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I've read that a way to construct supersymmetric invariant lagrangian could be either to integrate a superfield in the whole superspace, i.e. in all anticommuting coordinates (D-term), or in half of them (F-term).

Obviously I call F-term a lagrangian term that can't be written as D-term, because all D-term could be written trivially as integrals in half the superspace.

But now I can't understand why couldn't be supersymmetric invariant lagrangian terms that are not even F-term, but they are however invariant.

EDIT

I thought that the answer could be that given an ordinary lagrangian $F(x)$ term (dependent only on space-time coordinates) I can make it a part of a chiral superfields, as a coefficients of $\theta\theta$ in $y-\theta$ expansion

$$ \Phi(y,\theta) = \phi(y) + \sqrt{2} \theta \psi(y) - \theta\theta F(y)$$ $$ y^\mu = x^\mu + i\theta\sigma^\mu\bar{\theta}$$

choosing arbitrarily the $\phi$ and $\psi$ functions. The question now become: does it work?

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  • $\begingroup$ I'm pretty sure this is discussed in Weinberg's QFT, Vol.III. $\endgroup$ Apr 4, 2018 at 17:57
  • $\begingroup$ Following your suggestion I tried to find it in 26.3 and 26.6 paragraphs (I think that those are the only paragraphs where is possible to find this type of info) and I failed. Maybe it's my fault, and maybe I make a mistake eliminating the other paragraphs, if you can recall the page could be very useful (Weinberg III is not too short...) $\endgroup$
    – Annibale
    Apr 4, 2018 at 18:22

1 Answer 1

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This boils down to the fact that the supercharges are represented in superspace by

$$ Q_{\alpha}=\frac{\partial}{\partial\theta^{\alpha}}-i(\sigma^{m}\bar\theta)_{\alpha}\frac{\partial}{\partial x^{m}}=\left(\frac{\partial}{\partial\theta^{\alpha}}\right)_{y} $$

$$ Q^{\dot\alpha}=\frac{\partial}{\partial\theta_{\dot\alpha}}-i(\bar\sigma^{m}\theta)^{\dot\alpha}\frac{\partial}{\partial x^{m}}=\left(\frac{\partial}{\partial\theta_{\dot\alpha}}\right)_{y}-2i(\bar\sigma^{m}\theta)^{\dot\alpha}\frac{\partial}{\partial y^{m}} $$

And the D-terms and F-terms are written as fermionic integrals:

$$ \mathcal{L}_{D}=\int d^{2}\theta V(y,\theta),\,\,\,\,\,\,\,\,\,\,\, \mathcal{L}_{F}=\int d^2\theta d^{2}\bar\theta V(x,\theta,\bar\theta) $$

So if you act the supercharges in a D-term or a F-term, the first derivative of the supercharge, the $\theta$-derivative, will drop out because the fermionic integrals are only non-zero if it is saturated. This means that there will be only the $x$-derivatives, making a total derivative. This means that by doing a supersymmetric transformation in the Lagrangian we obtain a total derivative!, so the action is invariant under supersymmetry. Then we say that this Lagrangian is supersymmetric in a manifest way, since there is no need to check the invariance explicitly.

Now, for a given superfield, it is possible to write an action that is supersymmetric out of components that are not $F$-terms or $D$-terms, but this will be make the supersymmetry not manifest.

There is example for you. In pure spinor formalism for the second quantized superparticle in d=10 we can write the action as

$$ \int d^{10}x \langle \psi \,Q\psi\rangle $$

where $\psi(\lambda,\theta)$ is the pure spinor superfield, $Q=\lambda^{\alpha}D_{\alpha}$, and $\langle...\rangle$ is defined by picking just the $\langle\theta^5\lambda^3\rangle=1$, and zero otherwise. This means that it does not pick the last component of the superfields $\theta^{16}$. Nonetheless, this action describes linearized $d=10$ Super-Yang-Mills, so it is supersymmetric, although not in a manifest way.

Back to $d=4$, it is always possible to find an manifest supersymmetric formulation for a given non-manifest supersymmetric $N=1$ action. For large dimensions or extended supersymmetry things start to be more complicated, and the superspace should be generalized some how, e.g. harmonic superspace, pure spinors etc.

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  • $\begingroup$ Thank you for your answer, but my doubt was (with reference to your example) if I can construct a chiral superfield whose F component is the same $\rangle \psi Q\psi\langle$, and so also that is, not manifestly, an F-term. More clearly my question is are there any susy-invariant lagrangian that couldn't bring back to the F-term of some chiral superfield? $\endgroup$
    – Annibale
    Oct 4, 2018 at 10:01
  • $\begingroup$ For d=4 N=1 supersymmetry, the answer is no. All invariants in d=4 N=1 can be written as integrals over the whole superspace (F-term). $\endgroup$
    – Nogueira
    Oct 5, 2018 at 14:18
  • $\begingroup$ For change a chiral term $W$ into a F-term just solve the equation $\bar{D}^2 V= W$. An integral in the full superspace of $V$ gives the chiral integral of $W$. $\endgroup$
    – Nogueira
    Oct 5, 2018 at 14:26

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