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In the section 7.2.2 of Schwartz's QFT textbook, it says:

define the generation definition of time-evolution operators:

$$U_{21}\equiv U(t_2,t_1)=T{\exp[-i\int^{t_2}_{t_1} dt'V_I(t')]}\tag{7.46}$$

where $V_I$ is the interaction part of Hamiltonian in the interaction picture.

Then it has:

$$U_{21}U_{12}=1\tag{7.47}.$$

I don't understand how this relation is arrived. I tried to expand the definition directly, but to the second order of integral, it seems it cannot be canceled:

$$\int^{t_2}_{t_1} dt' dt'' V_I(t')V_I(t'')-\int^{t_2}_{t_1} dt' dt'' T[V_I(t')V_I(t'')]$$

which is obviously not zero.

Qmechanic
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Link
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  • I noted that it can be seen from $U_{21}^{\dagger}$ that when $t_1<t_2$, the $U_{12}$ is anti-time ordering. Or we can avoid defining $U_{12}$ and just use $U_{21}^{\dagger}$. – Link Jan 11 '23 at 11:56

2 Answers2

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Schwartz forgot to mention that the time-evolution operator $U(t_2,t_1)$ is antitime-ordered for $t_2<t_1$, cf. e.g. my Phys.SE answer here. Then eq. (7.47) becomes essentially just a telescoping product when discretized.

Qmechanic
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  • Thanks! I have read your answer you linked, but I wonder whether the definition of anti-time ordering is the same as the time ordering because they all put the later-time-operator on the left. If I expand the exponential in Taylor series, it seems that it is the same for both time-ordering. – Link Jan 06 '23 at 02:58
  • No, anti-time ordering put the later-time-operator on the right. – Qmechanic Jan 06 '23 at 05:01
  • But in the eq.(B) of your previous answer, it says when $t_1>t_2$, $U_{21}=lim_{N\to \infty}{ exp[-iH(t_1)\frac{t_2-t_1}{N}]...exp[-iH(t_2)\frac{t_2-t_1}{N}]}$. It seems that the later-time-operator with time $t_1$ is on the left. Do I understand it in a wrong way? – Link Jan 06 '23 at 05:37
  • I searched on the Internet, and looked up some QFT textbooks, but there is little information about the anti-time ordering. – Link Jan 06 '23 at 05:48
  • Eq. (B) had a typo. Corrected. Thanks for spotting it. – Qmechanic Jan 06 '23 at 06:09
  • I see. Thank you! – Link Jan 06 '23 at 08:08
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He must be tacitly using anti-time ordering when $t_2<t_1$. Then the cancellation works.

He should really use the path-ordering symbol "P" rather than "T".

mike stone
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