# 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. – user1620696 May 14 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. – Jahan Claes May 14 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. – Cosmas Zachos May 14 at 18:31
• How do we know that they create paricles? – Žarko Tomičić May 14 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. – Jerry Schirmer May 14 at 20:43