# Why do we need creation and annihilation operators in QFT?

2. Why do we need creation and annihilation operators?

Main point is that a particle can be created by creation operator and destroyed by the annihilation operator. But to destroy a particle is to change from one vector in Hilbert space to another, right? So these operators act on states and actually change them. In a definite way. So why do we need them? We can just write the states as we want...if I know I want a state with two particles of definite momenta I can just write it, why use these operators to create particles? Obviously, I am missing something here. What is the real purpose of creation and annihilation operators? Is it because through these operators, fields are represented better than through momentum or energy operators? So, combinations of creation and annihilation operators give us our observables?

• Give a look on Weinberg's "The Quantum Theory of Fields Vol. 1" and Duncan's "The Conceptual Framework of Quantum Field Theory" in what they both call "cluster decomposition principle". The basic idea they argue is that when you write down the interaction Hamiltonian in terms of creation and annihilation operators it becomes easier to find the condition for satisfying this cluster decomposition principle which, roughly speaking, means that experiments conducted far away have uncorrelated results, a property which you would like the dynamics of the theory to have.
– Gold
May 14, 2019 at 16:45
• I think you should read a little further. QFT is usually done in Heisenberg or interaction picture, so wiring down/manipulating states is less important than wiring down and manipulating operators. And creation/annihilation operators allow you to write down the operators you're interested in, and manipulating them is easy because they have simple commutation relations. May 14, 2019 at 16:45
• The most efficient accounting for oscillator eigenstates goes through creation and annihilation operators in a Fock space. Classical field theories are but an infinity of decoupled classical oscillators (normal modes). Quantizing those goes easiest through creation and annihilation operators. Your QFT text should detail that. May 14, 2019 at 18:31
• How do we know that they create paricles? May 14, 2019 at 18:41
• The action of the operator is the definition of the particle. The creation/annihilation operators have their appropriate commutation relations, and an associated number operator, and they form a basis for the space of states of the field, and so, they are used to define what a "particle state" is, BECAUSE they have the associated number operator. The field is the fundamental thing, not the particle. May 14, 2019 at 20:43

"Need them?" We don't need anything. They arise naturally. If you express the field equations in momentum space you get a set of harmonic oscillator equations and the algebra of that system applies to the field. These states form a basis for Hilbert space, but are not the only ones. Certainly one can use any number of orthonormal bases but the occupation number basis is the easiest one to use. The solution is easy to arrive at and easy to use and they map nicely to the processes we are trying to study. I wouldn't say we need them I would say the field equations have given them to us through the process of quantization.

A field theorist would start with a field equation for some A(x, t) as a scalar, vector, spinor, or tensor and then apply QM to it. I guarantee that if you do this for a "free field" you will arrive at a harmonic oscillator formalism. Particle theorists are more pragmatic. They tend to model processes using existing paradigms and then look for a field theory that fits that model. There is nothing wrong with this. this is how science works and is a pragmatic approach to particle phenomenology. If you are reading a text or paper by a particle phenomenologist they may start with the harmonic oscillator formalism and just start building up processes. This can look a little informal to a mathematician or pure theorist but it is just as valid or an approach as starting from a classical field theory and trying to quantize it.

• Tnx man. When i say need them of course we dont need them. So what quantum field does is it acts on hilbert spaces at space time points and makes it possible to conatruct observables? May 14, 2019 at 20:04
• The field operators act on Hilbert space states, yes. The "the field" as we observe it will not be either the operator of the state. The state describes the information content of the system and we measure things as expectation values <n|A|n> etc.
– user196418
May 14, 2019 at 21:56
• Don't confuse the three terms: (1) the classical field, (2) the operator, (3) the observed expectation value.
– user196418
May 14, 2019 at 21:57