The derivation steps of the LSZ reduction formula

Question

I am following the derivation steps of the LSZ reduction formula as done in the lecture notes by Timo Weigand the section on S matrix in the full interacting theory, here:

http://www.thphys.uni-heidelberg.de/~weigand/QFT1-13-14/SkriptQFT1.pdf

On page 50 the equation 2.56 below does not make sense to me.

$$ \langle p_1,\dots,p_n, out| a^{\dagger}_{out}(q_1)|q_2,\dots, q_r\rangle \sqrt(2E_{q_1})=$$ $$=\sum_{k=1}^{n}2E_{p_k}(2\pi)^3\delta^{3}(p_k-q_1)\langle p_1,\dots,\hat{p_k},\dots, p_n, out|q_2,\dots,q_r, in\rangle ,\tag{2.56}$$

where $\hat{p_k}$ has to be taken out.

For instance when I expand for n=2 the right side by replacing the delta function with commutation involving annihilation and creation of "out" operators I do not obtain the left side of the equation.

I mean using $$(2\pi)^3\delta^{3}(p_k-q_1)=[a(p_k), a^{\dagger}(q_1)], \qquad \langle p_1,p_2|/\sqrt(2E_{p2})=\langle p_1|a_{out}(p_2),$$

but this does not show that the two sides are equal.

Also do not see why this equation describes a process where one of the in- and outgoing states are identical and do not participate in scattering.

Answer 1

1. That this is the amplitude for a process where a particle with momentum $q_i$ does not participate in scattering is perhaps easier to see for the complex conjugate $$ \langle q,\text{in}\vert a_\text{out}(q_1)\vert p,\text{out}\rangle,$$ where $a_\text{out}(q_1)$ is the annihilation operator for $\lvert q_1,\text{out}\rangle$. Applying an annihilation operator to such a state gives zero unless the corresponding state exists at least once in the $\lvert p,\text{out}\rangle$, which is the origin of the $\delta$-function. If such a state is present, then the amplitude reduces to the amplitude where the $q_1$-state is taken out, i.e. does not participate in scattering.

2. If you do not get the r.h.s as a result of applying the definition of the $\lvert p_i,\text{out}\rangle$, then you are simply making a computational error. I can confirm that diligent application of the definitions and commutation relations yields the claimed result. Schematically, you start with $a(q_1) a^\dagger(p_1)\dots a^\dagger(p_n)$ and then move the $a(q_1)$ successively to the right. Each time you move it past an $a^\dagger$, you have to apply $aa^\dagger = a^\dagger a + (2\pi)^3 \delta$, and when it is completely to the right, it just annihilates the state, so you're left with the sum over the terms each involving a $\delta$ for a different $p_i$. The only potential tricky issue is that all the $\sqrt{E_{p_i}}$ are reabsorbed into the $\langle q,\text{in}\vert p,\text{out}\rangle$, except for the one that occurs in the delta, so you get $\sqrt{E_{q_1}E_{p_i}}$ terms, and the $E_{q_1}$ overall factor comes about by anticipating that the $\delta$ effectively sets $p_i = q_1$.