Supersymmetry and Supergravity

Abstract

In this chapter we will discuss the gravitational interactions of simple physical systems containing bosonic and fermionic degrees of freedom, taking into account the possibility that these different components are related among themselves by infinitesimal field transformations.

Appendices

Exercises Chap. 14

14.1

Commutation Properties of the Majorana Spinors Show that, given two Majorana spinors $\epsilon = \epsilon ^c$ and $\psi = \psi ^c$, they satisfy the property

$$\begin{aligned} \overline{\epsilon }\psi = {\overline{\psi }}\epsilon . \end{aligned}$$

(14.83)

14.2

On-Shell SUSY Transformations Derive the result (14.25) for the fields B and $\psi$ of the Wess–Zumino model, using the properties of the Majorana spinors and imposing that the equations of motion are satisfied .

14.3

Global SUSY for the Graviton-Gravitino System Compute the infinitesimal variation $\delta \mathcal{L}$ of the effective Lagrangian (14.42) under the global infinitesimal transformation (14.41), and show that the result is a total divergence, $\delta \mathcal{L}= \partial _\mu K^\mu$.

14.4

Local SUSY for ${{\varvec{N}}}\,{\varvec{=}}\,\mathbf{{1}}$ Supergravity Compute the infinitesimal variation $\delta \mathcal{L}$ of the effective Lagrangian (14.53) under the local transformation (14.55), and show that the result is a total divergence, $\delta \mathcal{L}= \partial _\mu K^\mu$, which is vanishing if the equations of motion of the gravitino field are satisfied.

Solutions

14.1

Solution From the Majorana condition (14.2) we have

$$\begin{aligned} C^{-1} \epsilon = \overline{\epsilon }^T, \end{aligned}$$

(14.84)

so that, using the properties (14.3) of the charge conjugation operator,

$$\begin{aligned} \overline{\epsilon }= \left( C^{-1} \epsilon \right) ^T= \epsilon ^T \left( C^{-1} \right) ^T= - \epsilon ^T C^{-1}. \end{aligned}$$

(14.85)

Hence:

$$\begin{aligned} \overline{\epsilon }\psi = - \epsilon ^T C^{-1} C{\overline{\psi }}^T =-\epsilon ^T {\overline{\psi }}^T= ( {\overline{\psi }}\epsilon )^T= {\overline{\psi }}\epsilon . \end{aligned}$$

(14.86)

The next to last step is due to the fact that the spinors $\epsilon ^A$ and $\psi ^A$ anticommute, and then

$$\begin{aligned} -\epsilon ^T {\overline{\psi }}^T= - (\epsilon ^A)^T \gamma _0 \psi _A^{*}= (\psi _A^{*T} \gamma _0\epsilon ^A) = ( {\overline{\psi }}\epsilon )^T. \end{aligned}$$

(14.87)

Finally, the result of the spinor product ${\overline{\psi }}\epsilon$ is actually a number, hence $( {\overline{\psi }}\epsilon )^T= {\overline{\psi }}\epsilon$.

14.2

Solution By applying to B the infinitesimal SUSY transformations (14.24), and computing the commutator, we obtain

$$\begin{aligned}&\delta _2 \delta _1 B=\delta _2 \left( i \overline{\epsilon }_1 \gamma ^5 \psi \right) = {1\over 2} \overline{\epsilon }_1 \gamma ^5 \gamma ^\mu \partial _\mu \left( A+i \gamma ^5B\right) \epsilon _2, \nonumber \\&[\delta _2, \delta _1] B= {1\over 2} \left( \overline{\epsilon }_1 \gamma ^5 \gamma ^\mu \epsilon _2\right) \partial _\mu A+{i\over 2} \left( \overline{\epsilon }_1 \gamma ^5 \gamma ^\mu \gamma ^5 \epsilon _2\right) \partial _\mu B -\{1 \leftrightarrow 2\}.\qquad \end{aligned}$$

(14.88)

The first term, proportional to $\partial _\mu A$, is symmetric in the exchange of the indices 1 and 2. In fact, by recalling Eqs. (14.3) and (14.85), and using the property

$$\begin{aligned} \{ \gamma ^5, \gamma ^\mu \}=0=[\gamma ^5,C], \end{aligned}$$

(14.89)

we have

$$\begin{aligned} \overline{\epsilon }_1 \gamma ^5 \gamma ^\mu \epsilon _2= & {} - \epsilon _1^T C^{-1} \gamma ^5 \gamma ^\mu C\, \overline{\epsilon }_2^T= \epsilon _1^T \gamma ^5 \gamma ^{\mu T} \overline{\epsilon }_2^T \nonumber \\= & {} -\left( \overline{\epsilon }_2 \gamma ^\mu \gamma ^5 \epsilon _1 \right) ^T = \overline{\epsilon }_2 \gamma ^5 \gamma ^\mu \epsilon _1. \end{aligned}$$

(14.90)

Hence, this type of term does not contribute to the commutator (14.88).

For the second term, proportional to $\partial _\mu B$, we note that

$$\begin{aligned} \gamma ^5 \gamma ^\mu \gamma ^5= - \gamma ^\mu , \end{aligned}$$

(14.91)

so that, by using the result (14.18), we finally obtain

$$\begin{aligned}{}[\delta _2, \delta _1] B=-i \left( \overline{\epsilon }_1 \gamma ^\mu \epsilon _2\right) \partial _\mu B, \end{aligned}$$

(14.92)

in agreement with Eq. (14.25).

Let us now consider the commutator of two infinitesimal transformations applied to $\psi$, starting from Eq. (14.24) and writing explicitly the spinor indices:

$$\begin{aligned} \delta _1 \psi _A= -{i\over 2}\partial _\mu A \left( \gamma ^\mu \epsilon _1\right) _A +{1\over 2} \partial _\mu B\left( \gamma ^\mu \gamma ^5 \epsilon _1\right) _A. \end{aligned}$$

(14.93)

Therefore:

$$\begin{aligned}{}[\delta _2 , \delta _1] \psi _A= -{i\over 2} \left( \overline{\epsilon }_2\partial _\mu \psi \right) \gamma ^\mu \epsilon _1 +{i\over 2} \left( \overline{\epsilon }_2 \gamma ^5 \partial _\mu \psi \right) \gamma ^\mu \gamma ^5 \epsilon _1 - \{ 1 \leftrightarrow 2\}. \end{aligned}$$

(14.94)

It is convenient, at this point, to use the Fierz identity (14.78) to re-arrange the right-hand side of Eq. (14.94), so as to move to the right $\partial _\mu \psi$. We obtain, in this way,

$$\begin{aligned}{}[\delta _2 , \delta _1] \psi _A= & {} {i\over 8} \sum _i \left( \overline{\epsilon }_2 \varGamma ^i \epsilon _1\right) \gamma ^\mu \varGamma _i \partial _\mu \psi \nonumber \\- & {} {i\over 8} \sum _i \left( \overline{\epsilon }_2 \varGamma ^i \epsilon _1\right) \gamma ^\mu \gamma ^5 \varGamma _i \gamma ^5 \partial _\mu \psi - \{ 1 \leftrightarrow 2\}, \end{aligned}$$

(14.95)

where $\varGamma ^i$ are the matrix operators defined in Eq. (14.79), and where the only nonzero contributions to this expression come from the operators $\varGamma ^i$ for which $\overline{\epsilon }_2 \varGamma ^i \epsilon _1$ is antisymmetric in the indices 1 and 2. Using the properties of the Majorana spinor we find that this is possible only if $\varGamma ^i$ corresponds to $\gamma ^\mu$ or to $\sigma ^{\mu \nu }$ (defined by Eq. (13.10)).

In the case of $\sigma ^{\mu \nu }$, however, we have $\gamma ^5 \sigma ^{\mu \nu } \gamma ^5= \sigma ^{\mu \nu }$, and the four terms of Eq. (14.95) identically cancel among each other. We are thus left with the contribution of $\gamma ^\mu$, which gives

$$\begin{aligned}{}[\delta _2 , \delta _1] \psi= & {} {i\over 2} \left( \overline{\epsilon }_2 \gamma ^\nu \epsilon _1\right) \gamma _\mu \gamma _\nu \partial ^\mu \psi \nonumber \\= & {} {i\over 2} \left( \overline{\epsilon }_2 \gamma ^\nu \epsilon _1\right) \left( - \gamma _\nu \gamma _\mu + 2 \eta _{\mu \nu }\right) \partial ^\mu \psi . \end{aligned}$$

(14.96)

The first term on the right-hand side is zero for the spinor equation of motion, which imposes $\gamma _\mu \partial ^\mu \psi =0$. The final result, using Eq. (14.18), can then be rewritten as follows:

$$\begin{aligned}{}[\delta _2 , \delta _1] \psi = -i \left( \overline{\epsilon }_1 \gamma ^\mu \epsilon _2\right) \partial _\mu \psi , \end{aligned}$$

(14.97)

in agreement with Eq. (14.25).

14.3

Solution Let us compute, first of all, the infinitesimal variation of the conjugate field ${\overline{\psi }}_\mu$. Starting from the definition (14.41) of $\delta \psi _\mu$, and using the result of Eq. (13.46), we obtain

$$\begin{aligned} \delta {\overline{\psi }}_\mu = \left( \gamma ^{[\alpha } \gamma ^{\beta ]}\epsilon \right) ^\dag \gamma ^0 \partial _\alpha h_{\mu \beta }=- \overline{\epsilon }\gamma ^{[\alpha } \gamma ^{\beta ]} \partial _\alpha h_{\mu \beta }. \end{aligned}$$

(14.98)

By varying the Lagrangian (14.42), and applying the infinitesimal transformations of h, $\psi$ and ${\overline{\psi }}$ we are then led to

$$\begin{aligned} \delta \mathcal{L}= & {} \partial ^\alpha h^{\mu \nu } \left( \overline{\epsilon }\gamma _\mu \partial _\alpha \psi _\nu \right) \nonumber \\&+{i\over 2} \epsilon _{\mu \nu \alpha \beta } \Big [ -\left( \overline{\epsilon }\gamma ^{[\rho } \gamma ^{\sigma ]} \gamma ^5 \gamma ^\nu \partial ^\alpha \psi ^\beta \right) \partial _\rho h^\mu \,_\sigma \nonumber \\&+ \left( {\overline{\psi }}^\mu \gamma ^5 \gamma ^\nu \gamma ^{[\rho } \gamma ^{\sigma ]} \epsilon \right) \partial ^\alpha \partial _\rho h^\beta \,_\sigma \Big ] \end{aligned}$$

(14.99)

(where we have enclosed in round brackets all terms containing spinor products).

Consider the third (and last) contribution to the above equation. By extracting a total divergence,

$$\begin{aligned} \partial ^\alpha V_\alpha \equiv \partial ^\alpha \left( {i\over 2} \epsilon _{\mu \nu \alpha \beta } {\overline{\psi }}^\mu \gamma ^5 \gamma ^\nu \gamma ^{[\rho } \gamma ^{\sigma ]} \epsilon \, \partial _\rho h^\beta {}_\sigma \right) , \end{aligned}$$

(14.100)

by exploiting the anticommuting properties of the Majorana spinor ${\overline{\psi }}^\mu$ and $\epsilon$, and by renaming the summation indices $\mu$ and $\beta$, such a last contribution can be rewritten as:

$$\begin{aligned} \partial ^\alpha V_\alpha - {i\over 2} \epsilon _{\mu \nu \alpha \beta } \left( \overline{\epsilon }\gamma ^5 \gamma ^\nu \gamma ^{[\rho } \gamma ^{\sigma ]} \partial ^\alpha \psi ^\beta \right) \partial _\rho h^\mu {}_\sigma . \end{aligned}$$

(14.101)

The variation (14.99) then reduces to

$$\begin{aligned} \delta \mathcal{L}= & {} \partial ^\alpha h^{\mu \nu } \left( \overline{\epsilon }\gamma _\mu \partial _\alpha \psi _\nu \right) + \partial ^\alpha V_\alpha \nonumber \\&-{i\over 2} \epsilon _{\mu \nu \alpha \beta } \,\overline{\epsilon }\gamma ^5 \Big [ \gamma ^{[\rho } \gamma ^{\sigma ]} \gamma ^\nu +\gamma ^\nu \gamma ^{[\rho } \gamma ^{\sigma ]} \Big ] \partial ^\alpha \psi ^\beta \partial _\rho h^\mu \,_\sigma . \end{aligned}$$

(14.102)

For the product of Dirac matrices we can use Eqs. (13.34), (13.36), (13.49), which imply

$$\begin{aligned} \gamma ^{[\rho } \gamma ^{\sigma ]} \gamma ^\nu +\gamma ^\nu \gamma ^{[\rho } \gamma ^{\sigma ]}=2 \gamma ^{[\nu } \gamma ^{\rho } \gamma ^{\sigma ]}=-2i \epsilon ^{\nu \rho \sigma \lambda } \gamma ^5 \gamma _\lambda . \end{aligned}$$

(14.103)

Inserting this result into Eq. (14.102), and using the product rule (3.39) for completely antisymmetric tensors, we arrive at:

$$\begin{aligned} \delta \mathcal{L}= \partial ^\alpha h^{\mu \nu } \left( \overline{\epsilon }\gamma _\mu \partial _\alpha \psi _\nu \right) + \partial ^\alpha V_\alpha - \delta ^{\rho \sigma \lambda }_{\mu \alpha \beta } \left( \overline{\epsilon }\gamma _\lambda \partial ^\alpha \psi ^\beta \right) \partial _\rho h^\mu {}_\sigma . \end{aligned}$$

(14.104)

We now notice that the last term of the above expression in nonvanishing only for $\mu \not = \rho$ e $\mu \not = \sigma$, due to the gauge conditions (14.39). The only contribution of the symbol $\delta ^{\rho \sigma \lambda }_{\mu \alpha \beta }$ thus comes from the term with $\mu =\lambda$, and the general symbol $\delta ^{\rho \sigma \lambda }_{\mu \alpha \beta }$ reduces to

$$\begin{aligned} \delta _\mu ^\lambda \left( \delta _\alpha ^\rho \delta _\beta ^\sigma - \delta _\alpha ^\sigma \delta _\beta ^\rho \right) . \end{aligned}$$

(14.105)

Using this condition into Eq. (14.104) we obtain

$$\begin{aligned} \delta \mathcal{L}= & {} \partial ^\alpha h^{\mu \nu } \left( \overline{\epsilon }\gamma _\mu \partial _\alpha \psi _\nu \right) + \partial ^\alpha V_\alpha \nonumber \\&- \left( \overline{\epsilon }\gamma _\mu \partial _\alpha \psi _\beta \right) \partial ^\alpha h^{\mu \beta } +\left( \overline{\epsilon }\gamma _\mu \partial _\alpha \psi _\beta \right) \partial ^\beta h^{\mu \alpha } . \end{aligned}$$

(14.106)

The first and third contributions to $\delta \mathcal{L}$ cancel among each other. The last contribution can be rewritten as a total divergence,

$$\begin{aligned} \partial _a W^\alpha \equiv \partial _\alpha \left( \overline{\epsilon }\gamma _\mu \psi _\beta \partial ^\beta h^{\mu \alpha } \right) , \end{aligned}$$

(14.107)

since $\partial _\alpha h^{\mu \alpha }=0$ in the gauge (14.39). The total infinitesimal variation of the Lagrangian can be then rewritten in the final form

$$\begin{aligned} \delta \mathcal{L}= \partial _\alpha \left( V^\alpha + W^\alpha \right) \equiv \partial _\alpha K^\alpha , \end{aligned}$$

(14.108)

where, according to the definitions (14.100) and (14.107),

$$\begin{aligned} K^\alpha = \left( \overline{\epsilon }\gamma _\mu \psi _\beta \right) \partial ^\beta h^{\mu \alpha } +{i\over 2} \epsilon ^{\mu \nu \alpha \beta } \left( {\overline{\psi }}_\mu \gamma _5 \gamma _\nu \gamma _{[\rho } \gamma _{\sigma ]} \epsilon \right) \partial ^\rho h_\beta {}^\sigma . \end{aligned}$$

(14.109)

Using Eq. (14.58) and the properties of the Majorana spinors the current K can also be recast in the form

$$\begin{aligned}&K^\alpha = {1\over 2} \left( \overline{\epsilon }\gamma _\mu \psi _\nu \right) \left( \partial ^\nu h^{\mu \alpha }+ \partial ^\alpha h^{\mu \nu } \right) +{i\over 2} \epsilon ^{\mu \nu \alpha \beta } \left( {\overline{\psi }}_\mu \gamma _5 \gamma ^\rho \epsilon \right) \partial _\nu h_{\beta \rho }. \end{aligned}$$

(14.110)

14.4

Solution As discussed in Sect. 14.3, it will be enough to compute the variation of the action induced by the supersymmetry transformations of the two independent varianles V and $\psi$. Hence, we must compute

$$\begin{aligned} \delta \mathcal{L}= \delta _V\mathcal{L}_2 + \delta _V \mathcal{L}_{3/2}+ \delta _\psi \mathcal{L}_{3/2}, \end{aligned}$$

(14.111)

where $\mathcal{L}_2$ and $\mathcal{L}_{3/2}$ denote, respectively, the gravitational and spinorial part of the Lagrangian (14.53). For simplicity, and for consistency with the transformations defined in (14.55), we will use everywhere the convention $\chi = 8 \pi G/c^4=1$.

For the gravitational action there is only the variational contribution of the vierbeins, so that, using the results (12.62), (12.75) we can immediately conclude that

$$\begin{aligned} \delta _V\mathcal{L}_2= \delta _V \left( -{V\over 2} R\right) = G^\mu {}_a \delta V_\mu ^a = \left( \overline{\epsilon }\gamma ^a \psi _\mu \right) G^\mu {}_a, \end{aligned}$$

(14.112)

where G is the Einstein tensor (12.64).

Let us now vary the Rarita–Schwinger Lagrangian with respect to the gravitino, using the infinitesimal transformations $\delta \psi _\mu = -2 D_\mu \epsilon$, $\delta {\overline{\psi }}_\mu = -2 D_\mu \overline{\epsilon }$. We obtain:

$$\begin{aligned} \delta _\psi \mathcal{L}_{3/2}= & {} - i \epsilon ^{\mu \nu \alpha \beta } \left( D_\mu \overline{\epsilon }\gamma _5 \gamma _\nu D_\alpha \psi _\beta + {\overline{\psi }}_\mu \gamma _5 \gamma _\nu D_\alpha D_\beta \epsilon \right) \nonumber \\= & {} -i \epsilon ^{\mu \nu \alpha \beta }\Big [ {\overline{\psi }}_\mu \gamma _5 \gamma _\nu D_{[\alpha } D_{\beta ]} \epsilon -\overline{\epsilon }\gamma _5 \gamma _\nu D_{[\mu } D_{\alpha ]} \psi _\beta \nonumber \\&-\left( \overline{\epsilon }\gamma _5 \gamma _a D_\alpha \psi _\beta \right) D_\mu V_\nu ^a \Big ] + \mathrm{total} ~ \mathrm{divergence}. \end{aligned}$$

(14.113)

We will first consider the joint contribution of the first two terms (which we will denote by C), containing the second derivatives of the spinor field.

By exploiting the result for the commutator of two covariant derivatives, see (14.70), we obtain

$$\begin{aligned} C= -{i\over 8} \epsilon ^{\mu \nu \alpha \beta } \left( {\overline{\psi }}_\mu \gamma _5 \gamma _\nu \gamma _{[a}\gamma _{b]} \epsilon R_{\alpha \beta }\,^{ab} - \overline{\epsilon }\gamma _5 \gamma _\nu \gamma _{[a}\gamma _{b]}\psi _\beta R_{\mu \alpha }\,^{ab} \right) . \end{aligned}$$

(14.114)

The combination of Dirac matrices appearing here has been already evaluated in Eq. (14.58). Inserting that result we get

$$\begin{aligned} C= & {} -{1\over 8} \epsilon ^{\mu \nu \alpha \beta }\epsilon _{abcd} V_\nu ^c \left( {\overline{\psi }}_\mu \gamma ^d \epsilon R_{\alpha \beta }\,^{ab}- \overline{\epsilon }\gamma ^d \psi _\beta R_{\mu \alpha }\,^{ab}\right) \nonumber \\&-{i\over 4} \epsilon ^{\mu \nu \alpha \beta } V_{\nu a} \left( {\overline{\psi }}_\mu \gamma _5 \gamma _b \epsilon R_{\alpha \beta }\,^{ab}- \overline{\epsilon }\gamma _5 \gamma _b \psi _\beta R_{\mu \alpha }\,^{ab}\right) . \end{aligned}$$

(14.115)

By recalling the commutation property of the Majorana spinors we can also write

$$\begin{aligned} {\overline{\psi }}_\mu \gamma ^d \epsilon =- \overline{\epsilon }\gamma ^d \psi _\mu , ~~~~~~~~~ {\overline{\psi }}_\mu \gamma _5 \gamma _b \epsilon =\overline{\epsilon }\gamma _5 \gamma _b \psi _\mu \end{aligned}$$

(14.116)

(see Exercises 14.1 and 14.2). The last two terms of Eq. (14.115) thus cancel among each other, while the first two terms give

$$\begin{aligned} C= {1\over 4} \epsilon ^{\mu \nu \alpha \beta }\epsilon _{abcd} V_\nu ^cR_{\mu \alpha }\,^{ab} \left( \overline{\epsilon }\gamma ^d \psi _\beta \right) =- G^\beta {}_d \left( \overline{\epsilon }\gamma ^d \psi _\beta \right) \end{aligned}$$

(14.117)

(we have used Eq. (12.75)).

Let us consider now the last term of Eq. (14.113), and eliminate $D_{[\mu } V_{\nu ]}^a$ through the torsion equation (14.60). Summing up all contributions, we then find that the variation of the gravitino Lagrangian (14.113) with respect to $\psi$ reduces to

$$\begin{aligned} \delta _\psi \mathcal{L}_{3/2} =- G^\mu {}_a \left( \overline{\epsilon }\gamma ^a \psi _\mu \right) -{i\over 4} \left( \overline{\epsilon }\gamma _5 \gamma _a D_\alpha \psi _\beta \right) \left( {\overline{\psi }}_\mu \gamma ^a \psi _\nu \right) \epsilon ^{\mu \nu \alpha \beta } \end{aligned}$$

(14.118)

(modulo a total divergence).

We still need the variation of $\mathcal{L}_{3/2}$ with respect to V, which gives

$$\begin{aligned} \delta _V \mathcal{L}_{3/2}= & {} {i\over 2} \epsilon ^{\mu \nu \alpha \beta } {\overline{\psi }}_\mu \gamma _5 \gamma _a D_\alpha \psi _b \delta V_\nu ^a \nonumber \\= & {} {i\over 2} \epsilon ^{\mu \nu \alpha \beta } \left( {\overline{\psi }}_\mu \gamma _5 \gamma _a D_\alpha \psi _b \right) \left( \overline{\epsilon }\gamma ^a \psi _\nu \right) . \end{aligned}$$

(14.119)

Operating on the spinors a Fierz rearrangement of type (14.78) we have

$$\begin{aligned} \delta _V \mathcal{L}_{3/2} = -{i\over 8} \epsilon ^{\mu \nu \alpha \beta }\left( \overline{\epsilon }\gamma ^a \varGamma ^i \gamma _5 \gamma _a D_\alpha \psi _\beta \right) \left( {\overline{\psi }}_\mu \varGamma _i \psi _\nu \right) , \end{aligned}$$

(14.120)

and we can note that the nonzero contribution, corresponding to a spinor current ${\overline{\psi }}_\mu \varGamma _i \psi _\nu$ antisymmetric in $\mu$ and $\nu$, comes from the matrix $\varGamma ^i = \gamma ^a$. The possible term ${\overline{\psi }}_\mu \sigma _{\alpha \beta } \psi _\nu$, which is also antisymmetric, must be excluded because $\gamma ^a \sigma _{\alpha \beta } \gamma _a \equiv 0$. Therefore:

$$\begin{aligned} \delta _V \mathcal{L}_{3/2}= & {} -{i\over 8} \epsilon ^{\mu \nu \alpha \beta }\left( \overline{\epsilon }\gamma _5 \gamma ^a \gamma ^b \gamma _a D_\alpha \psi _\beta \right) \left( {\overline{\psi }}_\mu \gamma _b \psi _\nu \right) \nonumber \\= & {} -{i\over 8} \epsilon ^{\mu \nu \alpha \beta } \overline{\epsilon }\gamma _5 \left( - \gamma ^b \gamma ^a + 2 \eta ^{ab} \right) \gamma _a D_\alpha \psi _\beta \left( {\overline{\psi }}_\mu \gamma _b \psi _\nu \right) \nonumber \\= & {} {i\over 4} \epsilon ^{\mu \nu \alpha \beta }\left( \overline{\epsilon }\gamma _5 \gamma ^b D_\alpha \psi _\beta \right) \left( {\overline{\psi }}_\mu \gamma _b \psi _\nu \right) . \end{aligned}$$

(14.121)

Adding all contributions—given by Eqs. (14.112), (14.118) and (14.121)—we obtain for $\delta \mathcal{L}$ a vanishing result, modulo the total divergence that we have neglected in Eq. (14.113), and which is given by

$$\begin{aligned} \partial _\mu K^\mu =-i D_\mu \left( \epsilon ^{\mu \nu \alpha \beta } \overline{\epsilon }\gamma _5 \gamma _\nu D_\alpha \psi _\beta \right) . \end{aligned}$$

(14.122)

We can immediately check that $\partial _\mu K^\mu =0$ if we impose the gravitino equations of motion $R^\mu =0$, together with the consistency condition $D_\mu R^\mu =0$ which is always satisfied, on-shell, as discussed in Sect. 14.3.1.