Confusion with $S$-matrix of Bhabha scattering

Question

Here are the two Feynman diagrams that corresponds to $e^- + e^+ \rightarrow e^- + e^+$ processes: electron positron scattering and electron positron annihilation respectively.

I can clearly see the final results should be in the form of

$$S_{fi} \propto (\bar{v}\gamma_{\nu}{v})(\bar{u}\gamma_{\mu}{u})$$ $$S_{fi} \propto (\bar{u}\gamma_{\mu}{v})(\bar{v}\gamma_{\nu}{u}).$$

however when I expand the field operators in terms of negative and positive energy fields (momenta and spin subscripts are omitted), I run into a conceptual problem.

$$S_{fi} = \langle{0 | d b : \bar{\psi_1}\gamma_{\mu}{\psi_1}\bar{\psi_2}\gamma_{\nu}{\psi_2}: \underbrace{A_{1}^{\mu}A_{2}^{\nu}} d^{\dagger}b^{\dagger}} |0 \rangle $$

$$S_{fi} = \langle{e^- e^+ | : (\bar{\psi^+_1}+\bar{\psi^-_1})\gamma_{\mu}({\psi^+_1}+{\psi^-_1})(\bar{\psi^+_2}+\bar{\psi^-_2})\gamma_{\nu}({\psi^+_2}+{\psi^-_2}): \underbrace{A_{1}^{\mu}A_{2}^{\nu}}} | e^- e^+ \rangle$$

$$ =\langle{e^- e^+ | (\bar{\psi^-_1})\gamma_{\mu}({\psi^-_1})(\bar{\psi^+_2})\gamma_{\nu}({\psi^+_2}) \underbrace{A_{1}^{\mu}A_{2}^{\nu}}} | e^- e^+ \rangle $$

which produces the correct result for the $s$-channel, while for the $t$-channel I am supposed to get the following based on what I expect from the first equation

$$\langle{e^- e^+ | (\bar{\psi^+_1})\gamma_{\nu}({\psi^-_1})(\bar{\psi^-_2})\gamma_{\mu}({\psi^+_2}) \underbrace{A_{1}^{\mu}A_{2}^{\nu}}} | e^- e^+ \rangle. $$

But I don't understand how $\bar{\psi^+_1}$ term with an annihilation operator can act on the final state after normal ordering.

Answer 1

Given the Dirac fields

$$ \psi(x) = \sum_{\vec p} \dfrac{1}{\sqrt{2V E_{\vec p}}} (b_\vec p u(\vec p)e^{-ip\cdot x} + c^\dagger_\vec p v(\vec p) e^{ip\cdot x}) = \psi^+ + \psi^- $$ $$ \bar\psi(x) = \sum_{\vec p} \dfrac{1}{\sqrt{2V E_{\vec p}}} (c_\vec p \bar v(\vec p)e^{-ip\cdot x} + b^\dagger_\vec p \bar u(\vec p) e^{ip\cdot x}) = \bar\psi^+ + \bar\psi^- $$ the matrix elements should be of the form $$S_{fi} \propto (\bar u \gamma_\alpha u)(\bar v\gamma^\alpha v)$$ $$S_{fi} \propto (\bar u \gamma_\alpha v)(\bar v \gamma^\alpha u).$$ Note that one indice is up and one down. Moreover, note that if your final state is $$\vert{f}\rangle = d^\dagger b^\dagger\vert 0 \rangle$$ then $$ \langle f \vert = \langle 0 \vert b d.$$ Assuming that your underbraces are indicating a Wick contraction, then $$\underbrace{A^{\mu}(x_1)A^{\nu}(x_2)}= -g^{\mu \nu} \lim_{m\rightarrow 0}\Delta(x_1 - x_2) = -g^{\mu \nu} \lim_{m\rightarrow 0} \int \dfrac{d^4k}{(2\pi)^4}\dfrac{e^{-ik\cdot(x_1 - x_2)}}{k^2 - m^2 + i\delta},$$ which you can pull out to the right. Writing out the fields in terms of creation and annihilation operator should yield $$\langle f \vert : b_\vec k ^\dagger b_\vec p c_{\vec p\,'} c^\dagger_{\vec k\,'}:\vert i \rangle$$ and similarly $$\langle f \vert : b^\dagger_\vec k c^\dagger_{\vec k\,'} c_\vec p b_{\vec p\,'}:\vert i \rangle.$$ Finally, normal ordering and using that $$\{b_\vec p, b^\dagger_{\vec p\,'}\} = \delta_{\vec p, \vec p\,'}$$ $$\{c_\vec p, c^\dagger_{\vec p\,'}\} = \delta_{\vec p, \vec p\,'}$$ (all other anticommutator vanish and interchanging operators thus gives a minus sign!) gives the desired result. That means, you have to apply the anticommutator relations a few times, pull out the Kronecker-Deltas and use that $b \vert 0 \rangle = 0$ respectively $c \vert 0 \rangle = 0$.

Edit: When calculating the $S$ matrix for $n = 2$ you apply Wick's theorem and its modified version where you do not take equal time contraction. Writing all those contractions out, only the term where you contract the 4-potential corresponds to the Bhaba scattering, because your inital and final state contain an electron and a positron. The term looks like this $$\underbrace{\not A_1 \not A_2}:\bar \psi_1 \psi_1 \bar \psi_2 \psi_2:. $$ Writing the fields in terms of positive/negative energy fields gives $$ :(\bar \psi^+ \gamma^\alpha \psi^+ + \bar \psi^+ \gamma^\alpha\psi^- + \bar \psi^- \gamma^\alpha\psi^+ + \bar \psi^- \gamma^\alpha \psi^-)_{x_1}\\\cdot (\bar \psi^+ \gamma^\beta\psi^+ + \bar \psi^+ \gamma^\beta\psi^- + \bar \psi^- \gamma^\beta\psi^+ + \bar \psi^- \gamma^\beta \psi^-)_{x_2}: i D_{F\alpha\beta}(x_1 - x_2)$$ where $D_{F\alpha\beta}(x_1 - x_2)$ is the propagator of the photon field. The process $e^+ e^- \rightarrow e^+ e^-$ is described by the two terms $$:(\bar\psi^- \gamma^\alpha \psi^+)_{x_1}(\bar \psi^+ \gamma^\beta\psi^-)_{x_2} + (\bar \psi^+ \gamma^\alpha\psi^-)_{x_1}(\bar\psi^- \gamma^\beta \psi^+)_{x_2}: i D_{F\alpha \beta}(x_1-x_2)$$ and $$:(\bar\psi^- \gamma^\alpha \psi^-)_{x_1}(\bar \psi^+ \gamma^\beta\psi^+)_{x_2} + (\bar \psi^+ \gamma^\alpha\psi^+)_{x_1}(\bar\psi^- \gamma^\beta \psi^-)_{x_2}: i D_{F\alpha \beta}(x_1-x_2)$$ since each contains only one of each creation and annihilation operator for the electron and positron repsectively. Note that after normal ordering the two summands of the terms are equivalent and give rise to a factor of $2$, canceling the $2!$ you get due to the $S$-matrix expansion. So, $$S^{(2)}(e^+e^- \rightarrow e^+ e^-) = -e^2\int d^4x_1\int d^4x_2 :(\bar\psi^- \gamma^\alpha \psi^+)_{x_1}(\bar \psi^+ \gamma^\beta\psi^-)_{x_2}: i D_{F\alpha \beta}(x_1-x_2)\\ -e^2\int d^4x_1\int d^4x_2 :(\bar\psi^- \gamma^\alpha \psi^-)_{x_1}(\bar \psi^+ \gamma^\beta\psi^+)_{x_2}:i D_{F\alpha \beta}(x_1-x_2).$$ One of the terms corresponds to the s-channel and one to the t-channel. Writing out the fields before normal ordering and only normal ordering the annihilation and creation operators at the end will give you the Feynman amplitudes you wrote down. From them you can construct the corresponding Feynman diagram by just applying the Feynman rules.