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In Sidney Coleman's Lecture Notes on Quantum Field Theory, under section 7.4, we have the following,

For a scattering of particles in a potential, we have a very simple formula for the S-matrix. We don’t have to worry about in states and out states, because the in states are the states in the far past, and the out states are the states in the far future: $$|\psi(-\infty)\rangle^{\text{in}}= \lim_{t\rightarrow -\infty} e^{iH_0 t} e^{-iHt}|\psi\rangle=\lim_{t\rightarrow -\infty} U_I(0,t)|\psi\rangle \tag{7.57} $$

I cannot derive the above equation - I have tried, but unsuccessfully. The same question was asked here, but the only answer the question has seems to solve the issue but is actually using a different notation to Coleman.

To start with, we have a Hilbert space for the scattering problem on which the complete Hamiltonian, $H$, (interaction included) acts upon, $\mathcal{H}$. We have another Hilbert space that is basically the free particle Hilbert space, $\mathcal{H}_{0}$, on which the free particle Hamiltonian, $H_{0} = \hat{p}^{2}/2m$, acts.

The free particle state kets (in $\mathcal{H}_{0}$) are defined as $|\psi (t)\rangle$ while the state kets in $\mathcal{H}$ that correspond to $|\psi (t)\rangle$ in the distant past (or, future) are denoted by $|\psi (t)\rangle^{in}$ (or, $|\psi (t)\rangle^{out}$). To express this distant past (future) correspondence in between the states, we can write,

$$ \lim_{t\rightarrow -\infty}\big| e^{-iHt}|\psi(0)\rangle^{in} - e^{-iH_{0}t}|\psi(0)\rangle\big| = 0 \tag{1}$$

(In the linked answer, this equation is written in a different notation that leads to the problem)

Now, Coleman defines $|\psi(0)\rangle = |\psi\rangle$ - this is the free particle state (the linked answer considers this to be the incoming state). The operator $U_{I}$ stands for, $$U_{I}(t, t') = e^{iH_{0}t} U(t, t') e^{-iH_{0}t'}$$ And it satisfies, $U_{I}(t, t') = U^{\dagger}_{I}(t', t) = U^{-1}_{I}(t', t)$.

Now, in the quoted equation, the third expression then means, $$ \lim_{t\rightarrow -\infty} U_I(0,t)|\psi\rangle = \lim_{t\rightarrow -\infty}U(0, t)e^{-iH_{0}t}|\psi\rangle = \lim_{t\rightarrow -\infty}e^{iHt}e^{-iH_{0}t}|\psi\rangle$$ This is evidently different from the middle term. Not only is the order different in the two equations, the signs differ as well. How are the two equal then and how do we get to these equations?

Any help will be appreciated.

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    $\begingroup$ Note: In the YouTube lectures made available, Coleman directly "adopts" a definition of the S-matrix without mentioning or computing the above steps. Link here. $\endgroup$
    – ShKol
    Oct 25, 2023 at 6:58
  • $\begingroup$ Around the 1 hour 20 minutes mark. $\endgroup$
    – ShKol
    Oct 25, 2023 at 7:27

1 Answer 1

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After looking into this a bit, I realised that there is a separate way to arrive at the final expression for the $S$ matrix without writing the states in this way. While this does not solve my original issue, it does make the question a bit redundant since these expressions are used to derive the $S$ matrix. I include the proof here in case someone happens to have the same issue (so that the expression won't seem arbitrary).

From the relation between the free states and the in (out) states, we have, $$ \lim_{t\rightarrow -\infty}\big| e^{-iHt}|\psi(0)\rangle^{in} - e^{-iH_{0}t}|\psi(0)\rangle\big| = 0 \tag{1}$$ This implies that, $$ |\psi(0)\rangle^{in} = \lim_{t\rightarrow -\infty} e^{iHt}e^{-iH_{0}t}|\psi(0)\rangle = \lim_{t\rightarrow -\infty} e^{iHt}e^{-iH_{0}t}|\psi\rangle = \lim_{t\rightarrow -\infty} U_{I}(0,t)|\psi\rangle \tag{2}$$ And, similarly, $$ |\phi(0)\rangle^{out} = \lim_{t\rightarrow \infty} e^{iHt}e^{-iH_{0}t}|\phi(0)\rangle = \lim_{t\rightarrow \infty} e^{iHt}e^{-iH_{0}t}|\phi\rangle = \lim_{t\rightarrow \infty} U_{I}(0,t)|\phi\rangle \tag{3}$$ Here, $|\phi\rangle^{out}$ is the state (belonging to $\mathcal{H}$) that resembles the free state $|\phi\rangle \in \mathcal{H}_{0}$ in the distant future. I write eqn. $3$ as, $$ |\phi(0)\rangle^{out} = \lim_{t\rightarrow \infty} U_{I}(0,t)|\phi\rangle = \lim_{t\rightarrow -\infty} U_{I}(0,-t)|\phi\rangle \tag{4}$$ The $S$ matrix is defined by Coleman as, $$ \langle \phi |S| \psi\rangle =\ ^{out}\langle \phi|\psi\rangle^{in} \tag{5}$$ As Coleman states,

Notice that we don’t have to put a $t$ in this expression because both $|\psi\rangle^{in}$ and $|\phi\rangle^{out}$ evolve according to the exact Schrodinger equation, and their inner product is independent of time.

This means that I might as well evaluate the inner product at $t = 0$ without a loss of generality (since these states in $\mathcal{H}$ evolve according to some unitary time evolution operator, the time does not matter for the inner product).

This leads to, $$ \langle \phi |S| \psi\rangle =\ ^{out}\langle \phi|\psi\rangle^{in} = \lim_{t \rightarrow -\infty} \langle \phi |U^{\dagger}_{I}(0,-t)U_{I}(0,t)|\psi\rangle = \lim_{t \rightarrow -\infty} \langle\phi|U_{I}(-t, t)|\psi\rangle$$ $$\Rightarrow S \equiv U_{I}(\infty, -\infty) = \mathrm{T}\Big(e^{-i\int_{-\infty}^{\infty}H_{I}(t)dt}\Big) \tag{6}$$

This is the required expression for the $S$ matrix. I would appreciate it if someone could check the correctness of the argument. Thanks!

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