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Reading Martin's SUSY Primer, section 4.4 on Chiral Superfields, he makes the statement that the SUSY chiral covariant derivatives

$$D_\alpha=\dfrac{\partial}{\partial\theta^\alpha}-i(\sigma^\mu\theta^\dagger)_\alpha\partial_\mu,\quad\bar{D}^\dot{\alpha}=\dfrac{\partial}{\partial\theta_\dot{\alpha}^\dagger}-i(\bar{\sigma}^\mu\theta)^\dot{\alpha}\partial_\mu,$$

under the change of coordinates

$$x^\mu\to y^\mu\equiv x^\mu+i\theta^\dagger\bar{\sigma}^\mu\theta,$$

become

$$D_\alpha=\dfrac{\partial}{\partial\theta^\alpha}-2i(\sigma^\mu\theta^\dagger)_\alpha\dfrac{\partial}{\partial y_\mu},\quad\bar{D}^\dot{\alpha}=\dfrac{\partial}{\partial\theta_\dot{\alpha}^\dagger}.$$

Question: could someone make explicit the derivation of these last two equations?


Note: I feel a bit more confident with another derivation of the sought-after result that this choice of coordinates automatically satisfies the condition $\bar{D}^\dot{\alpha}\Phi=0$ required for $\Phi$ to be a chiral superfield, but it don't think it reduces $\bar{D}$ to $\partial/\partial\bar{\theta}$:

$$\begin{align} \bar{D}^\dot{\alpha}\Phi(y^\mu,\theta)&=\left(\dfrac{\partial}{\partial\theta_\dot{\alpha}^\dagger}-i(\bar{\sigma}^\mu\theta)^\dot{\alpha}\partial_\mu\right)\Phi(y^\mu,\theta)\\ &=\left(i(\bar{\sigma}^\mu\theta)^\dot{\alpha}\dfrac{\partial}{\partial y^\mu}-i(\bar{\sigma}^\mu\theta)^\dot{\alpha}\dfrac{\partial}{\partial y^\mu}\right)\Phi(y^\mu,\theta)\\ &=0.\quad\blacksquare \end{align}$$

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  • $\begingroup$ Hint: expansion of $f(y+a)$ around $y$ is $f(y+a)=f(y)+f'(y)a+\frac{1}{2}f''(y)a^2$, and for $a=-i\theta^{\dagger}\bar{\sigma}^\mu\theta$ all higher (~$a^3$ etc.) order terms vanish. Try applying $\bar{D}^\dot{\alpha}$ to this, it should reduce to $\bar{\partial}^{\dot{\alpha}}$. If you won't get the result, I'll do it later, when I have time. $\endgroup$ – Kosm Jun 21 '17 at 16:55
  • $\begingroup$ @Kosm I'd like to see that, please :) because I'm not sure how the $\bar{\sigma}\theta$ terms cancel out... $\endgroup$ – Demosthene Jun 22 '17 at 10:07
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Ok, forget components, let's take your last equation, $$\bar{D}^{\dot{\alpha}}\Phi(y,\theta,\bar{\theta})=\left(\bar{\partial}^\dot{\alpha}-i(\bar{\sigma}^\mu\theta)^{\dot{\alpha}}\partial_\mu\right)\Phi(y,\theta,\bar{\theta})\\ =\left( \bar{\partial}^\dot{\alpha}+i(\bar{\sigma}^\mu\theta)^{\dot{\alpha}}\partial_\mu-i(\bar{\sigma}^\mu\theta)^{\dot{\alpha}}\partial_\mu \right)\Phi(y,\theta,\bar{\theta})=\bar{\partial}^\dot{\alpha}\Phi=0~, $$ where $\partial_\mu=\frac{\partial}{\partial y^\mu}$.

You missed the $~\bar{\partial}^\dot{\alpha}\equiv\frac{\partial}{\partial\theta^\dagger_{\dot{\alpha}}}$ term :) It's there because you don't know that $\Phi$ doesn't depend explicitly on $\bar{\theta}$ until the end. The last equality actually shows that.

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  • $\begingroup$ In my second derivation, I did $$\bar{\partial}\Phi(y,\theta)=\left(\dfrac{\partial\theta}{\partial\bar{\theta}}\dfrac{\partial}{\partial\theta}+\dfrac{\partial y}{\partial\bar{\theta}}\dfrac{\partial}{\partial y}\right)\Phi(y,\theta)$$ where the first term vanishes and the second one cancels out the other term in $\bar{D}$, once the substitution $\dfrac{\partial}{\partial x^\mu}\to\dfrac{\partial}{\partial y^\mu}$ is made. From what you wrote above, I don't see how $\bar{D}=\bar{\partial}$ is justified. $\endgroup$ – Demosthene Jun 22 '17 at 22:08
  • $\begingroup$ as I said, you start with $\Phi=\Phi(y,\theta,\bar{\theta})$, so that you have explicit $\bar{\theta}$ dependence (it's more general), and you cannot drop $\bar{\partial}^\dot{\alpha}$. Only after applying the operator $\bar{D}$, you find that there is no explicit $\bar{\theta}$. $\endgroup$ – Kosm Jun 22 '17 at 22:12

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