0
$\begingroup$

These questions arose while reading the paper ``Comments on Supercurrent Multiplets, Supersymmetric Field Theories and Supergravity" by Komargodski and Seiberg (arXiv:1002.2228)

The Ferrara-Zumino multiplet is defined by the equations

\begin{align} {\overline{D}}^{\dot{\alpha}} \mathcal{J}_{\alpha\dot{\alpha}} &= D_{\alpha}X\\ {\overline{D}}^{\dot{\alpha}} X &= 0 \end{align}

and is subject to the improvement transformations

\begin{align}\label{eq:improvement} \begin{split} \mathcal{J}_{\alpha\dot{\alpha}}' &= \mathcal{J}_{\alpha\dot{\alpha}} - i\partial_{\alpha\dot{\alpha}}(\Xi - \overline{\Xi}) = \mathcal{J}_{\alpha\dot{\alpha}} + [D_{\alpha}, \overline{D}_{\dot{a}}](\Xi + \overline{\Xi})\\ X' &= X + \frac{1}{2}{\overline{D}}^2\overline{\Xi}\\ \overline{D}_{\dot{\alpha}} \Xi &= 0 \end{split} \end{align}

The claim is that these transformations preserve the form of the Ferrara-Zumino equation. But I'm a bit puzzled by this:

\begin{align} \overline{D}^{\dot{\alpha}}\mathcal{J}_{\alpha\dot{\alpha}}' &= \overline{D}^{\dot{\alpha}}\left(\mathcal{J}_{\alpha\dot{\alpha}} + [D_{\alpha}, \overline{D}_{\dot{\alpha}}](\Xi + \overline{\Xi})\right)\\ &= D_{\alpha} X + \overline{D}^{\dot{\alpha}}[D_{\alpha}, \overline{D}_{\dot{\alpha}}](\Xi + \overline{\Xi})\\ &= D_{\alpha} X + \overline{D}^{\dot{\alpha}}(D_{\alpha} \overline{D}_{\dot{\alpha}}\overline{\Xi} - \overline{D}_{\dot{\alpha}}D_{\alpha}\Xi)\\ &= D_{\alpha} X + \overline{D}^{\dot{\alpha}}D_{\alpha} \overline{D}_{\dot{\alpha}}\overline{\Xi} - \overline{D}^{\dot{\alpha}}\overline{D}_{\dot{\alpha}}D_{\alpha}\Xi\\ &= D_{\alpha} X - \overline{D}_{\dot{\alpha}}D_{\alpha}\overline{D}^{\dot{\alpha}}\overline{\Xi} + \overline{D}_{\dot{\alpha}}\overline{D}^{\dot{\alpha}}D_{\alpha}\Xi\\ &= D_{\alpha} X - (\{\overline{D}_{\dot{\alpha}}, D_{\alpha}\}-D_{\alpha}\overline{D}_{\dot{\alpha}})\overline{D}^{\dot{\alpha}}\overline{\Xi} + \frac{1}{2}{\overline{D}}^2 D_{\alpha}\Xi\\ &= D_{\alpha} X - \{D_{\alpha}, \overline{D}_{\dot{\alpha}}\}\overline{D}^{\dot{\alpha}}\overline{\Xi} + \frac{1}{2}D_{\alpha}{\overline{D}}^2 \overline{\Xi} + \frac{1}{2}([{\overline{D}}^2, D_{\alpha}] + D_{\alpha}{\overline{D}}^2)\Xi\\ &= D_{\alpha}\left(X + \frac{1}{2}{\overline{D}}^2\overline{\Xi}\right) - \{D_{\alpha}, \overline{D}_{\dot{\alpha}}\}\overline{D}^{\dot{\alpha}}\overline{\Xi} + \frac{1}{2}(-2i\overline{D}^{\dot{\alpha}}\partial_{\alpha\dot{\alpha}})\Xi\\ &= D_{\alpha}X' - i\partial_{\alpha\dot{\alpha}}\overline{D}^{\dot{\alpha}}\overline{\Xi} - i \overline{D}^{\dot{\alpha}}\partial_{\alpha\dot{\alpha}}\Xi\\ &= D_{\alpha}X' - i\partial_{\alpha\dot{\alpha}}\overline{D}^{\dot{\alpha}}\overline{\Xi} - i \partial_{\alpha\dot{\alpha}}\overline{D}^{\dot{\alpha}}\Xi\\ &= D_{\alpha}X' - i\partial_{\alpha\dot{\alpha}}\overline{D}^{\dot{\alpha}}\overline{\Xi} \end{align}

whereas I also have

\begin{align} \overline{D}^{\dot{\alpha}}\mathcal{J}_{\alpha\dot{\alpha}}' &= \overline{D}^{\dot{\alpha}}\left(\mathcal{J}_{\alpha\dot{\alpha}} - i\partial_{\alpha\dot{\alpha}}(\Xi-\overline{\Xi})\right)\\ &= D_{\alpha}X - i\partial_{\alpha\dot{\alpha}}(\overline{D}^{\dot{\alpha}}\Xi - \overline{D}^{\dot{\alpha}}\overline{\Xi})\\ &= D_{\alpha}X -i\partial_{\alpha\dot{\alpha}}\overline{D}^{\dot{\alpha}}\overline{\Xi} \end{align}

One of these calculations has to be wrong: am I missing something?

$\endgroup$
0
$\begingroup$

The improvement should preserve the Ferrara-Zumino equation, meaning that you expect to get \begin{align} {\overline{D}}^{\dot{\alpha}} \mathcal{J}'_{\alpha\dot{\alpha}} & = D_{\alpha}X' \\ & = D_{\alpha} X + \frac{1}{2}D_{\alpha}{\overline{D}}^2\overline{\Xi}. \end{align}

The only identity that you need is $$ [ D_\alpha, \overline{D}^2 ] = -2i \partial_{\alpha\dot{\alpha}} \overline{D}^{\dot{\alpha}}, $$ then you can see right away that the second expression that you obtained is correct: \begin{align} {\overline{D}}^{\dot{\alpha}} \mathcal{J}'_{\alpha\dot{\alpha}} & = \ldots \\ & = D_{\alpha} X - i \partial_{\alpha\dot{\alpha}} \overline{D}^{\dot{\alpha}} \overline{\Xi}\\ & = D_{\alpha} X + \frac{1}{2}D_{\alpha}{\overline{D}}^2\overline{\Xi}. \end{align} There is certainly a mistake somewhere in your first derivation, even though I can't tell where it is.

$\endgroup$

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.