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Help to 2-point correlation function in $ \phi^{4} $-theory

  1. Jan 23, 2015 #1
    Hi everyone! After a few slow days in the office I thought I would like to derive the 2-point correlation function within statistical(Euclidian) ## \phi^{4} ##-theory but I ran into some problems. For the sake of clarity I will show you from where I startet and ask questions when I need help. Lets start!\\
    The energy functional is given by
    \begin{equation}
    E[\phi] = \int dx \frac{1}{2}\left( (\partial_{x}\phi(x) )^{2} + m^{2}\phi(x)^{2}\right) + V(\phi)
    \end{equation}
    Where ## x ## might as well be a n-dimensional vector and ## V(\phi) = \frac{\lambda}{4!}\phi^{4} ##. The partition function is then given by
    \begin{equation}
    Z[J] = \int D\phi \exp(-E[\phi] + J\phi) =\int D\phi \exp\left( -\int dx V(\frac{\delta}{\delta J(x)})\right) \exp(\frac{1}{2}JG_{0}(x,x')J)
    \end{equation}
    With ## J ## some current added for computational purposes. For the second equality sign I have rewritten the free part of the energy, completed the square and set the vacuum expectation energy to 1, and used(what in my notes is called) Wicks theorem on the interaction part. ## G_{0}(x,x') ## is the free propagator given by ## G_{0}(x,x') = \int \frac{dP}{(2\pi)^{D}} \frac{\exp(ip(x-x'))}{p^{2} + m^{2}}##. We now have the n-point correlation functions
    \begin{equation}
    G^{n}(x_{1},...,x_{n}) = \frac{\delta}{\delta J(x_{1})}...\frac{\delta}{\delta J(x_{n})} Z[J]\big|_{J=0}
    \end{equation}
    To calculate the 2-point function we expand equation 2 and performs the functional derivatives. To zeroth order in $ \lambda $ we (of cause) just gets
    \begin{equation}
    G^{2}_{0}=G_{0}
    \end{equation}
    To first order we get some 15 diagrams, some of which are tadpoles and some of which are disconnected. The first problem shows up now. I believe there is some easy argument as to why I can discard the disconnected diagrams, but I can not seem to remember what it is. Also is there any easy way to write the diagrams on this forum?\\
    Lets use my intuition and throw away the disconnected diagrams. To first order we now have the diagramatic expansion.\\
    C2_to_first_order.jpg

    Lets do the calculations in Fourier space. Using the usual Feynman rules the zeroth diagram becomes
    \begin{equation}
    I_{0} = \frac{1}{p^{2} + m^{2}}
    \end{equation}
    The first diagram hopefully becomes
    \begin{equation}
    I_{1} = \frac{\delta(p-p')}{p^{2} + m^{2}}\frac{\lambda}{(p')^{2}+m^{2}} \int \frac{dq}{(2\pi)^{D}} \frac{1}{q^{2} + m^{2}} = \frac{\delta(p-p')}{p^{2} + m^{2}}\frac{\lambda}{(p')^{2}+m^{2}} \frac{\Gamma (1-D/2)(m^{2})^{D/2-1}}{(4\pi)^{D/2}}
    \end{equation}
    So in three dimension I would be done by now?(besides having to Fourier transforming back again, and is that even possible?). Suppose 4 dimensions then. ## I_{1} ## diverges and we need to regularize. Using dimensional regularization we define ## \epsilon = 4-d ## and ## \tilde{\lambda} = \mu^{\epsilon}\lambda ##. The gamma-function part can now be written as
    \begin{equation}
    \frac{\lambda \Gamma(\epsilon/2 - 1)}{2(4\pi)^{2}} \left(\frac{4\pi \mu^{2}}{m^{2}}\right)^{\epsilon/2} = \frac{\lambda}{(4\pi)^{2}}m^{2}\left( \frac{1}{\epsilon} - \frac{1}{2}\log\left( \frac{m^{2}}{4\pi \mu^{2}}\right) + \frac{1}{2}\psi(2) + O(\epsilon) \right)
    \end{equation}
    With ## \psi(x) ## being the Digamma function.\\
    This is as far as I got. Now what? I suppose I have to use some renormalization to throw away the divergent terms? And how about the renormalization group flow?
     
  2. jcsd
  3. Jan 23, 2015 #2

    vanhees71

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  4. Jan 24, 2015 #3
    Thanks for the link :). You seem to be treating the same calculation in chapter 5 but I still have some questions. In eq (5,5) you calculate the tadpole diagram using a momentum cut off regulator and finds a constant which consists of a part diverging for ## \Lambda \rightarrow \infty ## and one going to zero. You then go on about how we should subtract the divergence plus some constant(5,8). What is this constant? Now doing the same with dimensional regularization you find a similar result to mine(5,80). This result however, seem to have a finite part in the ## \epsilon \rightarrow 0 ## limit. So what is this term? And why are there differences? I thought different regularization schemes should give the same result? ( I haven't gotten to RG-Flow yet).
     
  5. Jan 24, 2015 #4

    vanhees71

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    The self energy (or in case of bosons also called polarization piece, i.e., the trunctated one-particle irreducible two-point function) figures into the connected propagator according to the Dyson equation, which is easily algebraically resolved to
    $$G(s)=\frac{1}{s-m_0^2-\Pi(s)}.$$
    The polarization function is analytic everywhere in the complex ##s=p^2## (with ##p^2=p_0^2-\vec{p}^2##) plane except for a cut along the positive real axis. Your tadpole diagram is a special case, because it's effectively a one-point function and thus a quadratically divergent constant (i.e., independent of ##s##). The first diagram, where a cut occurs is the 2nd-order sunset diagram. The cut starts thus at ##s=(3m)^2##.

    Now to make sense out of this, you have to remember the meaning of physical parameters in the model: The pole of ##G(s)## denotes the physical mass, ##M^2##. Thus
    $$M^2-m^2-\Pi(M^2)=0$$
    defines the physical mass of the particle, provided this equation has a real solution ##M^2>0##.

    Now you can impose renormalization conditions. One scheme is the "on-shell scheme", where you identify ##M=m##, i.e., you say ##m## is the physical mass of the particle. This already fixes your tadpole diagram: You simply subtract it. So there is no correction to the two-point function in leading order of ##\phi^4## theory. For the higher-order cases you need one more renormalization condition. To this end you Taylor expand the self-energy around ##s=m^2##. Then you get
    $$\Pi(s)=\Pi(M^2)+(s-m^2) \Pi'(M^2)+\tilde{\Pi}(s).$$
    From power counting it's clear that ##\tilde{\Pi}## is finite, while the first two terms are quadratically and logarithmically divergent respectively. But these divergences you can subtract with a mass-counter term and a wave-function-renormalization-counter term, i.e., you result reads
    $$\Pi_s=A+B(s-M^2) + \tilde{\Pi}(s).$$
    Now you have to choose ##A## and ##B## according to your renormalization scheme. In the onshell scheme you set ##A=0##. Then the Green's function has a pole at ##s=M^2##. Now to the wave-function renormalization condition, which leads to the determination of the constant ##B##. First note that
    $$\tilde{\Pi}(s)=\mathcal{O}[(s-M^2)^2].$$
    From this you get that the propagator
    $$G(s)=\frac{1}{(1-B)(s-M^2) - \tilde{\Pi}(s)}$$
    has a first-order pole in ##s=M^2## with residuum ##1/(1+B)##. It's convenient to make this residuum to ##1##, i.e., to set ##B=0##.
    Thus your renormalization conditions for the on-shell scheme reads
    $$\Pi(s=M^2)=0, \quad \Pi'(s=M^2)=0.$$
    Of course, you can choose other schemes. Particularly useful are mass-independent renormalization schemes. Using dim. reg. you can just subtract the pieces diverging for ##\epsilon \rightarrow 0##. This is called the minimal-substraction or MS scheme. You should be aware that at higher orders you have to treat subdivergences properly to show that these subtractions are "allowed" in the sense that they occur from appropriate counter terms in the action. In this case of a Dyson-renormalizable theory. The counter terms are precisely of the same local form as the original Lagrangian, i.e., you have a finite number of wave-function, mass and coupling-constant renormalizations.

    Although the MS scheme is technically pretty simple, the BPHZ technique is conceptionally more clear, because it defines the parameters in a specific way at an appropriate renormalization point (energy-momentum scale).

    The RG you also find in my manuscript, using the BPHZ scheme.
     
  6. Feb 2, 2015 #5
    Okay I think I am almost understanding the basics now!
    After finding that the 2-point correlation function is divergent I introduce a mass constant ## \mu ##, and subtract some constant*divegence, to get a finit result( nice, since the 2-point function should in principle be measureable). The constant is dependent on the renormalization scheme. So I now have, ## L^{ren} = L{post 1} + L{c.t} ##.
    For ## \phi^4 ## theory the 2-point function only gives a mass renormalization, but in principle we could have gotten something proportional to ## \mu^{\epsilon} ## and ## p^2 ##. I suppose the ## p^2 ## dependence might be observable in an experiment. Is this correct?
    To define a renormalization scheme I now demand that the correlation function should obey some conditions in momentum space( ## G^2(m^2) = \infty ## for example). I suppose I need 3 conditions due to the 3 terms in the Lagrangian, the differential term, the mass term and the coupling term. The different schemes should be reachable through finite counter terms, as opposed to the counter term for the 2-point correlation function.
    From my new and renormalized Lagrangian I can find some running couplings and so on.

    Okay so a new question: For some momentum I measure the mass of a particle, ## m_1 = K## and the coupling constant( do not know how though), ## \lambda_1 = C ##. Now if I wanted to calculate the mass and coupling constant for some other momentum, I would find the beta-function from renormalization theory from which I could predict the "new" mass and coupling. Is this how calculations is done?

    I suppose this means that If I were to give you the ## \phi^4 ## Lagrangian above and tell you m=1 and ## \lambda = 0.1 ##( no momentum given) you would not be able to calculate the 2-point function?
     
  7. Feb 2, 2015 #6

    vanhees71

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    You get the physical (finite) parameters from fitting your predictions to appropriate scattering cross sections and then you can predict other cross sections and compare to experiments. You get the running of the parameters in terms of the renormalization scale by using the renormalization-group equations. Then you do experiments at another energy-momentum scale and compare the newly fitted constants and compare it with the predicted behavior from the RG calculation. In practice this has been done e.g., with the running of the electromagnetic coupling. At low momenta you get ##\alpha \simeq 1/137##. If measuring at momentum scales of the order of the ##Z##-boson mass (around 90 MeV) you get about ##1/128##.

    Another example is the running of the strong coupling constant ##\alpha_s##, which decreases with higher momenta ("asymptotic freedom"). The RG flow of QCD describes this running very well. For a plot see the end of the review article in the Review of Particle Physics:

    http://pdg.lbl.gov/2014/reviews/rpp2014-rev-qcd.pdf
     
  8. Feb 3, 2015 #7
    Thanks for the reply! For some reason the coupling between renormalization and experiments were never really brought up back when I had a course in field theory( or I have forgotten).
    Another question springs to mind:
    If I expand the beta function as a power series in the coupling constant, the coefficients at different order seems to be scheme dependent( mentioned on the page 3 in the review you just posted). How do I figure out which schemes will converge the fastest? As in this scheme the power series coefficients will fall off this fast, while in this other scheme they will fall off even faster.
     
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