CHAPTER 2 A Crash Course on Weyl Spinors
37
To be honest, SUSY afficionados usually don’t write any dot symbol (·) between
the spinors, but it makes expressions more clear to include it, so that’s what we will
do for the rest of this book. However, even the notation used here may cause some
confusion, so let us issue a few warnings.
First, note that the bar appearing over the Weyl spinors has nothing whatsoever
to do with the bar used to represent Dirac conjugation for a four-component spinor,
=
†
γ
0
. We will discuss the bar notation for Weyl spinors in more depth later.
For now, consider the bars in the second line of Eq. (2.50) to be simply a notation
to distinguish the two types of dot products.
Second, those “spinor dot products” have nothing to do with the usual dot prod-
ucts between four-vectors or three-vectors familiar from introductory physics. It
may be tempting to think of χ ·χ, say, as χ
2
1
+ χ
2
2
, but this, of course, would be
incorrect (and equal to zero!). If we want to calculate these dot products explicitly
in terms of components, we need to go back to their explicit definitions given in
Eq. (2.50).
Third, it’s very important to keep in mind that the difference between the two
types of dot products is not only that one involves the hermitian conjugate of the
spinors, whereas the other does not; there is also a sign difference between the two
definitions, as is shown explicitly in Eqs. (2.51) and (2.52).
However, the two types of dot products obviously can be written in terms of one
another using
¯χ · ¯χ =−χ
†T
· χ
†T
(2.53)
The right-hand side is admittedly rather ugly, which is why the bar notation of the
left-hand side is preferred. The point of showing this relation is that the bar notation
is redundant; we could write everything in terms of unbarred spinors, at the price
of sometimes getting awkward expressions.
From Eq. (2.51), we see that the product of the two components of a spinor is
related to the dot product of that spinor with itself in a very simple way:
χ
1
χ
2
=−χ
2
χ
1
=−
1
2
χ ·χ
whereas, of course, χ
2
1
= χ
2
2
= 0. These four relations can be summarized into a
very useful identity:
χ
a
χ
b
=−
1
2
(iσ
2
)
ab
χ ·χ (2.54)