# Scalar Yukawa theory derivation

I am using Tong's notes for QFT, and on page 59 there is a derivation for the scattering amplitude of $\psi\psi \rightarrow \psi\psi$ in Scalar Yukawa theory.

It goes from here:

$$\langle p_1',p_2'|:\psi(x_1)^\dagger\psi(x_1)\psi(x_2)^\dagger\psi(x_2):|p_1,p_2\rangle$$

to here:

$$\langle p_1',p_2'|\psi(x_1)^\dagger\psi(x_2)^\dagger|0\rangle\langle0|\psi(x_1)\psi(x_2)|p_1,p_2\rangle.$$

I honestly have no idea what the step is there.

Shoulnd't the identity be $\int |i\rangle \langle i|$ and not just the vacuum? How can he move the fields around?

I am not aware of an elegant answer to this question, but writing out the expansion of the fields $\psi$ in creation- and annihilation operators gives an answer. $\psi$ is a complex scalar with $$\psi(x)=\int\frac{d^3 k}{(2\pi)^3\sqrt{2E_{\vec{k}}}}\left(b_{\vec{k}}e^{-ikx}+c_{\vec{k}}^\dagger e^{ikx}\right)$$ The initial and the final state only contain the $b$-type particle, not its anti-particle. This is not apparent from your question, but in Tong's lecture notes you can see that \begin{align} |p_1,p_2\rangle &= \sqrt{2E_{\vec{p}_1}}\sqrt{2E_{\vec{p}_2}}b_{\vec{p}_1}^\dagger b_{\vec{p}_2}^\dagger|0\rangle\\ |p_1',p_2'\rangle &= \sqrt{2E_{\vec{p}_1'}}\sqrt{2E_{\vec{p}_2'}}b_{\vec{p}_1'}^\dagger b_{\vec{p}_2'}^\dagger|0\rangle \end{align} Since the normal ordering moves all annihilation operators to the right and the creation operators to the left and the $c_{\vec{k}}$, $c_{\vec{k}}^\dagger$ operators commute with $b_{\vec{p}}^\dagger$, all terms in the product $$\langle p_1',p_2'|:\psi(x_1)^\dagger\psi(x_1)\psi(x_2)^\dagger\psi(x_2):|p_1,p_2\rangle$$ containing either $c_{\vec{k}}$ or $c_{\vec{k}}^\dagger$ will vanish. That leaves us with an expression of the following form, leaving away all normalization factors and exponentials etc, just focusing on the ladder operator structure: $$\int d^3k_1d^3k_2d^3k_3d^3k_4 \langle 0| b_{\vec{p}_1'} b_{\vec{p}_2'} b_{\vec{k}_1}^\dagger b_{\vec{k}_2}^\dagger b_{\vec{k}_3} b_{\vec{k}_4} b_{\vec{p}_1}^\dagger b_{\vec{p}_2}^\dagger|0\rangle$$ The crucial point is, that this expression is equal to one where we insert a $|0\rangle\langle 0|$ into the middel in the following way: $$\int d^3k_1d^3k_2d^3k_3d^3k_4 \langle 0| b_{\vec{p}_1'} b_{\vec{p}_2'} b_{\vec{k}_1}^\dagger b_{\vec{k}_2}^\dagger |0\rangle\langle 0| b_{\vec{k}_3} b_{\vec{k}_4} b_{\vec{p}_1}^\dagger b_{\vec{p}_2}^\dagger|0\rangle$$ A quick way to see this is by looking at the annihilation operator $b_{\vec{p}_2'}$, the second from the left in the original expression. One can use the commutator twice to move it to the right. On the way, there are two $\delta$-functions arising from the commutator, but these equally arise in the expression with the inserted $|0\rangle\langle 0|$. The remaining term, in which the annihilation operator has been moved to the right by two positions is then $$\langle 0| b_{\vec{p}_1'} b_{\vec{k}_1}^\dagger b_{\vec{k}_2}^\dagger b_{\vec{p}_2'} b_{\vec{k}_3} b_{\vec{k}_4} b_{\vec{p}_1}^\dagger b_{\vec{p}_2}^\dagger|0\rangle$$ which vanishes, because there are three annihilation operators acting after two creation operators on the vacuum.
Of course the argument can be made more explicit by calculating the whole expression more carefully. It is much more easy to see that $$\langle p_1',p_2'|\psi(x_1)^\dagger\psi(x_2)^\dagger|0\rangle\langle0|\psi(x_1)\psi(x_2)|p_1,p_2\rangle$$ will lead to the same expression we have found for starting with the normal ordered expression.
Here's a quick way to see this: once we write down the mode expansions for $$\psi$$ and $$\psi^{\dagger}$$, we see immediately---since any term with $$c$$ or $$c^{\dagger}$$ gives us zero trivially---that the expression for the normal ordered product of the field operators is,$$\int \frac{dp^3dp^4dp^5dp^6}{(2\pi)^{12}}\langle p_1',p_2'|b^{\dagger}_{p_3}b^{\dagger}_{p_4}b_{p_5}b_{p_6}|p_1,p_2\rangle,$$where we have not explicitly written down the phase factors and stuff like that. Now note that the two $$b$$ operators on the right annihilate particles from the state $$|p_1,p_2\rangle$$, leaving us with the zero-particle state. Similarly, the action of the two $$b^{\dagger}$$ operators on the left give us the zero particle state. So, what we are left with is simply an overlap between two zero-particle states. So we are free to insert the identity operator of the zero particle sector, which is simply $$|0\rangle\langle0|$$.