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I saw the question What are field quanta? but it's a bit advanced for me and probably for some people who will search for this question.

I learned QM but not QFT, but I still hear all the time that "particles are the quanta of fields" and I don't really understand what it means.

Is there a simple explanation for people who know QM but not QFT?

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Wave-particle duality, so (for example) vibrations in materials can be considered "particles"... if you remember your relation between wavelength and energy in decay of atoms, then a wave corresponds to an energy difference.... these energy differences are the "field quanta", because they are discrete jumps. [I don't like the answer given below because he completely ignores your last sentence] –  Chris Gerig May 22 '12 at 2:19

4 Answers 4

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In terms of measurements of the field, we can regard the vacuum state as the state that has the least correlations between measurements of the field in different places. At an elementary level, we can call the vacuum state the zero-quantum field state. The statistics of field measurements in the vacuum state are also highly symmetric, being isotropic, homogeneous, and invariant under Lorentz boosts. As we add more quanta to the quantum field state, the correlations between field measurements in different places become more complex and less symmetrical.

There are many significant differences between a classical field and a quantum field, but the most basic is that instead of talking just about the field at a particular place, we talk about the probability that the field will take one value or another, at many different places all at once. As a result, instead of describing modulations of a classical field --how a particular configuration of the field is different from the zero field--, we have to describe modulations of all the correlations of the field --how a particular configuration of the quantum field is different from the vacuum state.

A classical field is a function of just one point in space time, something like $\phi(x)$, but a quantum field can best be understood (IMO) in terms of a function of many points, $W(x_1,x_2,...,x_n)$, that describes the correlations between the field at $n$ different places. For the vacuum state, these functions are called the Vacuum Expectation Values (VEVs), but we can equally well construct a similar function in any state. A quantum field operator $\hat\phi(x)$ describes how to construct correlation functions for all different $n$, in any state that we can construct in the Hilbert space of quantum field states. The exact changes that are made to the VEVs by the action of a single quantum field operator are in a mathematical sense elementary operations on the discrete structure that ultimately derives from the fact that we don't talk about 2½-point correlation functions, for example, we only talk about 2-point, 3-point, ..., $n$-point correlation functions.

A single quantum of the field is often taken to be associated with a given frequency, but it can in general be associated with an arbitrarily complicated modulation of the correlation functions, at many different frequencies. The significance of this is that a quantum field operator is not just "add one quantum" (though it is that), it also says where in space-time to add the quantum, $\hat\phi(x)$. A single quantum, however, may be described as a superposition of different frequencies, or it may equally well be described as a superposition of different positions. In particular, $\hat\phi(x_1)+\hat\phi(x_2)$ is in many ways as good as a single quantum field operator as is $\hat\phi(x)$.

The mathematics of creation and annihilation operators is given in answers to the question you linked to, in very bare form, so if that's not what you want then you must want something different. To put the above into two bullet points, $\bullet$ quantum field operators modulate the correlations that are present in the vacuum state, and $\bullet$ correlations are an intrinsically discrete concept because they are constructed as correlations between 2, 3, or any integer number of measurements.

There is quite a lot of detail missing from the above, of course; I would particularly single out the issue of how to understand the consequences of measurement incompatibility at time-like separation. I have attempted not to speak too much beyond your question. If you remember your QM well enough, BTW, the quantized simple harmonic oscillator (SHO) can be regarded as the mathematical foundation of QFT; the quantum field can be constructed as an infinite number of interacting SHOs (albeit non-rigorously). But you can get that from many of the textbooks.

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In a harmonic oscillator, the number operator $N=a^*a$ that takes the eigenvalue $n$ on the $n$th lowest eigenstate (counting the ground state as the $n=0$ baseline) counts the number of quanta of energy contained in a state: The $n$th lowest eigenstate is obtained by adding $n$ quanta of energy to the ground state, by applying $n$ times the creation operator $s^*$.

You may regard the harmonic oscillator as a toy model for a very degenerate quantum field theory where particles have no momenta, so that the number of particles is the only characteristics an energy eigenstate has. Then particle = quant, and the ground state without excitations = vacuum state.

In reality, each particle can be in infinitely many distinct energy modes characterized by the momentum $p$ of the particle, and the field is generated by a corresponding number of independent oscillators. Ignoring interactions, one can treat these oscillators as harmonic, and represent them by creationoperators $a^*(p)$. The number operator takes now the form of an integral over all $a^*(p)a(p)$, but still counts the number of quanta (elementary excitations of the ground state = vacuum). The eigenstates with $n$ quanta of energy are now much more varied, however, being given by all possible states $|p_1,\dots,p_n\rangle=a^*(p_1)\dots a^*(p_n)|vac\rangle$. These states describe $n$ particles with definite momenta $p_1,\dots,p_n$. In this sense, particles are the quanta of fields.

Physically realizes states are obtained by superimposing (typically infinitely) many of these states as the precise momenta (and even the number of quanta involved) cannot be determines with mathematical precision.

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The easiest way to learn field theory is to learn what a Schrodinger field is. This is a half-way house for relativistic quantum mechanics, it's field theory, but it's non-relativistic.

A Schrodinger field is a classical field that satisfies the Schrodinger equation. It's a number in space which satisfies the equation of motion:

$$ i \partial_t \phi = {\nabla^2\over 2m} \phi $$

You can classically measure this field, it's a supefluid density and phase, or a Bose-Einstein condensate density and phase.

In quantum mechanics, this field is an operator on states, and the operator obeys the same equation. The Fourier transform of the operator equation is:

$$ i \partial_t \phi(k) = - {k^2\over 2m} \phi(k) $$

Notice that the field $\phi$ has all negative frequencies. In quantum mechanics, a field with pure positive frequency raises the energy. If

$$ [a^\dagger,H] = i\omega a^\dagger$$

Then $a^\dagger$ only has matrix elements between states of energy E and energy $E+\omega$. There are many ways to understand this, you can prove it quickly from the commutation relation, but it is also just obvious from Heisenberg's intuitive understanding of off-diagonal matrix elements in the energy representation as those with a definite frequency.

So $\psi^\dagger(k)$ has the property that it adds ${k^2\over 2m}$ to the system. Further $\psi(k)$ removes ${k^2\over 2m}$ from the system, and is the complete left inverse to $\psi^\dagger$. Further, in the vacuum state, the operator $\psi(k)$ gives 0 for all k.

The Hamiltonian (with zero point energy subtracted out) is a sum of Harmonic oscillators, one at each k (the sum is an integral in infinite volume, I use sum to represent either):

$$ H = \sum_k \psi^\dagger(k)\psi(k)({k^2\over 2m}) $$

And the obvious interpretation is that the state of level n at k is the n-particle state, $\psi^\dagger(k)$ creates a particle, $\psi(k)$ annihilates a particle.

$\psi^\dagger(x)$ and $\psi(x)$ create and annihilate a particle at x, there is no ambiguity in localizing particles in space. This formalism describes N identical particles in a wavefunction $\eta(x_1,....,x_n)$ by acting with the operator:

$$ \eta(x_1,...,x_n) \psi^\dagger (x_1)\psi^\dagger(x_2) ... \psi^\dagger(x_n) $$

Notice that $\eta$ is automatically symmetric--- this field described bosons. The field representation is an alternate classical limit for bosonic particles, they can make a field if they superpose coherently in a wave.

The statement that "all particles are quanta of a field" is saying that this same construction works for everything. In relativity it is more central, because the field is the only causal entity, the particles go back in time.

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In simple words, the "quanta of field" is DEFINED as MOMENTUM EIGENSTATE. In terms of QM it can be interpreted as a wavefunction $\psi$ that is an eigenfunction of momentum operators:

$P^{\mu} \psi = p^{\mu}\psi$,

where $P^{\mu}$ is the (4d) momentum operator, and $p^{\mu}$ is a numeric eigenvalue.

In QFT the "Normal Product" of operators is used in hamiltonian to make sure that vacuum state $\Phi_0$ is a state with zero momentum: $P^{\mu}\Phi_0=0\Phi_0$. Vacuum is a "zero particle state".

If we then act on vacuum state with creation operator $a^+(p)$, we will obtain the momentum eigenstate corresponding to eigenvalue $p$:

$P^{\mu}a^+(p)\Phi_0=p^{\mu}a^+(p)\Phi_0$

Due to equations of motion in this case $p^{\mu}p_{\mu}=m^2$. This state is normally interpreted as "one particle state".

If we further act on $a^+(p)\Phi_0$ with $a^+(q)$, we will also obtain momentum eigenstate:

$P^{\mu}a^+(q)a^+(p)\Phi_0=(p^{\mu}+q^{\mu})a^+(q)a^+(p)\Phi_0$

but in this case, of course, $(p^{\mu}+q^{\mu})(p_{\mu}+q_{\mu})$ is not equal to $m^2$.

So, this new state cannot be interpreted as a "one particle state", but still can be interpreted as "two particles state". This is because two particles with momentums $p$ and $q$ ($p^{\mu}p_{\mu}=q^{\mu}q_{\mu}=m^2$) can have total momentum of $p+q$.

Similarly, "n-particle states" can be obtained from vacuum state by acting on it with creation operator. All these states are "quanta of field".

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