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I'm switching between QFT books all the time. From Maggiore's chapter 5 I got the impression that

$$ \langle \phi_1 \phi_2 \phi_3 \rangle \equiv \langle 0 |T\{\phi_1 \phi_2 \phi_3\}|0 \rangle =0 $$

because, on using Wick contraction to expand the time-ordered fields, there is always a spare un-contracted field which kills the term due to its action on the vacuum.

Now, following Schwartz I'm deriving the Schwinger-Dyson's equations in chapter 7. Here, he does not mention any such thing. In fact, he asks readers to calculate $$ \Box_x \langle\phi_1 \phi_2 \phi_x \rangle $$ Why isn't this object automatically zero?

knzhou
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topologically_astounded
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1 Answers1

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You have to be careful to compare the definitions between the books.

Maggiore is setting up perturbation theory in the usual way. To recap, the LSZ reduction theorem relates $S$-matrix elements to time-ordered correlators of Heisenberg fields. These may be related to time-ordered correlators of fields in the interaction picture by Maggiore's Eq. 5.67. And fields in the interaction picture always behave like free fields, so we may apply Wick's theorem to conclude that $$\langle 0 | T \phi_I(x_1) \phi_I(x_2) \phi_I(x_3) | 0 \rangle = 0.$$ Since interaction picture perturbation theory is so common, the $I$ subscript is usually dropped, as Maggiore warns in the beginning of section 5.4.

Schwartz eventually does the same thing, in section 7.2, but he first derives the Feynman rules using a completely different route, by the Schwinger-Dyson equation in section 7.1. Here we simply work with Heisenberg fields the entire time, so $$\langle 0 | T \phi_H(x_1) \phi_H(x_2) \phi_H(x_3) | 0 \rangle$$ where the $H$ subscript is dropped. To add to this confusion, he refers to these fields as "the interacting fields", but only to emphasize they we're no longer doing free field theory, not that we're in interaction picture. Somewhat paradoxically the point of going to interaction picture is to make the interacting fields look free instead.

knzhou
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  • Alright, this I follow. So when you'd expand the time-ordered fields in the heisenberg picture, you'd expand the fields in their normal modes? This expansion will eventually lead you to wick contractions, and using the definition of annihilation and creation on vaccum, you'd get zero for odd number of fields?

    Where does this line of reasoning go wrong?

     – topologically_astounded May 30 '18 at 13:54
  • @topologically_astounded If you define a mode expansion using the exact same equation you did for free fields, then the operators $a$, $a^\dagger$ in the expansion will not satisfy the commutation relations for creation and annilhilation operators, so none of the other reasoning goes through. This reflects the fact that nontrivial stuff can happen; if you create 8 particles you could, e.g. come back later and see there are now 12 particles. – knzhou May 30 '18 at 13:58
  • Wait a minute, I re-checked Schwartz properly now.

    The idea of Heisenberg picture or interaction picture is not in at the point where he derives the Schwinger-Dyson equation. Moreover, he explicitly mentions he is using free fields.

    Above all, he just used the annihilation-creation idea I mentioned for the same fields in a different paragraph. I don't see how your answer answer's my question.

     – topologically_astounded May 30 '18 at 14:06
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    @topologically_astounded The idea of Heisenberg picture has been there from the start; it is introduced in section 2.3.2 and used implicitly everywhere afterward. I don't see any indication that Schwartz is using free fields in section 7.1, can you directly quote him? – knzhou May 30 '18 at 14:29
  • Read 'Schwartz': pg-85, line above eq (7.31).

    "The interaction picture fields are just what we had been calling (and will continue to call) the free fields:"

    Your answer is in principle wrong!

     – topologically_astounded May 30 '18 at 14:49
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    @topologically_astounded I'm not sure what's bugging you, by knzhou's answer is correct. If $\Phi$ is a free field, or an interacting field in the interaction picture, then $\langle \Phi^1\cdots\Phi^{2n+1}\rangle\equiv 0$. On the other hand, if $\phi$ is an interacting field in the Heisenberg (or Schrödinger) picture, then $\langle \phi^1\cdots\phi^{2n+1}\rangle$ may in principle be different from zero. To see this, recall the Gell-Mann and Low formula, $\langle \phi^1\cdots\phi^{2n+1}\rangle=\langle \Phi^1\cdots\Phi^{2n+1}\mathrm e^{iS_\mathrm{int}[\Phi]}\rangle$ (1/2) – AccidentalFourierTransform May 30 '18 at 14:55
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    (2/2) (cf. e.g. this PSE post). If $S_\mathrm{int}[\Phi]$ only has even powers of $\Phi$, then $\langle \Phi^1\cdots\Phi^{2n+1}\mathrm e^{iS_\mathrm{int}[\Phi]}\rangle$ will contain an odd number of $\Phi$'s, and $\langle\phi^1\cdots\phi^{2n+1}\rangle\equiv 0$. If, on the other hand, $S_\mathrm{int}[\Phi]$ contains a term with an odd number of $\Phi$'s (e.g., $\phi^3$ theory), then $\langle \Phi^1\cdots\Phi^{2n+1}\mathrm e^{iS_\mathrm{int}[\Phi]}\rangle$ will contain an even number of $\Phi$'s, and $\langle\phi^1\cdots\phi^{2n+1}\rangle\neq 0$ – AccidentalFourierTransform May 30 '18 at 14:58
  • @topologically_astounded Note that you asked about section 7.1, where the fields are Heisenberg and interacting. In section 7.2, Schwartz switches to interaction picture to do what Maggiore is doing, and interaction picture fields are free. – knzhou May 30 '18 at 15:00
  • @AccidentalFourierTransform So this is the part I'm teaching myself right now, which is:

    Converting a correlation function from interaction picture to free-fields can induce an interaction part of the action in the correlation? And the number of fields coming from this interaction part now must be taken into account for doing even/odd counts. This is fine.

    Schwartz on pg-80, blatantly asks to compute $\Box<\phi_1 \phi_2 \phi_3>$. The fields are free and there is no interaction going on!

     – topologically_astounded May 30 '18 at 15:04
  • @topologically_astounded pg-80, which equation? Maybe we're reading different editions, but I don't see $\square\langle\phi_1\phi_2\phi_3\rangle$ anywhere on page 80! – AccidentalFourierTransform May 30 '18 at 15:09
  • Edition - 2014 (first published).

    After eqn(7.11), he asks us to verify that eqn in this special case:

    $$ \Box_x \langle\phi_x \phi_2 \phi_x \rangle$$

     – topologically_astounded May 30 '18 at 15:13
  • @topologically_astounded I see. Nowhere on that page, except for the paragraph around eq.7.9, are the fields free. Everywhere on that section the fields are interacting, and in the Heisenberg picture. – AccidentalFourierTransform May 30 '18 at 15:20
  • and that is 0! (hence, my confusion) – topologically_astounded May 30 '18 at 15:24