The two-component helicity eigenstates ${\displaystyle \xi _{\lambda }}$  satisfy

${\displaystyle \sigma \cdot {\hat {p}}\xi _{\lambda }\left({\hat {p}}\right)=\lambda \xi _{\lambda }\left({\hat {p}}\right)\,}$ 

where

${\displaystyle \sigma \,}$  are the Pauli matrices,

${\displaystyle {\hat {p}}\,}$  is the direction of the fermion momentum,

${\displaystyle \lambda =\pm 1\,}$  depending on whether spin is pointing in the same direction as ${\displaystyle {\hat {p}}\,}$  or opposite.

To say more about the state, ${\displaystyle \xi _{\lambda }\,}$  we will use the generic form of fermion four-momentum:

${\displaystyle p^{\mu }=\left(E,\left|{\vec {p}}\right|\sin {\theta }\cos {\phi },\left|{\vec {p}}\right|\sin {\theta }\sin {\phi },\left|{\vec {p}}\right|\cos {\theta }\right)\,}$ 

Then one can say the two helicity eigenstates are

${\displaystyle \xi _{+1}({\vec {p}})={\frac {1}{\sqrt {2\left|{\vec {p}}\right|\left(\left|{\vec {p}}\right|+p_{z}\right)}}}{\begin{pmatrix}\left|{\vec {p}}\right|+p_{z}\\p_{x}+ip_{y}\end{pmatrix}}={\begin{pmatrix}\cos {\frac {\theta }{2}}\\e^{i\phi }\sin {\frac {\theta }{2}}\end{pmatrix}}\,}$ 

and

${\displaystyle \xi _{-1}({\vec {p}})={\frac {1}{\sqrt {2|{\vec {p}}|(|{\vec {p}}|+p_{z})}}}{\begin{pmatrix}-p_{x}+ip_{y}\\\left|{\vec {p}}\right|+p_{z}\end{pmatrix}}={\begin{pmatrix}-e^{-i\phi }\sin {\frac {\theta }{2}}\\\cos {\frac {\theta }{2}}\end{pmatrix}}\,}$ 

These can be simplified by defining the z-axis such that the momentum direction is either parallel or anti-parallel, or rather:

${\displaystyle {\hat {z}}=\pm {\hat {p}}\,}$ .

In this situation the helicity eigenstates are for when the particle momentum is ${\displaystyle {\hat {p}}=+{\hat {z}}\,}$ 

${\displaystyle \xi _{+1}({\hat {z}})={\begin{pmatrix}1\\0\end{pmatrix}}\,}$  and ${\displaystyle \xi _{-1}({\hat {z}})={\begin{pmatrix}0\\1\end{pmatrix}}\,}$ 

then for when momentum is ${\displaystyle {\hat {p}}=-{\hat {z}}\,}$ 

${\displaystyle \xi _{+1}(-{\hat {z}})={\begin{pmatrix}0\\1\end{pmatrix}}\,}$  and ${\displaystyle \xi _{-1}(-{\hat {z}})={\begin{pmatrix}-1\\0\end{pmatrix}}\,}$ 

A fermion 4-component wave function, ${\displaystyle \psi \,}$  may be decomposed into states with definite four-momentum:

${\displaystyle \psi (x)=\int {{\frac {d^{3}p}{(2\pi )^{3}{\sqrt {2E}}}}\sum _{\lambda \pm 1}{\left({\hat {a}}_{p}^{\lambda }u_{\lambda }(p)e^{-ip\cdot x}+{\hat {b}}_{p}^{\lambda }v_{\lambda }(p)e^{ip\cdot x}\right)}}\,}$ 

where

${\displaystyle {\hat {a}}_{p}^{\lambda }\,}$  and ${\displaystyle {\hat {b}}_{p}^{\lambda }\,}$  are the creation and annihilation operators, and

${\displaystyle u_{\lambda }(p)\,}$  and ${\displaystyle v_{\lambda }(p)\,}$  are the momentum-space Dirac spinors for a fermion and anti-fermion respectively.

Put it more explicitly, the Dirac spinors in the helicity basis for a fermion is

${\displaystyle u_{\lambda }(p)={\begin{pmatrix}u_{-1}\\u_{+1}\end{pmatrix}}={\begin{pmatrix}{\sqrt {E-\lambda \left|{\vec {p}}\right|}}\chi _{\lambda }({\hat {p}})\\{\sqrt {E+\lambda \left|{\vec {p}}\right|}}\chi _{\lambda }({\hat {p}})\end{pmatrix}}\,}$ 

and for an anti-fermion,

${\displaystyle v_{\lambda }(p)={\begin{pmatrix}v_{+1}\\v_{-1}\end{pmatrix}}={\begin{pmatrix}-\lambda {\sqrt {E+\lambda \left|{\vec {p}}\right|}}\chi _{-\lambda }({\hat {p}})\\\lambda {\sqrt {E-\lambda \left|{\vec {p}}\right|}}\chi _{-\lambda }({\hat {p}})\end{pmatrix}}}$ 

Dirac matrices edit

To use these helicity states, one can use the Weyl (chiral) representation for the Dirac matrices.

The plane wave expansion is

${\displaystyle \psi (x)=\int {{\frac {d^{3}p}{(2\pi )^{3}{\sqrt {2E}}}}\sum _{\lambda =0}^{3}\left({\hat {a}}_{p,\lambda }\epsilon _{\lambda }(p)e^{-ip\cdot x}+{\hat {a}}_{p,\lambda }^{\dagger }\epsilon _{\lambda }^{*}(p)e^{ip\cdot x}\right)}\,}$ .

For a vector boson with mass m and a four-momentum ${\displaystyle q^{\mu }=(E,q_{x},q_{y},q_{z})}$ , the polarization vectors quantized with respect to its momentum direction can be defined as

${\displaystyle {\begin{aligned}\epsilon ^{\mu }(q,x)&={\frac {1}{\left|{\vec {q}}\right|q_{\text{T}}}}\left(0,q_{x}q_{z},q_{y}q_{z},-q_{\text{T}}^{2}\right)\\\epsilon ^{\mu }(q,y)&={\frac {1}{q_{\text{T}}}}\left(0,-q_{y},q_{x},0\right)\\\epsilon ^{\mu }(q,z)&={\frac {E}{m\left|{\vec {q}}\right|}}\left({\frac {\left|{\vec {q}}\right|^{2}}{E}},q_{x},q_{y},q_{z}\right)\end{aligned}}}$ 

where

${\displaystyle q_{\text{T}}={\sqrt {q_{x}^{2}+q_{y}^{2}}}\,}$  is transverse momentum, and

${\displaystyle E={\sqrt {|{\vec {q}}|^{2}+m^{2}}}\,}$  is the energy of the boson.