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Meson propagator development

  1. Nov 17, 2011 #1
    Could anyone please explain the sequence below taken from Mandl QFT textbook (p.53)?

    1. [itex]i\hbar c\Delta^+(x-x')=[\phi^+(x),\phi^-(x')][/itex]

    2. [itex]i\hbar c\Delta^+(x-x')=\langle 0|[\phi^+(x),\phi^-(x')]|0\rangle[/itex]

    3. [itex]i\hbar c\Delta^+(x-x')=\langle 0|\phi^+(x)\phi^-(x')|0\rangle[/itex]

    4. [itex]i\hbar c\Delta^+(x-x')=\langle 0|\phi(x)\phi(x')|0\rangle[/itex]

    From 1. to 2. does it mean that the vacuum expected value of the commutator is the commutator itself? How?

    From 2. to 3. does it mean that the term [itex]\langle 0|\phi^-(x')\phi^+(x)|0\rangle[/itex] is null? How?

    From 3. to 4. does it mean that the terms

    [itex]\langle 0|\phi^+(x)\phi^+(x')|0\rangle[/itex]

    [itex]\langle 0|\phi^-(x)\phi^+(x')|0\rangle[/itex]

    [itex]\langle 0|\phi^-(x)\phi^-(x')|0\rangle[/itex]

    are all null? How?

    Thank you for any help.
  2. jcsd
  3. Nov 17, 2011 #2


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    Yes to 2, 3 and 4. φ+ is an absorption operator, so φ+|0> = 0 since there is nothing to absorb. Likewise <0|φ- = 0.
  4. Nov 17, 2011 #3
    Thank you. What about the first transformation?
  5. Nov 17, 2011 #4


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    I don't have a copy of Mandl to compare, but I'm looking at Weinberg section 6.1 where he says something similar. He's evaluating the S-matrix as a sum of terms, <0| ... |0> where ... is a string of creation and annihilation operators, and he's talking about rearranging the order of the operators. Every time you switch the order of two of them you get a numerical factor.

    And on p262 he says: (f) Pairing of a field ψ with a field adjoint ψ in H(y) yields a factor -iΔ(x,y). (I'm leaving some subscripts out.) This is close to what you're saying. So I think the context is that Mandl's Eq (1) represents a subexpression that's eventually going to be placed between <0| |0>'s.
  6. Nov 18, 2011 #5


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    To go from eq.1 to eq.2, sandwich eq.1 between <0| and |0>. Use the fact that the left side of eq.1 is just a c-number (not an operator), and so on the left we just get that function times <0|0>=1.
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