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I'm reading introductory quantum field theory, from the book Relativistic Quantum Physics by T. Ohlsson. In the derivation of the unitary time-evolution operator, chapter 11.2, there is an equation that I get stuck on. Here it is, with a little bit of context:

... In the case of fermions, the shifted interaction Hamiltonian $H'_I$ always includes a product of an even number of fermions. Thus, for the ${H'_I}^2$ term, we can symmetrize the integrals in the following way $$\int_{t'}^t dt_1 \int_{t'}^{t_1} dt_2 T\left(H'_I(t_1)H'_I(t_2)\right) = \int_{t'}^t dt_2 \int_{t'}^{t_2} dt_1 T\left(H'_I(t_1)H'_I(t_2)\right)\\ = \frac{1}{2} \int_{t'}^t dt_1 \int_{t'}^{t} dt_2 T\left(H'_I(t_1)H'_I(t_2)\right).$$

I have been trying to verify this by using the definition of the time ordered product for fermionic fields, $$T\left(\psi_1(x_1)\psi_2(x_2)\right) = \theta(t_1 - t_2)\psi_1(x_1)\psi_2(x_2) - \theta(t_2 - t_1)\psi_2(x_2)\psi_1(x_1),$$ but I'm not really getting anywhere. The way it is introduced makes it seem like it should be obvious, and maybe it is, but I'm not getting it, so I would greatly appreciate if anyone could show me how the quoted equalities are derived.

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    $\begingroup$ Peskin and Schroeder Figure 4.1 should answer your question $\endgroup$ Mar 9, 2021 at 3:05
  • $\begingroup$ This seems promising, but they state that "$T{H_I(t_1)H_I(t_2)}$ is symmetric about the line $t_1=t_2$". But would it not be antisymmetric for fermionic fields? Or is this where the statement "$H'_I$ always includes a product of an even number of fermions" from the book I quoted comes in? Does that somehow make $H'_I$ work like a bosonic field with respect to time-ordering? $\endgroup$
    – ummg
    Mar 9, 2021 at 3:56
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    $\begingroup$ Exactly, if the time-ordering dictates that $H'_I(t_1)$ and $H'_I(t_2)$ need to be swapped, then you need to make sure (even within the time ordering) that you add the minus signs for anticommuting Grassman numbers. But when an even number of fermions are pushed past an even number of fermions, you get $(-1)^{2n}=1$, which is symmetric $\endgroup$ Mar 9, 2021 at 4:05
  • $\begingroup$ Okay, I think I'm with you. Thank you for answering! $\endgroup$
    – ummg
    Mar 9, 2021 at 4:33

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$$\int_{t'}^t dt_1 \int_{t'}^{t_1} dt_2 \ T\{H'_I(t_1)H'_I(t_2)\} \overset{(1)}{=} \int_{t'}^t dt_2 \int_{t'}^{t_2} \ dt_1 T\{H'_I(t_1)H'_I(t_2)\} \\ \overset{(2)}{=} \frac{1}{2} \int_{t'}^t dt_1 \int_{t'}^{t} dt_2 T\left(H'_I(t_1)H'_I(t_2)\right).$$

Where in the first equality we use time ordering symmetry, and in the second we use an integration trick which, at second order, is summed up well in this diagram: Peskin and Schroeder Diagram [Peskin and Schroeder, An Introduction to Quantum Field Theory, Figure 4.1]

and it's not the greatest leap of faith to see how this holds at order $N$, with the time-ordered product of $N$ Hamiltonian operators.

Now in P&S, this expression is derived for Hamiltonians of bosonic fields, since it relies on the symmetry of the time-ordered expression about $t_1=t_2...=t_n$. For fermionic fields, recall that the time ordering operation is not a blanket command to be able to shuffle fields within it willy-nilly - one must remember to include minus signs when a Grassman-odd fermion is anticommuted past another. However, when the interaction part of the Hamiltonian contains products of an even number of fermionic fields, then you will pick up factors of $(-1)^{2n}=1$ for moving $H'_I(t_l)$ past $H'_I(t_m)$, so this symmetry is retained.

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  • $\begingroup$ Again, thank you. I don't really know about Grassman numbers though. Am I correct that it could be equally well stated in terms of permutations? That is, when time-ordering $H'_I(t_1)H'_I(t_2)$ we may do as follows: First time-order $H'_I(t_1)$ and $H'_I(t_2)$ individually. This will be an even permutation, since each "swap" is performed twice; once in $H'_I(t_1)$ and once in $H'_I(t_2)$. Then swap the $2n$ fermions in $H'_I(t_1)$ with the $2n$ fermions in $H'_I(t_2)$, which is of course also an even permutation. $\endgroup$
    – ummg
    Mar 9, 2021 at 14:10
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    $\begingroup$ @ummg yep, that's exactly it (and also essentially the premise behind Grassman numbers) $\endgroup$ Mar 9, 2021 at 14:11

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