Question 3.3: The Nucleonic Scattering Tensor with Weak Interaction Repeat......

Have a look at Fig. 3.2. We are again discussing the elastic process described by that figure, but allow for non-parity-conserving currents. In this case the transition current is of the general form

with real functions A, B, C, \ldots and with

In (1) we have assumed that the transition current is real, i.e.

\left(\bar{u}\left(P^{\prime}\right) \Gamma_{\mu} u(P)\right)^{\dagger}=\bar{u}\left(P^{\prime}\right) \Gamma_{\mu} u(P)

and therefore

u^{\dagger}(P) \gamma_{0} \gamma_{0} \Gamma_{\mu}^{\dagger} \gamma_{0} u\left(P^{\prime}\right)=\bar{u}(P) \Gamma_{\mu} u\left(P^{\prime}\right).

Here the relation \gamma^{0} \Gamma_{\mu}^{\dagger} \gamma^{0}=\Gamma_{\mu} was used, which can easily be verified in the standard representation of the \gamma_{\mu} and \sigma_{\mu \nu}=\frac{\mathrm{i}}{2}\left[\gamma_{\mu}, \gamma_{\nu}\right]. Obviously, the exchange P \leftrightarrow P^{\prime} should have the same effect as complex conjugation. In other words, the right-hand side must therefore be invariant under the transformation

\left.(\cdots)^{*}\right|_{P_{\mu}^{\prime} \leftrightarrow P_{\mu}}.     (3)

Time inversion yields (see Table 2.1)

S \rightarrow S, \quad P \rightarrow-P, \quad V_{\mu} \rightarrow V^{\mu}, \quad A_{\mu} \rightarrow A^{\mu}, \quad T_{\mu \nu} \rightarrow-T^{\mu \nu},     (4)

and

{P} \leftrightarrow-{P}^{\prime}, \quad P^{0} \leftrightarrow P^{\prime 0}, \quad \text { i.e. } \quad P_{\mu} \leftrightarrow P^{\prime \mu} \text {. }    (5)

The transition {P} \leftrightarrow-{P}^{\prime}, \quad P^{0} \leftrightarrow P^{\prime 0} is due to the complex conjugation of \bar{u}\left(P^{\prime}\right) \Gamma_{\mu} u(P), which replaces the momentum P_{\mu}^{\prime} by the negative value of P_{\mu} and vice versa, i.e., P_{\mu}^{\prime} \leftrightarrow-P_{\mu}. Because t \rightarrow-t there is an additional change in the sign of the zero component. This is easily understood, because under time reversal the direction of motion changes and initial and final states are exchanged. The energies, however, remain positive.

Under combined transformations (4) and (5), (1) assumes the form

Only the spatial components of the current vector should change sign under time reversal. In order to conserve T invariance (6) must therefore be equal to \bar{u}\left(P^{\prime}\right) \Gamma^{\mu} u(P). For the \varepsilon tensor the relation \varepsilon_{\mu \nu \alpha \beta}=-\varepsilon^{\mu \nu \alpha \beta} holds. This can be verified from the definition -4 \mathrm{i} \varepsilon^{\mu \nu \alpha \beta}=\operatorname{tr}\left[\gamma_{5} \gamma_{\mu} \gamma_{\nu} \gamma_{\alpha} \gamma_{\beta}\right]. The indices \mu, \nu, \alpha, \beta have to be 0,1,2,3, and different from each other. Now -4 \mathrm{i} \varepsilon \varepsilon^{0,1,2,3}= \operatorname{tr}\left[\gamma_{5} \gamma^{0} \gamma^{1} \gamma^{2} \gamma^{3}\right]=-\operatorname{tr}\left[\gamma_{5} \gamma_{0} \gamma_{1} \gamma_{2} \gamma_{3}\right]=+4 \mathrm{i} \varepsilon_{0,1,2,3}. For different permutations of 0,1,2,3 the analogous relation holds. Employing now \varepsilon_{\mu \nu \alpha \beta}=-\varepsilon^{\mu \nu \alpha \beta} we have

C=D=B^{\prime}=E^{\prime}=0 .   (7)

Because of the Gordon decomposition (see earlier text and (3.10)),

\bar{u}\left(P^{\prime}\right) \Gamma_\mu\left(P^{\prime}, P\right) u(P)=\bar{u}\left(P^{\prime}\right)\left[A\left(q^2\right) \gamma_\mu+\mathrm{i} B\left(q^2\right) q^ν \sigma_{\mu \nu}\right] u(P)    (3.10)

E can again be replaced by A and B. In an analogous way D^{\prime} can also be eliminated using A^{\prime} and C^{\prime}. In order to derive this identity we consider the expression

Differentiating this relation with respect to a^{\mu} yields

\frac{\partial}{\partial a^{\mu}} a^{\alpha} \gamma_{\alpha}=\delta_{\mu}^{\alpha} \gamma_{\alpha}=\gamma_{\mu}

and we obtain

With (1/2) \left\{\gamma_{\lambda}, \gamma_{\mu}\right\}=g_{\lambda, \mu} and (1/2) \left[\gamma_{\lambda}, \gamma_{\mu}\right]=-\mathrm{i} \sigma_{\lambda, \mu} and with (8) we therefore conclude that

which represents the desired result. Consequently we have reduced (1) to

\bar{u}\left(P^{\prime}\right) \Gamma_{\mu} u(P)=B\left(P_{\mu}^{\prime}+P_{\mu}\right) S+A V_{\mu}+\mathrm{i} C^{\prime}\left(P_{\mu}^{\prime}-P_{\mu}\right) P+A^{\prime} A_{\mu}.   (10)

Here the letters B, A, C^{\prime}, and A^{\prime} denote constants while S, V_{\mu}, P, and A_{\mu} stand for the various currents denoted in (2). Similarly, as in (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)

W_{\mu \nu} then becomes

Note again that we have used the normalization for the u and ν spinors expressed in (2.50). The first two traces do not have to be evaluated explicitly. It is sufficient to know that because P_{\mu}^{\prime}=q_{\mu}+P_{\mu} their contribution can only be of the form

g_{\mu \nu} V_{1}+P_{\mu} P_{\nu} V_{2}+\left(P_{\mu} q_{\nu}+P_{\nu} q_{\mu}\right) V_{3}+q_{\mu} q_{\nu} V_{4} .      (12)

In the case of the last trace the identity

\operatorname{tr}\left\{\gamma_{5} \not a \not b \not c \not d\right\}=4 \mathrm{i} \varepsilon_{\mu \nu \alpha \beta} a^{\mu} b^{ν} c^{\alpha} d^{\beta}

holds. Therefore the hadronic tensor is of the form

W_{\mu \nu}=g_{\mu \nu} V_{1}+P_{\mu} P_{\nu} V_{2}+\left(P_{\mu} q_{\nu}+P_{\nu} q_{\mu}\right) V_{3}+q_{\mu} q_{\nu} V_{4}+\mathrm{i} \varepsilon_{\mu \nu \alpha \beta} P^{\alpha} q^{\beta} V_{5} .      (13)

We have derived (13) for elastic lepton-nucleon scattering. In the case of inelastic processes, including particle creation, q_{\mu}=P_{\mu}^{\prime}-P_{\mu} is no longer valid. q \cdot P^{\prime}, q^{2} and q \cdot P are then independent quantities. However, the results again are quite simple if one sums over all possible final states, i.e., over the final states of the reactions

Therefore the inclusive inelastic scattering tensor for parity-violating leptonnucleon scattering is