# Can creation and annihilation operators be used in interaction case?

1. Jul 14, 2011

### Nixom

In the free case,we decomposite the free Hamiltonian into the creation and annihilation operators, i just wonder why this ad hoc method can not be used to the interaction theory?

2. Jul 15, 2011

### tom.stoer

What you are simply doing is to re-write field operators in position space via a Fourier transformation. You can then reformulate all derived operators (Hamiltonian, interaction, charge, Poincare generators, ...) in terms of creation and annihilation operators. You can (at least in simple cases) regularize these operators.

3. Jul 28, 2011

### Nixom

OK,if I originally take the a+ and a- as the Fourier coefficients, can they be interpreted as creation and annihilation operators definitely, or it is just depend on some specific form in Hamiltonion, e.g. the quadratic form?

4. Jul 28, 2011

### blechman

You can (and we do!) expand fields out in terms of "creation" and "annihilation" operators. We do this even in ordinary SHO quantum mechanics. Consider the Hamiltonian

$$H=\frac{p^2}{2m}+\frac{1}{2}m\omega^2x^2+\lambda x^4$$

We still expand $x\propto(a+a^\dagger)$ and expand out the anharmonic term. We can then apply perturbation theory to get answers as long as $\lambda$ is not too large.

However, doing this will only work in perturbation theory. This is fine for physics.

This generalized almost exactly for field theory.

5. Jul 28, 2011

### tom.stoer

The interpretation as creation and annihilation operators works simply b/c of their (purely algebraic) commutation relations; it has nothing to do with a specific interaction term. The quantization is non-perturbative, that means perturbation theory is possible but not mandatory.

The problem when you not use perturbation theory is that the creation operators do not create physical particles. Consider QCD: You can write down a physical, non-perturbative QCD Hamiltonian with creation and annihilation operators for quarks and gluons. The problem is that these operators create "free" = colored plane wave particles, but of course we know that these particles do not exist in nature. Instead the quarks and gluons do only exist in confined, color-neutral tsingulet states. That means that a one-particle Fock state (one quark) is physically nonsense (this problem is hidden when you do perturbation theory). What you would need instead are creation and annihilation operators for color-neutral pions, nucleons etc. Unfortunately nobody is able to write down a mathematical expression for these physical particles. This is not a problem for the definition of QCD, but "only" for the solution :-)

6. Jul 29, 2011

### bobloblaw

There are somewhat analogous operators in the interacting theory
They are called in and out operators and they create "free" particles in the interacting theory that are eigenstates of the full hamiltonian. Look in Weinberg vol. 1

7. Jul 29, 2011

### tom.stoer

Yes, but they are formal objects which cannot be constructed explicitly.

8. Aug 4, 2011

### Nixom

If I expand $$x\propto(a+a^\dagger)$$ in the Hamiltonian $$H=\frac{p^2}{2m}+\frac{1}{2}m\omega^2x^2+\lambda x^4$$, I can't get the results such as $$[H,a^\dagger]=\omega a^\dagger$$ or $$[H,a]=-\omega a$$, so how can I interpret them as creation and annihilation operators?

The commutation relations of these operators derived from the algebraic relations of the field(coordinate) and the conjugate momentum, so dose it have some relationship with the symplectic constructure of the Hamiltonian?

How do we interpret the in and out state? I can't understand that since they are the eigenstate of the full Hamiltonian, why they are used to describe the "free" particles?

9. Aug 5, 2011

### bobloblaw

"How do we interpret the in and out state? I can't understand that since they are the eigenstate of the full Hamiltonian, why they are used to describe the "free" particles?"

Saying they are not interacting was misleading. Rather in and out states appear to be eigenstates of the free hamiltonian to observers sitting at t= plus or minus infinity. To observers not sitting at infinity they will appear as interacting states. The formula relating free particle states to in and out states is called the lippmann-schwinger equation: http://en.wikipedia.org/wiki/Lippmann–Schwinger_equation

Wikipedia has an ok discussion of in and out operators in reference to the LSZ reduction formula: http://en.wikipedia.org/wiki/LSZ_reduction_formula

The LSZ reduction formula article talks about operators in position space whereas the lippmann-schwinger equation deals with operators in momentum space.

10. Aug 5, 2011

### bobloblaw

"I can't get the results such as
[H,a†]=ωa†
or
[H,a]=−ωa
, so how can I interpret them as creation and annihilation operators?"

I think part of the answer to this question is this: The creation and annhilation operators create multi-particle states in both the free and the interacting case. In the interacting case the creation operators do NOT create eigenstates of the Hamiltonian. So that is why you can't derive

[H,a†]=ωa†
or
[H,a]=−ωa

This is actually a good thing because if multi-particle states were energy eigenstates then multi-particle states would be orthogonal to each other and we could not get any multi-particle -> multi-particle transitions, where the particle number is different. Ie particle number would be conserved which we know is not the case in reality.

The in and out states i spoke of only look like they have definite particle content to people sitting at infinity. To every other observer they will have components along other multiparticle states.

11. Aug 19, 2011

### inempty

In interaction picture, we can actually do this. However, the creation/annhilation operators are different from ones in the free-particle-case. In other words, if we act a interacion-picture creation operator on a vacuum in-state, we couldn't get a one-particle in-state, Which makes the external line factors of Feynman diagrams be not usual ones. This is a reason of field operator renormalization.

12. Aug 20, 2011

### K^2

Interaction Lagrangians in QFT are written in terms of creation and annihilation operators for the fields involved in the interaction. So yeah, that's exactly what we do.

13. Aug 20, 2011

### inempty

I think it should be interaction Hamiltonians instead of Lagrangians, since fields in Lagrangians aren't field operators and we can only write field operators in terms of creation and annihilation operators. In addition, to remember the difference between creation/annihilation operators in in/out field operators (Heisenberg picture) and them in field operators of interaction picture is important.