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Calculating weights of Feynman Diagrams

  1. Apr 26, 2010 #1
    I have a two point Green's Function with a phi^4 interacting Hamiltonian and am considering the second order of pertubation, and must work out the following:

    [tex]T[\phi(x_1) \phi(x_2) \phi(y_1)^4 \phi(y_2)^4] = T[\phi(x_1) \phi(x_2) \phi(y_1) \phi(y_1) \phi(y_1) \phi(y_1) \phi(y_2) \phi(y_2) \phi(y_2) \phi(y_2)][/tex]

    where [tex]\phi(x_1)[/tex]and [tex]\phi(x_2)[/tex] are two interacting fields.

    Now I am trying to work out the weight of a diagram, but I am not sure if I am doing it right?

    This is our diagram:

    x1___y1____y2____x2

    with one loop at y1 and one loop at y2.

    -For the line joining x1 to y1, there are 4 ways of contracting x1 with y1 (there are three y1's left after doing this)
    -For the loop at y1, there are 2 ways of contracting the next y1 with another y1 (there is one y1 left after doing this)
    -For the line joining y1 and y2, there are 4 ways of contracting the remaining y1 with a y2
    -For the loop at y2, there are 2 ways of contracting the next y2 with another y2 (there is one y2 left after doing this)
    -Finally there is one way of contracting the remaining y2 with x2.

    Hence the weight for this diagram is 4x2x4x2=64. Is this right?

    Thanks.
     
  2. jcsd
  3. Apr 26, 2010 #2
    It should be 4*3*4*3 not 4x2x4x2. You can select y1 with y1 in 3 different ways (3 choose 2), not 2 ways and similarly with y2 with y2.

    Usually there is a 1/4! for each quartic interaction, and instead of weights you find symmetry factors, and for the diagram

    J(r1)_y1_y2_J(r2)

    you can have:

    J(r2)_y2_y1_J(r1)

    and each of the loops attached to y1 and y2 has a degeneracy of 2 corresponding to the fact that switching the ends of the propagator can be achieved be rearranging the derivatives at each vertex.

    So the total symmetry factor would be 1/8, but then you select either r1 or r2 for x1 and x2, and that multiplies it by 2 to get 1/4.

    This 1/4 is what you would get if you multiply 4*3*4*3/(4!)^2

    So however you want to do it, but I prefer the symmetry factor way because you don't have to go through those calculations.
     
  4. Apr 27, 2010 #3
    Thanks RedX - good explanation:)

    I was infact trying to work out the symettry factor by working out the weight and dividing by the 4!^2 (the coefficient of the second pertubative Hamiltonian) - I was under the impression this is the definition of the symettry factor?

    Is there another way to work out the symettry factor (as I think that's what you're suggesting) without having to work out the weight?

    Thanks.
     
  5. Apr 27, 2010 #4
    Symmetry factors result when some rearrangement of external sources in a diagram can also be achieved by just rearrangement of vertices, or by the rearrangement of the multiple derivatives within each vertex. Also, symmetry can occur within loops subdiagrams if switching propagators can be achieved by swapping derivatives at each of the two endpoint vertices, or in the case of tadpole diagrams where swapping the endpoints of the propagator can be achieved by swapping the derivatives at a vertex. Just multiply all these symmetries together to get the symmetry factor.

    This might sound confusing, but the best I can do is refer you to page 74 of Srednicki's most excellent free draft version of his book:

    http://www.physics.ucsb.edu/~mark/qft.html

    Here he talks about symmetry factors in the context of a phi^3 theory, and there are plenty of diagrams from pages 73 to 78 with the symmetry factor listed. These symmetry factors are fairly easy to calculate using the rules I mentioned above.
     
  6. Apr 28, 2010 #5
    Thanks for the info, and the link!
     
  7. Apr 28, 2010 #6
    It's a good book.

    Symmetry factors are for the most part easy, but some can be tricky, as in the last diagram in Figure 9.2. on page 73, compared to the first diagram of Figure. 9.9 on page 78. That one I still can't figure out.

    The way I like to do it is to first only switch propagators, and then see if I could achieve that with just a swap of vertices or derivatives at vertices. So for example on the 3rd row left column of Figure 9.7, the following symmetries are:

    swap the left circle propagator with the right circle one, along with the stems of the circle propagators (for a total of 4 propagators swapped). This can be achieved just by rearranging the derivatives at the middle vertex where the stems meet.

    swap the left source (solid circle) propagator with the right source one. This can be achived by changing the derivative at the vertex where the two sources meet.

    and finally, each of the circle propagators has a factor of 2 as changing the endpoints of the circle propagator can be achieved by changing the derivatives at the vertex.

    so 2^4.

    So I can derive almost all the symmetry factors with this rule, but every now and then there comes something like the last diagram in Fig. 9.2. that beats me up. Luckily, I rarely have to calculate such weird diagrams.

    good luck.
     
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