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This relates to Peskin & Schroeder's QFT book, equation 4.70 on page 104.

To define in and out states we take our initial state and evolve it far into the past, and do the same for our final state. Peskin & Schroeder write this:

$$_{out}\langle p_1p_2...|k_\mathcal Ak_\mathcal B\rangle_{in}=\lim_{T\rightarrow \infty}\langle p_1p_2...|e^{-iH(2T)}|k_\mathcal A k_\mathcal B\rangle \tag{4.70},$$

in which the time evolution operator has a negative sign. However, in my attempt to explain this I have written the following:

$$|k_\mathcal Ak_\mathcal B\rangle_{in}=\lim_{T\rightarrow\infty}e^{iHT}|k_\mathcal Ak_\mathcal B\rangle,$$

since $U(T,0)=e^{-iHT}$ is the forward time evolution operator and $U(0,T)=e^{iHT}$ is the backward time evolution operator (or rather, the inverse), since we are evolving the initial state back to the far past this (perhaps naively) seems like the correct choice. I would then expect an analogous statement to hold for the out state, and combining the two would give:

$$_{out}\langle p_1p_2...|k_\mathcal Ak_\mathcal B\rangle_{in}=\lim_{T\rightarrow \infty}\langle p_1p_2...|e^{+iH(2T)}|k_\mathcal A k_\mathcal B\rangle,$$

in which there is a $+$ sign in the exponential. What have I done wrong?

Qmechanic
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Charlie
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1 Answers1

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The 'in' and 'out' states are the (asymptotically free) initial and final states of particals interacting at $t=t_0$: $$|k_1k_2...\rangle_{in/out}=|k_1k_2...(t=\pm\infty)\rangle_{interacting}$$ The 'in' and 'out' states are independent on time, however the interacting states are not. $$ |k_1k_2...(t)\rangle_{interacting} = e^{-iH(t-t_0)}|k_1k_2...(t_0)\rangle_{interacting} $$ What we are interested in is e.g. an initial state $|k_1k_2...(t=-T)\rangle_{interacting}$ long before ($-T\ll t_0$) the particles in it interact with each other and we want to know how this states looks like long after ($T\gg t_0$) the interaction $|p_1p_2...(t=+T)\rangle_{interacting}$. The overlap is $$ S_{fi} = \langle p_1p_2...(t_0)|k_1k_2...(t_0)\rangle = \langle p_1p_2...(T)|e^{-iH2T}|k_1k_2...(-T)\rangle $$ If we take the limit $T\rightarrow\infty$, we can relate these interacting states to the 'in' and 'out' states. $$ S_{fi}\rightarrow {}_{out}\langle p_1p_2...|k_1k_2...\rangle_{in} $$ The reason for wanting to relate the transition amplitude of interacting states to those of free states is, that we know the explicit form of the creation and annihilation operators only for free theories. Another important subtlety is that $a_{in}(p)\neq a_{out}(p)$, which means, that the overlap of the asymptotically free states does not simply produce a bunch of delta functions, but is instead more complicated and can be calculated e.g. using the LSZ reduction formula.

KilianM
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  • I had a quick think about what you've written, thanks for your answer. Just to be sure I've understood, we define our states in the asymptotic past and future (because that way we can consider them essentially non-interacting fields), but the actual amplitude we want is that between the states when they are very close (at $T=0$), so we evolve the asymptotic initial state forward and the asymptotic final states backwards until they are both at some reference time (like $T=0$), at which point we overlap them, and the evolution operator that does this is the S-matrix? – Charlie Dec 25 '20 at 23:24
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    Yes, that is basically it! – KilianM Dec 25 '20 at 23:39
  • Great, thank you! – Charlie Dec 25 '20 at 23:40
  • @KilianM what I don't understand is that Peskin explicitly states that "in the last line, states are defined at any common reference time". This seems to be in contrast with your answer. – Quantumwhisp Jun 28 '21 at 08:21
  • @Quantumwhisp Good point. I have edited/ corrected my answer to hopefully make it more clear. – KilianM Jun 28 '21 at 09:01
  • @KilianM I think you wrote one thing that gave me the right clue: the states $|k\rangle$ evolve by $e^{iHt}$, not by $e^{-iHt}$, because they are not states in the schroedinger picture, but eigenstates of the time evolving operators in the heisenberg picture. This explaines why the "-" sign is there. – Quantumwhisp Jun 28 '21 at 15:21
  • Now (after the edit), all the states are in the schroedinger picture? – Quantumwhisp Jun 28 '21 at 17:55
  • The states in the interacting theory are in the Schrödinger picture, while the asymptotic 'in' and 'out' states are time-independent free particle states. – KilianM Jun 29 '21 at 13:20
  • @KilianM I'm confused because Peskin states he works in the Heisenberg Picture. – Quantumwhisp Jun 30 '21 at 23:01
  • Ok I see, the states $|k_1k_2...(t_0)\rangle_{interacting}$ are states in the Heisenberg picture since they are defined at a fixed time $t_0$. On the other hand $|k_1k_2...(t)\rangle_{interacting}$ are defined for any time $t$, i.e. are states in the Schrödinger picture. I should have been more clear there. – KilianM Jul 01 '21 at 08:08