What do the commutation/anti-commutation relations mean in QFT?

[moss]

Hello everybody,
In all QFT courses one starts very early with commutation and anti-commutation relation. My main question is why do we do this and what is the motivation?

I have already asked few people including our professor but could not get a clear answer.
I am talking about the commutation relations of the scaler fields and the creation/annihilation operators.

My next question is that what does the negative sign means in anti-commutation relations? OR what do anti-commuting fields means?

Then my last Qs. is, what is the form of creation and annihilation operator in QFT? In quantum mechanics harmonic oscillator we write them in terms of x and p but in QFT?

many thanks in advance for any help or hint.

PS: I have read the previous posts on this topic but I want a concise answer please.

 

[DarMM]

In all QFT courses one starts very early with commutation and anti-commutation relation. My main question is why do we do this and what is the motivation?

What is unclear about commutation and anti-commutation in QFT since it already appears in QM? Just to help form an answer.

 

[moss]

For example, Peskin&Schroeder says in 2.3 that field quantities goes to operators and suitable commutation relation are imposed, so what is suitable? and are there any other motives than just to quantize the theory?

 

[DarMM]

For example, Peskin&Schroeder says in 2.3 that field quantities goes to operators and suitable commutation relation are imposed, so what is suitable? and are there any other motives than just to quantize the theory?

What I more mean is that similar relations are imposed on $q$ and $p$ (position and momentum) in quantum mechanics. Do you have a similar question with why we do this in the non-relativistic QM case?

 

[moss]

Of course, like in Sakurai where he said that we impose commutation relations on compatible observable but what is compatible in QFT? and why commutation relation, why not anti-commutation relation?
Further 2 operators can commute maybe due to some missing info, redundant dof?
My knowledge is not as deep as yours and that is why i need help to understand this.

 

[meopemuk]

I prefer the explanation given in Weinberg's book

S. Weinberg, The Quantum Theory of Fields, Vol. 1, (University Press, Cambridge, 1995).

His basic idea is particle-centric rather than the usual field-centric approach:

Traditionally in quantum field theory one begins with such field equations, or with the Lagrangian from which they are derived, and then uses them to derive the expansion of the fields in terms of one-particle annihilation and creation operators. In the approach followed here we start with the particles, and derive the fields according to the dictates of Lorentz invariance, with the field equations arising almost incidentally as a byproduct of this construction.

Basically, Weinberg wants to build a relativistic interacting Hamiltonian (and interacting boost operator) in the Fock space occupied by electrons, positrons, photons, etc. This construction is both necessary and sufficient for having a complete quantum relativistic theory. This is a technical problem and Weinberg solves it by a series of technical steps:

1. Define (free) quantum fields as certain linear combinations of particle creation and annihilation operators and satisfying postulated (anti)commutator relations. Weinberg explicitly refuses to assign any physical meaning to these (anti)commutators:

The point of view taken here is that Eq. (5.1.32) is needed for the Lorentz invariance of the S-matrix, without any ancillary assumptions about measurability or causality.

2. Build Hamiltonian density as a polynomial in these fields.

3. Define interacting Poincare generators of time translations (H) and boosts ($\boldsymbol{K}$) as space integrals of the Hamiltonian density.

4. Prove that the above generators together with trivial generators of space translations ($\boldsymbol{P}$) and rotations ($\boldsymbol{J}$) satisfy all commutators of the Poincare Lie algebra. This means that the theory is relativistically invariant.

5. Insert the above Hamiltonian H in the perturbation formula for the S-matrix, perform renormalization, and check experimental predictions of QFT.

The beauty of this approach is that it is based on the familiar and transparent idea of interacting particles. Quantum fields and their (anti)commutators are introduced only as formal technical devices. Moreover, you don't even have to worry about interacting quantum fields. They are not present in this picture.

Eugene.

 

[Demystifier]

In all QFT courses one starts very early with commutation and anti-commutation relation. My main question is why do we do this and what is the motivation?

A possible, in a sense more physical, way to understand this is the following. In nonrelativistic QM of many identical particles you know that the wave function is either symmetric or antisymmetric under the exchange of all particles. But to write the (anti)symmetrized wave function explicitly is a bit clumsy. To avoid that clumsiness, there is a mathematical trick: one introduces the creation operators that act on vacuum, such that action with N operators creates the N-particle state. Requiring that these states should be (anti)symmetric, one finds that the operators must satisfy certain (anti)commutation relations. For more details see e.g.
http://de.arxiv.org/abs/quant-ph/0609163 Secs. 8.1 and 8.2.

 

[A. Neumaier]

Of course, like in Sakurai where he said that we impose commutation relations on compatible observable but what is compatible in QFT? and why commutation relation, why not anti-commutation relation?

The commutation or anticommutation relations are dictated by the spin-statistics theorem, or equivalently the need for cluster decomposition (i.e., that objects far away hardly interact). See Chapter 4 in Weinberg's book 'The quantum theory of fields, Vol. 1).

The beauty of this approach is that it is based on the familiar and transparent idea of interacting particles.

But he only considers noninteracting in/out particles and scattering theory (i.e., no finite time dynamics) - see Chapter 3.2 where he considers the asymptotic Fock space, and he does so to motivate the topic of his book: The quantum theory of fields!

In Section 4.1 (p.169 bottom) he explicitly says that according to his experience, relativistic many-particle theories ''have always run into trouble in sectors with more than two particles: either the three-particle S-matrix is not Lorentz invariant, or else it violates the cluster decomposition principle''. Your dressed particle approach seems to suffer from the first defect. Did you ever compute the 3-particle S-matrix in your approach and check its Lorentz invariance? Did you check the validity of the cluster decomposition principle for your version of QED?

 

[meopemuk]

For more details see e.g. http://de.arxiv.org/abs/quant-ph/0609163 Secs. 8.1 and 8.2.

Thank you, @Demystifier, for mentioning this.

We have to distinguish two kinds of (anti)commutation relations in QFT. One kind is the (anti)commutation relations for particle creation $a^{\dagger}_p$ and annihilation $a_p$ operators. They are usually written in the momentum representation, e.g.

$[a_p,a^{\dagger}_p]_{\pm} = \delta(p-p')$

These operators have purely technical meaning (they formally add/subtract one particle to/from a general multiparticle Fock state). There is no physics associated with creation/annihilation operators. They are just convenient tools, a useful notation. One can easily check that by definition these operators have to satisfy the above (anti)commutators for (fermions)bosons. No controversy.

The controversy begins when one introduces "quantum fields" $\psi(x,t)$ and "field momenta" $\pi(x,t)$. One school of thought claims that these are quantum representatives of certain all-penetrating physical substances -- fields of photons, electrons, quarks, etc. These fields interact with each other and obey certain field equations. Everything we experience in this world is the result of mutual interactions of these fields. The fields are postulated to satisfy specific (anti)commutation relations, e.g.

$[\psi(x,t),\pi(x',t)]_{\pm} = i \hbar \delta(x-x')$ (1)

The justification of this postulate was always a mystery to me.

In my previous post #6, I just tried to explain that there exists an alternative point of view, in which quantum fields are not regarded as fundamental ingredients of matter. They are regarded instead as formal technical objects (similar to creation-annihilation operators in multiparticle Fock spaces), which are useful building blocks for constructing relativistically covariant interaction operators in the Fock space. Then (anti)commutators (1) are just parts of this formal definition a la Weinberg. No specific physical meaning.

Eugene.

 

[meopemuk]

But he only considers noninteracting in/out particles and scattering theory (i.e., no finite time dynamics) - see Chapter 3.2 where he considers the asymptotic Fock space, and he does so to motivate the topic of his book: The quantum theory of fields!

In section 4.4 Weinberg writes the interacting Hamiltonian as a polynomial in particle creation and annihilation operators. Although he doesn't mention the Fock space by name, I understand his eq. (4.4.1) as an admission that interaction operators act in the Fock space.

In Section 4.1 (p.169 bottom) he explicitly says that according to his experience, relativistic many-particle theories ''have always run into trouble in sectors with more than two particles: either the three-particle S-matrix is not Lorentz invariant, or else it violates the cluster decomposition principle''. Your dressed particle approach seems to suffer from the first defect. Did you ever compute the 3-particle S-matrix in your approach and check its Lorentz invariance? Did you check the validity of the cluster decomposition principle for your version of QED?

10 Poincare generators of the dressed particle approach are obtained by a unitary transformation of the traditional QED generators. So, all commutators between the generators are preserved by this transformation, and the relativistic invariance of QED implies the relativistic invariance of the dressed theory.

According to Weinberg's sections 4.3 - 4.4, a Fock space Hamiltonian is cluster separable if coefficients in front of products of creation and annihilation operators are smooth functions of particle momenta (apart from a single delta function that is needed to conserve the total momentum). When doing the dressing transformation in section 2.3 of volume 3 of my https://www.researchgate.net/publication/326264565_Elementary_Particle_Theory_Vol_3_Relativistic_Quantum_Dynamics, I take special care to ensure that the Hamiltonian of the dressed theory has smooth and finite coefficient functions, i.e., separable.

Eugene.

 

[A. Neumaier]

In section 4.4 Weinberg writes the interacting Hamiltonian as a polynomial in particle creation and annihilation operators. Although he doesn't mention the Fock space by name, I understand his eq. (4.4.1) as an admission that interaction operators act in the Fock space.

Yes, of course, but all this is before renormalization. The terms don't make sense as operators but only formally, hence the need for renormalization. Renormalization necessarily destroys the Fock space structure by Haag's theorem - the number operator cannot be renormalized.

10 Poincare generators of the dressed particle approach are obtained by a unitary transformation of the traditional QED generators. So, all commutators between the generators are preserved by this transformation, and the relativistic invariance of QED implies the relativistic invariance of the dressed theory.

Your unitary transformations proceed order by order and have errors at each order. This invalidates the claimed equivalence.

 

[vanhees71]

Haag's theorem is not too much to worry about. Just put everything in a finite volume with periodic boundary conditions, and all is fine. Of course, mathmaticians have to worry about the mathematical details of how to take the infinite-volume limit. AFAIK nobody has found a rigorous formulation for QFTs in (1+3) dimensions which are applicable to what's observed in nature. The Standard Model of course works well for physicists, but it's not well-defined for mathematicians.

A very nice book, complementary to Weinberg's QT of Fields, dealing with these questions in a pragmatic physicist's way is

A. Duncan, The conceptual framework of quantum field
theory, Oxford University Press, Oxford (2012).

 

[A. Neumaier]

Haag's theorem is not too much to worry about. Just put everything in a finite volume with periodic boundary conditions, and all is fine.

The problem with this Procrustean solution is that it destroys Poincare invariance. Restoring it by taking the infinite volume limit brings back the issues in Haag's theorem.

 

[meopemuk]

Yes, of course, but all this is before renormalization. The terms don't make sense as operators but only formally, hence the need for renormalization. Renormalization necessarily destroys the Fock space structure by Haag's theorem - the number operator cannot be renormalized.

There could be problems with renormalization and Haag's theorem in the field-based theory, but everything is fine in the dressed theory. The trick is to fix our Fock space structure from the beginning and never allow it to change. So, we define our electrons, protons, photons as physical particles with measurable masses and charges, and these definitions remain valid throughout the renormalization and dressing procedures.

The renormalization is performed by adding certain (cutoff-dependent) counterterms to the Hamiltonian. This results in a good S-matrix comparable with experiment, as described in every QFT textbook. However, at the end of the renormalization procedure the Hamiltonian is not acceptable. First, it is cutoff-dependent and diverges in the limit of the lifted cutoff. Second, the vacuum and one-particle Fock states are not eigenstates of this Hamiltonian. This is what Haag's theorem complained about.

This set of problems is fixed by dressing. Again, we do not touch the definitions of particles. The dressing is performed by a unitary transformation of the Hamiltonian (and the boost operator). This transformation is designed so that the transformed interaction operator does not act on the vacuum and one-particle states. So, the Fock vacuum state is a true eigenstate of the dressed Hamiltonian with eigenvalue zero. Haag should be happy about that. This also means that the theory avoids having self-interactions, which are the true sources of QFT divergences. At the end we have a finite cutoff-independent Hamiltonian, whose S-matrix agrees with experiment and which can be used directly as a generator of time translations.

Of course, what I described above is an ideal construction. In practice, we have to do this order-by-order in perturbation theory. Moreover, instead of doing a honest unitary transformation, we often simply fit interaction terms in the dressed Hamiltonian to Feynman loop diagrams calculated from the field theory. No theory is perfect. Everyone has to use approximations, perturbation theory, etc. But it is OK as long as the theoretical foundation is solid.

Eugene.

 

[A. Neumaier]

everything is fine in the dressed theory. The trick is to fix our Fock space structure from the beginning and never allow it to change. So, we define our electrons, protons, photons as physical particles with measurable masses and charges

The problem is already there. Fixing this the way you do it produces a 2-point function for the electron inconsistent with standard QED, as mentioned in my post #39 in another thread.

 

[vanhees71]

The problem with this Procrustean solution is that it destroys Poincare invariance. Restoring it by taking the infinite volume limit brings back the issues in Haag's theorem.

Why then works the "robust math" of practitioners (partially to 12 or more digits of precision)?

 

[A. Neumaier]

Why then works the "robust math" of practitioners (partially to 12 or more digits of precision)?

Because these calculations are not done with the Procrustean method you suggested. To my knowledge nobody ever used periodic boundary conditions to do any sort of high precision calculations in QED.

 

[vanhees71]

The periodic boundary conditions and the finite-volume regularization is just envoked once to derive the relation of the S-matrix to observable cross sections ("Fermi's trick"). You can even forget about it there and argue with wave packets in the initial state (see, e.g., Peskin and Schroeder). Everything else is calculated with the usual Feynman rules leading to manifestly covariant S-matrix elements and then you square them.

You need the finite quantization volume also to argue about thermodynamics like the derivation of the black-body spectrum in thermal field theory, and the "thermodynamic limit" can also get tricky like in the case of charged bosons to get the correct Bose-Einstein condensate at low temperatures. Also there there are other means to deal with the mathematically ill-defined expressions like the Schwinger proper time, heat-kernel, or $\zeta$-function regularization methods.

I think it's indeed a non-trivial question, how to deal with this more rigorously, but at the end it boils down to the naive results physicists get for almost 80 years using perturbative QFT.

 

[A. Neumaier]

I think it's indeed a non-trivial question, how to deal with this more rigorously, but at the end it boils down to the naive results physicists get for almost 80 years using perturbative QFT.

These results are not naive (where one just gets infinities) but need substantial dirty trickery to get the final good, renormalized and RG-corrected results. Haag's theorem is just a rigorous expression of this mess...