In a multiple particle Hilbert space (any space of any multi-particle system), is it sufficient to define creation and annihilation operators by their action (e.g. mapping an n-particle state to an n+1-particle state) or does one have to do anything else, like "proving existence". Intuitively, one can apriori say it exists if you can write it down, but I donno.


3 Answers 3


If the action is given, the existence is obvious (unless the definition is faulty). But one would usually want to verify that all creation operators commute, all annihilation operators commute, and the commutator of a creation operator and an annihilation operator is a c-number.

Moreover, for a field theory, one would usually want to verify that this c-number transforms covariantly under Poincare transformations, and vanishes at spacelike separation. This applies for free field theories. For interactive field theories, there are no intrinsic c/a operators, as their properties are destroyed by renormalization. But one can associate (according to Haag--Ruelle theory) to each bound state a family of creation and annihilation operators parameterized by momentum, describing the free asymptotic motion in a scattering process.


Providing the action of an operator on the vectors of a basis is exactly the same as "writing it down" since it extends to all other vectors by linearity. This is the same as writing the matrix of a linear operator between vector spaces. Thus creation and annihilation operators automatically exist when they are defined.

  • $\begingroup$ This is true, modulo infrared problems as in QED or more sever infrared problems such as in QCD. The demonstration is simply from the noninteracting nature of plane waves. $\endgroup$
    – Ron Maimon
    Aug 5, 2012 at 19:52
  • $\begingroup$ This is true in the fermionic case where the operators are bounded but not in the bosonic case. On rather defines the operators on vectors with finite particle number which are just dense in the Hilbert (Fock) space. I refer to the second volume of the books of Reed, Simon. $\endgroup$
    – Marcel
    Aug 11, 2012 at 14:32
  • $\begingroup$ Yes, that is correct, and it is fundamentally due to bosonic operators being unbounded - just like position and momentum in the SHO. Whatever mathematical difficulties one encounters for boson creation and annihilation has an exact counterpart in the SHO formalism. $\endgroup$ Aug 11, 2012 at 23:20

For a theory with a mass gap, it is sufficient to define the creation and annihilation operators by their action, but these operators will not have local properties in general, they won't be the Fourier transform of fields that obey microcausality or come from a local Lagrangian.

When particles are spread out over all space, they have no interaction. Their S-matrix element is

$$ S = I + i A $$

Where I is the identity, so has delta-functions making the incoming k's equal to the outgoing k's, while A has only one overall delta function for energy momentum conservation (A might be the identity on some of the particles and not the identity on others, but this is still fewer delta functions). This means that two infinite plane waves have delta-function scattering in the forward direction, but only a smooth distribution in the off-forward direction. The interpretation is just that as you make the beam of particles more tenuous, the number of collisions goes to zero.

This argument fails to a certain extent in infrared divergent theories (like quantum electrodynamics), because you need to make a nonlocal field with every charged particle. In this case, you need to prove existence, since it isn't clear that the same Hilbert space includes both a zero electron state and a one electron state with an infinite range field (although the zero electron state must contain electron-positron states which are net neutral, and so have a shorter range field).

But in any relativistic field theory with a mass gap, if you have a vacuum and a one particle state, you can define creation and annihilation operators which make n-particle (necessarily noninteracting) states. Then you can define a quantum field from these operators. This quantum field will not have local interactions in general, since the particles can be H-atoms, or pions, they don't have to be pointlike photons.

The issue with constructing quantum field theories is to make sure that the field for the elementary particles is local, and this sometimes requires different degrees of freedom, like quarks, which are not asymptotic. In this case, you can't define a quark creation operator, because the quark's infinite range strong field is an infinite mass string, and is definitely outside the Hilbert space of the theory.


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.