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


under the change of coordinates

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


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

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 $\partial/\partial\bar{\theta}$:

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

  • $\begingroup$ 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

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.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.