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Just assume that I understand that a field in quantum field theory is an operator-valued distribution. For simplicity, forget about the distribution and think about a function

$\varphi:M \rightarrow L(H)$

that assigns an operator to each point of spacetime.

Can someone explain to me what (mathematically speaking) physicists mean by "quanta" of this field?

EDIT: (follow up question): if one fixes an inertial system and a point $t_0$ in time (say 4:00pm), then there is "space" and a map

$a^\dagger:\mathbb R^3 \rightarrow L(H) \rightarrow H$

$(x,y,z)\mapsto a^\dagger(t_0,x,y,z) \mapsto a^\dagger(t_0,x,y,z) |0>$

, which creates "quanta". Would it be correct to think about "particles at the point (x,y,z)" then one thinks about this map?

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4 Answers 4

up vote 6 down vote accepted

some authors of the questions expect a less mathematical answer than what they are given at the end. In a sense, you are the opposite example because you expect a more mathematical explanation than what the right explanation actually is.

The term "quantum of a field" is not representing a particular state or operator; it is meant to describe a particular "object" that an experimenter may measure. In the theoretical language, you could say that a quantum of a field is the difference between the states $$|\psi\rangle\mbox{ and } a^\dagger|\psi\rangle$$ Well, the difference is multiplicative in a sense - it's what the creation operator $a^\dagger$ created. But note that the state $|\psi\rangle$ may be anything - a quantum of a field may be added to any configuration of other (or the same) particles. Most typically, we talk about quanta in momentum eigenstates, so $a^\dagger$ is usually expressed by $$a^\dagger(p) = C \cdot \int d^3x\,\exp(ipx) \phi_i(x)$$ where $p$ is the momentum of the quantum - which usually allows us to calculate the energy as well - and $\phi_x(x)$ is a component of the quantum field (which may be both bosonic or fermionic).

So in the "intuitive" or "experimental" language, quanta of quantum fields are the actual particles (such as photons, Higgs bosons, or electrons). They're what the quantum fields are able to create (and annihilate).

Best wishes Lubos

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I always suspected that. I think there is a ontological quagmire here... –  foobar Jan 16 '11 at 18:37
@foobar: No quagmire--- the quantum is a state description, the field is an operator. The operators create states. –  Ron Maimon May 22 '12 at 3:38

It goes more or less like this. You define what is known as a creation operator:

$a^{\dagger}(\vec{k}) = \int \varphi(x)i\overleftrightarrow{\partial}_0e^{-ikx}d\vec{x}$

and when this is applied on a distinguished vector of the Hilbert space called "vacuum" $|0\rangle$, which describes the ground state of the system, you create a quanta of the field:

$|k\rangle = a^{\dagger}(\vec{k})|0\rangle $

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The operators assigned to your field usually are creation operators. Quanta are the things these creation operators create. This can be photons, phonons, electron-hole-pairs etc.

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I realize that my answers will never be to par as some of the other brains at this sight, but I want to throw my two cents in.

If we can imagine the situation in ordinary quantum mechanics, we understand that there are quantum states that a particle can be in that are described by a complete set of commutating observables.


Quantum field theory evolved to answer questions about the number of particles that may be in a closed system at a certain energy level. Since relativity equates particle mass to energy, we run into the situation where the number and types of particles will change over time, just like the state of a system might change over time.

So QFT is largely concerned with the evolution of systems of particles, where the observables are the particles themselves. So by analogy, the goal is to find a complete set of commutating observables (particles) that can be used to determine the exact quantum state of the closed system in question.

The quantization of a classical field into one described by particles is called the second quatization.


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