory. The delta(E) just says that there exists an uncertain energy, QFT creates particle pairs within that uncertainty.
Oct 27, 2012 at 6:37 comment added John Rennie You're quite correct that $t$ isn't usually treated as an operator so there isn't a direct analogy between the two forms of the HUP. Over the years I've seen many papers on this subject but I don't think there's any explanation that triggers your "oh yes, of course" detectors. Most of us just shrug and accept that it works.
Oct 27, 2012 at 5:09 comment added contrariwise If I understand correctly, the similarity between momentum-position uncertainty and energy-time uncertainty isn't that straightforward. This is because time isn't an operator like momentum or position. It is an external parameter. So, a standard deviation for t doesn't make sense. Which is why one has to interpret what one means by $\Delta t$ in the uncertainty relation. Interpretations similar to the one I have referred to are the only ones I have seen that sound reasonable to me. But, I don't understand how that is related to what I mention in the second paragraph. Thanks.
Oct 27, 2012 at 4:53 history answered anna v CC BY-SA 3.0