# Why is free field called free ?

Why is free field called "free"?

For example the Klein-Gordon field, consider if we have no spatial dimension(0-dimensional QFT should be identical to QM, is it correct?), then the field hamiltonian reduces to something like p^2+ m^2q^2, which is a hamiltonian for a harmonic oscillator in QM, not a free particle, then why we call it a free field?

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Free field simply means there is no interactions terms in the Lagrangian, which means the fields do not interact with each other. The quanta associated with such fields simply pass through each other. If we start with the KG field and add something like a phi^3 term however this will lead to vertices where 3 particles can interact.

Normally the KG equation is introduced by the analogy of an infinite number of simple harmonic oscillatiors, which I guess gives the impression of a potential being present, and perhaps where the confusion is arising from? but this isn't what free means in this context.

But emmm, suppose we really want to quantize a string made up by infinite number of particles, wouldn't we arrive at the same quantum field? But you can't say particles in the string are free, can you?

No, but it's a different kind of free, when we say a quantum field is a free field, it means non interacting, with other quantum fields, not free from a potential function as in ordinary QM.

A. Neumaier

For example the Klein-Gordon field, consider if we have no spatial dimension(0-dimensional QFT should be identical to QM, is it correct?), then the field hamiltonian reduces to something like p^2+ m^2q^2, which is a hamiltonian for a harmonic oscillator in QM, not a free particle, then why we call it a free field?

A field does not describe a particle but an arbitrary number of identical particles.

The harmonic oscillator is the 0-dimension free field (not a free particle). The ground state is the vacuum, and the k-th excited state describes a field in which k indistinguishable free particles are present.

Emm, ok, I guess it's just semantics? BTW is it correct to say "0-dimensional QFT reduces to QM"

A. Neumaier

Emm, ok, I guess it's just semantics? BTW is it correct to say "0-dimensional QFT reduces to QM"

No.

QM in the Heisenberg picture is 1-dimensional QFT.

The 1-dimension refers to time dimension? I was thinking about spatial dimensions.

Well, i guess I'm too much impressed by the resemblance between the harmonic oscillator system in QM and QFT, all the stuff like the raising and lowering operators.....I'm still confused in QM harmonic oscillator raising and operators only change the energy level, and no mention of creation or annilation of particles...so do they really have some relation or just the similar appearance?

A. Neumaier

Actually, you were right. I wasn't numbering the dimensions consistently.
To be unambiguous:

Single-particle QM in the Heisenberg picture is 1+0-dimensional field theory,
and the harmonic oscillator is the 1+0-dimensional free field.

But note that the harmonic oscillator has two different interpretations:
(i) as a single particle in 1+1 dimensions (which is how it is introduced in QM), where it describes a particle in 1-dimensional R^1 space with states in L^2(R^1), bound in an external field and oscillating in time,
(ii) as a free field in 1+0-dimensions, where it describes an arbitrary number of noninteracting particles in 0-dimensional R^0={0} with states in C^1=L^2(R^0).

Mathematically, it is precisely the same - physically, the interpretation is radically different!

Well, i guess I'm too much impressed by the resemblance between the harmonic oscillator system in QM and QFT, all the stuff like the raising and lowering operators.....I'm still confused in QM harmonic oscillator raising and operators only change the energy level, and no mention of creation or annilation of particles...so do they really have some relation or just the similar appearance?

They resemble each other because one is indeed a special case of the other, if interpreted correctly.

For the harmonic oscillator, a^* creates one particle at a fixed (unmentioned) time. Since there is no space, there is no momentum to distinguish single particle states.
|0>=|> is the vacuum (ground state),
|1>=a^*|0> is the 1-particle state,
|2>=a^*|1> the 2-particle state, etc..
The frequency omega of the harmonic oscillator is the particle mass (setting c=1 and hbar=1): m=omega, and the N-particle state has the mass E_N=N*omega of N particles.
The Hamiltonian is H=omega a^*a.

In higher dimension, each particle comes together with its quantum numbers (for a scalar field just the momentum p); thus there is one creation operator a^*(p) for each allowed quantum number p, and there are many 1-particle states |p>=a^*(p)|>, and even more 2-particle states |p',p>=a^*(p')|p>.

Thus the notation is a bit different. To make the analogy perfect, one should assign the 1-particle state in dimension 0 a zero momentum and write a^*(0) for a^*, |> for the vacuum, |0> for the 1-particle state, |0,0> for the 2-particle state, etc., and the Hamiltonian as
$$H = \sum_{p \in R^{0}}m a^*(p)a(p),$$
which indeed equals omega a^*a, since R^{0} contains only one element.

G01
Homework Helper
Gold Member

Actually, you were right. I wasn't numbering the dimensions consistently.
To be unambiguous:

Single-particle QM in the Heisenberg picture is 1+0-dimensional field theory,
and the harmonic oscillator is the 1+0-dimensional free field.

But note that the harmonic oscillator has two different interpretations:
(i) as a single particle in 1+1 dimensions (which is how it is introduced in QM), where it describes a particle in 1-dimensional R^1 space with states in L^2(R^1), bound in an external field and oscillating in time,
(ii) as a free field in 1+0-dimensions, where it describes an arbitrary number of noninteracting particles in 0-dimensional R^0={0} with states in C^1=L^2(R^0).

Mathematically, it is precisely the same - physically, the interpretation is radically different!

They resemble each other because one is indeed a special case of the other, if interpreted correctly.

For the harmonic oscillator, a^* creates one particle at a fixed (unmentioned) time. Since there is no space, there is no momentum to distinguish single particle states.
|0>=|> is the vacuum (ground state),
|1>=a^*|0> is the 1-particle state,
|2>=a^*|1> the 2-particle state, etc..
The frequency omega of the harmonic oscillator is the particle mass (setting c=1 and hbar=1): m=omega, and the N-particle state has the mass E_N=N*omega of N particles.
The Hamiltonian is H=omega a^*a.

In higher dimension, each particle comes together with its quantum numbers (for a scalar field just the momentum p); thus there is one creation operator a^*(p) for each allowed quantum number p, and there are many 1-particle states |p>=a^*(p)|>, and even more 2-particle states |p',p>=a^*(p')|p>.

Thus the notation is a bit different. To make the analogy perfect, one should assign the 1-particle state in dimension 0 a zero momentum and write a^*(0) for a^*, |> for the vacuum, |0> for the 1-particle state, |0,0> for the 2-particle state, etc., and the Hamiltonian as
$$H = \sum_{p \in R^{0}}m a^*(p)a(p),$$
which indeed equals omega a^*a, since R^{0} contains only one element.

I think the OP is possibly getting confused about the use of the word "particle" in this context.

Just to avoid any more possible confusion:

In the above post, A.N. uses the word particle to mean "field quantum."

i.e.

|1> stands for the state with one particle (field quantum) present.

|2> stands for the state with two field quanta, etc.

In the 1-D mechanical quantum oscillator picture, the analog to the "particle" in the field theory would be the number of quanta of energy present, not the mass on the spring. i.e. |1> is the state with one quantum of energy, |2> with two quanta, etc.

A. Neumaier

In the above post, A.N. uses the word particle to mean "field quantum."

Yes, like everywhere in quantum field theory. Except that there the name ''particle'' is customary. Essentially nobody speaks there today about quanta.

kof9595995 I had a course at ens paris this autumn and this was a point of confusion for us too. so taking from my lecture notes that have as reference basically the first book on qft of Weinberg it goes like this:

following the viewpoint in Weinberg's first 7-8 chapters (landau viewpoint, symmetries first) we construct our fields as fourier transforms of annihilation and creation operators and determine the coefficients depending on the representation that this field will transform under, say scalar. To be precise it is the annihilation and creation operators by which it is constructed that transform under a given representation. Now the thing is that this operators are acting on the states in our Hilbert space that represent non interacting multiparticle states.

That is when we say $$a(q_2)\Psi_{q_1}=\Psi_{q_1 q_2}$$, we mean by $$\Psi_{q_1 q_2}$$ the state of two particles that are described by the quantum numbers q1 and q2 (e.g. momentum,spin or helicity,other stuff) in which these two particles are not interacting between them. This base of multi particle states is called the base of non interacting particles in french at least. So the creation and annihilation operators are defined in this base.

It turns out that following the construction of the fields having in mind the scalar interaction hamiltonian (one admits the splitting H=Hfree + Hinteraction ) which is an operator operator in the interaction picture the fields are too in this picture. This means for any operator in the interaction picture and specifically for the fields that $$\psi_I(t)=exp(iH_0 t) \psi_H exp(-iH_0 t)$$ where psi_H is the field in the heisenberg picture. So the temporal evolution of the field is governed by the free hamiltonian and it is in this sense that it is called a free field. Note that an operator in the heisenberg picture has no dependence on time, and that the fields are operators too.

So this is a different viewpoint from the ones taken before although I am actually putting this up for debate, it is interesting to me too.

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G01
Homework Helper
Gold Member

Yes, like everywhere in quantum field theory. Except that there the name ''particle'' is customary. Essentially nobody speaks there today about quanta.

Yes, of course. But, the following post led me to think that the OP might not be aware of this terminology.

Well, i guess I'm too much impressed by the resemblance between the harmonic oscillator system in QM and QFT, all the stuff like the raising and lowering operators.....I'm still confused in QM harmonic oscillator raising and operators only change the energy level, and no mention of creation or annilation of particles...so do they really have some relation or just the similar appearance?

I thought that the bolded statement necessitated pointing out that the "energy quanta" in the QSHO are the (1+0)D analogs to the particles in QFT, and not the hypothetical "mass on the spring" that he may have been picturing while thinking about the QSHO.

I also wondered about the point raised by the OP before.

Free fields are ones that 'wiggle', composed of infinitely many coupled harmonic oscillators.

But static fields, e.g. the Coulomb field in EM, are still classical fields in QFT, since the number of photons is indefinite. Correct?

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In reply to the last, the resemblance of fields propagating say uniformely in space like one created by an infinite charged plane wiggled right and forth, are in field theory the states that are eigenstates of the momentum operator. The fourier transform states if you think of it as waves. But they are waves in a quantum mecanical (probabilistic) sense it is a quantum theory after all. What oscillates is probability amplitudes given by the field at every point in spacetime.

Well, what I do not get

- a free field has a definite number of photons, which implies that the expectation value of the free field is zero

- a static field has a indefinite number of photons (or none at all, rather), but its (classical) field strength can be measured precisely

How are free fields and static fields related in QFT then? They seem somewhat the opposite if what I just wrote is true. (Which I'm not sure of!)

Wow,that's a lot to digest, I need some time to think about these and then ask again.

In the 1-D mechanical quantum oscillator picture, the analog to the "particle" in the field theory would be the number of quanta of energy present, not the mass on the spring. i.e. |1> is the state with one quantum of energy, |2> with two quanta, etc.

Yes, I guess that's part of my confusion. If I understand you correctly, we can't use 0+1 dimensional free field to really describe a physical system like a mass on the spring. So is it possible to use QFT to really describe a system like a mass on the spring(or mass in an external potential)? What kind of theory should we use?

Well the thing is in my opinion that the mass you are referring to would be a localized particle. When we are talking about particle fields in qft we have the free states that correspond to the quanta of the theory i.e. solutions of the type |p,n,s>. By that I mean a state vector in Hilbert space (not a field) that describes (contains all the possible information) a free particle (quanta of theory) with momentum p of particle type n and of spin s for example (that is by assumption).

In order to actually describe a real particle that propagates and is localized one has to use a number of these states and have them superimposed (that is one cannot have only a state with momentum p describing the particle due to Heisenberg principle and also exp(ipx) on its own is an oscillation through all of space i.e nowhere to be found since it has energy p0 definite). That is writing down a field in this sense $$\psi(x)=\int dx^{D+1} W(p) a(p)$$. For simplicity I do not include annihilation operators. This field acting on vacuum would create a localized particle (of type n of spin s implied) from a continuum of momentum eigenstates whose contribution would be wighted by the coefficients W. These are determined by the way the field transforms under lorentz transformations (they could be of tensorial or spinorial nature but for a scalar they are a simple function of the energy of the state) and it comes out they will always include a exp(ipx) factor (that is the solution of the klein gordon describing "free" fields, that is the solution of a wave equation, also that is a solution of a 1-d harmonic oscillator) . In that sense it is a fourier transform. The thing that has definite energy in qft is the states of the hilbert space not the actual particles we are creating (they do in their "free" form but if they are localized i.e. what we literally mean particles, only within the uncertainty principle ).

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Yes, I guess that's part of my confusion. If I understand you correctly, we can't use 0+1 dimensional free field to really describe a physical system like a mass on the spring. So is it possible to use QFT to really describe a system like a mass on the spring(or mass in an external potential)? What kind of theory should we use?

well in order to use qft to describe a mass attached on a spring thinking of this macroscopically it would be a hassle. One would use newton's theory which works fine in this case. Maby what you mean is a potential of the form -kx^2 used in qm for the harmonic oscillator that gives definite energy states. Again this applies for definite energy solutions of schrodinger's equation this time (not free because there is a potential, but that is not the problem in thinking of the particle as actually going back and forth).

A note here to make it a bit more clear, I hope I am helping. In the potential well you would imagine the sinoids with two or three or four etc nodes (zero probability amplitude points, you wouldn't find the particle there). There is no point of thinking to localize the particle in these states since the energy is very well defined. That is the same thing as saying not to think of the atom as in the planetary model, meaning the electron swirling around the nucleus. The electron in the atom is confined in a potential similar in low energies to the harmonic oscillator (in 3d) and we think of the probability of finding it somewhere in terms of the harmonic functions or the orbitals as chemists put it, but not of it actually being somewhere and propagating around the nucleus. Sorry if I am blubbering i'll stop here!

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As someone mentioned before, free just means non-interacting.

In this sense, an ideal gas is just as free as an idealized paramagnet in an external field, or an einstein solid, or the modes of radiation in a box. The gas molecules, the paramagnets, the oscillators, the standing wave modes - they do not interact with each other.

The mass term (that looks like a harmonic oscillator term) is part of the nature of the particle. You can't take it away like you can take away a spring attached to a box, leaving only the box. How can you take away the mass of a particle?

In that sense that potential is the freest you can get.

G01
Homework Helper
Gold Member

Yes, I guess that's part of my confusion. If I understand you correctly, we can't use 0+1 dimensional free field to really describe a physical system like a mass on the spring. So is it possible to use QFT to really describe a system like a mass on the spring(or mass in an external potential)? What kind of theory should we use?

No. That is not what I am saying. The 1D QSHO is described correctly by the (0+1) dimensional free scalar field in QFT. The only problem is semantic: The "particles" the theory describes represent quanta of energy in the system, not the number of masses attached to springs.

In fact, this is one reason why the QSHO problem is stressed as being very important when one learns quantum mechanics. As the lowest dimensional form of a scalar field, it is the starting point for quantum field theory.

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A. Neumaier

kof9595995 I had a course at ens paris this autumn and this was a point of confusion for us too. so taking from my lecture notes that have as reference basically the first book on qft of Weinberg [...]
It turns out that following the construction of the fields having in mind the scalar interaction hamiltonian (one admits the splitting H=Hfree + Hinteraction ) which is an operator operator in the interaction picture the fields are too in this picture.

We are currently discussing the relation between the free and the interacting representation in the thread https://www.physicsforums.com/showthread.php?t=388556 ,
also following Weinberg.

A. Neumaier

I also wondered about the point raised by the OP before.

Free fields are ones that 'wiggle', composed of infinitely many coupled harmonic oscillators.

But static fields, e.g. the Coulomb field in EM, are still classical fields in QFT, since the number of photons is indefinite. Correct?

In QED, the Coulomb field is an interaction term in the Hamiltonian, written in terms of the electron field rather than the photon field. But electron fields of course also wiggle (de Broglie waves)!