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I Two point correlation function in Peskin book

  1. Jul 9, 2017 #1
    Hello! 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_0,\mathbf{x})=\int{\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_\mathbf{p}}}(a_\mathbf{p}e^{i\mathbf{p}\mathbf{x}}+a_\mathbf{p}^\dagger e^{-i\mathbf{p}\mathbf{x}})}##. 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? Thank you!
     
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  3. Jul 10, 2017 #2

    vanhees71

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    I don't understand this claim by Peskin&Schroeder either. Interacting fields in the Heisenberg picture cannot be defined in terms of creation and annihilation operators in this kind, and that makes it impossible to find a proper particle interpretation for states that are not asymptotically free.

    To derive the Feynman rules of perturbation theory the field operators, however, evolve according to the free Hamiltonian and that's why the propagator lines in these Feynman diagrams stand for the free not for the interacting propagator. The full propagator is calculated with help of the self energy (represented by one-particle irreducible diagrams with two (truncated) external legs). In terms of the self-energy ##\Pi(p)## the full propagator is given by Dyson's equation
    $$G(p)=\Delta(p)+\Delta(p) \Pi(p) G(p) \; \Rightarrow \; G(p)=\frac{1}{p^2-m^2-\Pi(p)+\mathrm{i} 0^+}.$$
    The self-energy can of course only calculated approximately in perturbation theory as a (formal) power series in the coupling or ##\hbar##.
     
  4. Jul 10, 2017 #3
    Thank you for your reply. But still, Peskin obtains the right answer in the end, so I assume that the statement he makes must be right (unless he makes another mistake that compensate that one, which I highly doubt).
     
  5. Jul 10, 2017 #4

    ChrisVer

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    Isn't it that the creation and annihilation operators don't depend on time?
     
  6. Jul 10, 2017 #5
    I am not sure. But anyway, if you chance the interactions (i.e. add an extra potential), they shouldn't be the same as in the free case, whether or not they depend on time, right?
     
  7. Jul 10, 2017 #6

    vanhees71

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    Well, looking on page 83, it's in fact correct what he writes: As usual he made the interaction and Heisenberg pictures coincide at ##t=t_0##. This is always possible, because changing the initial time means just a unitary time-independent transformation between the pictures, which doesn't lead to any observable consequences.

    You should note that in (4.15) he's writing the solution for the free-field equation in terms of annihilation and creation operators wrt. the single-particle momentum basis. That's fine, because the field in the interaction picture in fact is by definition the free-field solution. The rest is a pretty good presentation of the standard derivation of the Dyson series (which is from a strict point of view wrong due to Haag's theorem, but that's swept under the carpet from a practical point of view ;-0).
     
  8. Jul 10, 2017 #7
    Wait, sorry, I am still a bit confused. If you want the Heisenberg picture, you need to take the field at an (arbitrary) initial time, in this case ##\phi(t_0,x)## and evolve it using H (and not ##H_0## in this case), right (please let me know if I say something wrong)? But that initial ##\phi(t_0,x)##, is initial just because we chose it like that, but it still should satisfy the KG equation with an interaction (i.e. instead of ##=0## we have ##=\frac{\lambda}{3!}\phi^3##). So the solution in terms of ladder operators, satisfies the free equation, so how can it satisfy the interaction equation at the same time, such that we can write ##\phi(t_0,x)## in terms of ladder operators, just as in the free case. Because Peskin says we can do that for any fixed ##t_0## so what I understand from this is that at any instant, the field in the interaction picture can be written in terms of ladder operators exactly as the free field.
     
  9. Jul 10, 2017 #8

    MathematicalPhysicist

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    I should learn Haag's book and his theorem more patiently and rigoroursly.

    I tried once reading his book and the proof of the theorem (he refers you to some reference) but I think I first need to learn other more advanced mathematical tools. (Before reading this book, I didn't know SCV (Several Complex Variables) is being used in QFT).

    Is there some mathematical tool not being used in physics?! :-D
     
  10. Jul 10, 2017 #9

    vanhees71

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    The KG equation as an initial-value problem depends on two arbitrary functions, ##\phi(t_0,\vec{x})## and ##\dot{\phi}(t_0,\vec{x})##. The "hand-waving argument", ignoring Haag's theorem, is that you start with asymptotic free particles at ##t=t_0 \rightarrow -\infty##.

    How this is to be understood is also discussed in Peskin Schröder, but a much more careful description is given in the old textbook by Bjorken and Drell (Vol. 2 of course, because you shouldn't read Vol. 1 in the 21st century anymore, except you are interested in old-fashioned ideas).
     
  11. Jul 10, 2017 #10
    Ok, so just to make sure I understand by ##t_0## he means an initial time when the field was free and from which point on, the field starts to interact? And if this is the case why does he say that we can do that at any fixed ##t_0##? Because taking the limit ##t_0 \to \infty## isn't actually any ##t_0##.
     
  12. Jul 10, 2017 #11

    vanhees71

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    I guess, that's a very formal book. If you don't want to do research in the mathematical rigorous formulation of QFTs, maybe it's not necessary to read it. A very nice review for the "standard bread-and-butter QFT user" is

    https://link.springer.com/article/10.1007/s10670-005-5814-y?LI=true

    In googling for this link, I've also found Fraser's PhD thesis (which is most probably not behind a paywall, which I can't check, because I'm at the university right now):

    http://d-scholarship.pitt.edu/8260/
     
  13. Jul 10, 2017 #12

    vanhees71

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    Good question. That's what I said in my first answer: I don't understand, why P&S are writing it in this way. I'd stick to the interaction picture from the very beginning and express everything in the interaction picture. Then you get to the Dyson series by considering the time evluation of the state vectors, which is generated with the time-dependent (!) interaction Hamiltonian in the interaction picture right away.
     
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