Prof. Dr. Stefan Floerchinger

Researcher and professor for quantum field theory.

Mode expansion for Dirac fields

We will also need a mode expansion for free Dirac fields in order to describe asymptotic (incoming and outgoing) fermion states. We write the fields as \[\begin{split} \Psi(x) = & \sum_{s=1}^2 \int \frac{d^3 p}{(2\pi)^3}\frac{1}{\sqrt{2E_\mathbf{p}}}\left\{b^s_{\mathbf{p}}\; u_s(p)\; e^{ipx} + d^{s\dagger}_{\mathbf{p}}\;v_s(p)\;e^{-ipx}\right\}, \\ \bar \Psi(x) = & \sum_{s=1}^2 \int \frac{d^3 p}{(2\pi)^3}\frac{1}{\sqrt{2E_\mathbf{p}}}\left\{ d^s_{\mathbf{p}}\;\bar{v}_s(p)\;e^{ipx} + b^{s\dagger}_{\mathbf{p}}\; \bar{u}_s(p)\; e^{-ipx} \right\}. \end{split}\] Here, \(b^s_{\mathbf{p}}\), \(d^s_{\mathbf{p}}\) etc. can be seen as expansion coefficients and become operators in the operator picture. Because of the fermionic or Grassmann exchange symmetry, they fulfill actually anti-commutation relations, \[\{ b_\mathbf{p}^r, b^{s\dagger}_\mathbf{q} \} = \{ d_\mathbf{p}^r, d^{s\dagger}_\mathbf{q} \} = (2\pi)^3 \delta^{rs} \delta^{(3)}(\mathbf{p} - \mathbf{q}),\] with all other anti-commutators vanishing. The index \(s\) sums over independent spin states. One can interpret \(b_\mathbf{p}^{s\dagger}\) as a creation opertors for fermionoc particles (such as electrons) with momentum \(\mathbf{p}\) and spin \(s\), and similarly \(d_\mathbf{p}^{s\dagger}\) as the corresponding creation operator for anti-particles (such as positrons). The fact that there are particles and anti-particles arises from the “complex” nature of Dirac spionors; for Majorana fermions there would be only one kind of creation operator.

Solutions of Dirac equation

The Dirac equation \[(\gamma^{\mu} \partial_{\mu} + m)\psi(x) = 0,\] becomes for the plane waves \[\begin{split} (i\slashed{p} + m)\;u_s(\mathbf{p}) & =0, \\ (-i\slashed{p} + m )\; v_s(\mathbf{p}) & =0, \end{split}\] with the Dirac slash notation \(\slashed{p} = \gamma^{\mu} p_{\mu}\). We consider this first in the frame where the spatial momentum vanishes, \(\mathbf{p}=0\), such that \(p^{\mu}=(m,0,0,0)\), \[\slashed{p} = -\gamma^{0}m = im \begin{pmatrix} & \mathbb{1} \\ \mathbb{1} & \end{pmatrix}.\] The last equation holds in the chiral basis where \[\gamma^{\mu} = \begin{pmatrix} 0 & -i \sigma^{\mu} \\ -i \bar\sigma^{\mu} & 0 \end{pmatrix}.\] with \(\sigma^{\mu} = (\mathbb{1},\vec{\sigma})\) and \(\bar\sigma^{\mu} = (\mathbb{1},-\vec{\sigma}).\) For the spinor \(u_s(\mathbf{0})\) one has the equation \[(i \slashed{p}+m) u_s(\mathbf{0}) = m \begin{pmatrix} +\mathbb{1} & -\mathbb{1} \\ -\mathbb{1} & +\mathbb{1} \end{pmatrix} u_s(\mathbf{0}) = 0.\] The two independent solutions are \[u_1(\mathbf{0})=\sqrt{m}\begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix},\quad\quad\quad u_2(\mathbf{0}) = \sqrt{m}\begin{pmatrix} 0 \\ 1 \\ 0 \\ 1 \end{pmatrix}.\] The normalization has been chosen for later convenience. Similarly \[(-i\slashed{p}+m) v_s(\mathbf{0}) = m \begin{pmatrix} \mathbb{1} & \mathbb{1} \\ \mathbb{1} & \mathbb{1} \end{pmatrix} v_s(\mathbf{0}) = 0,\] has the two independent solutions \[v_1(\mathbf{0})=\sqrt{m}\begin{pmatrix} 0 \\ +1 \\ 0 \\ -1 \end{pmatrix},\quad\quad\quad v_2(\mathbf{0}) = \sqrt{m}\begin{pmatrix} -1 \\ 0 \\ +1 \\ 0 \end{pmatrix}.\] We see here that the Dirac equation has two independent solutions (for spin up and and down with respect to some basis) for particles and two more for anti-particles. One can now go to an arbitrary reference frame by performing a Lorentz transformation to obtain \(u_s(\mathbf{p})\), \(v_s(\mathbf{p})\) and their conjugates \(= u^\dagger_s(\mathbf{p})\beta\) as well as \(\bar v_s(\mathbf{p}) = v^\dagger_s(\mathbf{p})\beta\). We derive more identities about these objects when we need them later on.

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 \[\langle\mathbf{p}|\mathbf{q}\rangle = (2\pi)^3 \delta^{(3)}(\mathbf{p}-\mathbf{q}).\] For relativistic physics this has the drawback that it is not Lorentz invariant. To see this let us consider a boost in \(z\)-direction \[\begin{split} E^\prime =& \gamma(E+\beta p^3), \\ p^{1\prime} = & p^{1}, \\ p^{2\prime} = & p^2, \\ p^{3\prime} = & \gamma(p^3+\beta E). \end{split}\] Using the identity \[\delta\left(f(x) - f(x_0) \right) = \frac{1}{|f'(x_0)|}\delta(x-x_0),\] one finds \[\begin{split} \delta^{(3)}(\mathbf{p}-\mathbf{q}) &= \delta^{(3)}(\mathbf{p}^\prime-\mathbf{q}^\prime) \frac{dp^{3\prime}}{dp^3} = \delta^{(3)}(\mathbf{p}-\mathbf{q}) \gamma \left(1+\beta \frac{dE}{dp^3} \right)\\ &= \delta^{(3)}(\mathbf{p}'-\mathbf{q}') \frac{1}{E}\gamma \left(E+\beta p^3 \right)\\ &= \frac{E^\prime}{E} \delta^{(3)} (\mathbf{p}^\prime- \mathbf{q}^\prime). \end{split}\] This shows, however, that \(E\;\delta^{(3)} (\mathbf{p}-\mathbf{q})\) is in fact Lorentz invariant.

This motivates to change the normalization such that \[|p; \text{in}\rangle = \sqrt{2E_p} a^\dagger_{\mathbf{p}}(-\infty) |0\rangle = \sqrt{2E_{\mathbf{p}}} \, |\mathbf{p}; \text{in} \rangle.\] 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 \[\langle p; \text{in}| q; \text{in} \rangle = 2E_p (2\pi)^3 \delta^{(3)}(\mathbf{p}-\mathbf{q}).\] With this normalization we must divide by \(2E_p\) at the same places. In particular the completeness relation for single particle incoming states is \[\mathbb{1}_{1-\text{particle}} = \int \frac{d^3p}{(2\pi)^3} \frac{1}{2E_{\mathbf{p}}}|p;\text{in}\rangle\langle p; \text{in}|.\] In fact, what appears here is a Lorentz invariant momentum measure. To see this consider \[\int \frac{d^4 p}{(2\pi)^4}\;(2\pi)\;\delta(p^2+m^2)\;\theta(p^0) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{2E_{\mathbf{p}}}.\] 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 \[S_{\beta\alpha} = \langle \beta; \text{out}| \alpha; \text{in}\rangle = \delta_{\beta\alpha}+ i \, \mathcal{T}_{\beta\alpha}(2\pi)^4 \delta^{(4)}(p^\text{in}-p^\text{out}).\] But now we take elements with relativistic normalization, e.g. for \(2 \to 2\) scattering \[S_{q_1q_2, p_1p_2} = \langle q_1,q_2; \text{out}| p_1,p_2 ; \text{in}\rangle.\] We can calculate these matrix elements as before using the LSZ reduction formula to replace \(\sqrt{2E_p} a^\dagger_{\mathbf{p}}(-\infty)\) by fields. For example, for relativistic scalar fields \[\sqrt{2 E_{\mathbf{p}}}\; a^\dagger_{\mathbf{p}}(-\infty) = \sqrt{2E_{\mathbf{p}}} \; a^\dagger_{\vec p}(\infty) + i\left[-(p^0)^2+ \mathbf{p}^2 + m^2\right] \phi^{*}(p).\] 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 \[P = \frac{|\langle \beta; \text{out}| \alpha; \text{in}\rangle|^2}{\langle \beta; \text{out}| \beta;\text{out}\rangle \langle \alpha; \text{in}| \alpha; \text{in}\rangle}.\] 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 \[\begin{split} \langle p| p \rangle &= \lim_{q \to p} \langle p|q \rangle\\ &= \lim_{q \to p} 2E_p (2\pi)^3 \delta^{(3)}(\mathbf{p}-\mathbf{q})\\ &= 2E_p (2\pi)^3 \delta^{(3)}(0)\\ &= 2E_p V \end{split}\] One has thus for the incoming state of two particles \[\langle \alpha; \text{in}| \alpha; \text{in}\rangle = 4E_1 E_2 V^2.\] For the outgoing state of \(n\) particles one has instead \[\langle \beta; \text{out}| \beta; \text{out}\rangle = \prod_{j=1}^n \{2q^0_j V\}.\] The product goes over final state particles which have the four-momentum \(q_j^n\). So, far we have thus \[\dot{P} = \frac{V\;(2\pi)^4\;\delta^{(4)}(p^\text{in} - p^\text{out})|\mathcal{T}|^2}{4E_1 E_2 V^2\; \prod_{j=1}^n \{2q_j^0 V\} }.\]

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 \[\sum_{\vec{n}_j} \to V\;\int\frac{d^3 q}{(2\pi)^3}\] for each final state particle. The transition rate becomes \[\dot{P} = \frac{|\mathcal{T}|^2}{4E_1 E_2 V} \left[(2\pi)^4\;\delta^{(4)}{\Big (}p^\text{in}-\sum_j q_j {\Big )} \prod_{j=1}^n \left\{\frac{d^3 q_j}{(2\pi)^3 2q^0_j}\right\}\right]\] 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 \[\mathcal{F} = \frac{|v|}{V} = \frac{|v_1-v_2|}{V} = \frac{\left|\frac{p_1^3}{p_1^0}-\frac{p_2^3}{p_2^0}\right|}{V}.\] In the last equality we chose the beam axis to coincide with the \(z\)-axis. For the differential cross section we obtain \[d\sigma = \frac{|\mathcal{T}|^2}{4E_1 E_2 |v_1-v_2|}\; [\text{LIPS}].\] The expression in the prefactor can be rewritten like \[\frac{1}{E_1 E_2 |v_1- v_2|} = \frac{1}{p_1^0 p_2^0 \left|\frac{p_1^3}{p_1^0}-\frac{p_2^3}{p_2^0}\right|} = \frac{1}{| p_2^0 p_1^3 - p_1^0 p_2^3 |} = \frac{1}{|\epsilon_{\mu x y \nu} p_2^\mu p_1^\nu|}.\] 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 \(p_2^3 = -p_1^3 = \pm |\mathbf{p}_1|\) and \[\frac{1}{|p_2^0 p_1^3 -p_1^0 p_2^0|} = \frac{1}{|\mathbf{p}_1|(p_1^0 + p_2^0)} = \frac{1}{|\mathbf{p}_1|_\text{COM}\sqrt{s}}\] This leads finally to the result for the differential cross section \[d\sigma = \frac{|\mathcal{T}|^2}{4 |\mathbf{p}_1|_{\text{COM}}\sqrt{s}} \left[ (2\pi)^4 \delta^{(4)}{\Big(}p^\text{in}-\sum_j q_j {\Big )} \prod_{j=1}^n \left\{\frac{d^3 q_j}{(2\pi)^3 2q_j^0}\right\}\right].\]

\(2 \to 2\) scattering

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

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 \[\langle \alpha; \text{in}|\alpha; \text{in}\rangle = 2E_1 V.\] We find then for the differential transition or decay rate \(d\Gamma = \dot{P}\) \[d\Gamma = \frac{|\mathcal{T}|^2}{2E_1}\left[(2\pi)^4 \delta^{(4)}{\Big (}p^{\text{in}}-\sum_j q_j {\Big )} \prod_{j=1}^n \left\{\frac{d^3 q_j}{(2\pi)^3 2q_j^0}\right\}\right]\] 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 \[d\Gamma = \frac{|\mathcal{T}|^2 |\mathbf{q}_1|}{32 \pi^2 m_1^2} d\Omega.\]

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