In QM, one takes the inner product $(\phi|\hat x|\phi)$ and $(\phi|\hat p|\phi)$ to compute the position and momentum expected values, but what does one do in QFT? What is the relationship between the wavefunction in QM and the QFT field?
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asked Dec 28, 2015 at 14:53
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2$\begingroup$ Related: physics.stackexchange.com/q/223783/2451 $\endgroup$– Qmechanic ♦Dec 28, 2015 at 16:04
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1 Answer
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For 1-particle states you can do exactly the same in (nonrelativistic) QFT, with position operator $X=\int dx a^*(x)xa(x)$. For multiparticle states, the notion of position makes no sense.
In quantum field theory, particle collision experiments are interpreted in terms of the S-matrix, which describes the transition from the far past, where particles are still described by separate, single-particle states, to the far future, where particles are again described by separate, single-particle states. In the intermediate time, close to the collision, one cannot identify individual in/out particles, and the system dynamics is that of a quantum field, not of particles.
In quantum field theory, the S-matrix is approximately computable in renormalized perturbation theory.
answered Dec 30, 2015 at 11:35
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$\begingroup$ so the wavefunction is just the same as non-relativistic qm? $\endgroup$ Dec 30, 2015 at 15:05
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$\begingroup$ @MarcusQuinnRodriguezTenes: Yes, in the 1-particle sector. Only the notation has changed. $\endgroup$ Dec 30, 2015 at 15:08
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$\begingroup$ in the multiparticle scenario, could you explain why position makes no sense? how is one going to calculate probabilities of trajectories then? $\endgroup$ Dec 30, 2015 at 17:05
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$\begingroup$ @MarcusQuinnRodriguezTenes: Well, what is the position of a multiparticle system? Even classically, there is none. You can ask instead about the position of its center of mass. - Trajectories are in phase space, which is not a space of positions, because of indistinguishability of the particles. And if you have a syuperposition of states with different particle numbers even this is no longer meaningful. - In field theory, one never thinks in terms of particles but of currents! $\endgroup$ Dec 30, 2015 at 17:33
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$\begingroup$ so then how do you make the transition from a current, to understanding the trajectories of a particle collision observed in experiment? experimentally you get information on position and momentum of some particles after the scattering, so how do you get this from the current? $\endgroup$ Dec 30, 2015 at 17:49