I am reading Peskin's book on QFT and I reached a part (in chapter 4) where he is analyzing the two-point correlation function for $\phi^4$ theory. At a point he wants to find the evolution in time of $\phi$, under this Hamiltonian (which is basically the Klein-Gordon - $H_0$ - one plus the interaction one). Anyway, when he begins his derivation he says that for a fixed time $t_0$ we can still expand $\phi$ in terms of ladder operators in the same way as we did in the free (non-interaction) case (this is on page 83), i.e. $$\phi(t,\mathbf{x})=\int{\frac{d^3p}{(2\pi)^3}\frac{1}{E_\mathbf{p}}(a_\mathbf{p}e^{i\mathbf{x}\mathbf{p}}+a_\mathbf{p}^\dagger e^{-i\mathbf{x}\mathbf{p}})}.$$ I am not sure I understand why can we do this, for a fixed time. When we wrote this in term of ladder operators for the free case, we used the KG equation in the free case, which resembled to a harmonic oscillator in the momentum space, and hence we got the ladder operators. But now, the equation of motion is different (it has $\phi^3$ term, instead of 0, as before), so can someone explain to me why we can still use the same formula as before, even if the equation of motion is different?
3 Answers
Klein-Gordon equation only determines the dispersion relation between the energy and the momentum $(p^0)^2-(\vec p)^2=0$ for free scalar field. When the interaction exists in the theory, at any time, the field $\phi$ can be Fourier expanded to the momentum space, with the operators $a$ and $a^\dagger$ explicitly dependent on time $t$ without requiring the above free scalar dispersion relation. This is similar to the situation in the quantum mechanics: any wave function can be expanded using momentum eigenstates, with time-dependent coefficients. The point here is that $a(t)$ and $a^\dagger(t)$ also satisfy the commutation relation at any time $[a_{\vec p}(t),a^\dagger_{\vec p'}(t)]\sim \delta^3(\vec p-\vec p')$ which follows from $[\dot \phi(\vec x ,t),\phi(\vec x',t)]\sim \delta^3(\vec x-\vec x')$, if the interaction involves no derivative of $\phi$.
- 151
I am now very sure that Peskin made a mistake here. Please check the lecture Notes by Weigand
on page 43, where it is said clearly that The crucial difference to the free theory is, though, that $\phi(x)$ cannot simply be written as a superposition of its Fourier amplitudes $a(\vec{p})$ and $a^{\dagger}(\vec{p})$ because it does not obey the free equation of motion...
- 1,110
I think the question is related with the answer given by Valter Moretti from
Clearly, you cannot say that the field $\phi(x)$ can be decomposed as creating and annihilation part.
- 1,110