Time dependence of ladder operators in QFT

Question

I'm currently going through Matthew D. Schwartz book Quantum Field Theory and the Standard Model, p. 23. For free (non interacting) field theories we are able to quantise the field by expanding our field operator as a Fourier transform of ladder operators for each mode, i.e.

$$\phi_0(x) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}}(a_pe^{-ip.x} + a_p{^\dagger}e^{ip.x}).\tag{2.78}$$

For our free theories this leads to the Hamiltonian

$$H_0 \propto \int d^3p \space \omega_p a_p^{\dagger}a_p $$ with $$[a_k,a^\dagger_p]=(2\pi)^3\delta^3(\vec{k}-\vec{p}).\tag{2.69}$$

This gives us a clear physical interpretation. The ladder operators, say $a^\dagger_p$, adds a 'quanta' to the mode $\omega_p$ in a similar way to the simple harmonic oscillator due to analogous commutation relationships between the ladder operators and the Hamiltonian. This I am happy with.

However the problem arises for me when trying to interpret the quantum field operator under a general interacting theory. In the Heisenberg picture the field operator must obey

$$i\partial_t\phi(x)=[\phi,H].\tag{2.80}$$

It is then said this can be solved if $a_p \rightarrow a_p(t)$ such that the interacting field operator is given by $$\phi(x) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}}[a_p(t)e^{-ip.x} + a_p{^\dagger}(t)e^{ip.x}].\tag{2.81}$$

My problem is the interpretation of $a^\dagger_p(t)$ as an operator that creates particles in the general interacting theory and how it says in Matthew's book that these time dependent ladder operators satisfy the same algebra as the free theory ones. In our free theory the commutator $[H_0,a^\dagger_p]=+\omega_pa^\dagger_p$ is what lead to the interpretation of the ladder operators adding or removing particles to the system, it is not clear to me that this would hold for $[H,a^\dagger_p(t)]=+\omega_p(t)a^\dagger_p(t)$ in the interacting theory.

Answer 1

"Satisfy the same algebra" means that the equal time commutation relations $$ [a_p(t), a^\dagger_{p'}(t)]= 2E_p (2\pi)^3 \delta^3(p-p') $$ continue to hold. The $a_p(t)$ are no longer ladder operators for the interacting-theory Hamiltonian though. There is no equation like $$ [H, a^\dagger_p(t)]= E_p a^\dagger_p(t). $$

The best that one can hope for is that $\phi$ couples the vacuum to a single-particle state, so that $$ \langle p |\phi(x)|{\rm vac} \rangle= \sqrt Z e^{-ip\cdot x} $$ The $\sqrt Z$ is there because $\phi$ can also connect the vacuum to many-particle states. If so, then $Z<1$. It is also possibe that there is no particle in the interacting system that has the quantum numbers of $\phi$. For example there are no charge +2/3 particles in the spectrum of QCD so the state resulting from acting on the vacuum with an up-quark field operator has no overlap with any QCD eigenstate, as is the result of acting with one of its constituent $a^\dagger$'s

Answer 2

QFT (quantum field theory) describes the scattering of incoming states $\vert i \rangle$ to outgoing states $\vert f \rangle$ in terms of asymptotic in- and out-states.

In the asymptotic past, $t \to −\infty$, the in-states $\vert i \rangle$ are described as distinct wave-packets corresponding to well-separated single particle states. Being far apart for $t \to −\infty$ they travel freely as individual states. For $t \to +\infty$ the out-states $\vert f \rangle$ are again asymptotically free and well-separated single particle states.

The in-state $\phi_{in}$ has $E = \sqrt{\vec p^2 + m^2}$ where $m$ is the 1-particle pole in the Feynman propagator of the full interacting theory. Therefore, $\phi_{in}$ is a free field obeying the free Klein-Gordon-equation, but with the full mass $m \ne m_0$, where $m_0$ is the pole of the free theory
$(\partial^2 + m^2) \phi_{in} = 0$

It is thus possible to expand $\phi_{in}$ in terms of $a_{in} (\vec p)$ and $a_{in}^\dagger (\vec p)$, respectively annihilation and creation operators.

In the Heisenberg picture
$\phi_{in} (x) = \int \frac{d^3p}{(2 \pi)^3} \frac{1}{\sqrt{2 E_p}} (a_{in} (\vec p) e^{-ipx} + a_{in}^\dagger (\vec p) e^{ipx})$

A similar expression holds for $\phi_{out} (x)$.