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Crossing Symmetry and Mandelstam Variables; and (unrelated) external photons

  1. Jan 4, 2005 #1
    Greetings--a few further questions and thoughts from Peskin and Schroeder An Introduction to Quantum Field Theory. (I'll do my best to keep the discussion self-contained, but I'll provide references as well.)

    Is there an easy way to produce Feynman diagrams in LaTeX? (Maybe Daniel has a trick? :smile: ) In the mean while I'll resort to describing them.

    In section 5.4 (p.155-157) P&S introduce the principle of crossing symmetry and Mandelstam variables and proceed to use these tricks to easily compuse the squared, spin-avereaged amplitude for [tex]e^-\mu^- \rightarrow e^-\mu^-[/tex] by exploiting a previous result for [tex]e^+e^- \rightarrow \mu^+\mu^-[/tex].

    P&S say (p. 156) "it is conventional to define [tex]t[/tex] as the squared difference of the initial and final momenta of the most similar particles."

    • First of all, is there some standard criteria for what constitutes a "similar" particle? i.e. are an electron and a positron more similar than an electron and a muon (charge versus family)?

    This seems to lead to some ambiguity in the example:

    Following P&S's notation for [tex]e^+e^- \rightarrow \mu^+\mu^-[/tex] as follows: the momentum of the incoming [tex]e^-[/tex] is [tex]p[/tex] and incoming [tex]e^+[/tex] is [tex]p'[/tex], while the momentum of the outgoing [tex]\mu-[/tex] is [tex]k[/tex] and outgoing [tex]\mu^+[/tex] is [tex]k'[/tex]. (That is, particles are unprimed while antiparticles are primed and incoming particles are labelled with [tex]p[/tex], outgoing particles are labelled with [tex]k[/tex]). ((See, that wasn't too hard to visualize, was it? o:) ))

    The result (p. 156, eq. 5.70) is: [tex]\frac{1}{4}\sum_{spins}|M|^2 = \frac{8e^4}{s^2}\left[ \left( \frac{t}{2} \right)^2 + \left( \frac{u}{2} \right)^2\right][/tex]

    Now we use this result to compute the corresponding value for [tex]e^-\mu^- \rightarrow e^-\mu^-[/tex]. We first turn the photon propagator sideways (which doesn't change the topology or physics). Then we apply crossing symmetry twice: (1) the incoming [tex]e^+[/tex] with momentum [tex]p'[/tex] becomes an outgoing [tex]e^-[/tex] with momentum [tex]-p'[/tex], (2) the outgoing [tex]\mu^+[/tex] with momentum [tex]k'[/tex] becomes an incoming [tex]\mu^-[/tex] with momentum [tex]-k'[/tex].

    Let us call the corresponding Mandelstam variables for this process: [tex]\~{s}, \~{t}, \~{u}[/tex] to avoid confusion with the Mandelstam variables for [tex]e^+e^- \rightarrow \mu^+\mu^-[/tex], which we write without squigglies.

    Now all that remains to be done is to figure out what the old Mandelstam variables turn into in terms of the new Mandelstam variables after we employ crossing symmetry: [tex]p', k' \rightarrow -p', -k'[/tex].

    First of all, [tex]s=(p+p')^2 = (k+k')^2 \rightarrow (p-p')^2 = (k-k')^2 = \~{t} [/tex], which is written out explicitly at the bottom of page 156. However, the text says that [tex]t \rightarrow \~{s}, u\rightarrow \~{u}[/tex], which appears to be a typo--or perhaps I'm confused.

    • I would think that [tex]t=(k-p)^2=(k'-p')^2 \rightarrow (k-p)^2=(k'-p')^2=\~{u}[/tex] and [tex]u=(k'-p)^2=(k-p')^2 \rightarrow (p+k')^2 = (k+p')^2 = \~s[/tex]. Is this correct? (Either way yields the same result for the spin-averaged amplitude-squared.)

    Ok, now an unrelated question: in chapter 5.5 P&S discuss Compton Scattering and introduce photon polarization sums. Their goal is to motivate the replacement [tex]\sum_{polarizations} \epsilon^*_\mu \epsilon_\nu \rightarrow -g_{\mu \nu}[/tex]. They define [tex]M^\mu(k)[/tex] as the part of the total amplitude that does not depend on [tex]\epsilon^*_\mu[/tex] such that [tex]M(k)=M^\mu(k)\epsilon^*_\mu(k)[/tex].

    On p. 160, however, I am confused by their motivation for the form of this [tex]M^\mu(k)[/tex]: "Now recall from Chapter 4 that external photons are created by the interaction term [tex]\int d^4x e j^\mu A_\mu[/tex], where [tex]j^\mu = \overline{\psi} \gamma^\mu \psi[/tex] is the Dirac vector current. Therefore we expect M to be given by the matrix element of the Heisenberg field [tex]j^\mu[/tex]: "

    [tex]M^\mu(k)=\int d^4xe^{ik\cdot x} \langle f | j^\mu (x) i| \rangle[/tex]

    Where the sum is over the inital and final states of all particles except the photon in question.

    • How are external photons created from the interaction term [tex]\int d^4x e j^\mu A_\mu[/tex] and how does this motivate the form for [tex]M^\mu(k)[/tex]: ?

    Thanks very much,
    easily bewildered student
  2. jcsd
  3. Jan 4, 2005 #2

    The reason for this is that certain possible polarizations of photons (more specifically the longitudinal polarization) are NOT physical and need to be eliminated from the physical model. This modification is done by Gupta and Bleuler. Not-physical means that if you were to use this polarization then you would acquire a negative expectation value for the photons and that is ofcourse not possible. So basically the Matrix M describes or contains only the physical polarizations and therefore it describes the photons. The interaction term can be interpreted in several ways if which you already gave one. Now you can also see the j-term as some disturbance of the system that leads to the generation of particles, of which the interactions are mediated by the A-tensor. in this case the A-tensor are photons...

    But this is not important, the point is that M must be written in terms of jA because these are real and physical photons...

    As an addendum : There is another way in order to eliminate unphysical degrees of freedom. Just add the opposite degrees of freedom to your model so that the "sum" yields a model that is liberated of these unphysical things. the opposite degrees of freedom (opposite to the unphysical ones that is) are called the fadeev Poppov gohsts and they are very famous in QFT. Besides in order to make sure that this system works you need to write it down in Grassmann variables which i explained in your other post...


    ps : let me get back on the Mandelstamm variables tomorrow...They are a renormalization thing...
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