Timeline for Is measuring energy with arbitrary precision inherently impossible?
Current License: CC BY-SA 4.0
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| Apr 3 at 19:47 | comment | added | just a phase | Thanks for pointing this out. I agree about QFT, though it needn't be relativistic. Anyway, I understand how a signal $f(t)$ or its Fourier transform $\hat{f}(\omega)$ cannot both be sharply peaked. Does this apply to photon emission events? I would have thought that fluorescence, e.g., would emit a photon with very low energy uncertainty, but without the need for near-infinite uncertainty in when the emission occurred? I don't see where the Fourier analysis comes in. I'd like to understand this better so I can update the answer. | |
| Apr 3 at 8:16 | comment | added | AwkwardWhale | If you bring up photons you're talking about QFT now. Regardless, ordinary Fourier Analysis uncertainty (the Gabor limit) still limits how precisely you can measure and/or localise the momentum of a photon: any detection process involves scattering with well localised detectors, and so involves interactions that are localised in the EM field. The more localised the detector response is, the more momentum components of the EM field will couple and thus the less certain the measurement. | |
| Apr 3 at 1:32 | comment | added | just a phase | You did not "lose" a pair of canonical position and momentum operators because you never had one to begin with for photons, which is the only setting in which I made that claim, and it was shown by Dirac in the reference I provided. The uncertainty principle doesn't come from Fourier analysis but noncommuting operators, so I don't understand your last sentence. And I'm talking about QM not QFT btw. | |
| Apr 2 at 23:19 | comment | added | AwkwardWhale | I don’t agree at all with the statement that you can measure the momentum with arbitrary position while still being somewhat localised. In fact in moving to a field theory we have lost a canonical pair of position and momentum operators, and need to be far more careful with how we define a position or momentum measurement. Indeed for any field theory an uncertainty relation holds in the ordinary Fourier analysis sense, in this sense the notion of an energy time uncertainty has as much weight as a position momentum uncertainty when it comes to a field | |
| Apr 2 at 17:54 | history | edited | just a phase | CC BY-SA 4.0 |
Attempted to clarify, added references to earlier parts
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| Apr 2 at 17:15 | comment | added | just a phase | I've added the reference to Dirac's book, where he shows that photons don't have a position operator. This means there is no momentum/position uncertainty principle. As I also explain in my answer, the uncertainty principle for $a$ and $a^\dagger$ has nothing to do with $x$ and $p$ for photons (that's only for the harmonic oscillator). For photons, $a$ and $a^\dagger$ are instead related to the $E$ field and vector potential $A$, which indeed have an uncertainty relation. But there is no position operator, per Dirac's argument. | |
| Apr 2 at 17:13 | history | edited | just a phase | CC BY-SA 4.0 |
added reference to Dirac's book for absence of a position operator for photons
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| Apr 2 at 13:00 | comment | added | Paradoxy | The unsettling part in your answer is"there is no barrier to measuring the particle's momentum and knowing the particular location where this occurred". While it does answer the question(probably correctly so),I can't help but think that your argument is controversial. I would argue that, for the second quantization we demand commutator relationships for creation and annihilation of photons in momentum Fock space hold. This immediately gives uncertainty principle, so things are not this simple, even in QFT.Although I will become more convinced if you provide a creditable source for your claim. | |
| Apr 1 at 15:36 | comment | added | just a phase | Thanks, though I think you've asked some good questions and would like to have satisfactorily answered them :). My answer was actually meant to be about QM! I understand measurements there much better. But the things I said about photon position operators still hold: the QM of photons is still QFT-like in that position / momentum is a label, rather than an operator. Do you have lingering questions at this point? | |
| Apr 1 at 15:33 | history | edited | just a phase | CC BY-SA 4.0 |
deleted 2 characters in body
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| Apr 1 at 3:47 | history | bounty ended | Paradoxy | ||
| Apr 1 at 3:47 | comment | added | Paradoxy | Yup, with QFT we have no problem at all. I was wondering about QM. Nevertheless you have put too much effort in the answer, so I will accept yours | |
| Mar 31 at 16:03 | comment | added | just a phase | @Paradoxy I've updated my answer to address the "edit" part, hopefully this addresses your question? | |
| Mar 31 at 16:02 | history | edited | just a phase | CC BY-SA 4.0 |
updated based on comment so as to answer EDIT part of question
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| Mar 30 at 18:55 | comment | added | just a phase | Oh sorry, I misunderstood the "edit" part. But I don't see how measuring the energy of a battery is a quantum measurement? But also there are subtleties to defining position operators for photons due to gauge redundancy, etc. There might not be canonically conjugate position and momentum operators for photons (an operator of the form $a + a^\dagger$ corresponds to the $E$ field, not position, e.g.). | |
| Mar 30 at 18:15 | comment | added | Paradoxy | Thanks. I didn't understand your answer for Edit part fully. It seems that you are suggesting $\Delta t$ refers to the time that it takes for the photon to reach the detector. But this is not what I was asking. We measure battery's energy projectively over and over again. due to the conservation of energy, any energy shift observed in the battery corresponds to photon's energy, from which we find photon's momentum. So If nothing stops me from making $\Delta p = 0$(knowing photon's momentum exactly), we have a violation of uncertainty because $\Delta x = length of system$ which is not infinite. | |
| Mar 28 at 23:46 | history | answered | just a phase | CC BY-SA 4.0 |