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?
-
5$\begingroup$ Possible duplicate of Interacting Fields in QFT $\endgroup$– AccidentalFourierTransformCommented Jul 17, 2017 at 20:39
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 $\phi$.
-
$\begingroup$ Thank you for your reply. I understand the procedure he uses in the case of a free field. But (assuming that $t_0$ is not the moment where the perturbation is just started, as suggested above) the ladder operator should not be the same in the interaction case as in the free case i.e. in the interaction case you can still expand $\phi$ in momentum space, but the ladder operator would not be the same as in the free case (which is what Peskin implies if I understood it right). $\endgroup$– SilviuCommented Jul 9, 2017 at 18:36
-
$\begingroup$ Obviously, in my answer the ladder operators are different from the free field, because in free field case $a(t)\sim e^{-iEt}a$. and $E^2-\vec p^2=0$ is only used here. $\endgroup$– XiaoaiXCommented Jul 9, 2017 at 23:45
-
$\begingroup$ With interactions turned on, you do not know the energy eigenfunction like in Klein-Gordon, but you still know the momentum eigenfunctions and can expand any wave function in terms of momentum eigenfunctions. $\endgroup$– XiaoaiXCommented Jul 10, 2017 at 0:03
-
$\begingroup$ So what is the meaning of the ladder operators in the interaction picture? Do they still create a particle of momentum p and energy $E_p$? $\endgroup$– SilviuCommented Jul 10, 2017 at 7:02
-
$\begingroup$ What I don't understand is that when he does the derivation in the free case, he gets this equation in momentum space: $[\frac{\partial^2}{\partial t^2}+(|p|^2+m^2)]\phi(p,t)=0$, which resemble a harmonic oscillator and this is how he gets to ladder operator. However in the interaction case that =0 is replaced with =$\frac{\lambda}{3!}\phi^3$, so I am not sure how can you write the function in the same way as before (even with different coefficients) when you don't have a harmonic oscillator anymore $\endgroup$– SilviuCommented Jul 10, 2017 at 7:09
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...
-
2$\begingroup$ Weigand's statement is about expanding the field as a function of time, while this question is about expanding it at a fixed time. There is no contradiction, both are correct. The point is that for a free field the expansion is valid at every time with time-independent $a_p$, which is not the case for an interacting field. So you'd get some $a_p(t)$, which are not really all that useful since they have no clear interpretation. $\endgroup$– ACuriousMind ♦Commented Jul 19, 2017 at 10:07
-
$\begingroup$ How is $a_{p}(t)$ defined? I assume that $a^{\dagger}_{p}(t)$ is defined as creating a particle with momentum $p$ out of the perturbed vacuum $|\Omega>$. But this is not possible because the interacting field does not satisfy that the free EOM. physics.stackexchange.com/questions/274921/… $\endgroup$ Commented Jul 20, 2017 at 10:57
-
1$\begingroup$ It's simply a Fourier coefficient, as explained in this answer. It has no direct meaning of creating a "particle", which is why it's not as useful as the free field expansion. $\endgroup$– ACuriousMind ♦Commented Jul 20, 2017 at 11:02
-
$\begingroup$ What do you think of the answer given by Valter Moretti? $\endgroup$ Commented Jul 20, 2017 at 11:07
-
1$\begingroup$ It is certainly correct, but has nothing to do with the answer to this question - note that this question is simply about the possibility of the expansion, and does not attribute any particular properties to the $a_p(t)$. If you look at the corresponding derivation in P&S, you'll note that they proceed to do the usual interaction picture shenanigans to make the $a_p(t_0)$ at a fixed time into the modes of a free field, etc. There are certainly technical problems (Haag's theorem) with this from a rigorous mathematical standpoint, but, again, that is not what this question asks about. $\endgroup$– ACuriousMind ♦Commented Jul 20, 2017 at 11:14
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.