Why is the second quantization formulation important in quantum field theory?

  • Thread starter Thread starter thegreenlaser
  • Start date Start date
  • Tags Tags
    Field Quantum
thegreenlaser
Messages
524
Reaction score
16
I'm just starting to dip my toes into quantum field theory using Klauber's "Student friendly quantum field theory." I just got through the section on free spin-0 waves and it mostly makes sense: we solve the Klein-Gordon equation to get an operator which is a function of space and time, i.e., a quantum field. The math steps make sense to me. What's confusing me is, why should I care about this quantum field we just worked so hard to find? How is it connected to reality? In non-relativistic QM, the wavefunction we solve for has a very definite meaning in terms of experiment: it gives the probability of different outcomes. But this quantum field is made up of raising and lowering operators, and it doesn't even need to be Hermitian. Is it observable, or is it some sort of handy mathematical tool, or what?
 
Physics news on Phys.org
The quantum fields correspond to particles. In general the reason that you work with them, is because you want to get results from interactions. The interactions happen in the scheme of the quantum fields, and the particles that are observable correspond to the asymptotic states.
From those field interactions you can calculate the cross sections for example, which is the observable.
 
ChrisVer said:
The quantum fields correspond to particles. In general the reason that you work with them, is because you want to get results from interactions. The interactions happen in the scheme of the quantum fields, and the particles that are observable correspond to the asymptotic states.
From those field interactions you can calculate the cross sections for example, which is the observable.
I'm having trouble understanding what you mean; as I said, I'm very new to this. Are you saying that quantum fields are just a tool for calculating the real observables? Perhaps I'll just have to be patient until I get to the section on interactions.
 
Yes they are. Nobody ever sees the quantum fields, nor the interactions. What they see are the asymptotic states [which are the physical particles] and what someone measures are the cross-sections for the interactions.
QFT can give you the amplitudes via with you can measure the cross sections.

Without going into the scheme of calculating cross sections, everything you are doing is just mathematics and you can figure out good tools to give you the answer.
 
  • Like
Likes thegreenlaser
Before you study relativistic quantum-field theory, have a brief look in a good non-relativistic quantum-mechanics textbook on the topic of identical particles and then the socalled "second-quantization formulation", which is a misnomer, because there is only one non-relativistic quantum theory, and thus at most one "quantization", but the historical name stuck.

The "second quantization formulation" is nothing else than quantum-field theory for non-relativistic particles. You'll see that it is completely equivalent to the many-body Schrödinger wavefunction formalism but much more convenient, because with it's creation and annihilation operators it automatically keeps track of the symmetrization/anti-symmetrization of the many-body states, describing bosons and fermions, respectively (in the latter case implementing also the Pauli exclusion principle).

More importantly, the quantum-field formulation of many-body quantum mechanics makes it very easy to extent quantum theory to processes, where particles can be destroyed and created, i.e., when the particle number is not conserved. In non-relativistic quantum theory this necessity only occurs when you use the quasi-particle method to describe collective modes of many-body systems like phonons (i.e., the quantized vibrations of crystal lattices and the collision of electrons with those, etc.).

Now, if you try to make sense of relativistic wave mechanics, you'll find out that the Schrödinger one-particle or fixed-number-of-particle formalism doesn't work for various formal reasons, usually treated in modern introductions of realtivistic quantum theory (don't waste your time with old-fashioned books, where they make the attempt to go through relativistic quantum mechanics, where this failing concept of a "first-quantization formulation" of relativistic quantum theory is followed). Nowadays the argument, why relativisitic quantum theory is most easily formulated in terms of quantum field theory from the very beginning is the observation that as soon as relativistic collision energies are involved (i.e., center-mass kinetic energies that are large compared to the typical mass scale of the lightest particles involved, like in the most simple version of QED, where you have electrons, positrons, and photons as field degrees of freedom), new particles can be created (as long as the conservation laws of energy, momentum and conserved charges are obeyed) and others destroyed. That's exactly what quantum-field theory makes easy to describe. As soon as you learn perturbation theory and Feynman diagrams with some exercise this will occur quite natural to you, the Feynman diagrams being quite intuitive in depicting what's going on in what's called "elementary reactions". This has to be taken, however with a large grain of "quantum salt". Strictly speaking Feynman diagrams are a very clever way to write down the equations for transition-probability rates (S-matrix elements) used to calculate scattering cross sections within perturbation theory!
 
  • Like
Likes thegreenlaser
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
Back
Top