Timeline for What happens when one "observes' a quantum field, and how do particles get involved?
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11 events
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| Jul 17, 2017 at 19:40 | comment | added | Cosmas Zachos | I can't be sure, but a glaring handle on the almost-but-not-quite localized QF wavepackets might be 344853#344853. | |
| Jul 17, 2017 at 1:35 | comment | added | Cosmas Zachos | Related 54603. The "analogy" between quantum fields and wavefunctions is at best counterproductive and at worst confusing. One rarely observes the x of a field in the lab: normally one observes momentum eigenstates. (Energies, momenta, and their angles, virtually decohered semiclassical pellets--except for neutrino or flavor oscillation phenomena.) | |
| Jul 16, 2017 at 15:13 | comment | added | Keith McClary | ... that factor involving $(k^2 + m^2)^{1/2}$. EDIT: I mean there are not vectors in Fock space representing localised position states. | |
| Jul 16, 2017 at 5:11 | comment | added | Keith McClary | Think of a (say, 1 space +1 time dimensional) free (bosonic) QFT "in a box" with momentum cutoff. This is just a multidimensional harmonic oscillator, with each dimension corresponding to a momentum state. The "particles" are the eigenstates of these oscillators. These "particles" are not distinguishable however - if you want that you need a tensor product of two different such theories. As for measuring the position, a very vague statement, you can't really define position states in relativistic QFT because the expression for $\phi(x)$ in terms of $a(k)$ is not a Fourier transform, it has ... | |
| Jul 16, 2017 at 4:05 | comment | added | Dragonsheep | Thanks for the response! That makes a lot of sense. For 2 particles distinguished by different, say, spin or momenta, are there operators we can use to observe 1 particle while keeping the other particle untouched, or do our operators only give us information about the system as a whole? I ask because most second-quantized operators I've seen involve some kind of sum over all momentum states. Despite this, though, if my particles are distinguishable, I intuitively should be know the position of one and momenta of the other. How this translates to second-quantized operators is unclear. | |
| Jul 16, 2017 at 3:58 | comment | added | Keith McClary | I suppose you could say in QFT you are measuring the number operator for each momentum mode whereas in QM you are measuring the momentum operator. But these are far removed from what we actually do in the lab. | |
| Jul 16, 2017 at 2:38 | comment | added | Dragonsheep | I think right now I'm just trying to grasp how I can recover non-QFT effects from QFT. E.g., creation operators apply to momentum eigenstates, and with addition of the right momentum eigenstates we can create a wavepacket that "looks like" a particle. Now suppose we use creation operators such that we have 2 wave packets/ particles in our system. Now, if I want to observe some observable of this particle, can I hit the resultant field constructed from the creation operators with my usual QM matrices and get my observables, treating the field as though it were a wave function? | |
| Jul 16, 2017 at 2:18 | comment | added | Keith McClary | Besides scattering, there are shifts in energy levels (or masses) and perhaps in decay rates due to QFT effects. What other experimental scenarios are you thinking of? | |
| Jul 15, 2017 at 22:18 | history | edited | Dragonsheep | CC BY-SA 3.0 |
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| Jul 15, 2017 at 21:32 | history | edited | Dragonsheep | CC BY-SA 3.0 |
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| Jul 15, 2017 at 21:26 | history | asked | Dragonsheep | CC BY-SA 3.0 |