Spinor-Helicity Formalism

It is a new language to make on-shell properties manifest.

Consider on-shell momentum $p^{\mu }$. Define

$p^{\dot{\alpha }\alpha }=p_{\mu }({\bar{\sigma }}^{\mu }{)}^{\dot{\alpha }\alpha }=\left(\begin{array}{cc}{p}^0+p^3& p^1-i{p}^2\\ p^1+i{p}^2& p^0-p^3\end{array}\right)$

where ${\bar{\sigma }}^{\mu }=(1_z,-\vec{\sigma })$

with $1_z=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$ and Pauli Matrices as always.

Locally, the Lorentz Group can be written as $SU(2)\times SU(2)$ and $\alpha ,\dot{\alpha }$ are the indices for the fundamental representation of each $SU(2)$.

Then, $p^2=p_{\mu }{p}^{\mu }=det(p^{\dot{\alpha }\alpha })=(p^0{)}^2-{\sum }_i(p^i{)}^2=m^2$.

For the case m=0, $det(p^{\dot{\alpha }\alpha })=0⟹\text{rank }1$.

Then, we can write $\boxed{p^{\dot{\alpha }\alpha }={\lambda }^{\alpha }{\tilde{\lambda }}^{\dot{\alpha }}}$, i.e., the bispinor form, where $\lambda$ and $\tilde{\lambda }$ are spinors.

If p is real, $\tilde{\lambda }=\pm {\lambda }^{\ast }$.

If p is complex, $\lambda$ and $\tilde{\lambda }$ are independent.

Now, $p^{\dot{\alpha }\alpha }$ is invariant under

$\{\begin{array}{l}\lambda \to e^{-i\varphi }\lambda \\ \tilde{\lambda }\to e^{i\varphi }\tilde{\lambda }\end{array}$.

The generator for this symmetry is calledhelicity and is defined as

$h=-\frac{1}{2}(-{\lambda }^{\alpha }\frac{\partial }{\partial {\lambda }^{\alpha }}+{\tilde{\lambda }}^{\dot{\alpha }}\frac{\partial }{\partial {\tilde{\lambda }}^{\dot{\alpha }}})$.

When helicity is operated on the spinor, we get an eigenfunction-eigenvalue equation,

$h\lambda =-\frac{1}{2}\lambda ,h\tilde{\lambda }=\frac{1}{2}\tilde{\lambda }$.

The U(1) group generated by the helicity corresponds to rotations in the 2D plane orthogonal to the spatial momentum, called the Little Group.

To raise and lower indices, we use the Levi-Civita tensor,

${\lambda }_{\alpha }={ϵ}_{\alpha \beta }{\lambda }^{\beta },{\tilde{\lambda }}_{\dot{\alpha }}={ϵ}_{\dot{\alpha }\dot{\beta }}{\tilde{\lambda }}^{\dot{\beta }}$

where

${ϵ}_{\alpha \beta }={ϵ}_{\dot{\alpha }\dot{\beta }}=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right),{ϵ}^{\alpha \beta }={ϵ}^{\dot{\alpha }\dot{\beta }}=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, so that

${ϵ}_{\alpha \beta }{ϵ}^{\beta \gamma }=+{\delta }_{\alpha }^{\gamma }$.

Inner Products

Now, we define the inner products (with particle labels i and j) as:

1. $⟨ij⟩={\lambda }_i^{\alpha }{\lambda }_{j\alpha }$
2. $[ij]={\tilde{\lambda }}_{i\dot{\alpha }}{\tilde{\lambda }}_j^{\dot{\alpha }}$

Note: $⟨ij⟩={ϵ}^{\alpha \beta }{\lambda }_{i\beta }{\lambda }_{j\alpha }=-{ϵ}^{\beta \alpha }{\lambda }_{i\beta }{\lambda }_{j\alpha }=-{ϵ}^{\beta \alpha }{\lambda }_{j\alpha }{\lambda }_{i\beta }=-⟨ji⟩$

Similarly, $[ij]=-[ji]$ and $⟨ii⟩=0=[ii]$.

Now, $p_{\alpha \dot{\alpha }}={ϵ}_{\dot{\alpha }\dot{\beta }}{ϵ}_{\alpha \beta }{p}^{\dot{\beta }\beta }={\lambda }_{\alpha }{\tilde{\lambda }}_{\dot{\alpha }}$

$={ϵ}_{\dot{\alpha }\dot{\beta }}{ϵ}_{\alpha \beta }{p}_{\mu }({\bar{\sigma }}^{\mu }{)}^{\dot{\beta }\beta }\equiv p_{\mu }({\sigma }^{\mu }{)}_{\alpha \dot{\alpha }}$

where $({\sigma }^{\mu }{)}_{\alpha \dot{\alpha }}=(1,\vec{\alpha })={ϵ}_{\alpha \beta }{ϵ}_{\dot{\alpha }\dot{\beta }}({\bar{\sigma }}^{\mu }{)}^{\dot{\beta }\beta }$

Then, p_{i\alpha \dot{\alpha}}p_{j}^{\dot{\alpha} \alpha}=\lambda_{i\alpha}\tilde{\lambda}_{i\dot{\alpha}}\tilde{\lambda}^{\dot{\alpha}}_{j}\lambda_{j}^\dot{\alpha}=\langle j\;i\rangle[i\;j]=-\langle i\; j\rangle[i\;j]

But, $p_{i\alpha \dot{\alpha }}{p}_j^{\dot{\alpha }\alpha }=p_i^{\mu }{p}_j^{\nu }\underset{Tr({\sigma }_{\mu }\overline{{\sigma }_{\nu }})=2{\eta }_{\mu \nu }}{\underbrace{{\sigma }_{\mu \alpha \dot{\alpha }}{\bar{\sigma }}_{\nu }^{\dot{\alpha }\alpha }}}=2p_i.p_j=(p_i+p_j{)}^2\text{for m=0}$

Hence, $\boxed{(p_i+p_j{)}^2=-⟨ij⟩[ij]}$

Polarisation

Massless Dirac Equation: $\mathcal{L}=i\bar{\psi }{\gamma }^{\mu }{\partial }_{\mu }\psi ,\text{with }\bar{\psi }={\psi }^{+}{\gamma }^0,{\gamma }^{\mu }=\left(\begin{array}{cc}0& {\sigma }^{\mu }\\ {\bar{\sigma }}^{\mu }& 0\end{array}\right)$

But, ${\gamma }^{\mu }{\partial }_{\mu }\psi =0$. Multiplying by ${\gamma }^{\nu }{\partial }_{\nu }$, ${\gamma }^{\mu }{\gamma }^{\nu }{\partial }_{\mu }{\partial }_{\nu }\psi =0={\eta }_{\mu \nu }{\partial }_{\mu }{\partial }_{\nu }\psi$.

Hence, ${\partial }^2\psi =0$. This means that $\psi$ has a plane wave solution of the form

$\psi \sim u(p)e^{-ipx}+v(p)e^{i}px$.

In the massless case, $\gamma .pu=\gamma .pv=0$.

For $u_{+}=\left(\begin{array}{c}{x}_{\alpha }\\ 0\end{array}\right),v_{+}=\left(\begin{array}{c}0\\ {\tilde{x}}^{\dot{\alpha }}\end{array}\right)$,

$\sigma .p\tilde{x}=0\text{ and }\bar{\sigma }.px=0$.

Recall that $(\sigma .p{)}_{\alpha \dot{\alpha }}={\lambda }_{\alpha }{\tilde{\lambda }}_{\dot{\alpha }}\text{ and }(\bar{\sigma }.p{)}^{\dot{\alpha }\alpha }={\tilde{\lambda }}^{\dot{\alpha }\alpha }$.

Then, we can choose x_{\alpha}=\lambda_{\alpha}\text{ and }\tilde{x}^\dot{\alpha}=\tilde{\lambda}^\dot{\alpha}, up to an overall constant.

Now, we can denote solutions by

\lambda_{\alpha}\\0 \end{pmatrix},\;|p] =\begin{pmatrix} 0\\\tilde{\lambda}^\dot{\alpha} \end{pmatrix}$$ $$\bra{p}=\begin{pmatrix} \lambda^\alpha\\ 0 \end{pmatrix},\;[p|=\begin{pmatrix} 0\\\tilde{\lambda}^\dot{\alpha} \end{pmatrix}$$. We consider all the particles to be ongoing and | Helicity | $+\frac{1}{2}$ | $-\frac{1}{2}$ | | ---------- | -------------- | -------------- | | Quark | $[p\|$ | $\bra{p}$ | | Anti-quark | $\|p]$ | $\ket{p}$ | For ingoing particles, $p_{\mu}\to-p_{\mu}$ and $h\to-h$. <u>Note</u>: All spinors above are commuting spinors. To construct a Grassman spinor, write $$\psi_{G}=\sum_{s=\pm}\int d\tilde{p}\;[b_{s}(p)u_{s}(p)e^{-ipx}+d^+_{s}(p)v_{s}(p)e^{ipx}]$$, where $b_{s}(p)$ and $d_{s}^+(p)$ are the fermionic creation and annihilation operators respectively.