94
4. Lorentz Transformations and Discrete Symmetries
with E = m, p = 0, and with spin up along the z-axis; in S
′ , the particle has
′
energy E
′ , momentum p = (0, 0, p
′ ), and spin up along the z-axis. If we apply
′
a boost such that S
′ has velocity (0, 0, −v
′ ) relative to S, where v = p
′ /E
′ ,
then E and p become
E
′
= coshϑ
′ E = mγ(v
′ )
(4.52)
′
p = sinh ϑ
′ E = mv
′ γ(v
′ )
(4.53)
as required. Now consider the forms of the 4-spinors in S and S
′ . In S,
from (4.14) and (4.15) we have simply φ = χ, and if we normalize such that
¯ = 2m we may take
uu
(
)
(
)
√
φ +
1
u S = m
,
φ + =
.
(4.54)
φ +
0
In S
′ the spinor is
(
)
(
)
φ +
φ +
(
)
(
)
u S ' = N
E
' −σz p
'
= N
E
' −p
'
(4.55)
φ +
φ +
m
m
where the normalization N is determined (since ¯
uu is invariant) from the
condition ¯
u S ' u S ' = 2m to be N = (E
′ + p
′ )
1/2 , giving
(
)
′ )
1/2
(E
′ + p
φ +
u S ' =
.
(4.56)
′ )
1/2
(E
′
− p
φ +
But we can also calculate u S ' by applying the transformation (4.49) with
′
tanh ϑ
′ = −v to u S . Then the upper two components become
ϑ
'
ϑ
'
φ
′ =
√
m e
σz /2 φ + =
√
m e
/2 φ + ,
(4.57)
while the lower two components become
−ϑ
'
χ
′ =
√
m e
/2 φ + .
(4.58)
Now we can write
(
) 1/2
′
ϑ
'
E
′ + p
e
/2 = (e
ϑ
'
)
1/2 = (cosh ϑ
′ + sinh ϑ
′ )
1/2 =
(4.59)
m
and
(
) 1/2
′
−ϑ
' /2
E
′
− p
e
=
;
(4.60)
m
and so we recover (4.56).