Why is free field called free ?

[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.