Why is free field called "free"?

[kof9595995]

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?

[LAHLH]

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.

[kof9595995]

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?

[LAHLH]

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.

[kof9595995]

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.

[kof9595995]

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

[kof9595995]

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]

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.

[christodouloum]

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.

[G01]

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.

[Lapidus]

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?

[christodouloum]

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.

[Lapidus]

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!)

[kof9595995]

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

[kof9595995]

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?

[christodouloum]

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 ).

[christodouloum]

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!

[RedX]

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]

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.

[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 http://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)!

[A. Neumaier]

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

No. One cannot assign photons to a quantum field, except in a very loose sense.
It is the state that may or may not have a definite number of photons. The field has different expectation values in different states, and in most states, it is not zero.

[A. Neumaier]

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?

0+1 dimensions means no time and one space dimension. Thus it is a static view. No springs, except in equilibrium where it is uninteresting.

The harmonic oscillator with the Hamiltonian p^2/2m +k q^2/2 is in 1+1 dimensions,
and describes an oscillating mode in a single direction - of anything, be it a spring or a laser beam. If for the sake of spectral analysis you rewrite it in terms of ladder operators, you get a new representation in a new basis consisting of eigenstates.

It just happens that this second representation has another interpretation as a field theory in 1+0 dimensions (one time dimension and space 0-dimensional - a single point).
where what were excited states in the oscillator representation turn into states counting the number of particles.

[A. Neumaier]

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.

The same abstract model in an abstract Hilbert space can have several interpretations, each with its own semantics. This is the case for the harmonic oscillator in the ladder representation.

The confusion gets away if one doesn't try to see both semantics in the same picture. Draw two pictures, one for each semantics!

[kof9595995]

0+1 dimensions means no time and one space dimension. Thus it is a static view. No springs, except in equilibrium where it is uninteresting.

The harmonic oscillator with the Hamiltonian p^2/2m +k q^2/2 is in 1+1 dimensions,
and describes an oscillating mode in a single direction - of anything, be it a spring or a laser beam. If for the sake of spectral analysis you rewrite it in terms of ladder operators, you get a new representation in a new basis consisting of eigenstates.

It just happens that this second representation has another interpretation as a field theory in 1+0 dimensions (one time dimension and space 0-dimensional - a single point).
where what were excited states in the oscillator representation turn into states counting the number of particles.
Ok so I should've used 1+0 actually, now I fully agree with you.

[kof9595995]

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.

I'm not sure if I really understand you, so I'll just elaborate what I meant and see if clarifies anything: If we look at the lagrangian q^2+p^2(omitting all the coefficients), we can either interpret it as a 1+0 QFT or a 1+1 QSHO. However, when we interpret it as a 1+0 QFT, we can't say it describes a QSHO, or a mass on a spring. It's easy to imagine to describe a QSHO fully in terms of QFT, you need a field to describe the existence of the particle(i.e. the mass on the spring), and some fields to describe the external field(i.e. the spring), so it's probably a interacting QFT rather than free.

[A. Neumaier]

I'm not sure if I really understand you, so I'll just elaborate what I meant and see if clarifies anything: If we look at the lagrangian q^2+p^2(omitting all the coefficients), we can either interpret it as a 1+0 QFT or a 1+1 QSHO. However, when we interpret it as a 1+0 QFT, we can't say it describes a QSHO, or a mass on a spring. It's easy to imagine to describe a QSHO fully in terms of QFT, you need a field to describe the existence of the particle(i.e. the mass on the spring), and some fields to describe the external field(i.e. the spring), so it's probably a interacting QFT rather than free.

A real spring is described by a (nonrelativistic) matter field in 1+3 dimensions.
The 1+1D picture is a simplification where you reduce space to the direction in which the spring can oscillate, the particle number is 1, and the interaction is given by the stiffness of the spring.

In a 1+0D picture, there is no direction to oscillate and the only oscillations that can happen are unobservable oscillations of the phase of the wave function (for the single particle state) or oscillations in the number of particles (for a general state).

[kof9595995]

In a 1+0D picture, there is no direction to oscillate and the only oscillations that can happen are unobservable oscillations of the phase of the wave function (for the single particle state) or oscillations in the number of particles (for a general state).

That's why I think there's no exact (both mathematical and physical at the same time) analogy between QM oscillator and quantum free field.

[A. Neumaier]

That's why I think there's no exact (both mathematical and physical at the same time) analogy between QM oscillator and quantum free field.

Analogies always only capture part of the whole.

Mathematically (i.e., in essence), the free oscillator and the free 1+0-D quantum field are the same - they behave the same in every respect; you can translate everything from one to the other and back. Even physicists call them by the same name. It is the same mathematics applied to different problems.

[kof9595995]

Ok, I guess these clarify all my confusions.

[G01]

Ooops. Yeah I seem to have perpetuated the dimension typo in all of my posts above. I meant (1+1 )field theory in all my above posts, NOT 0+1. Very sorry!

I read (0+1) in the thread and it stuck in my mind.

[kof9595995]

Today I tried to do the same to Dirac field--reduce it to 1+0 dimension, but I could get nothing like [x,p]=i, is it that Dirac field can't be reduce to a QM analog?

[A. Neumaier]

Today I tried to do the same to Dirac field--reduce it to 1+0 dimension, but I could get nothing like [x,p]=i, is it that Dirac field can't be reduce to a QM analog?

Only the harmonic oscillator can easily be reduced to 1+0D. This is exceptional - it plays a double role since the same math describes two intrinsically different situation.

The Dirac equation describes a single electron or positron, and cannot be interpreted as a multi-particle theory. (Though it can be generalized to do so. This then gives the QFT of a free fermion.)

[kof9595995]

I see, thank you.

[kof9595995]

BTW it'd be really nice to have a teacher like you : )

[A. Neumaier]

BTW it'd be really nice to have a teacher like you : )

Well, come to Vienna and study mathematics!

[kof9595995]

Actually, if Dirac field can't be reduced to QM form in simple manner, would QFT give consistent result with QM result of Dirac equation? We know single-particle interpretation of Dirac equation still gives a lot of meaningful result, does it?

[A. Neumaier]

Actually, if Dirac field can't be reduced to QM form in simple manner, would QFT give consistent result with QM result of Dirac equation? We know single-particle interpretation of Dirac equation still gives a lot of meaningful result, does it?

The Dirac field and the Dirac equation are very different things.

The Dirac field is the quantized Dirac equation and descrbes an arbitrary number of noninteracting spinor particles. But it has a perfect quantum mechanical representation in a Fermion Fock space.

The Dirac equation describes only superpositions of the electron and positron state of a single particle, and also has a quantum mechanical representation, though the Hamiltonian looks nice only in the momentum representation.

[kof9595995]

Then I'm worried, can we really recover results QM from QFT? Especially for fermion fields, I don't even know where to start, how can we construct something like [x,p] from anti-commutation relations of fields?

[dextercioby]

Then I'm worried, can we really recover results QM from QFT? Especially for fermion fields, I don't even know where to start, how can we construct something like [x,p] from anti-commutation relations of fields?

A heuristic way to pass from QFT back to <ordinary QM> is to remove the so-called 2nd quantization, which (very) roughly speaking means leaving Fock space aside. The Dirac equation in the <ordinary QM> is a wave equation (though imprecise, the terminology has historical value) for a specially-relativistic wavefunction describing a particle of spin 1/2.

For the fermion part of fields, the passage from QFT back to QM makes no sense, as the fundamental propery of anticommutation is lost. Nonetheless, the Dirac equation for the H-atom brings huge improvements on Schroedinger's early results.

[A. Neumaier]

Then I'm worried, can we really recover results QM from QFT? Especially for fermion fields, I don't even know where to start, how can we construct something like [x,p] from anti-commutation relations of fields?

This is called second quantization, and explained in most books on statistical mechanics.

The free Dirac equation is the 1-particle sector of the quantum field theory of a free electron field. Operators A(p,q) on the 1-particle Hilbert space extend to corresponding field operators by means of the functor that maps A to a bilinear form N(A) in the c/a operators, which in the simplest case of a scalar field takes the form
N(A)=\int dp a^*(p)A(p,q)a(p).
[for some unknown reason, the dp prints as dx on my screen. Note:q is the differential operator in the momentum representation, not a variable.]


For the fermion part of fields, the passage from QFT back to QM makes no sense, as the fundamental property of anticommutation is lost.

Of course it does. Only the formula for N(A) gets more complicated since A is now a matrix and you need to sum over four spinor indices.

[dextercioby]

Of course it does. Only the formula for N(A) gets more complicated since A is now a matrix and you need to sum over four spinor indices.

Your statement doesn't really address mine.

The point I was making was that a correct theory of fermions (anticommuting particles with positive semi-integer spin) makes sense only in the realms of QFT on a Fock space, because, as far as I'm aware (here I may wrong and hopefully someone would correct me), the spin-statistics theorem is a result of QFT, not of Galilean or specially relativistic quantum mechanics of systems (I call the latter <normal QM>).

Unfortunately I cannot back my statement by quoting this page http://en.wikipedia.org/wiki/Spin-statistics_theorem, because it's not writtem in the language of axiomatical QFT. Rigorous treatments of the theorem can be found of course in the books by Streater & Wightman, Bogolubov (1975 one), Lopuszanski and probably many others.

So anticommutation required by the spin-stat. thm. is forced by existence of quantum fields (or Wightman functionals if you prefer), which means QFT. Dropping QFT, that is going back to the 1-particle Hilbert space of <normal QM>, means dropping the notion of fermion, dropping anticommutation relations.

[Careful]

Your statement doesn't really address mine.

The point I was making was that a correct theory of fermions (anticommuting particles with positive semi-integer spin) makes sense only in the realms of QFT on a Fock space, because, as far as I'm aware (here I may wrong and hopefully someone would correct me), the spin-statistics theorem is a result of QFT, not of Galilean or specially relativistic quantum mechanics of systems (I call the latter <normal QM>).

Unfortunately I cannot back my statement by quoting this page http://en.wikipedia.org/wiki/Spin-statistics_theorem, because it's not writtem in the language of axiomatical QFT. Rigorous treatments of the theorem can be found of course in the books by Streater & Wightman, Bogolubov (1975 one), Lopuszanski and probably many others.

So anticommutation required by the spin-stat. thm. is forced by existence of quantum fields (or Wightman functionals if you prefer), which means QFT. Dropping QFT, that is going back to the 1-particle Hilbert space of <normal QM>, means dropping the notion of fermion, dropping anticommutation relations.
It is even a bit more subtle than that, people always rely upon a statistics theorem (that only bose and fermi occur in nature, if you move away from Hilbert space that might evaporate too). But your general comment is correct, that is why the natural construction for a relativistic dynamics is one in which free field theory applies locally (see my book for that statement) so that spin statistics holds, but globally particle notions as well as the spin statistics relation become dependent upon the way you coarse grain and an effective quantum group description may be appropriate here.

Careful

[A. Neumaier]

Your statement doesn't really address mine.

The point I was making was that a correct theory of fermions (anticommuting particles with positive semi-integer spin) makes sense only in the realms of QFT on a Fock space, because, as far as I'm aware (here I may wrong and hopefully someone would correct me), the spin-statistics theorem is a result of QFT, not of Galilean or specially relativistic quantum mechanics of systems (I call the latter <normal QM>).

The point you make now is unrelated to the passage between single-particle and Fock representation. There is no spin-statistic theorem in nonrelativistic QFT, but the relation between 1-particle Hilbert spaces and second quantization (in both directions) is precisely the one I described, no matter whether you require commutation or anticommutation relations.

A fermion is a particle with nonintegral spin. The relativistic spin 1/2 case is described by the Dirac equation in quantum mechanics, and by anticommuting fields in quantum field theory. And the passage in both direction goes via the functor A--> N(A) and its inverse.


So anticommutation required by the spin-stat. thm. is forced by existence of quantum fields (or Wightman functionals if you prefer), which means QFT. Dropping QFT, that is going back to the 1-particle Hilbert space of <normal QM>, means dropping the notion of fermion, dropping anticommutation relations.

True. But a spin-statistic theorem makes sense only in quantum field theory since there is no concept of statistics in a single-particle theory. To keep up your argument

For the fermion part of fields, the passage from QFT back to QM makes no sense, as the fundamental property of anticommutation is lost.
you'd also have to admit that for the boson part of fields, the passage from QFT back to QM makes no sense, as the fundamental property of commutation is lost. But this is nonsense.

[kof9595995]

The free Dirac equation is the 1-particle sector of the quantum field theory of a free electron field. Operators A(p,q) on the 1-particle Hilbert space extend to corresponding field operators by means of the functor that maps A to a bilinear form N(A) in the c/a operators, which in the simplest case of a scalar field takes the form
N(A)=\int dp a^*(p)A(p,q)a(p).
[for some unknown reason, the dp prints as dx on my screen. Note:q is the differential operator in the momentum representation, not a variable.]

How does this come about? Do a(p) and A(p,q) commute? Anyway you help me realize that I had a wrong impression that all QM operators can somehow be express by field operators phi.

[A. Neumaier]

How does this come about? Do a(p) and A(p,q) commute? Anyway you help me realize that I had a wrong impression that all QM operators can somehow be express by field operators phi.

They can be expressed as integrals over products of field operators and their derivatives.
In the particular expression, q is the differential operator, q= i partial/hbar (apart perhaps from a minus sign) so A(p,q) a(p) is a linear combination of derivatives of a(p) multiplied by appropriate functions of p.

[kof9595995]

Why can't we just use the QM operator directly,but have to map it to a new object? You mean the 1-particle state in QFT is different from the 1- particle state in QM? How?

[A. Neumaier]

Why can't we just use the QM operator directly,but have to map it to a new object? You mean the 1-particle state in QFT is different from the 1- particle state in QM? How?

The 1-particle state spaces are isomorphic.

But second quantization allows the 1-particle operators to act on arbitrary N-particle states, and on states of indefinite particle number.

[kof9595995]

About second quantization, I guess what really confused me is why the field operator equation in QFT look so much like the equation for wavefunction in QM, despite that we interpret them very differently.
To make the question clearer:
A generic one-particle state can be written as
|\Phi > = \int {\Phi (x,t)|x > dx}
So Phi(x) is the wavefunction
With field operators we can write it as
|\Phi > = \int {\Phi (x,t){\psi _{op}}(x)dx} |0 >
Psi_op is the field operator which creates a particle at x.
Now QM requires some equation to be satified (Schrodinger for example):
- \frac{\hbar }{{2m}}\frac{{{\partial ^2}}}{{\partial {x^2}}}\Phi (x,t) = i\hbar \frac{\partial }{{\partial t}}\Phi (x,t)............(1)
But in the corresponding field theory we also have
- \frac{\hbar }{{2m}}\frac{{{\partial ^2}}}{{\partial {x^2}}}{\psi _{op}}(x,t) = i\hbar \frac{\partial }{{\partial t}}{\psi _{op}}(x,t)............(2)
It confused me that they look so much alike, I know the way we construct psi_op naturally makes it satisfy (2), but this only makes me feel like it's a pure coincidence. I would be happy if we can somehow show (1) and (2) is indeed logically connected.

[kof9595995]

Operators A(p,q) on the 1-particle Hilbert space extend to corresponding field operators by means of the functor that maps A to a bilinear form N(A) in the c/a operators, which in the simplest case of a scalar field takes the form
N(A)=\int dp a^*(p)A(p,q)a(p).
[for some unknown reason, the dp prints as dx on my screen. Note:q is the differential operator in the momentum representation, not a variable.]

I used this prescription and try to calculate commutator [x,p], but it seems to lead me nowhere:
\begin{array}{l}
[x,p]=\int {dpdk\{ {a^\dag }(p)\frac{\partial }{{\partial p}}} a(p)k{a^\dag }(k)a(k) - k{a^\dag }(k)a(k){a^\dag }(p)\frac{\partial }{{\partial p}}a(p)\} ......................(1) \\
= i\int {dpdk\{ k{a^\dag }(p)\frac{\partial }{{\partial p}}} [{a^\dag }(k)a(p) + \delta (p - k)]a(k) - k{a^\dag }(k)a(k){a^\dag }(p)\frac{\partial }{{\partial p}}a(p)\} .............(2) \\
= i\int {dpdk\{ k{a^\dag }(k){a^\dag }(p)\frac{\partial }{{\partial p}}} a(p)a(k) + k{a^\dag }(p)\frac{\partial }{{\partial p}}\delta (p - k)a(k) - k{a^\dag }(k)a(k){a^\dag }(p)\frac{\partial }{{\partial p}}a(p)\} .........(3) \\
= i\int {dpdk\{ k{a^\dag }(k){a^\dag }(p)a(k)\frac{\partial }{{\partial p}}} a(p) + k{a^\dag }(p)a(k)\frac{\partial }{{\partial p}}\delta (p - k) - k{a^\dag }(k)a(k){a^\dag }(p)\frac{\partial }{{\partial p}}a(p)\} .........(4) \\
= i\int {dpdk\{ k{a^\dag }(k)[a(k){a^\dag }(p) - \delta (k - p)]\frac{\partial }{{\partial p}}} a(p) + k{a^\dag }(p)a(k)\frac{\partial }{{\partial p}}\delta (p - k) - k{a^\dag }(k)a(k){a^\dag }(p)\frac{\partial }{{\partial p}}a(p)\} .......(5) \\
= i\int {dpdk\{ - k{a^\dag }(k)\delta (k - p)\frac{\partial }{{\partial p}}} a(p) + k{a^\dag }(p)a(k)\frac{\partial }{{\partial p}}\delta (p - k)\} ...................(6) \\
\end{array}
Did I do something wrong?(Sorry, I know it looks horribly tedious)

[A. Neumaier]

I used this prescription and try to calculate commutator [x,p], (Sorry, I know it looks horribly tedious)

In the definiition of q you forgot a factor of =i/hbar (or -i/hbar).

Ding these calculations by brute force is tedious indeed. But it can be done in a smarter way:

First compute for general A the commutator [N(A),a(p)]. Then conjugate the resulting formula to get the commutator [N(A),a^*(p)]. Then use this to compute the commutator formula [N(A),N(B)]=N([A,B]), using the rule [a,bc]=[a,b]c+b[a,c]. Finally specialize.