I’m stuck on a basic introductory point regarding twistors. I understand the mapping of 4-vectors to hermitian matrices, and the incidence relation defined by pairs of spinors. And I understand that a null vector will result in the matrix having determinant zero, therefore being of rank one. Then it can be regarded as the outer product of a spinor and its complex conjugate (so I'm told). But I can’t find a single example of such a spinor-conjugate pair, dotted and undotted; and I can’t see how the outer product of such a pair would produce the matrix one gets by taking the inner product of a vector with Pauli matrices.


1 Answer 1


The problem was to find two complex numbers (a,b) whose outer product with their complex conjugates (aa*,ab*,ba*,bb*) encodes a null spacetime 4-vector (T,X,Y,Z) in a Hermitian 2x2 matrix (T+Z, X-iY, X+iY, T-Z).

This is an answer I found in Steane’s “Intro to Spinors”. It uses spherical coordinates r,𝛉,𝚽 plus α for the “flag” angle.

a=√r cos(𝛉/2)e^i(-α-𝚽)/2

b=√r sin(𝛉/2)e^i(-α+𝚽)/2

If we work out ab* = rsincos(𝛉/2)e^-i𝚽 we get

(rsin𝛉cos𝚽 - irsin𝛉sin𝚽)/2

which in rectangular coordinates is


as desired.

Similarly aa* = rcos^2(𝛉/2)(e^i(-α-𝚽)/2)(e^-i(-α-𝚽)/2)

= rcos^2(𝛉/2)

= r(1+cos𝛉)/2

= (r + Z)/2

in this case we identify r with T (the radius of the “light cone” at T, because T^2=X^2+Y^2+Z^2 for a null 4-vector).

The same calculations give bb* = (r-Z)/2 and bb*=(X+iY)/2.


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