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Ryder QFT 2nd ed. Page 192

  1. Sep 3, 2007 #1
    [SOLVED] Ryder QFT 2nd ed. Page 192

    1. The problem statement, all variables and given/known data
    equation 6.51. This equation is actually 2 equations separated by a comma. I don't understand either one and would appreciate any help to get me started. For the time being, I would like to concentrate on the first one.
    [tex]\frac{1}{i}\frac{\delta}{\delta J(x)}exp[-\frac{i}{2}\int dx_1dx_2J(x_1)\Delta_F(x_1 - x_2)J(x_2)][/tex]
    [tex]= -\int\Delta_F(x - x_1)j(x_1)dx_1exp[-\frac{i}{2}\int dx_1dx_2J(x_1)\Delta_F(x_1 - x_2)J(x_2)][/tex]



    2. Relevant equations
    None



    3. The attempt at a solution
    I don't have so much trouble with the form of the factor
    [tex]\int\Delta_F(x - x_1)J(x_1)dx_1[/tex]
    as I do with the placement of x and x_1. x doesn't appear in the exponential on the lhs and [itex]x_1[/itex] appears with a plus sign on the lhs, and a minus sign on the rhs.
     
  2. jcsd
  3. Sep 3, 2007 #2

    Avodyne

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    Does it help to know that [tex]\Delta_F(x)[/tex] is an even function of x? That is, [tex]\Delta_F(x - x_1)=\Delta_F(x_1 - x)[/tex]
     
  4. Sep 4, 2007 #3
    Thanks for taking a look at this Avodyne. I see that [itex]\Delta_F(x)[/itex], defined in eqn (6.14) on page 184, is even:
    [tex]\Delta_F(x) = \frac{1}{(2\pi)^4}\int d^4k\frac{e^{-ikx}}{k^2 - m^2 + i\epsilon}[/tex]
    because k can be replaced with -k without changing the value of the integral. And this helps a great deal. I'm still fuzzy on the idea of differentiating wrt J(x) when J(x) doesn't appear in the thing being differentiated.
     
  5. Sep 4, 2007 #4
    Moreover, [tex]\Delta_F(x) [/tex] is a Green Function, as defined in equation (6.10), which is always symmetric (between x and x').

    As for the differentiation with respect to J(x), take in mind that it's a functional differentiation (check out the section 5.4), and the result yields as follows:

    [tex]\frac{1}{i}\frac{\delta}{\delta J(x)}exp[-\frac{i}{2}\int dx_1dx_2J(x_1)\Delta_F(x_1 - x_2)J(x_2)]=[/tex]

    [tex]=\frac{1}{i}\left[-\frac{i}{2}\int dx_1dx_2\delta_{x_1x}\Delta_F(x_1 - x_2)J(x_2)\right]exp\left[-\frac{i}{2}\int dx_1dx_2J(x_1)\Delta_F(x_1 - x_2)J(x_2)\right]+[/tex]

    [tex]+\frac{1}{i}\left[-\frac{i}{2}\int dx_1dx_2\delta_{x_2x}\Delta_F(x_1 - x_2)J(x_1)\right]exp\left[-\frac{i}{2}\int dx_1dx_2J(x_1)\Delta_F(x_1 - x_2)J(x_2)\right][/tex]

    Integrate with respect to x_1 and to x_2 with the delta functions, and you got your result!

    Edit:You can treat functional differentiation almost like in simple calculus, putting delta functions where is need for. One can always use the Functional differentiation definition (5.55), but it's a pretty longer road!
     
    Last edited by a moderator: Sep 4, 2007
  6. Sep 4, 2007 #5

    Avodyne

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    J(x) *does* appear in the exponential, because x is in the integration range of x1 and x2.

    Think of x like an index; suppose we had a vector [tex]J_i[/tex] and a matrix [tex]\Delta_{ij}[/tex], and we wanted to differentiate [tex]\exp[-{\textstyle{1\over2}\sum_{jk}}J_j\Delta_{jk}J_k][/tex] with respect to [tex]J_i[/tex].
     
  7. Sep 4, 2007 #6
    Thanks to both of you, PhysiSmo and Avodyne. When you boil it down, the point I was missing, was this:
    [tex]\frac{\delta J(y)}{\delta J(x)} = \delta (x - y).[/tex]

    This is in analogy with the discrete case:
    [tex]\frac{dx^i}{dx^j} = \delta_{ij}[/tex]

    It looks like both of you were trying to tell me this each in your own way.
     
  8. Sep 4, 2007 #7

    dextercioby

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    Even if the sources are fermionic variables, the argument of the Pauli-Jordan functional [itex] \Delta (x) [/itex] is still symmetric due to its origin, i.e. appearing in a complex exponential.
     
  9. Sep 4, 2007 #8

    Avodyne

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    Yes, exactly. This definition, and the usual rules for derivatives (chain rule, product rule, etc) are all you need.

    This is not correct. The Dirac propagator (appropriate for complex fermionic sources) has no definite symmetry, while the Majorana propagator (appropriate for real fermionic sources) is antisymmetric on swapping both the position labels and the spin indices; see Srednicki eq.(42.24).
     
  10. Sep 5, 2007 #9

    dextercioby

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    I don't think you read my answer very well. The Pauli-Jordan functional is nothing but the coordinate space representation of the propagator for bosonic scalar field. That's why it's a simple Delta, with no spinor indices.
     
  11. Sep 5, 2007 #10

    Avodyne

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    Your terminology is not correct. The Pauli-Jordan function (not functional) is the vacuum expectation value of the commutator of two free scalar fields (see, e.g., Bogoliubov & Shirkov or Greiner & Reinhardt), and it is an odd function of its argument. On the other hand, the Feynman propagator, which is what appears in the path integral for vacuum boundary conditions, is the vacuum expectation value of the time-ordered product of two free scalar fields, and it is an even function of its argument.

    Your orginal comment was about fermionic sources. In relativistic theories, where the spin-statistics theorem applies, fermionic sources are used only with fields of half-odd-integer spin.
     
  12. Sep 5, 2007 #11

    dextercioby

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    Yes, pardon me, it was a confusion. Of curse, i meant the Feynman propagator.
     
  13. Sep 5, 2007 #12
    ps. On page 199, there are two unnumbered equations below eqn (6.77) that I wish I had encountered earlier in the book. It might have obviated the need for this thread.
     
  14. Sep 5, 2007 #13

    dextercioby

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    Since the fields are operator-valued distributions, their commutator is still an operator-valued distribution and by taking the VEV, we obtain a distribution. Any distribution is a functional, so we speak about the Pauli-Jordan functional.
     
  15. Sep 5, 2007 #14

    Avodyne

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    "Pauli-Jordan function" gets 398 hits on google, and appears in the index of Bogoliubov & Shirkov and of Greiner & Reinhardt. "Pauli-Jordan functional" gets none, and does not. Using terminology that no one else uses will definitely lead to confusion.

    Physicists usually use "functional" to mean a function on a vector space of functions, e.g., the path integral Z(J) is a functional of the source J(x), and "function" to mean a function or distribution on Rn; e.g., the Dirac delta function.
     
  16. Sep 5, 2007 #15
    to the reals or complex plane?
    Ryder muddies the waters on page 172.

    Quantities like the propagator
    [tex]<x_ft_f|x_it_i> = \int \mathcal{D}x exp[\frac{i}{\hbar}\int_{t_i}^{t_f}L(x, \dot{x})dt][/tex]
    are functional integrals: the integration is taken over all functions x(t). The left hand side is a number, so the integral associates with each function x(t) a number.

    It seems to me that these two sentences are contradictory. What's worse, there are two integrals in the equation. The inner one is a functional.
     
    Last edited: Sep 5, 2007
  17. Sep 5, 2007 #16

    Avodyne

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    Yes; see http://mathworld.wolfram.com/Functional.html

    He sure does. [tex]\langle x_ft_f|x_it_i\rangle[/tex] is a function (not a functional) of the four numbers xf, tf, xi, and ti. It's not a functional of x(t), just like an integral over x is not a function of x.

    If we add a source term J(t)x(t) to L, then the result is a function of xf, tf, xi , and ti, and a functional of J(t).
     
  18. Sep 5, 2007 #17
    Interesting. The article only allows functionals to map to the reals. But if you click on the link "functional analysis", it allows both real and complex numbers.
     
  19. Sep 6, 2007 #18

    dextercioby

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    Function instead of functional is unfortunately a misuse of terminology. It's definitely confusing and that's why i find it misfortunate. Anyway, i don't use such terminology, for me there's always the "Dirac functional" or the "Dirac delta" and likewise it's the "Pauli-Jordan functional", the Feynman functional a.s.o.a.s.f.

    Anyway, I think this terminology issue is not really important. Bringing "google hits" as an argument is not valid for me, since for me the terminology issue is: use whatever words you can, just don't create confusion. "Function" instead of "functional" is definitely misleading.

    To me this is "case closed".
     
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