# Interacting QFT - construction of states

#### paweld

Let's consider self-interacting scalar field described by action:
$$S = \int d^4 x \frac{1}{2} \left[ \partial_\mu \phi \partial^\mu \phi - m^2 \phi^2 - \frac{\lambda}{4!}\phi^4\right]$$
Using relation between Poisson bracket in classical mechanics and
comutators of quantum operators we have:
$$[\partial_0\hat{\phi}(t,x_1),\hat{\phi}(t,x_2)] = - i \delta(x_1-x_2)$$
Unfortunatelly in case of interacting field we cannot decompose field in terms
of basic solution (modes) and we don't have operators which create and anihilate
modes. So in case of interaction field construction of Hilbert space on which
$$\hat{\phi}$$ acts in not straightforward. What is the Hilbert space
on which field operator acts and how the action of it on some basic states
looks like.

Some people claim that in case of interaction we have to assume that the spectrum
of full hamiltonian is the same as spectrum of hamiltonian without interaction
(so that the theory poses particle interpretation). However even if this assumption
is correct we still don't know how $$\hat{\phi}$$ on these states, and whole
theory is underdetermined.

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#### TriTertButoxy

Just because you don't know how $\hat\phi$ act on interacting states doesn't mean your whole theory is undermined.

Here is one possible solution to your dilemma:
You can still use the normal mode decomposition of the scalar field,
$$\hat\phi(x)=\int\frac{d^3\mathbf{p}}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{\mathbf{p}}}} (\hat{a}_{\mathbf{p}}e^{-ipx}+\hat{a}^\dag_{\mathbf{p}}e^{ipx})$$,​
but now, the single-particle states are no longer created by $\hat{a}^\dag_\textbf{p}$. Instead, the physical, one particle state in the presence of interactions may be written in terms of Fock states as follows
$$|\mathbf{p}\rangle = c_0 a^\dag_{\mathbf{p}}|0\rangle + c_1 a^\dag_{\mathbf{p}/2} a^\dag_{\mathbf{p}/2}|0\rangle + \ldots$$​
where there are an infinity of terms in the Fock expansion and the coefficients are, in principle, calculable in field theory. Note that you know how $\hat{\phi}$ acts on each term individually. Your problem should be solved.

#### paweld

Thanks for answer. I'm afraid that your decomposition of $$\phi$$ does not fulfill
the Euler-Lagrange equation derived form my action:
$$\square \phi + m^2 \phi = \frac{\lambda}{3!}\phi^3$$
(or probably the momentum in exponens is not on shell).
The decomposition of state in terms of Fock states looks interesting.
It means that the state of momentum p in case of interacting theory
needs to be a superposition of multiparticles states.

#### A. Neumaier

Let's consider self-interacting scalar field described by action:
$$S = \int d^4 x \frac{1}{2} \left[ \partial_\mu \phi \partial^\mu \phi - m^2 \phi^2 - \frac{\lambda}{4!}\phi^4\right]$$
Unfortunatelly in case of interacting field we cannot decompose field in terms
of basic solution (modes) and we don't have operators which create and anihilate
modes. So in case of interaction field construction of Hilbert space on which
$$\hat{\phi}$$ acts in not straightforward. What is the Hilbert space
on which field operator acts and how the action of it on some basic states
looks like.
In 4-dimensional space-time, this is one of the unsolved problems. In lower dimensions it is solved bot the answer cannot be explained in few words. For 2 dimensions, see, e.g.,

James Glimm and Arthur Jaffe
The λ(φ4)2 quantum field theory without cutoffs III
Acta Mathematica 125, 1970, 203-267

#### TriTertButoxy

Ok, I made a slight typo: the mode expansion should have been in the Schrodinger picture (not the Heisenberg as I have written it)

$$\hat\phi(\mathbf{x})=\int\frac{d^3\mathbf{p}}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{\mathbf{p}}}} (\hat{a}_{\mathbf{p}}e^{i\mathbf{p}\mathbf{x}}+\hat{a}^\dag_{\mathbf{p}}e^{-i\mathbf{p}\mathbf{x}})$$

The time evolution of the field operator will in general be complicated -- again, this necessitates an expansion similar to the one I described for the physical particle.

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