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In 'classical' quantum mechanics, a wave packet is a (more or less) localized particle. The wave packet can be expanded in a superposition of plane waves, each with a defined momentum and energy. This superposition is again still a wavefunction of one particle, with its physical interpretation being the probability amplitudes in space.

If we move up to quantum field theory, the quantized field is also a superposition of plane waves, which each represent a possible excitation (particle) of the field with well-defined momentum and energy. So let's say the electromagnetic field has many different quanta, created through several creation operators acting on the vacuum of the E-M field. They interfere with each other and again form a total 'wave packet' in configuration space. Should I interpret this as multiple photons, or can I think about it as if there is only one photon, more localized in space, but less localized in momentum space?

hft
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Remind
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2 Answers2

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First, just to clear up some confusion perhaps. A superposition of one-particle states $$ |\psi\rangle = \sum_n |\phi_n\rangle c_n , $$ is still a one-particle state. To get a state with multiple particles from one particle states one needs to form a tensor product$^*$ $$ |n\rangle = \frac{1}{n!}|1\rangle|1\rangle|1\rangle ...|1\rangle . $$

Now, what is a wave packet and how does it relate to particles? In general, a wave packet is the same thing as a wave function. However, the notion of a wave packet is related to the idea that it is almost like a particle in that it is somewhat localized.

If we now think of a wave function, then one can think of a complex field with the physical meaning that it can tell us what the probability is to observe a particle. In that sense, the wave functions (or wave packet) represents a single excitation (a single particle).

When one allows creation operators to act multiple times, then one will end up with multiple particles. If they are all in the same state (assuming they are bosons) then one can use the same wave function to represent the multi-particle state. Otherwise, the wave function would need to be a function of multiple sets of variables, one for each of the particles.

Different particles don't actually interfere with one another in the same way that photons would interfere in an optical interferometer. However, one can get quantum interference where the different terms in the expansion of a multi-particle state can cancel one-another, as in the Hong-Ou-Mandel effect.

Hope this helps.


$^*$ Often people don't like to think of it like this, but rather as multiple excitations in some Hilbert space, which is generated by multiple applications of creation operators.

flippiefanus
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An experimental particle physicists answer, who has been measuring particles for decades:

Quantum field theory is used to get numbers to be compared with experiments, (cross-sections, decay rates, mass distributions), the calculations using Feynman diagrams. The particle fields are assumed to be plane wave solutions of the corresponding quantum mechanical equations (Dirac, Klein Gordon, quantized Maxwell). These wavefunctions enter the integrals for the calculations.

It is not necessary to model a single elementary particle running along in order to validate the theory (standard model in this case). Nevertheless, as is well known that a plane wave cannot be localized, it goes from -infinity to +infinity in space time, so the need of a wave packet solution arises. This solution is consistent with the Heisenberg uncertainty principle which has to hold when one measures a single particle of a given momentum, and the probabilistic nature of quantum mechanical entities.

The wave packet solution describes a single particle when measured by itself in a consistent quantum mechanical probabilistic manner. It is one electron in the picture: turning in the magnetic field of the chamber

electron

If one wants to describe its wavefunction in QFT it will be a wave packet, interacting and ionizing consecutively the bubble chamber medium.

hft
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anna v
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  • "The particle fields are assumed to be plane wave solutions of the corresponding quantum mechanical equations" - i hope you don't mind me asking here instead of another question, and I hope I'm not "going all philosophical" here, but... in classical wave mechanics we can "consider" a wave to be the superposition of many plane waves. We can also consider it to be an infinity of spherical waves. We don't consider either "to be the wave", do we? But that seems to be what is happening here, it seems we are ascribing a mathematical short cut to "be real"? – Maury Markowitz Aug 24 '18 at 14:33
  • @MauryMarkowitz In quantum field theory each particle has a field on which a creation operator creates a particle and an annihilation operator annihilates it, and theoretically it should propagate.BUT it is plane waves that are used which have no meaning for a path in space time as in the picture. Thus the wavepacket. In quantum mechanics the waves are probability waves . A plane wave's Ψ*Ψ (probability distribution) has the information that the particle is somewhere from -infitnity to +infinity in space time when the creation operator works on it. – anna v Aug 24 '18 at 15:35
  • This is not what is happening in the picture. Fortunately there is the Heisenberg uncertainty principle and the wavepacket formalism to solve the conuundrum. – anna v Aug 24 '18 at 15:35