Question 3.4: The Inclusive Weak Lepton–Nucleon Scattering The coupling of......
The Inclusive Weak Lepton-Nucleon Scattering
The coupling of the charged vector bosons W_{\mu}^{ \pm}of the weak interaction to leptons is described by the interaction Lagrangean (4.23),
L_{\mathrm{int}}=\frac{+g}{\sqrt{2}} \bar{\Psi}_{\mathrm{f}}\left(\hat{T}_{-} W_{\mu}^{+}+\hat{T}_{+} W_{\mu}^{-}\right) \gamma^{\mu} \Psi_{\mathrm{f}}, (1)
where \Psi_{\mathrm{f}}=\left(\begin{array}{c}\nu_{\mathrm{e}} \\ \mathrm{e}\end{array}\right)_{\mathrm{L}} denotes the doublet of the left-handed electron-neutrino field: e_{\mathrm{L}}=\frac{1-\gamma_{5}}{2} e_{\mathrm{L}}. Writing (1) we have used \hat{T}_{ \pm}=\frac{1}{\sqrt{2}}\left(\frac{\lambda_{1}}{2} \pm \mathrm{i} \frac{\lambda_{2}}{2}\right). With \hat{T}_{-}= \left(\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right) and \hat{T}_{+}=\left(\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right) we get for the fundamental (e, \bar{\nu}) interaction
\frac{+g}{\sqrt{2}} \bar{\nu}(x) \gamma_{\mu} \frac{1-\gamma_{5}}{2} \mathrm{e}(x) W^{\mu}(x)+\text { h.c. . } (1a)
The corresponding processes are depicted in the figure. Derive from this equation the leptonic scattering tensor L_{\mu \nu}. Employ the result of Exercise 3.3 to evaluate the differential cross section for the inclusive weak lepton-nucleon scattering.
Since the neutrino is massless and the electron mass can be neglected, it is helpful to choose the following spinor normalization.
\begin{aligned} u_{\alpha}^{\dagger}(p) u_{\beta}(p) & =2 E \delta_{\alpha \beta}, \\ \sum_{\text {spin }} u(p) \bar{u}(p) & =\not p+m=\not p. & (2) \end{aligned}
The leptonic scattering tensor is then
\begin{aligned} L_{\mu \nu} & =\frac{g^{2}}{8} \sum_{s} \operatorname{tr}\left\{\bar{\nu}(p) \gamma_{\mu}\left(1-\gamma_{5}\right) \mathrm{e}\left(p^{\prime}, s\right) \overline{\mathrm{e}}\left(p^{\prime}, s\right) \gamma_{\nu}\left(1-\gamma_{5}\right) \nu(p)\right\} \\ & =\frac{g^{2}}{8} \operatorname{tr}\left\{\not p \gamma_{\mu}\left(1-\gamma_{5}\right) \not p^{\prime} \gamma_{\nu}\left(1-\gamma_{5}\right)\right\} \quad \text { (utilizing (2))) } \end{aligned}
\begin{aligned} & =\frac{g^{2}}{4}\left(\operatorname{tr}\left\{\not p \gamma_{\mu} \not p^{\prime} \gamma_{\nu}\right\}-\operatorname{tr}\left\{\gamma_{5} \not p^{\prime} \gamma_{\nu} \not p \gamma_{\mu}\right\}\right) \\ & =g^{2}\left(p_{\mu}^{\prime} p_{\nu}+p_{\mu} p_{\nu}^{\prime}-p^{\prime} \cdot p g_{\mu \nu}-\mathrm{i} \varepsilon_{\mu \nu \alpha \beta} p^{\alpha} p^{\prime \beta}\right) . & (3) \end{aligned}
Since the neutrino is always left-handed, there is no averaging of the initial spin. Nevertheless, in order to obtain the trace in the second line of (3) we have summed over all neutrino spins and made use of (2). This procedure is correct, because the operator \left(1-\gamma_{5}\right) cancels the contribution of the nonphysical righthanded neutrino state. In analogy to (3.22)
\frac{\mathrm{d}^2 \sigma}{\mathrm{d} E^{\prime} \mathrm{d} \Omega}=\frac{E^{\prime} \alpha^2}{E Q^4} L^{\mu \nu} W_{\mu \nu} . (3.22)
the differential cross section then becomes
\frac{\mathrm{d}^{2} \sigma}{\mathrm{d} E^{\prime} \mathrm{d} \Omega}=\frac{E^{\prime}}{(4 \pi)^{2} E}\left(\frac{1}{q^{2}-M_{\mathrm{W}}^{2}}\right)^{2} L^{\mu \nu}\left(\frac{g}{\sqrt{2}}\right)^{2} W_{\mu \nu} . (4)
The differences can be explained as follows.
(1) The propagator for massive \mathrm{W} bosons differs from that for massless photons. For \left|q^{2}\right| \ll M_{\mathrm{W}}^{2} we can use
\left(\frac{1}{q^{2}-M_{\mathrm{W}}^{2}}\right)^{2} \approx \frac{1}{M_{\mathrm{W}}^{4}}. (5)
This yields Fermi’s theory of weak interaction.
(2) The coupling constant of the \mathrm{W} bosons to the nucleon is effectively chosen to be G_{\mathrm{F}} / \sqrt{2} where G_{\mathrm{F}}=g^{2} /\left(4 \sqrt{2} M_{\mathrm{W}}^{2}\right) is the Fermi constant of weak interaction. Since we have already absorbed a factor of g^{2} / 8 (see (3)) in the definition of the leptonic tensor we are left with a factor of g^{2} /(4 \cdot 2) in front of the hadronic tensor W_{\mu \nu}, which according to (3.11)
W_{\mu \nu}=\frac{1}{2} \sum_{\mathrm{spin}}\left[\bar{u}\left(P^{\prime}\right) \Gamma_\mu u(P)\right]^*\left[\bar{u}\left(P^{\prime}\right) \Gamma_\nu u(P)\right] (3.11)
does not contain any coupling constant. (3) The factor 1 /(4 \pi)^{2} arises by replacing \alpha^{2}=e^{4} /(4 \pi)^{2} \rightarrow g^{4} /(4 \pi)^{2}. One factor g^{2} is contained in L_{\mu \nu}, the other one appears explicitly in (4).
In order to evaluate (4) we first show that q^{\mu} L_{\mu \nu} and q^{\nu} L_{\mu \nu} vanish:
\begin{aligned} q^{\mu} L_{\mu \nu} & =g^{2}\left(q \cdot p^{\prime} p_{\nu}+q \cdot p p_{\nu}^{\prime}-p^{\prime} \cdot p q_{\nu}-\mathrm{i} \varepsilon_{\mu \nu \alpha \beta} q^{\mu} p^{\alpha} p^{\prime \beta}\right) \\ & =g^{2}\left[-\frac{q^{2}}{2} p_{\nu}+\frac{q^{2}}{2} p_{\nu}^{\prime}+\frac{q^{2}}{2}\left(p_{\nu}-p_{\nu}^{\prime}\right)\right]=0 & (6) \end{aligned}
Here we have employed
\begin{gathered} q_{\mu}=p_{\mu}-p_{\mu}^{\prime} \quad \Rightarrow \quad p \cdot p^{\prime}=-\frac{q^{2}}{2}, \\ p \cdot q=\frac{q^{2}}{2} \quad \text { and } \quad p^{\prime} \cdot q=-\frac{q^{2}}{2}, & (7) \end{gathered}
which all hold in the high-energy limit m^{2} \ll q^{2}. With help of 15 from Exercise 3.3 it therefore follows that
\begin{aligned} L^{\mu \nu} W_{\mu \nu}= & L^{\mu \nu}\left(g_{\mu \nu} V_{1}+P_{\mu} P_{\nu} V_{2}+\mathrm{i} \varepsilon_{\mu \nu \alpha \beta} P^{\alpha} q^{\beta} V_{5}\right) \\ = & g^{2}\left[\left(2 p^{\prime} \cdot p-4 p^{\prime} \cdot p\right) V_{1}+\left(2 p^{\prime} \cdot P p \cdot P-M_{N}^{2} p \cdot p^{\prime}\right) V_{2}\right. \left.+\varepsilon_{\alpha \beta}^{\mu \nu} \varepsilon_{\mu \nu \delta \gamma} p^{\delta} q^{\gamma} P^{\alpha} q^{\beta} V_{5}\right] . & (8) \end{aligned}
With
p=(E, {p}), \quad p^{\prime}=\left(E^{\prime}, {p}^{\prime}\right), \quad P=(M, \mathrm{0}) (9)
and ^{8}
\varepsilon_{\alpha \beta}^{\mu \nu} \varepsilon_{\mu \nu \delta \gamma}=-2\left(g_{\alpha \delta} g_{\beta \gamma}-g_{\alpha \gamma} g_{\beta \delta}\right) (10)
and, also, utilizing relations (7) this assumes the form
L^{\mu \nu} W_{\mu \nu}=g^{2}\left[q^{2} V_{1}+M_{\mathrm{N}}^{2}\left(2 E E^{\prime}+\frac{1}{2} q^{2}\right) V_{2}-2\left(p \cdot P q^{2}-P \cdot q p \cdot q\right) V_{5}\right]
=g^{2}\left\{q^{2} V_{1}+2 M_{\mathrm{N}}^{2}\left(E E^{\prime}+\frac{1}{4} q^{2}\right) V_{2}-2 M_{\mathrm{N}}\left[E q^{2}-\left(E-E^{\prime}\right) \frac{q^{2}}{2}\right] V_{5}\right\} . (11)
Again, using (3.30),
Q^{2}=2\;p\cdot p^{\prime}=2(E E^{\prime}-p\cdot p^{\prime})=4E E^{\prime}\sin^{2}\left({\frac{\theta}{2}}\right)\;. (3.30)
we find that
q^{2}=-Q^{2}=-4 E E^{\prime} \sin ^{2}\left(\frac{\vartheta}{2}\right) (12)
leads to
\begin{aligned} L^{\mu \nu} W_{\mu \nu}= & g^{2} 2 E E^{\prime}\left[-2 \sin ^{2}\left(\frac{\vartheta}{2}\right) V_{1}+M_{\mathrm{N}}^{2} \cos ^{2}\left(\frac{\vartheta}{2}\right) V_{2}\right. \left.+2 M_{\mathrm{N}} \sin ^{2}\left(\frac{\vartheta}{2}\right)\left(E+E^{\prime}\right) V_{5}\right] . \end{aligned} (13)
In order to provide all structure functions with the same dimension we define
W_{1}=-V_{1}, \quad W_{2}=V_{2} M_{\mathrm{N}}^{2}, \quad W_{3}=-2 V_{5} M_{\mathrm{N}}^{2} (14)
and obtain
\begin{aligned} \frac{\mathrm{d}^{2} \sigma}{\mathrm{d} E^{\prime} \mathrm{d} \Omega}= & \frac{E^{\prime 2} g^{4}}{4 M_{\mathrm{W}}^{4}(4 \pi)^{2}}\left[2 \sin ^{2}\left(\frac{\vartheta}{2}\right) W_{1}\left(Q^{2}, \nu\right)+\cos ^{2}\left(\frac{\vartheta}{2}\right) W_{2}\left(Q^{2}, \nu\right)\right. \left.-\frac{E+E^{\prime}}{M_{\mathrm{N}}} \sin ^{2}\left(\frac{\vartheta}{2}\right) W_{3}\left(Q^{2}, \nu\right)\right] . \end{aligned} (15)
Finally we substitute the so-called Fermi constant of weak interaction:
\begin{aligned} G_{\mathrm{F}}= & \frac{g^{2}}{4 \sqrt{2} M_{\mathrm{W}}^{2}} ;& (16) \\ \frac{\mathrm{d}^{2} \sigma}{\mathrm{d} E^{\prime} \mathrm{d} \Omega}= & \frac{G_{\mathrm{F}}^{2} E^{\prime 2}}{2 \pi^{2}}\left[2 \sin ^{2}\left(\frac{\vartheta}{2}\right) W_{1}\left(Q^{2}, \nu\right)+\cos ^{2}\left(\frac{\vartheta}{2}\right) W_{2}\left(Q^{2}, \nu\right)\right. \left.-\frac{E+E^{\prime}}{M_{\mathrm{N}}} \sin ^{2}\left(\frac{\vartheta}{2}\right) W_{3}\left(Q^{2}, \nu\right)\right] . & (17) \end{aligned}
This is identical to (3.37).
\frac{\mathrm{d}^{2}\sigma}{\mathrm{d}E^{\prime}\mathrm{d}\Omega}\bigg|_{\nu\mathrm{N}}=\frac{G_{\mathrm{F}}^{2}}{2\pi^{2}}E^{\prime2}\left[ 2\sin^{2}\left(\frac{\theta}{2}\right)W_{1}^{\nu\mathrm{N}}\left(Q^{2},\nu\right) +\cos^{2}\left({\frac{\theta}{2}}\right)W_{2}^{\nu N}\left(Q^{2},\nu\right)-\;\frac{E+E^{\prime}}{M_{\mathrm{N}}}\sin^{2}\left(\frac{\theta}{2}\right)W_{3}^{\nu\mathrm{N}}\left(Q^{2},\nu\right) \right] (3.37)
In the reaction \bar{\nu}+\mathrm{N} \rightarrow \mathrm{e}^{+}+\mathrm{X} the incoming antineutrino corresponds to the outgoing electron in the reaction \nu+\mathrm{N} \rightarrow \mathrm{e}^{-}+\mathrm{X}, i.e., p and p^{\prime} are exchanged. Since the third term in L_{\mu \nu} is antisymmetric under this exchange, the sign of W_{3} changes.
{ }^{8} The proof of this relation can be found in Exercise 2.4 of W. Greiner and B. Müller: Gauge Theory of Weak Interactions, 3rd ed. (Springer, Berlin, Heidelberg 2000).