Naber G.L. The Geometry of Minkowski Spacetime. An Introduction to the Mathematics of the Special Theory of Relativity
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164 3 The Theory of Spinors
We derive a somewhat less obvious identity by beginning with (3.3.8)and
first raising C.
AB
CD
+
AC
DB
+
AD
BC
=0,
AB
(
CE
ED
)+(
CE
AE
)
DB
+
AD
(
CE
BE
)=0,
AB
C
D
+
A
C
DB
+
AD
B
C
=0,
−
AB
δ
C
D
+ δ
C
A
DB
+
AD
δ
C
B
=0.
Now, raise D.
−
AB
DE
δ
C
E
+ δ
C
A
(
DE
EB
)+(
DE
AE
)δ
C
B
=0,
−
AB
DC
+ δ
C
A
D
B
+
A
D
δ
C
B
=0,
−
AB
DC
− δ
C
A
δ
D
B
+ δ
D
A
δ
C
B
=0.
Using
DC
= −
CD
we finally obtain
AB
CD
= δ
C
A
δ
D
B
− δ
D
A
δ
C
B
, A,B,C,D =1, 0. (3.3.14)
It will also be useful to introduce, for each [G
A
B
]=
*
G
1
1
G
1
0
G
0
1
G
0
0
+
in SL(2, C),
an associated matrix [G
A
B
]=
*
G
1
1
G
0
1
G
1
0
G
0
0
+
,where
G
A
B
=
AA
1
G
A
1
B
1
B
1
B
,A,B=1, 0.
Exercise 3.3.4 Show that
#
G
1
1
G
0
1
G
1
0
G
0
0
$
=
#
−G
0
0
G
1
0
G
0
1
−G
1
1
$
= −
#
G
1
1
G
0
1
G
1
0
G
0
0
$
(3.3.15)
and that
G
A
A
1
G
B
B
1
A
1
B
1
=
AB
,A,B=1, 0. (3.3.16)
As usual, all of these have obvious barred and dotted versions.
We shall denote by ß
rs
mn
the set of all spinors of valence
rs
mn
.Beinga
collection of multilinear functionals, ß
rs
mn
admits a natural “pointwise” vector
space structure. Specifically, if ξ, ζ ∈ ß
rs
mn
and
1
φ,...,
r
φ ∈ ß,
1
¯
ψ,...,
s
¯
ψ ∈
¯
ß,
1
μ
,...,
m
μ
∈ ß
∗
and
1
¯ν,...,
n
¯ν ∈
¯
ß
∗
,then
(ξ + ζ)
1
φ,...,
n
¯ν
= ξ
1
φ,...,
n
¯ν
+ ζ
1
φ,...,
n
¯ν
3.3 Spinor Algebra 165
and, for α ∈ C,
(αξ)
1
φ,...,
n
¯ν
= α
ξ
1
φ,...,
n
¯ν
.
If {s
1
,s
0
} is a spin frame, {s
1
,s
0
} its dual basis for ß
∗
, {¯s
˙
1
, ¯s
˙
0
} the corre-
sponding conjugate basis for
¯
ßand{¯s
˙
1
, ¯s
˙
0
} its dual, we define s
A
1
⊗...⊗s
A
r
⊗
¯s
˙
X
1
⊗···⊗¯s
˙
X
s
⊗s
B
1
⊗···⊗s
B
m
⊗¯s
˙
Y
1
⊗···⊗¯s
˙
Y
n
, abbreviated s
A
1
⊗···⊗¯s
˙
Y
n
,by
s
A
1
⊗···⊗¯s
˙
Y
n
1
φ,...,
n
¯ν
= s
A
1
1
φ
···¯s
˙
Y
n
n
¯ν
=
1
φ
A
1
···
n
¯ν
˙
Y
n
.
Thus, for example, in ß
11
10
we have s
1
⊗¯s
˙
0
⊗s
0
defined by s
1
⊗¯s
˙
0
⊗s
0
(φ,
¯
ψ,μ)
= s
1
(φ)¯s
˙
0
(
¯
ψ)s
0
(μ)=φ
1
¯
ψ
˙
0
μ
0
,whereφ = φ
A
s
A
,
¯
ψ =
¯
ψ
˙
X
¯s
˙
X
and μ = μ
A
s
A
.
For ξ ∈ ß
rs
mn
we define
ξ
A
1
···
˙
X
s
B
1
···
˙
Y
n
= ξ
s
A
1
,...,¯s
˙
X
s
,s
B
1
,...,¯s
˙
Y
n
(3.3.17)
for A
i
,B
i
=1, 0and
˙
X
i
,
˙
Y
i
=
˙
1,
˙
0.
Exercise 3.3.5 Show that the elements s
A
1
⊗···⊗¯s
˙
Y
n
of ß
rs
mn
are linearly
independent and that any ξ ∈ ß
rs
mn
can be written
ξ = ξ
A
1
···
˙
X
s
B
1
···
˙
Y
n
s
A
1
⊗···⊗¯s
˙
X
s
⊗ s
B
1
⊗···⊗¯s
˙
Y
n
.
From Exercise 3.3.5 we conclude that the s
A
1
⊗···⊗ ¯s
˙
Y
n
form a basis for
ß
rs
mn
which therefore has dimension 2
r+s+m+n
.Foreachξ ∈ ß
rs
mn
the numbers
ξ
A
1
···
˙
X
s
B
1
···
˙
Y
n
defined by (3.3.17)arethecomponentsofξ relative to the basis
{s
A
1
⊗···⊗¯s
˙
Y
n
} and, in terms of them, the linear operations on ß
rs
mn
can be
expressed as
(ξ + ζ)
A
1
···
˙
X
s
B
1
···
˙
Y
n
= ξ
A
1
···
˙
X
s
B
1
···
˙
Y
n
+ ζ
A
1
···
˙
X
s
B
1
···
˙
Y
n
and
(αξ)
A
1
···
˙
X
s
B
1
···
˙
Y
n
= αξ
A
1
···
˙
X
s
B
1
···
˙
Y
n
.
The next algebraic operation on spinors that we must consider is a general-
ization of the procedure we just employed to construct a basis for ß
rs
mn
from a
spin frame. Suppose ξ is a spinor of valence
r
1
s
1
m
1
n
1
and ζ is a spinor of va-
lence
r
2
s
2
m
2
n
2
.Theouter product of ξ and ζ is the spinor ξ ⊗ ζ of valence
166 3 The Theory of Spinors
r
1
+r
2
s
1
+s
2
m
1
+m
2
n
1
+n
2
defined as follows. If
1
φ,...,
r
1
+r
2
φ ∈ ß,
1
¯
ψ,...,
s
1
+s
2
¯
ψ ∈
¯
ß,
1
μ
,...,
m
1
+m
2
μ
∈ ß
∗
and
1
¯ν,...,
n
1
+n
2
¯ν ∈
¯
ß
∗
,then
(ξ ⊗ ζ)
1
φ,...,
r
1
+r
2
φ,
1
¯
ψ,...,
s
1
+s
2
¯
ψ,
1
μ
,...,
m
1
+m
2
μ
,
1
¯ν,...,
n
1
+n
2
¯ν
= ξ
1
φ,...,
r
1
φ,
1
¯
ψ,...,
s
1
¯
ψ,
1
μ
,...,
m
1
μ
,
1
¯ν,...,
n
1
¯ν
× ζ
r
1
+1
φ ,...,
r
1
+r
2
φ,
s
1
+1
¯
ψ ,...,
s
1
+s
2
¯
ψ,
m
1
+1
μ
,...,
m
1
+m
2
μ
,
n
1
+1
¯ν ,...,
n
1
+n
2
¯ν
.
It follows immediately from the definition that, in terms of components,
(ξ ⊗ ζ)
A
1
···A
r
1
+r
2
˙
X
1
···
˙
X
s
1
+s
2
B
1
···B
m
1
+m
2
˙
Y
1
···
˙
Y
n
1
+n
2
=
ξ
A
1
···A
r
1
˙
X
1
···
˙
X
s
1
B
1
···B
m
1
˙
Y
1
···
˙
Y
n
1
ζ
A
r
1
+1
···A
r
1
+r
2
˙
X
s
1
+1
···
˙
X
s
1
+s
2
B
m
1
+1
···B
m
1
+m
2
˙
Y
n
1
+1
···
˙
Y
n
1
+n
2
.
Moreover, outer multiplication is clearly associative ((ξ ⊗ζ)⊗v = ξ ⊗(ζ ⊗v))
and distributive (ξ ⊗(ζ + v)=ξ ⊗ζ + ξ ⊗v and (ξ +ζ)⊗v = ξ ⊗v + ζ ⊗v), but
is not commutative. For example, if {s
1
,s
0
} is a spin frame, then s
1
⊗s
0
does
not equal s
0
⊗ s
1
since s
1
⊗ s
0
(φ
∗
,ψ
∗
)=φ
1
ψ
0
, but s
0
⊗ s
1
(φ
∗
,ψ
∗
)=φ
0
ψ
1
and these are generally not the same.
Next we consider a spinor ξ of valence
rs
mn
and two integers k and l
with 1 ≤ k ≤ r and 1 ≤ l ≤ m. Then the contraction of ξ in the indices A
k
and B
l
is the spinor C
kl
(ξ) of valence
r−1 s
m−1 n
whose components relative
to any spin frame are obtained by equating A
k
and B
l
in those of ξ and
summing as indicated, i.e., if
ξ = ξ
A
1
···A
k
···A
r
˙
X
1
···
˙
X
s
B
1
···B
l
···B
m
˙
Y
1
···
˙
Y
n
s
A
1
⊗···⊗¯s
˙
Y
n
,
then
C
kl
(ξ)=ξ
A
1
···A···A
r
˙
X
1
···
˙
X
s
B
1
···A···B
m
˙
Y
1
···
˙
Y
n
s
A
1
⊗···⊗¯s
˙
Y
n
, (3.3.18)
where, in this last expression, it is understood that s
A
k
and s
B
l
are missing
in s
A
1
⊗···⊗¯s
˙
Y
n
.Thus,forexample,ifξ is of valence
11
10
with
ξ = ξ
A
1
˙
X
1
B
1
s
A
1
⊗ ¯s
˙
X
1
⊗ s
B
1
3.3 Spinor Algebra 167
and k = l =1,thenC
11
(ξ) is the spinor of valence
01
00
given by
C
11
(ξ)=ξ
A
˙
X
1
A
¯s
˙
X
1
=
ξ
1
˙
X
1
1
+ ξ
0
˙
X
1
0
¯s
˙
X
1
.
Unlike our previous definitions that were coordinate-free, contractions are
defined in terms of components and so it is not immediately apparent that
we have defined a spinor at all. We must verify that the components of C
kl
(ξ)
as defined by (3.3.18) transform correctly, i.e., as the components of a spinor
of v
ale
nce
r−1 s
m−1 n
. But this is clearly the case since, in a new spin frame,
ˆ
ξ
A
1
···A···A
r
˙
X
1
···
˙
X
s
B
1
···A···B
m
˙
Y
1
···
˙
Y
n
= G
A
1
C
1
···G
A
C
k
···G
A
r
C
r
¯
G
˙
X
1
˙
U
1
···
¯
G
˙
X
s
˙
U
s
G
B
1
D
1
···G
A
D
l
···
G
B
m
D
m
¯
G
˙
Y
1
˙
V
1
···
¯
G
˙
Y
n
˙
V
n
ξ
C
1
···C
k
···C
r
˙
U
1
···
˙
U
s
D
1
···D
l
···D
m
˙
Y
1
···
˙
Y
n
=
G
A
C
k
G
A
D
l
G
A
1
C
1
···
¯
G
˙
Y
n
˙
V
n
ξ
C
1
···C
k
···C
r
˙
U
1
···
˙
U
s
D
1
···D
l
···D
m
˙
Y
1
···
˙
Y
n
= δ
D
l
C
k
G
A
1
C
1
···
¯
G
˙
Y
n
˙
V
n
ξ
C
1
···C
k
···C
r
˙
U
1
···
˙
U
s
D
1
···D
l
···D
m
˙
Y
1
···
˙
Y
n
= G
A
1
C
1
···
¯
G
˙
Y
n
˙
V
n
ξ
C
1
···A···C
r
˙
U
1
···
˙
U
s
D
1
···A···D
m
˙
Y
1
···
˙
Y
n
.
One can, in the same way, contract a spinor ξ of valence
rs
mn
in two
dotted indices
˙
k and
˙
l, one upper and one lower, to obtain a spinor C
˙
k
˙
l
(ξ)of
valence
rs−1
mn−1
. Observe that the processes of raising and lowering indices
discussed earlier are actually outer products (with an spinor) followed by a
contraction.
Exercise 3.3.6 Let φ be a spinor of valence
00
20
and denote its compo-
nents in a spin frame by φ
AB
. Show that
1. φ
1
1
= −φ
10
,φ
0
0
= φ
01
,φ
1
0
= φ
11
,φ
0
1
= −φ
00
,
2. φ
11
= φ
00
,φ
00
= φ
11
,φ
10
= −φ
01
,φ
01
= −φ
10
,
3. φ
AB
φ
AB
=2det
*
φ
11
φ
10
φ
01
φ
00
+
=2det
*
φ
1
1
φ
1
0
φ
0
1
φ
0
0
+
,
4. φ
AC
φ
B
C
=
⎧
⎨
⎩
0,A= B
det[φ
AB
],A=0,B=1
−det[φ
AB
],A=1,B=0.
Let ξ denote a spinor with the same number of dotted and undotted indices,
say, of valence
*
rr
00
+
. We define a new spinor denoted
¯
ξ and called the
168 3 The Theory of Spinors
conjugate of ξ by specifying that its components
¯
ξ
˙
A
1
···
˙
A
r
X
1
···X
r
in any spin
frame are given by
¯
ξ
˙
A
1
···
˙
A
r
X
1
···X
r
= ξ
A
1
···A
r
˙
X
1
···
˙
X
r
(here we must depart from our habit of selecting dotted/undotted indices
from the end/beginning of the alphabet). Thus, for example, if ξ has compo-
nents ξ
A
˙
X
, then the components of
¯
ξ are given by
¯
ξ
˙
01
= ξ
0
˙
1
,
¯
ξ
˙
11
= ξ
1
˙
1
,etc.
Exercise 3.3.7 Show that we have actually defined a spinor of the required
type by verifying the appropriate transformation law, i.e.,
¯
ˆ
ξ
˙
A
1
···
˙
A
r
X
1
···X
r
=
¯
G
˙
A
1
˙
C
1
···
¯
G
˙
A
r
˙
C
r
G
X
1
U
1
···G
X
r
U
r
¯
ξ
˙
C
1
···
˙
C
r
U
1
···U
r
.
Entirely analogous definitions and results apply regardless of the positions
(upper or lower) of the indices, provided only that the number of dotted
indices is the same as the number of undotted indices. We shall say that
such a spinor ξ is Hermitian if
¯
ξ = ξ.Thus,forexample,ifξ is of va-
lence
rr
00
,thenitisHermitianif
¯
ξ
A
1
···A
r
˙
X
1
···
˙
X
r
= ξ
A
1
···A
r
˙
X
1
···
˙
X
r
for all
A
1
,...,A
r
,
˙
X
1
,...,
˙
X
r
, i.e., if
ξ
˙
A
1
···
˙
A
r
X
1
···X
r
= ξ
A
1
···A
r
˙
X
1
···
˙
X
r
,
e.g., if r =1,
ξ
˙
01
= ξ
0
˙
1
, ξ
0
˙
0
= ξ
˙
00
,etc.
As a multilinear functional a spinor ξ operates on four distinct types of
objects (elements of ß,
¯
ß, ß
∗
and
¯
ß
∗
) and, if the valence is
rs
mn
,has
r + s + m + n “slots” (variables) into which these objects are inserted for
evaluation, each slot corresponding to an index position in our notation
for ξ’s components. If ξ has the property that, for two such slots of the
same type, ξ(...,p,...,q,...)=ξ(...,q,...,p,...) for all p and q of the
appropriate type, then ξ is said to be symmetric in these two variables (if
ξ(...,p,...,q,...)=−ξ(...,q,...,p,...), it is skew-symmetric). It follows at
once from the definition that ξ is symmetric (skew-symmetric) in the vari-
ables p and q if and only if the components of ξ in every spin frame are
unchanged (change sign) when the corresponding indices are interchanged.
We will be particularly interested in the case of spinors with just two indices.
Thus, for example, a spinor φ of valence
00
20
is symmetric (in its only two
variables) if and only if, in every spin frame, φ
BA
= φ
AB
for all A, B =1, 0; φ
is skew-symmetric if φ
BA
= − φ
AB
for all A and B. On the other hand, an
arbitrary spinor ξ of valence
00
20
has a symmetrization whose components
in each spin frame are given by
ξ
(AB)
=
1
2
(ξ
AB
+ ξ
BA
)
3.4 Spinors and World Vectors 169
and a skew-symmetrization given by
ξ
[AB ]
=
1
2
(ξ
AB
− ξ
BA
).
The symmetrization (skew-symmetrization) of ξ clearly defines a spinor, also
of valence
00
20
, that is symmetric (skew-symmetric).
Exercise 3.3.8 Let α and β be two spin vectors. The outer product α ⊗ β
is a spinor of valence
00
20
whose components in any spin frame are given
by α
A
β
B
,A,B =1, 0. Let φ be the symmetrization of α ⊗ β so that
φ
AB
= α
(A
β
B)
=
1
2
(α
A
β
B
+ α
B
β
A
).
Show that φ
AB
=
1
2
(α
A
β
B
+ α
B
β
A
)andthat
φ
AB
φ
AB
= −
1
2
<α,β>
2
.
3.4 Spinors and World Vectors
In this section we will establish a correspondence between spinors of valence
11
00
and vectors in Minkowski spacetime (also called world vectors
or 4-vectors). This correspondence, which we have actually seen before
(in Section 1.7), is most easily phrased in terms of the Pauli spin matrices.
Exercise 3.4.1 Let σ
1
=
*
01
10
+
,σ
2
=
*
0 i
−i 0
+
,σ
3
=
*
10
0 −1
+
,andσ
4
=
*
10
01
+
. Verify the following commutation relations:
σ
1
2
= σ
2
2
= σ
3
2
= σ
4
2
= σ
4
,
σ
1
σ
2
= −σ
2
σ
1
= −iσ
3
,
σ
1
σ
3
= −σ
3
σ
1
= iσ
2
,
σ
2
σ
3
= −σ
3
σ
2
= −iσ
1
.
For what follows it will be convenient to introduce a factor of
1
√
2
and some
rather peculiar looking indices, the significance of which will become clear
shortly. Thus, for each A =1, 0and
˙
X =
˙
1,
˙
0, we define matrices
σ
a
A
˙
X
=
#
σ
a
1
˙
1
σ
a
1
˙
0
σ
a
0
˙
1
σ
a
0
˙
0
$
,a=1, 2, 3, 4,
170 3 The Theory of Spinors
by
σ
a
A
˙
X
=
1
√
2
σ
a
,a=1, 2, 3, 4.
Thus,
σ
1
A
˙
X
=
1
√
2
*
01
10
+
,σ
2
A
˙
X
=
1
√
2
*
0 i
−i 0
+
,
σ
3
A
˙
X
=
1
√
2
*
10
0 −1
+
,σ
4
A
˙
X
=
1
√
2
*
10
01
+
.
We again adopt the convention that the relative position of dotted and undot-
ted indices is immaterial so σ
a
A
˙
X
= σ
a
˙
X
A
. Undotted indices indicate rows;
dotted indices number the columns. Observe that each of these is a Hermitian
matrix, i.e., equals its conjugate transpose (Section 1.7).
Now we describe a procedure for taking a vector v ∈M, an admissible
basis {e
a
} for M and a spin frame {s
A
} and constructing from them a spinor
V of valence
11
00
. We do this by specifying the components of V in every
spin frame and verifying that they have the correct transformation law. We
begin by writing v = v
a
e
a
. Define the components V
A
˙
X
of V relative to
{s
A
} by
V
A
˙
X
= σ
a
A
˙
X
v
a
,A=1, 0,
˙
X =
˙
1,
˙
0. (3.4.1)
Thus,
V
1
˙
1
=
1
√
2
(v
3
+ v
4
),
V
1
˙
0
=
1
√
2
(v
1
+ iv
2
),
V
0
˙
1
=
1
√
2
(v
1
− iv
2
), (3.4.2)
V
0
˙
0
=
1
√
2
(−v
3
+ v
4
)
(cf. Exercise 1.7.1). Now, suppose {ˆs
1
, ˆs
0
} is another spin frame, related to
{s
A
} by (3.2.1)(s
B
= G
A
B
ˆs
A
)and(3.2.6)(ˆs
A
= G
A
B
s
B
). We define the
components
ˆ
V
A
˙
X
of V relative to {ˆs
A
} as follows: Let Λ = Λ
G
= Spin(G)be
the element of L that G maps onto under the spinor map and ˆv
a
=Λ
a
b
v
b
,a=
1, 2, 3, 4. Now let
ˆ
V
A
˙
X
= σ
a
A
˙
X
ˆv
a
,A=1, 0,
˙
X =
˙
1,
˙
0. (3.4.3)
3.4 Spinors and World Vectors 171
That we have actually defined a spinor of valence
11
00
is not obvious, of
course, since it is not clear that the V
A
˙
X
transform correctly. To show this
we must prove that
ˆ
V
A
˙
X
= G
A
B
¯
G
˙
X
˙
Y
V
B
˙
Y
,A=1, 0,
˙
X =
˙
1,
˙
0. (3.4.4)
For this we temporarily denote the right-hand side of (3.4.4)by
˜
V
A
˙
X
, i.e.,
˜
V
A
˙
X
= G
A
B
¯
G
˙
X
˙
Y
V
B
˙
Y
. Writing this as a matrix product gives
⎡
⎢
⎢
⎢
⎢
⎢
⎣
˜
V
1
˙
1
˜
V
1
˙
0
˜
V
0
˙
1
˜
V
0
˙
0
⎤
⎥
⎥
⎥
⎥
⎥
⎦
=
⎡
⎢
⎢
⎢
⎢
⎢
⎣
G
1
1
¯
G
˙
1
˙
1
G
1
1
¯
G
˙
1
˙
0
G
1
0
¯
G
˙
1
˙
1
G
1
0
¯
G
˙
1
˙
0
G
1
1
¯
G
˙
0
˙
1
G
1
1
¯
G
˙
0
˙
0
G
1
0
¯
G
˙
0
˙
1
G
1
0
¯
G
˙
0
˙
0
G
0
1
¯
G
˙
1
˙
1
G
0
1
¯
G
˙
1
˙
0
G
0
0
¯
G
˙
1
˙
1
G
0
0
¯
G
˙
1
˙
0
G
0
1
¯
G
˙
0
˙
1
G
0
1
¯
G
˙
0
˙
0
G
0
0
¯
G
˙
0
˙
1
G
0
0
¯
G
˙
0
˙
0
⎤
⎥
⎥
⎥
⎥
⎥
⎦
⎡
⎢
⎢
⎢
⎢
⎢
⎣
V
1
˙
1
V
1
˙
0
V
0
˙
1
V
0
˙
0
⎤
⎥
⎥
⎥
⎥
⎥
⎦
. (3.4.5)
But if we let
G =
#
G
1
1
G
1
0
G
0
1
G
0
0
$
=
*
αβ
γδ
+
,
then (3.4.5) becomes
⎡
⎢
⎢
⎢
⎣
˜
V
1
˙
1
˜
V
1
˙
0
˜
V
0
˙
1
˜
V
0
˙
0
⎤
⎥
⎥
⎥
⎦
=
⎡
⎢
⎢
⎢
⎣
α¯αα
¯
β ¯αβ β
¯
β
α¯γα
¯
δβ¯γβ
¯
δ
¯αγ
¯
βγ ¯αδ
¯
βδ
γ¯γγ
¯
δ ¯γδ δ
¯
δ
⎤
⎥
⎥
⎥
⎦
⎡
⎢
⎢
⎢
⎣
V
1
˙
1
V
1
˙
0
V
0
˙
1
V
0
˙
0
⎤
⎥
⎥
⎥
⎦
. (3.4.6)
Now, using (3.4.2) and the corresponding equalities for
ˆ
V
A
˙
X
it follows from
Exercise 1.7.2 (with the appropriate notational changes) that the right-hand
side of (3.4.6)isequalto
1
√
2
⎡
⎢
⎢
⎣
ˆv
3
+ˆv
4
ˆv
1
+ iˆv
2
ˆv
1
− iˆv
2
−ˆv
3
+ˆv
4
⎤
⎥
⎥
⎦
=
⎡
⎢
⎢
⎢
⎣
ˆ
V
1
˙
1
ˆ
V
1
˙
0
ˆ
V
0
˙
1
ˆ
V
0
˙
0
⎤
⎥
⎥
⎥
⎦
,
where the ˆv
a
are the images of the v
a
under Λ = Λ
G
. Substituting this into
(3.4.6)thengives[
˜
V
A
˙
X
]=[
ˆ
V
A
˙
X
] and this proves (3.4.4). Observe that, since
#
G
1
1
G
1
0
G
0
1
G
0
0
$
= −
#
G
1
1
G
1
0
G
0
1
G
0
0
$
,
(3.4.4) can also be written as
ˆ
V
A
˙
X
= G
A
B
¯
G
˙
X
˙
Y
V
B
˙
Y
,A=1, 0,
˙
X =
˙
1,
˙
0. (3.4.7)
172 3 The Theory of Spinors
We conclude then that the procedure we have described does indeed define a
spinor of valence
11
00
which we shall call the spinor equivalent of v ∈M
(somewhat imprecisely since V depends not only on v, but also on the initial
choices of {e
a
} and {s
A
}). Observe that the conjugate
¯
V of V has components
¯
V
A
˙
X
= V
˙
AX
= σ
a
˙
AX
v
a
= σ
a
˙
AX
v
a
= σ
a
A
˙
X
v
a
= V
A
˙
X
since the matrices
σ
a
A
˙
X
are Hermitian and the v
a
are real. Thus, the spinor equivalent of any
world vector is a Hermitian spinor.
With (3.4.4) we can now justify the odd arrangement of indices in the
sym
b
ols σ
a
A
˙
X
by showing that the σ
a
A
˙
X
are constant under the combined
effect of a G∈SL(2, C) and the corresponding Λ = Λ
G
in L, i.e., that
Λ
a
b
G
A
B
¯
G
˙
X
˙
Y
σ
b
B
˙
Y
= σ
a
A
˙
X
,a=1, 2, 3, 4,A=1, 0,
˙
X =
˙
1,
˙
0 (3.4.8)
(one might say that the σ
a
A
˙
X
are the components of a constant “spinor-
covector”). To see this we select an arbitrary admissible basis and spin frame.
Fix A and
˙
X.Nowletv = v
a
e
a
be an arbitrary vector in M.ThenV
A
˙
X
=
σ
a
A
˙
X
v
a
. In another spin frame, related to the original by G,wehave
ˆ
V
A
˙
X
= σ
a
A
˙
X
ˆv
a
.
But also,
ˆ
V
A
˙
X
= G
A
B
¯
G
˙
X
˙
Y
V
B
˙
Y
= G
A
B
¯
G
˙
X
˙
Y
σ
b
B
˙
Y
v
b
= G
A
B
¯
G
˙
X
˙
Y
σ
b
B
˙
Y
δ
b
c
v
c
= G
A
B
¯
G
˙
X
˙
Y
σ
b
B
˙
Y
Λ
a
b
Λ
a
c
v
c
=Λ
a
b
G
A
B
¯
G
˙
X
˙
Y
σ
b
B
˙
Y
(Λ
a
c
v
c
)
=Λ
a
b
G
A
B
¯
G
˙
X
˙
Y
σ
b
B
˙
Y
ˆv
a
.
Thus,
Λ
a
b
G
A
B
¯
G
˙
X
˙
Y
σ
b
B
˙
Y
ˆv
a
= σ
a
A
˙
X
ˆv
a
.
But v was arbitrary so we may successively select v’s that give (ˆv
1
, ˆv
2
, ˆv
3
, ˆv
4
)
equal to (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0) and (0, 0, 0, 1) and thereby obtain
(3.4.8)fora =1, 2, 3
and 4
, respectively. Since
G
A
B
= −
G
A
B
we again
find that (3.4.8) can be written
Λ
a
b
G
A
B
¯
G
˙
X
˙
Y
σ
b
B
˙
Y
= σ
a
A
˙
X
,a=1, 2, 3, 4,A=1, 0,
˙
X =
˙
1,
˙
0. (3.4.9)
Exercise 3.4.2 Show that
G
A
B
¯
G
˙
X
˙
Y
σ
a
B
˙
Y
=Λ
α
a
σ
α
A
˙
X
,a=1, 2, 3, 4,A=1, 0,
˙
X =
˙
1,
˙
0. (3.4.10)
3.4 Spinors and World Vectors 173
From all of this we conclude that the σ
a
A
˙
X
behave formally like a combined
world covector and spinor of valence
11
00
. Treating them as such we raise
the index a and lower A and
˙
X, i.e., we define
σ
a
A
˙
X
= η
ab
σ
b
B
˙
Y
BA
¯
˙
Y
˙
X
for a =1, 2, 3, 4,A=1, 0and
˙
X =
˙
1,
˙
0. Thus, for example, if a =1,σ
1
A
˙
X
=
η
1b
σ
b
B
˙
Y
BA
¯
˙
Y
˙
X
= η
11
σ
1
B
˙
Y
BA
¯
˙
Y
˙
X
= σ
1
B
˙
Y
BA
¯
˙
Y
˙
X
.IfA =1,this
becomes σ
1
1
˙
X
= σ
1
B
˙
Y
B1
¯
˙
Y
˙
X
= σ
1
0
˙
Y
01
¯
˙
Y
˙
X
= σ
1
0
˙
Y
¯
˙
Y
˙
X
= σ
1
0
˙
1
¯
˙
1
˙
X
+
σ
1
0
˙
0
¯
˙
0
˙
X
.Thus,for
˙
X =
˙
1,σ
1
1
˙
1
= σ
1
0
˙
1
¯
˙
1
˙
1
+ σ
1
0
˙
0
¯
˙
0
˙
1
= σ
1
0
˙
0
= 0 and, for
˙
X =
˙
0,σ
1
1
˙
0
= σ
1
0
˙
1
¯
˙
1
˙
0
+ σ
1
0
˙
0
¯
˙
0
˙
0
= −σ
1
0
˙
1
= −
1
√
2
. Similarly, σ
1
0
˙
1
= −
1
√
2
and σ
1
0
˙
0
=0so
σ
1
A
˙
X
=
*
σ
1
1
˙
1
σ
1
1
˙
0
σ
1
0
˙
1
σ
1
0
˙
0
+
= −
1
√
2
*
01
10
+
= −σ
1
A
˙
X
.
Exercise 3.4.3 Continue in this way to prove the remaining equalities in
σ
1
A
˙
X
= −σ
1
A
˙
X
= −
1
√
2
*
01
10
+
,
σ
2
A
˙
X
= σ
2
A
˙
X
=
1
√
2
*
0 i
−i 0
+
,
σ
3
A
˙
X
= −σ
3
A
˙
X
= −
1
√
2
*
10
0 −1
+
,
σ
4
A
˙
X
= −σ
4
A
˙
X
= −
1
√
2
*
10
01
+
.
(3.4.11)
We enumerate a number of useful properties of these so-called Infeld-van der
Waerden symbols σ
a
A
˙
X
and σ
a
A
˙
X
.
σ
a
A
˙
X
= η
ab
¯
˙
X
˙
Y
AB
σ
b
B
˙
Y
, (3.4.12)
σ
a
A
˙
X
σ
b
A
˙
X
= −δ
b
a
, (3.4.13)
σ
a
A
˙
X
σ
a
B
˙
Y
= −δ
A
B
δ
˙
X
˙
Y
, (3.4.14)
σ
a
A
˙
X
σ
a
B
˙
Y
σ
b
B
˙
Y
= −σ
b
A
˙
X
. (3.4.15)
For the proof of (3.4.12)weinsertσ
b
B
˙
Y
= η
bc
(σ
c
C
˙
Z
CB
)¯
˙
Z
˙
Y
into the right-
hand side to obtain
η
ab
¯
˙
X
˙
Y
AB
σ
b
B
˙
Y
= η
ab
¯
˙
X
˙
Y
AB
η
bc
σ
c
C
˙
Z
CB
¯
˙
Z
˙
Y
=(η
ab
η
bc
)(¯
˙
X
˙
Y
¯
˙
Z
˙
Y
)(
AB
CB
)σ
c
C
˙
Z