# Quantum field theory 1, lecture 24

Electron-Muon Scattering. We can also consider the scattering process $$e^{-} \mu ^{-} \to e^{-}\mu ^{-}$$,

\begin{equation*} i \mathcal{T} = \bar{u}(q_3) (-e\gamma ^\mu ) iu(q_1) \left (-i\frac{\eta _{\mu \nu } -\frac{k_\mu k_\nu }{k^2}}{k^2}\right )\bar{u}(q_4)(-e\gamma ^\nu ) iu(q_2). \end{equation*} By a similar argument as before the term $$\sim k_\mu k_\nu$$ drops out, \begin{equation*} \mathcal{T} = \frac{e^2}{(q_1-q_3)^2} \bar{u}(q_3) \gamma ^\mu u(q_1) \bar{u}(q_4) \gamma _\mu u(q_2)\quad \quad \quad (e^{-}\mu ^{-} \to e^{-}\mu ^{-}). \end{equation*}

Comparison to electron to muon scattering. Compare this to what we have found for $$e^{-} e^{+} \to \mu ^{-}\mu ^{+}$$ \begin{equation*} \mathcal{T} = \frac{e^2}{(p_1+p_2)^2} \bar{v}(p_2) \gamma ^\mu u(p_1) \bar{u}(p_3) \gamma _\mu v(p_4), \end{equation*} where the conventions were according to

There is a close relation and the expressions agree if we put \begin{equation*} \begin{split} q_1 = +p_1, \quad \quad \quad & u(q_1) = u(p_1), \\ q_2 = -p_4, \quad \quad \quad & u(q_2) = u(-p_4) \to v(p_4), \\ q_3 = -p_2, \quad \quad \quad & \bar{u}(q_3) = \bar{u}(-p_2) \to \bar{v}(p_2), \\ q_4 = +p_3, \quad \quad \quad & \bar{u}(q_4) = \bar{u}(p_3). \end{split} \end{equation*}

Crossing symmetry. Recall that \begin{equation*} (ip \cdot \gamma+m)\;u(p) = 0 \quad \quad \quad \text{but} \quad \quad \quad (-ip \cdot \gamma+ m)\;v(p) = 0. \end{equation*} However one sign arises from the spin sums \begin{equation*} \begin{split} \sum _{s=1}^2 \; u_s(p)\bar{u}_s(p) & =-ip \cdot \gamma+m, \\ \sum _{s=1}^2 \; v_s(p)\;\bar{v}_s(p) & = -ip \cdot \gamma-m = -\sum _s u_s(-p)\;\bar{u}_s(-p). \end{split} \end{equation*} Because it appears twice, the additional sign cancels for $$|\mathcal{T}|^2$$ after spin averaging and one finds indeed the same result as for $$e^{-}e^{+} \to \mu ^{-}\mu ^{+}$$ but with \begin{equation*} \begin{split} s_q = & -(q_1+q_2)^2 = -(p_1-p_4)^2 = u_p, \\ t_q = & -(q_1 -q_3)^2 = -(p_1+p_2)^2 = s_p,\\ u_q = & -(q_1 -q_4)^2 = -(p_1 -p_3)^2 =t_p. \end{split} \end{equation*} We can take what we had calculated but must change the role of $$s$$, $$t$$and $$u$$ ! This is an example of crossing symmetries.

Electron-muon scattering in the massless limit. Recall that we found for $$e^{-}e^{+} \to \mu ^{-}\mu ^{+}$$ in the massless limit $$m_e = m_\mu = 0$$ simply \begin{equation*} \frac{1}{4}\sum _\text{spins} |\mathcal{T}|^2 = \frac{2e^4}{s^2} \left [t^2+u^2 \right ]. \end{equation*} For $$e^{-}\mu ^{-} \to e^{-}\mu ^{-}$$ we find after the replacements $$u \to s$$, $$s\to t$$, $$t \to u$$, \begin{equation*} \frac{1}{4} \sum _\text{spins}|\mathcal{T}|^2=\frac{2e^4}{t^2}\left [u^2+ s^2\right ]. \end{equation*}

More on Mandelstam variables. To get a better feeling for $$s$$, $$t$$ and $$u$$, let us evaluate them in the center of mass frame for a situation where all particles have mass $$m$$.

\begin{equation*} \begin{split} p_1^\mu & =(E,\vec{p}),\quad \quad p_2^\mu =(E,-\vec{p}), \\ p_3^\mu & =(E,\vec{p}'), \quad \quad p_4^\mu =(E, -\vec{p}'). \\ \end{split} \end{equation*} While $$s$$ measures the center of mass energy, $$t$$ is a momentum transfer that vanishes in the soft limit $$\vec{p}^2 \to 0$$ and in the colinear limit $$\theta \to 0.$$ Similarly, $$u$$ vanishes for $$\vec{p}^2 \to 0$$ and for backward scattering $$\theta \to \pi .$$

$$s$$-, $$t$$- and $$u$$-channels. One speaks of interactions in different channels for tree diagrams of the following generic types,

Electron-muon scattering. For the cross section we find for $$e^{-}\mu ^{-} \to e^{-}\mu ^{-}$$ in the massless limit \begin{equation*} \frac{d\sigma }{d\Omega } = \frac{1}{64\pi ^2 s} \frac{1}{4} \sum _{\rm spins} |\mathcal{T}|^2 = \frac{\alpha ^2[4+(1+\cos \theta )^2]}{2s(1-\cos \theta )^2} \end{equation*} This diverges in the colinear limit $$\theta \to 0$$ as we had already seen for Yukawa theory in the limit where the exchange particle becomes massless.

Note that by the definition $$s \geq 0$$ while $$u$$ and $$t$$ can have either sign. Replacements of the type used for crossing symmetry are in this sense always to be understood as analytic continuation.

#### 10.6 Relativistic scattering and decay kinematics

Covariant normalization of asymptotic states. For non-relativistic physics this we have used a normalization of single particle states in the asymptotic incoming and out-going regimes such that \begin{equation*} \langle \vec{p}|\vec{q}\rangle = (2\pi )^3 \delta ^{(3)}(\vec{p}-\vec{q}). \end{equation*} For relativistic physics this has the drawback that it is not Lorentz invariant. To see this let us consider a boost in $$z$$-direction \begin{equation*} \begin{split} E^\prime =& \gamma (E+\beta p^3), \\ p^{1\prime } = & p^{1}, \\ p^{2\prime } = & p^2, \\ p^{3\prime } = & \gamma (p^3+\beta E). \end{split} \end{equation*} Using the identity \begin{equation*} \delta \left (f(x) - f(x_0) \right ) = \frac{1}{|f'(x_0)|}\delta (x-x_0), \end{equation*} one finds \begin{equation*} \begin{split} \delta ^{(3)}(\vec{p}-\vec{q}) &= \delta ^{(3)}(\vec{p}^\prime -\vec{q}^\prime ) \frac{dp^{3\prime }}{dp^3} = \delta ^{(3)}(\vec{p}-\vec{q}) \gamma \left (1+\beta \frac{dE}{dp^3} \right )\\ &= \delta ^{(3)}(\vec{p}'-\vec{q}') \frac{1}{E}\gamma \left (E+\beta p^3 \right )\\ &= \frac{E^\prime }{E} \delta ^{(3)} (\vec{p}^\prime - \vec{q}^\prime ). \end{split} \end{equation*} This shows, however, that $$E\;\delta ^{(3)} (\vec{p}-\vec{q})$$ is in fact Lorentz invariant.

This motivates to change the normalization such that \begin{equation*} |p; \text{in}\rangle = \sqrt{2E_p} a^\dagger _{\vec{p}}(-\infty ) |0\rangle = \sqrt{2E_{\vec{p}}} \, |\vec{p}; \text{in} \rangle . \end{equation*} Note the subtle difference in notation between $$|p; \text{in}\rangle$$ (relativistic normalization) and $$|\vec p; \text{in}\rangle$$ (non-relativistic normalization). This implies for example \begin{equation*} \langle p; \text{in}| q; \text{in} \rangle = 2E_p (2\pi )^3 \delta ^{(3)}(\vec{p}-\vec{q}). \end{equation*} With this normalization we must divide by $$2E_p$$ at the same places. In particular the completeness relation for single particle incoming states is \begin{equation*} \mathbb{1}_{1-\text{particle}} = \int \frac{d^3p}{(2\pi )^3} \frac{1}{2E_{\vec{p}}}|p;\text{in}\rangle \langle p; \text{in}|. \end{equation*} In fact, what appears here is a Lorentz invariant momentum measure. To see this consider \begin{equation*} \int \frac{d^4 p}{(2\pi )^4}\;(2\pi )\;\delta (p^2+m^2)\;\theta (p^0) = \int \frac{d^3p}{(2\pi )^3} \frac{1}{2E_{\vec{p}}}. \end{equation*} The left hand side is explicitly Lorentz invariant and so is the right hand side.

Covariantly normalized S-matrix. We can use the covariant normalization of states also in the definition of S-matrix elements. The general definition is as before \begin{equation*} S_{\beta \alpha } = \langle \beta ; \text{out}| \alpha ; \text{in}\rangle = \delta _{\beta \alpha }+ i \, \mathcal{T}_{\beta \alpha }(2\pi )^4 \delta ^{(4)}(p^\text{in}-p^\text{out}). \end{equation*} But now we take elements with relativistic normalization, e.g. for $$2 \to 2$$ scattering \begin{equation*} S_{q_1q_2, p_1p_2} = \langle q_1,q_2; \text{out}| p_1,p_2 ; \text{in}\rangle . \end{equation*} We can calculate these matrix elements as before using the LSZ reduction formula to replace $$\sqrt{2E_p} a^\dagger _{\vec{p}}(-\infty )$$ by fields. For example, for relativistic scalar fields \begin{equation*} \sqrt{2 E_{\vec{p}}}\; a^\dagger _{\vec{p}}(-\infty ) = \sqrt{2E_{\vec{p}}} \; a^\dagger _{\vec p}(\infty ) + i\left [-(p^0)^2+ \vec{p}^2 + m^2\right ] \phi ^{*}(p). \end{equation*} This allows to calculate S-matrix elements through correlation functions.

Cross sections for $$2\to n$$ scattering. Let us now generalize our discussion of $$2 \to 2$$ scattering of non-relativistic particles to a scattering $$2 \to n$$ of relativistic particles. The transition probability is as before \begin{equation*} P = \frac{|\langle \beta ; \text{out}| \alpha ; \text{in}\rangle |^2}{\langle \beta ; \text{out}| \beta ;\text{out}\rangle \langle \alpha ; \text{in}| \alpha ; \text{in}\rangle }. \end{equation*} Rewriting the numerator in terms of $$\mathcal{T}_{\beta \alpha }$$ and going over to the transition rate we obtain as before $$\dot{P} = \frac{V (2\pi )^4 \delta ^{(4)}(p^\text{out}-p^\text{in})|\mathcal{T}|^2}{\langle \beta ; \text{out}| \beta ;\text{out}\rangle \langle \alpha ; \text{in}| \alpha ; \text{in}\rangle }. \label{eq:PDot}$$ But now states are normalized in a covariant way \begin{equation*} \begin{split} \langle p| p \rangle &= \lim _{q \to p} \langle p|q \rangle \\ &= \lim _{q \to p} 2E_p (2\pi )^3 \delta ^{(3)}(\vec{p}-\vec{q})\\ &= 2E_p (2\pi )^3 \delta ^{(3)}(0)\\ &= 2E_p V \end{split} \end{equation*} One has thus for the incoming state of two particles \begin{equation*} \langle \alpha ; \text{in}| \alpha ; \text{in}\rangle = 4E_1 E_2 V^2. \end{equation*} For the outgoing state of $$n$$ particles one has instead \begin{equation*} \langle \beta ; \text{out}| \beta ; \text{out}\rangle = \prod _{j=1}^n \{2q^0_j V\}. \end{equation*} The product goes over final state particles which have the four-momentum $$q_j^n$$. So, far we have thus \begin{equation*} \dot{P} = \frac{V\;(2\pi )^4\;\delta ^{(4)}(p^\text{in} - p^\text{out})|\mathcal{T}|^2}{4E_1 E_2 V^2\; \prod _{j=1}^n \{2q_j^0 V\} }. \end{equation*}

Lorentz invariant phase space. To count final state momenta appropriately we could go back to finite volume and then take the continuum limit. This leads to an additional factor \begin{equation*} \sum _{\vec{n}_j} \to V\;\int \frac{d^3 q}{(2\pi )^3} \end{equation*} for each final state particle. The transition rate becomes \begin{equation*} \dot{P} = \frac{|\mathcal{T}|^2}{4E_1 E_2 V} \left [(2\pi )^4\;\delta ^{(4)}{\Big (}p^\text{in}-\sum _j q_j{\Big )} \prod _{j=1}^n \left \{\frac{d^3 q_j}{(2\pi )^3 2q^0_j}\right \}\right ] \end{equation*} The expression in square brackets is known as the Lorentz-invariant phase space measure (sometimes ”LIPS”).

Flux and differential cross section. To go from there to a differential cross section we need to divide by a flux of particles. There is one particle per volume $$V$$ with velocity $$v=v_1 - v_2$$, so the flux is \begin{equation*} \mathcal{F} = \frac{|v|}{V} = \frac{|v_1-v_2|}{V} = \frac{\left |\frac{p_1^3}{p_1^0}-\frac{p_2^3}{p_2^0}\right |}{V}. \end{equation*} In the last equality we chose the beam axis to coincide with the $$z$$-axis. For the differential cross section we obtain \begin{equation*} d\sigma = \frac{|\mathcal{T}|^2}{4E_1 E_2 |v_1-v_2|}\; [\text{LIPS}]. \end{equation*} The expression in the prefactor can be rewritten like \begin{equation*} \frac{1}{E_1 E_2 |v_1- v_2|} = \frac{1}{p_1^0 p_2^0 \left |\frac{p_1^3}{p_1^0}-\frac{p_2^3}{p_2^0}\right |} = \frac{1}{| p_2^0 p_1^3 - p_1^0 p_2^3 |} = \frac{1}{|\epsilon _{\mu x y \nu } p_2^\mu p_1^\nu |}. \end{equation*} This is not Lorentz invariant in general but invariant under boosts in the $$z$$-direction. In fact it transforms as a two-dimensional area element as it should.

Differential cross section in the centre of mass frame. In the center of mass frame one has $$p_2^3 = -p_1^3 = \pm |\vec{p}_1|$$ and \begin{equation*} \frac{1}{|p_2^0 p_1^3 -p_1^0 p_2^0|} = \frac{1}{|\vec{p}_1|(p_1^0 + p_2^0)} = \frac{1}{|\vec{p}_1|_\text{COM}\sqrt{s}} \end{equation*} This leads finally to the result for the differential cross section \begin{equation*} d\sigma = \frac{|\tau |^2}{4 |\vec{p}_1|_{\text{COM}}\sqrt{s}} \left [ (2\pi )^4 \delta ^{(4)}{\Big (}p^\text{in}-\sum _j q_j{\Big )} \prod _{j=1}^n \left \{\frac{d^3 q_j}{(2\pi )^3 2q_j^0}\right \}\right ]. \end{equation*}

$$2 \to 2$$scattering. For the case of $$n=2$$ one can write the Lorentz invariant differential phase space element in the center of mass frame (exercise) \begin{equation*} \left [(2\pi )^4 \; \delta ^{(4)} (p^{\text{in}} -q_1 -q_2) \frac{d^3 q_1}{(2\pi )^3 2q_1^0}\; \frac{d^3 q_2}{(2\pi )^3 q_2^0}\right ] = \frac{|\vec{q}_1|}{16 \pi ^2 \sqrt{s}}d\Omega \end{equation*} such that \begin{equation*} \frac{d\sigma }{d\Omega } = \frac{1}{64\pi ^2 s} \frac{|\vec{q}_1|}{|\vec{p}_1|} |\mathcal{T}|^2. \end{equation*}

Decay rate. Let us now consider the decay rate of a single particle, i. e. a process $$1 \to n$$. We can still use equation \eqref{eq:PDot}, but now the initial state is normalized like \begin{equation*} \langle \alpha ; \text{in}|\alpha ; \text{in}\rangle = 2E_1 V. \end{equation*} We find then for the differential transition or decay rate $$d\Gamma = \dot{P}$$ \begin{equation*} d\Gamma = \frac{|\mathcal{T}|^2}{2E_1}\left [(2\pi )^4 \delta ^{(4)}{\Big (}p^{\text{in}}-\sum _j q_j{\Big )} \prod _{j=1}^n \left \{\frac{d^3 q_j}{(2\pi )^3 2q_j^0}\right \}\right ] \end{equation*} In the center of mass frame one has $$E_1 = m_1$$. For the special case of $$1\to 2$$ decay one finds in the center of mass frame or rest frame of the initial particle \begin{equation*} d\Gamma = \frac{|\mathcal{T}|^2 |\vec{q}_1|}{32 \pi ^2 m_1^2} d\Omega . \end{equation*}

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