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I've been trying for some time to find the expressions for conformal generators of Witten's paper in perturbative Yang-Mills.

Given $P_{\alpha \dot{\alpha}} = \lambda_{\alpha} \overline{\lambda}_{\dot{\alpha}}$, for example, I want to find the Lorentz generator $$J_{\alpha \beta} = \lambda_{\alpha} \frac{\partial}{\partial \lambda^{\beta}} + \lambda_{\beta} \frac{\partial}{\partial \lambda^{\alpha}}.$$

What I've been trying to do is, first, go to momentum space:

$$M_{mn} = x_{m} \partial_{n} - x_{n} \partial_{m} \longmapsto p_{n} \frac{\partial}{\partial p^{m}} - p_{m} \frac{\partial}{\partial p^{n}},$$

and then split this $M_{mn}$ into its self-dual and anti-self-dual parts:

$$ M_{mn} = J_{\alpha \beta} \epsilon_{\dot{\alpha} \dot{\beta}} + \overline{J}_{\dot{\alpha} \dot{\beta}} \epsilon_{\alpha \beta},$$

where, for instance,

$$J_{\alpha \beta} = p^{\dot{\gamma}}_{\,\, (\beta} \frac{\partial}{\partial p^{\alpha) \dot{\gamma}}}.$$

From here, I would apply the transformation $P = \lambda \overline{\lambda}$. The problem is that I don't know then how to deal with this kind of object:

$$\frac{\partial \lambda_{\beta}}{\partial p^{\alpha \dot{\alpha}}} = \,?$$

It was brought to my attention that $\partial/ \partial p^{m}$ doesn't commute with the constraint $p^{2} = 0$, and so it would not be properly defined.

Then, how people obtain these expressions? Every reference seems to state that it is obvious. Unfortunately, it is not to me.

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So, first let me point out an important reference which is John Terning's book "Modern Supersymmetry". I will work in the conventions of his Appendix A, which I think are pretty standard.

Are you familiar with the usual way to decompose $SO(4)$ into $(SU(2) \times SU(2))/\mathbb{Z}_2$? Because that is essentially what we are going to do, but with the Lorentz group $SO(3,1)$. Perhaps the 2-component spinor language is obscuring this a little. But in $SO(4)$ this is pretty straightforward to see: You can decompose every $SO(4)$ matrix into a product of two "isoclinic" rotations. An isoclinic rotation is one that rotates, e.g., by the same angle in the $xy$ plane as it does in the $zw$ plane. Combine this with an anti-isoclinic rotation (which rotates the planes by the same angle, but in opposite directions), and you can split the 6 generators of $SO(4)$ into two sets of 3, which each generate an $SU(2)$. In any case, that is all this fancy helicity spinor stuff is doing, except in Lorentzian signature.

Now, you have your Lorentz generators, in momentum space, given by

$$M_{\mu\nu} = p_\mu \partial_\nu - p_\nu \partial_\mu,$$

and you should be able to check that these do in fact preserve your momentum condition $p^2 = 0$, so no worries about that. Next you have your expression for $p_\mu$ in terms of (for +--- signature) a Weyl spinor and its conjugate:

$$\sigma^\mu_{\alpha \dot \alpha} \, p_\mu = \lambda_\alpha \bar \lambda_{\dot \alpha}$$

It will also be helpful to invert this relationship using

$$\sigma^\mu_{\alpha \dot \alpha} \bar \sigma_\nu^{\dot \alpha \alpha} = 2 \delta^\mu_\nu$$

By the way, the matrix $\sigma^\mu$ is called an intertwiner and describes the map between two different representations of $SO(3,1)$. It is useful to define the antisymmetric combination

$$\sigma^{\mu\nu} \equiv \frac{i}{4} \Big( \sigma^\mu \bar \sigma^\nu - \sigma^\nu \bar \sigma^\mu \Big),$$

because it will in fact describe how the intertwiner acts on our Lorentz generators. That is, it describes the particular linear combinations of generators to take in order to effect the notion of "isoclinic" Lorentz transformations.

So, let's define

$$J_{\alpha \beta} = \frac12 (\sigma^{\mu\nu})_{\alpha \beta} \, M_{\mu\nu}$$

as the particular linear combinations of $M_{\mu\nu}$ we are interested in. It is painful to write out all the indices, but we'll do it once:

$$(\sigma^{\mu\nu})_{\alpha \beta} = \frac{i}{4} \epsilon_{\beta \gamma} \Big( \eta^{\nu \rho} \sigma^\mu_{\alpha \dot \delta} \bar \sigma_\rho^{\dot \delta \gamma} - \eta^{\mu \rho} \sigma^\nu_{\alpha \dot \delta} \bar \sigma_\rho^{\dot \delta \gamma} \Big)$$

Now, just contract this with $M_{\mu\nu}$, and use the relation $\sigma^\mu_{\alpha \dot \alpha} \, p_\mu = \lambda_\alpha \bar \lambda_{\dot \alpha}$. You should wind up with

$$J_{\alpha \beta} = \frac{i}{4} \Big( \lambda_\alpha \bar \lambda^{\dot \gamma} \sigma^\mu_{\beta \dot \gamma} \, \partial_\mu + \lambda_\beta \bar \lambda^{\dot \gamma} \sigma^\mu_{\alpha \dot \gamma} \, \partial_\mu \Big)$$

Notice that I still have the derivatives $\partial_\mu \equiv \partial / \partial p^\mu$ in there. We need to work out what to do with them. To figure this out, just observe the following (notice the upper indices on $p^\mu$ here):

$$\frac{\partial p^\mu}{\partial \lambda^\alpha} = \frac{\partial}{\partial \lambda^\alpha} \Big( \frac12 \sigma^\mu_{\beta \dot \beta} \lambda^\beta \bar \lambda^{\dot \beta} \Big) = \frac12 \sigma^\mu_{\alpha \dot \alpha} \bar \lambda^{\dot \alpha}$$

Using this, we can see that

$$\frac12 \bar \lambda^{\dot \gamma} \sigma^\mu_{\beta \dot \gamma} \frac{\partial}{\partial p^\mu} = \frac{\partial p^\mu}{\partial \lambda^\beta} \frac{\partial}{\partial p^\mu} = \frac{\partial}{\partial \lambda^\beta}$$

Finally, plugging this into $J_{\alpha \beta}$, we get

$$J_{\alpha \beta} = \frac{i}{2} \Big( \lambda_\alpha \frac{\partial}{\partial \lambda^\beta} + \lambda_\beta \frac{\partial}{\partial \lambda^\alpha} \Big)$$

as desired. A similar sequence of steps can produce the complex conjugate $J_{\dot \alpha \dot \beta}$.

(I am fairly sure I have all the factors of 2 right, but be wary of minus signs!)

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