The Schrodinger equation: $$ i \frac{d}{d t}\left|\psi_{t}\right\rangle=H\left|\psi_{t}\right\rangle $$ and solutions are given by
$$\left|\psi_{t}\right\rangle=U(t)|\psi_\text{in}\rangle \equiv e^{-\imath H t}|\psi_\text{in}\rangle$$

In scattering experiment we are interested in solution with the following asymptotic condition \begin{gathered} U(t)|\psi\rangle \underset{t \rightarrow-\infty}{\longrightarrow} U^{0}(t)\left|\alpha\right\rangle \\ \qquad \qquad \underset{t \rightarrow+\infty}{\longrightarrow} U^{0}(t)\left|\beta\right\rangle \end{gathered} where $ U^0(t)=e^{-\imath H_0 t}$ with $H_0$ the Hamiltonian of a free theory.

From the expression above we can show that

$$ |\psi_\text{in}\rangle=\lim _{t \rightarrow-\infty} U(t)^{\dagger} U^{0}(t)\left|\alpha\right\rangle \equiv \Omega_{-}\left|\alpha \right\rangle $$ and $$ |\psi_\text{in}\rangle=\lim _{t \rightarrow+\infty} U(t)^{\dagger} U^{0}(t)\left|\beta\right\rangle \equiv \Omega_{+}\left|\beta\right\rangle $$

We are intereste in the probability that a particle that entered the collision with in asymptote $|\phi\rangle$ will be observed to emerge with out asymptote $|\chi\rangle .$ To evaluate this probability we note that the actual state at $t=0$, which will evolve from the in asymptote $|\phi\rangle$ is $|\phi_\text{in}\rangle=\Omega_{-}|\phi\rangle$, while the actual state at $t=0$, which would evolve into the out asymptote $|\chi\rangle$ is $|\chi_ \text{out} \rangle=\Omega_{+}|\chi\rangle .$

This probability is given by \begin{aligned} w(\chi \leftarrow \phi) &=|\langle\chi(t) \mid \phi(t)\rangle|^{2} \\ &=|\langle\chi_\text{out}\mid \phi_\text{in}\rangle|^{2} \\ &=\left|\left\langle\chi\left|\Omega_{-}^{\dagger} \Omega_{+}\right| \phi\right\rangle\right|^{2} \\ &=|\langle\chi|S| \phi\rangle|^{2} \end{aligned}

Now in this notes Quantum Field Theory by Mark Srednicki at page 51 they have that the scattering Amplitude from a initial state $\mid i\rangle$ to a final state $\mid f\rangle$ is given by

$$\langle f \mid i\rangle=\left\langle 0\left|a_{1^{\prime}}(+\infty) a_{2^{\prime}}(+\infty) a_{1}^{\dagger}(-\infty) a_{2}^{\dagger}(-\infty)\right| 0\right\rangle \tag{5.13}$$

From expression $(5.13)$ for example if the initial state is orthogonal to the final state than the scattering amplitude is zero.

My question is shouldn't this scattering amplitude be given by $\langle f \mid S\mid i\rangle$ where $S$ is the $S$ matrix?


1 Answer 1


The transformation from Interaction picture to Heisenberg picture is given by

$$ O_{H}(t)=U^{\dagger}(t, 0) O_{\mathrm{ip}}(t) U(t, 0) $$ with $U(t, 0)=e^{i H_{0} t} e^{-i H t} .$

Suppose we have $\left|\psi_{t}\right\rangle=U(t)|\alpha\rangle_\text{in} \equiv e^{-\imath H t}|\alpha\rangle _\text{in}$ with asymptote $|\phi\rangle$ so we have $$|\alpha\rangle_{\text {in }}=\lim _{t \rightarrow-\infty} U^{\dagger}(t, 0)|\alpha\rangle$$

so \begin{aligned} \lim _{t \rightarrow-\infty} a_{H}^{\dagger}(k ; t)|\alpha\rangle_{\text {in }} &=\lim _{t \rightarrow-\infty} U^{\dagger}(t, 0) a^{\dagger}(k) U(t, 0)|\alpha\rangle_{\text {in }} \\ &=\lim _{t \rightarrow-\infty} U^{\dagger}(t, 0)|k, \alpha\rangle \\ &=|k, \alpha\rangle_{\text {in }} \\ &=a_{\text {in }}^{\dagger}(k)|\alpha\rangle_{\text {in }} \end{aligned} From above we have that $$\lim _{t \rightarrow-\infty} a_{H}^{\dagger}(k ; t)=a_\text{in}^{\dagger}(k)$$

Similarly we can show that $$\lim _{t \rightarrow +\infty} a_{H}(k ; t)=a_\text{out}(k)$$

Now since $$\mid i\rangle=a_\text{in}^{\dagger}(k_1)a_\text{in}^{\dagger}(k_2)\mid 0\rangle$$ and $$\mid f \rangle=a_\text{out}^{\dagger}(k'_1)a_\text{out}^{\dagger}(k'_2)\mid 0\rangle$$

we obtain the result.


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