Why is it so that because particles can be destroyed and recreated we introduce QFT? I read at the begining of some textbook that this is so. My main problem is not the rest of the book but the first motivation for introducing QFT for modeling. My thinking...till now every quantum operator id est observable was attached to a particle in question but when you have variable number of particles you can not do this so you imagine that there is a more fundamental thing which we observe and one observable is also the number of particles. Also, somewhere I read that because we are now in relativistic regime we have to define observables that are spacelike separated to commute. And because of that also we have to define observables as functions of spacetime points.I dont see it.

  • 1
    $\begingroup$ The very first chapter of Peskin & Schroeder has an extremely clear summary of this. I might write an answer later when I have time. $\endgroup$
    – SuperCiocia
    Sep 25, 2019 at 18:22
  • 2
    $\begingroup$ "My main problem is not the rest of the book" Does this mean that you have read and understood the entire textbook? $\endgroup$
    – my2cts
    Sep 26, 2019 at 15:24
  • 1
    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/415175/2451 and links therein. $\endgroup$
    – Qmechanic
    Sep 26, 2019 at 15:50
  • 2
    $\begingroup$ As with any physical model, the real reason we are confident in it is that it works like a clock. QFT is exceptionally successful experimentally. Motivations are more like teasers or appetizers – they are usually given to spark your excitement about the subject before you dive into (arguably pretty convoluted) math. $\endgroup$ Sep 27, 2019 at 11:47
  • 4
    $\begingroup$ IMHO the best answer to this is given in the first five chapters of Weinberg's The Quantum Theory of Fields. Basically you want to construct interaction Hamiltonians which give rise to one Lorentz invariant $S$-matrix which further obeys the cluster decomposition principle (roughly the statement that experiments conducted far apart yield uncorrelated results). These two conditions imply constraints on the interaction Hamiltonian and Weinberg shows that the simplest way to construct such interactions is by employing quantum fields $\endgroup$
    – Gold
    Nov 11, 2020 at 22:12

2 Answers 2


To give a short answer, hoping it will be of some use to some other poor soul roaming the physics land. Lets look at it this way: In reinterpreting electrodynamics as a quantum theory we arrive at real relativistic quantum theory of fields. This is done from the obvious reason to explain the fact that EM radiation comes in quanta. This was known from black body radiation and from photoelectric effect. So there is no mystery in trying to quantize EM field theory. This gave good predictions and was a success. On the other hand, trying to make single fermionic particle wave equation relativistic turns out to be hard. All sorts of problems come out. So, it is not good. Also, we cant explain creation and destruction of the particles. SO, we would need some multiparticle theory. Knowing about EM fields we naturaly think of field theory. So we try to model these processes with quantum fields. Main connection here is, I think, quantization of EM field which assures us that idea of field quantization is good.


Quantum Mechanics is about mechanics, and Quantum Field Theory is about fields. Given that all the forces in nature are described by fields, this would mean that QFT is the more fundamental theory. In fact, we can describe QM as a zero dimensional QFT. Zero dimensional as particles are zero dimensional.

It turns out that QFT requires particle creation and annhilation, hence it's also called the many particle theory.

QFT is usually motivated in most textbooks as the unification of the relativity principle from Einsteinian mechanics and QM. Hence it is a partial unification of dream theory that physicists are busy searching for, that is a full quantisation of General Relativity and which would require quantising the metric.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.