Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Genereral:Questions about Srednicki's QFT

  1. Aug 17, 2009 #1

    haushofer

    User Avatar
    Science Advisor

    **********************
    Hi,

    I did some searching and found quite some questions about the Srednicki book on QFT, so apparently there are more people working with it. I thought maybe it would be a nice idea to have some sort of "questions about QFT encountered while reading Srednicki's book"-topic, so I hope I'm being appropriate here. If not, let me know.
    **********************



    I'm still a little confused about how the Feynman diagrams are generated with the functional Z. Just like you can define [itex]\Pi(k^2)[/itex] as the sum of all one-particle irreducible diagrams (1PI's), you can define [itex]V_n(k_1, \ldots, k_n)[/itex] as the sum of all 1PI's with n external lines.

    Now Srednicki claims that there is no tree-level contribution to [itex]V_{n \geq 4} [/itex] in [itex]\phi^3[/itex]-theory. The connected diagram of V=1, P=3 is a tree diagram, right? (tree external lines coming together at a single vertex). So does he basically mean that "you don't have E=4,P=4,V=1 diagrams in [itex]\phi^3[/itex] theory and all the other tree diagrams are not 1PI"?

    Also, a question about regularization which I already posed, but I'm still confused (but RedX, thanks for your efforts!) ;)

    I have another small question about Srednick's book;it's about ultraviolet cutt-off. In eq. (9.22) Srednicki makes the replacement

    [tex]
    \Delta(x-y) \rightarrow \int\frac{d^4k}{(2\pi)^4}\frac{e^{ik(x-y)}}{k^2 + m^2 - i\epsilon} \Bigl(\frac{\Lambda^2}{k^2 + \Lambda^2 - i\epsilon}\Bigr)^2
    [/tex]
    instead of cutting the integral explicitly of at [itex]\Lambda[/itex]. Are there any arguments besides Lorentz invariance why such a particular convergent replacement makes sense?
     
  2. jcsd
  3. Aug 17, 2009 #2

    haushofer

    User Avatar
    Science Advisor

    I think I got the first question; In the considered [itex]\phi^3[/itex] theory we have P = (E+ 3V)/2, (where E = # external lines, P = # propagators, V = # vertices) so for E=4 we have to start with V=1,P=5, and this diagram is not 1PI.
     
  4. Aug 17, 2009 #3

    haushofer

    User Avatar
    Science Advisor

    I'm also having troubles with the "skeleton expansion" described in chapter 19 (perturbation theory to all orders).

    First of all: if we are interested in a certain proces, then we fix our E, right? We know how many particles come in and out, and we want to calculate the cross-section of that process. So I'm not sure why we have to sum over all n-point vertices n=3,4,...,E.

    The expansion itself is described as:

    This means that we draw all the contributing 1PI , but omit diagrams that include either propagator or 3-point vertex corrections. That is, we omit any 1PI diagram that contains a subdiagram with two or three external lines that is more complicated than a single tree-level propagator (for a subdiagram with two external lines) or tree-level vertex (for a subdiagram with three external lines).

    Can someone elaborate on this?
     
  5. Aug 17, 2009 #4

    haushofer

    User Avatar
    Science Advisor

    Ok, to take a concrete example (like in chapter 20): elastic 2-particle scattering. I take E=4 here. Because P=(E+3V)/2, I get

    [tex]
    P = 2 + \frac{3}{2}V
    [/tex]

    This gives me the following list:

    V: |1 | 2 |3 |4 |5 |6
    P: |x |5 |x |8 |x |11

    where an x means that this particular combination of V and P is not possible in our [itex]\phi^3[/itex] theory. So, to start at lowest order in V, we get

    V=2,P=5
    V=4,P=8
    V=6,P=11

    The first one,V=2 and P =5, is a diagram with 2 external lines coming in at a vertex, and this vertex is connected with an internal line to another vertex. This last vertex is connected to, ofcourse, again 2 external lines.

    The second, V=4 and P=8, is a square where every edge is connected to an external line.

    The third, V=6, P = 11 is, I think, the same diagram as the second with an extra internal line in the square.

    Ofcourse, every diagram can be obtained in different ways (the first one for instance at 3!!=3 different ways etc). Is this skeleton expansion then the idea that:

    You take these first three orders in V, insert for the internal lines the exact propagator, and for the vertices the exact 3-point functions? What about the exact propagators of the external lines? Can I find some book/link where this is explained in detail and up to a reasonable order in V?

    I hope my question is a bit clear :)
     
  6. Aug 17, 2009 #5

    Avodyne

    User Avatar
    Science Advisor

    Yes.
    Yes, but after the expression for the diagram is put into the LSZ formula, the external propagators get replaced by the residue of the pole at the physical mass, which (at least at this stage of the book) has been set equal to 1.
    Srednicki appears to be following the program outlined by 't Hooft, http://arxiv.org/abs/hep-th/0405032, which seems to me to be different than the standard BPHZ procedure. For this, see Sterman or Kaku.

    As for your question about the cutoff procedure, the whole idea is that the details of the procedure should not matter (in a renormalizable theory), so we can use whatever is most convenient.
     
  7. Aug 18, 2009 #6

    haushofer

    User Avatar
    Science Advisor

    Great! It's good to hear I get a hang of it! Indeed, I forgot the LSZ-formula, I see the point now! ;) When I asked,

    "We know how many particles come in and out, and we want to calculate the cross-section of that process. So I'm not sure why we have to sum over all n-point vertices n=3,4,...,E."

    the point is then that IN the diagram there are n-point functions which we want to evaluate. For instance, in the E=4, P=8 case we want to evaluate the vertices exactly which are 3-point functions, and if we go higher in order in our skeleton we will encounter higher n-point functions, right?

    So I could use

    [tex]

    \Delta(x-y) \rightarrow \int\frac{d^4k}{(2\pi)^4}\frac{e^{ik(x-y)}}{k^2 + m^2 - i\epsilon} \Bigl(\frac{\Lambda^2}{k^2 + \Lambda^2 - i\epsilon}\Bigr)^n

    [/tex]

    for an arbitrary, finite n?
     
  8. Aug 18, 2009 #7

    Avodyne

    User Avatar
    Science Advisor

    Right!

    Yes.
     
  9. Aug 21, 2009 #8

    haushofer

    User Avatar
    Science Advisor

    Another question :)

    Chapter 25 talks about decay. A Lagrangian of 2 different particles [itex]\phi[/itex] and [itex]\chi[/itex] is written down, and in this Lagrangian an interaction between [itex]\phi[/itex] and [itex]\chi[/itex] is included. Then it is stated that

    For [itex]m_{\phi}>2m_{\chi}[/itex] it is kinematically possible for the [itex]\phi[/itex]particle to decay in two [itex]\chi[/itex] particles.

    My question is: does this follow directly from Z(J)? Ofcourse, I'm familiar with relativistic kinematics and decays and all that, but I'm wondering where and how exactly this possibility slips in our definition of Z(J).
     
  10. Aug 21, 2009 #9
    haushofer -> It's built in Z(J) through translation invariance. You use Z(J) to generate the n-point functions of the theory, and these are translation invariant. In momentum space this means you will end up with a momentum conservation delta. For the decay of a particle you'll have [tex]\delta(k_1 + k_2 - k)[/tex]. (See Srednicki's eq. 25.4) Now, Dirac's delta is only non zero when the argument is zero. And if [tex]m_\phi < 2 m_\chi[/tex] then it will always be [tex]k_1 +k_2 - k \ne 0[/tex], so the momentum delta is identically zero, giving you an overall vanishing probability for the process to occur.
     
    Last edited: Aug 21, 2009
  11. Aug 21, 2009 #10

    haushofer

    User Avatar
    Science Advisor

    Hey DrFaustus!

    I also traced down what happens if I plug in the Lagrangian, and also came down to the Dirac delta functions. It's nice to see how energy conservation is expressed in that way in QFT.

    (I often tend to throw down questions here before intensive investigation, because this really helps me to come closer to an answer, and it's very good to see how other people think about it. So sometimes I find an answer some time after posing the question ;) Thanks for your answer! :) )
     
  12. Aug 21, 2009 #11

    turin

    User Avatar
    Homework Helper

    Shouldn't Re{n}>1? (So, "no, not completely arbitrary.")
     
  13. Aug 22, 2009 #12

    haushofer

    User Avatar
    Science Advisor

    Yes, I implicitly assumed that n was real. :)
     
  14. Sep 2, 2009 #13

    haushofer

    User Avatar
    Science Advisor

    Okido, another question which I encountered in chapter 27. It's about equation 27.11 and 27.12. We have

    [tex]
    m^2_{ph} = m^2[1+ \frac{5}{12}\alpha(\ln{\frac{\mu^2}{m^2}} + c') + O(\alpha^2)]
    [/tex]

    He takes the log on both sides and divides by 2. So that leaves us with

    [tex]
    \ln{m_{\ph}} = 2\ln{m} + \ln{[\ldots]}
    [/tex]

    Now I don't understand how he gets equation 27.12. He obviously does a Taylor expansion

    [tex]
    \ln(1+x) = x - \frac{x^2}{2} + \ldots
    [/tex]

    and implicitly assumes that alpha is very small, or m>>mu. Or can we sweep the corrections under the rug of [itex]O(\alpha^2)[/itex]? Can someone comment on this? :)

    edit: I see the point; you expand in x, and x is of at least order alpha.
     
    Last edited: Sep 2, 2009
  15. Sep 25, 2009 #14

    haushofer

    User Avatar
    Science Advisor

    Another question about Srednicki, chapter 34. It's about Lorentz representations.

    The Lorentz representation is described by (2n+1,2m+1), which gives the dimensions of both SU(2) algebra's to which SO(3,1) is isomorphic to.

    In ordinary QM we can add to electronspins together and obtain a singlet (which is antisymmetric) and a triplet (which is symmetric). This is written as

    [tex]
    2 \otimes 2 = 1_A \oplus 3_S
    [/tex]

    At the bottom of page 211 the book mentions "For the Lorentz group, the relevant equation is

    [tex]
    (2,1) \otimes (2,1) = (1,1)_A \oplus (3,1)_S
    [/tex]

    "

    Why is this exactly? Another question arises at page 213, "For example, we can deduce the existence of [itex]g_{\mu\nu}=g_{\nu\mu}[/itex] from

    [tex]
    (2,2) \otimes (2,2) = (1,1)_S \oplus (1,3)_A \oplus (3,1)_A \oplus (3,3)_S
    [/tex]
    ", and the following "another invariant symbol..." Frankly, I couldn't derive these results by my own, so could some-one elaborate on this or give a link/book/whatever in which this is properly explained?
     
  16. Sep 25, 2009 #15

    samalkhaiat

    User Avatar
    Science Advisor

  17. Sep 26, 2009 #16

    haushofer

    User Avatar
    Science Advisor

    Great Sam, I've never been properly exposed to Clebsch-Gordon, so now it's the time to be properly exposed :) I'll look at your posts!
     
  18. Sep 26, 2009 #17

    turin

    User Avatar
    Homework Helper

    I think that you need to translate between yours and sam's notation. Basically, your notation specifies the dimensionality of the representation, whereas sam's notation specifies the "total angular momentum" (in multiples of hbar). To put it simply, your values of n and m are twice those in sam's notation.
     
  19. Sep 27, 2009 #18
    Almost certain this integral converges (as a distribution) even for n=0. It's the Feynman propagator - it doesn't need to be regularised.

    I'm not familiar with Srednicki's book - why did s/he regularise this?

    Cheers

    Dave
     
  20. Oct 31, 2009 #19

    haushofer

    User Avatar
    Science Advisor

    I missed this one. But it's a good point, and I have to say that his treatment of this isn't very clear to me at all. So if anyone can comment on it, I'm curious!

    But I came with another computational question which I encountered in Chapter 4, eqn(4.7).
    For a proper orthochronous Lorentz Transformation we have

    [tex]
    U(\Lambda)^{-1}\phi(x)U(\Lambda)=\phi(\Lambda^{-1}x) \ \ \ \ \ (1)
    [/tex]
    which means for the annihilation operator (the creation operator goes the same)

    [tex]
    U(\Lambda)^{-1}a(\bold{k})U(\Lambda) = a(\Lambda^{-1}\bold{k}) \ \ \ \ \ (2)
    [/tex]

    I thought the following: Expand phi in creation and annihilation operators and plug this into (1). On the right hand side we then have in the exponential of the expansion an inner product between [itex]\Lambda^{-1}x[/itex] and [itex]k[/itex]. This basically is the same as (k and x transform oppositely) the inner product between [itex]\Lambda k[/itex] and [itex]x[/itex]. If we then change variables in the expansion,
    [tex]
    k \rightarrow \Lambda^{-1}k
    [/itex]
    and use that the measure [itex]\tilde{dk}[/itex] doesn't change, we arrive at the result.

    However, in (1) our Lorentz transformation acts on the four-vector x,while in (2) it acts on the three-vector [itex]\bold{k}[/itex]. So is my computation valid? What does it mean for this Lorentz transformation to act on a three-vector in (2)?
     
  21. Nov 1, 2009 #20

    turin

    User Avatar
    Homework Helper

    Yeah, that's just a kind of cheap notation. You can just as well index the mode operator with the 4-momentum, subject to the mass-shell constraint. Using the 3-momentum as the index just helps to remind you that the index space is (homeomorphic? to) R3, not R4. The meaning of the transform of the 3-momentum is precisely the resulting value of the 3-momentum components after the transform of the 4-momentum.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Genereral:Questions about Srednicki's QFT
  1. Qft question (Srednicki) (Replies: 14)

Loading...