7. Relativity Wave Equations and Field Theories
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Transcript 7. Relativity Wave Equations and Field Theories
7. Relativity Wave Equations and Field Theories
7.1.
The Klein-Gordon Equation
7.2.
Scalar Field Theory for Free Particles
7.3.
The Dirac Equation and spin-1/2 Particles
7.4.
Spinor Field Theory
7.5.
Weyl and Marjorana Spinors
7.6.
Particles of Spin 1 and 2
7.7.
Wave Equations in Curved Spacetime
Natural Units
c
1
c=1
→
[L]=[T]
=1
→
[ E ] = [ T ]1
Choosing [ E ] = MeV, we have
• [ M ] = [ P ] = [ E ] = MeV
• [ L ] = [ T ] = [ E ] 1 = (MeV) 1
c = 2.997 108 m/s
= 6.582 1022 MeV s
1 MeV = 1.602 1013 J
T( s ) = 6.582 1022 T( MeV 1 )
→
L( m ) = 1.937 1013 L( MeV 1 )
M( kg ) = 1.783 1030 M( MeV )
7.1.
p i
→
The Klein-Gordon Equation
p p E 2 p p m2
2
2 2 m2
t
W m 0
2
Klein-Gordon equation
2
W 2 2
t
D’Alembertian
μμ is a Lorentz scalar → must be a Lorentz tensor
Only case = scalar considered.
Conserved Current Density
Statistical interpretation of quantum mechanics requires
existence of a conserved probability density .
4-current density: j , j
m2 0
m2 * 0
j
→
Equation of continuity: j 0 t j
0 * * * *
i
* *
2m
i
*
2m
is conserved.
a b ab a b
However, jμ cannot be the 4-current probability density because
j0
i
* 0 0 *
2m
is not positive definite.
Plane Wave Solutions
Plane wave with 4-wavevector k = (k0 , k ):
k x exp ik x exp i k 0t i k x
→
k ik k
k k kk k k k
ϕ is a solution to the K-G eq only if
→ energy of the particle is
k k m
2
k 0 k 2 m2
2
E k 0 k 2 m 2 k
E < 0 solutions are unphysical since the vaccuum is at E = 0.
If E < 0 solutions are allowed, ground state is E → .
However, E < 0 solutions are needed to form a complete set of basis functions.
These problems are resolved through the concept of anti-particles.
Precursor to a Quantum Field
Problems of the K-G eq can be fully resolved only by switching to QFT.
Consistent 1-particle relativistic quantum theory does not exist
( particle- antiparticle pairs always emerge spontaneously at sufficiently high E )
Action for the K-G eq:
Conjugate momenta :
L
0 *
0
General solution:
x
1
x xj
f x
df
j
dx
L * m 2 *
S d 4x L
d 4k
2
*
L
0
*
0
2
2
k
m
k exp ik x
3
k k 2 m 2
x x j
1
k 0 k k 0 k
k 2 m 2 k 0 k 2 m 2 k 0 k 2 m 2
2k
x
where
d 3k
2 2k
3
a k exp ik t i k x b k exp ikt i k x
a k k k0
b k k k 0
k
k k :
x
k
d 3k
2 2k
3
d 3k
2 2k
3
a k exp ik x b k exp ik x k 0
k
a k exp ik x c* k exp ik x 0
k k
c* k b k
The following aims to find a and c :
x
d 3k
2 2k
i x
0
3
i k x
d
x
e
b :
3
i k x
d
x
e
a :
k
Add:
3
d 3k
2
3
2
a k exp ik x c* k exp ik x 0
k k
a k exp ik x c* k exp ik x 0
k k
(a)
(b)
1
a k exp ik t c* k exp ikt
2
1
3
*
d
x
exp
i
k
x
x
a
k
exp
i
t
c
k exp ikt
k
k
2
i d 3 x exp i k x 0 x
a k exp ik t d 3 x exp i k x i 0 x k x
0
a k d x exp ik x i x k x i d x exp ik x x
3
0
3
d 3 x exp ik x i * x k x
Subtract:
c* k exp ik t d 3 x exp i k x i 0 x k x
0
*
c k d x exp ik x i x k x i d x exp ik x x
3
0 *
*
d 3 x exp ik x i x k * x
3
7.2.
Scalar Field Theory for Free Particles
Scalar field → particles are bosons.
Quantization is expressed by equal-time commutators :
ˆ x, t i x x
ˆ x, t ,
ˆ x, t i x x
ˆ x, t ,
ˆ
ˆ
ˆ x, t , ˆ x, t
x, t , x , t 0
ˆ x, t ,
ˆ x, t 0
ˆ x, t , ˆ x, t
ˆ x, t ,
ˆ x, t 0
ˆ x, t , ˆ x, t
ˆ x , ˆ x i
ˆ x
aˆ k , aˆ k d 3 x d 3 x eik xik x k ˆ x i
k
'
t t'
d 3 x d 3 x eik x ik x k k ' x x
d x e
3
i k k ' t i k k x
e
cˆ k , cˆ k 2 2k k k
k k '
t t '
2 2k k k
3
3
c kills anti-particle of E > 0.
Covariant Normalization
aˆ k , aˆ k 2 2k k k
changes the normalization of the 1-P eigenstates.
Factor (2π)3 2ωk in
3
2 2k k k aˆ k aˆ k aˆ k aˆ k
3
where
k aˆ k
k k
aˆ k 0
For a system with exactly one “particle”, the states { | k } are complete :
k k
d k 2
3
| k is a free particle state →
x
d 3k
3
2 2k
3
2k
I
k x x k Ck exp i k x
k k x
d 3k
3
2 2k
k Ck* exp i k x
x x x x
d 3k
d 3k
3
3
2 2k 2 2k
Ck Ck* exp i k x i k x k k
2
d 3k
Ck
*
C
C
exp
i
k
x
i
k
x
x x
k k
3
2k
2 2k
Ck 2k
→
→
x
d 3k
2
3
2k
k exp i k x
k x x k 2k exp i k x
P x k x
2
2k
transforms like k0
2 2k k k k k
3
is a Lorentz covariant normalization.
Hamiltonian
0 *
j *
2 *
H 0 *0* L 0 j m
*
* m2 * * * m 2 *
t t
ˆ ˆ ˆ ˆ
Ignoring the possible ambiguity that may arise from
ˆ ˆ
Hˆ d x
ˆ ˆ m 2ˆ ˆ
t t
we write
3
ˆ ˆ
d 3k
3
d x t t d x 2 2 3
3
d 3k
2 2 aˆ k e
3
aˆ k e i k ' x cˆ k e i k ' x
i k k x
3
d
x
e
cˆ k e i k x
k 0 k
k '0 k '
2 k k exp i k 0 k 0 t
3
k k , k k
0
0
ˆ ˆ
d 3k
d x t t 4 2 3
3
ik x
k k , k k
aˆ k aˆ k cˆ k cˆ k cˆ k aˆ k e
2 k k
3
0
0
2 i k t
aˆ k cˆ k e 2ik t
ˆ ˆ d 3 x
d
x
3
d 3k
d 3k
2k 2
3
2 2
k k aˆ k e i k x cˆ k e i k x
3
k'
aˆ k e i k ' x cˆ k e i k ' x
d 3k
4k 2
d x m ˆ ˆ m
3
2
d x
3
d 3k
d 3k
2k 2
3
2 2
aˆ k e
3
→
4k2 2
3
2
k
k 0 k
k '0 k '
aˆ k aˆ k cˆ k cˆ k
d 3k
2k 2
Vacuum energy:
cˆ k e i k x
k 2 aˆ k aˆ k cˆ k cˆ k cˆ k aˆ k e 2ik t aˆ k cˆ k e 2i k t
d 3k
ˆ
H
3
2 2
ik x
k'
aˆ k e i k ' x cˆ k e i k ' x
d 3k
k '0 k '
k 2 aˆ k aˆ k cˆ k cˆ k cˆ k aˆ k e 2ik t aˆ k cˆ k e 2i k t
3
2
k 0 k
3
k aˆ k aˆ k cˆ k cˆ k 2 2k 0
3
E0 d 3k k 0 0 d 3k k
Normal Ordering
Determination of structure of spacetime by matter distribution requires E0 = 0.
This is accomplished by writing H in normal ordering : … : ,
which means all creation operators are to the left of all annihilation operators.
Hˆ
d 3k
2k 2
k : aˆ k aˆ k cˆ k cˆ k :
3
d 3k
2k 2
3
k aˆ k aˆ k cˆ k cˆ k
The technique should be applied to all “total” operators.
Total number operator:
3
d
k
i
3
0 ˆ
3
0
ˆ
ˆ
ˆ
d x : :
N d x : j :
2 2 3
2m
k
aˆ k aˆ k cˆ k cˆ k
j0 is the net probability density so that net number of “particles” is conserved
Some neutral particles are identical to their anti-particles, e.g., γ, π0…
→ Number of γ is not conserved ( necessary if they’re the quanta of EM fields )
Summary
x
d 4k
2
k m k exp ik x
3
2
2
a k d 3 x exp ik x i 0 x k x
ˆ x, t i x x
ˆ x, t ,
3
2 2k k k
d 3k
2 2k
3
c k d 3 x exp ik x i 0 * x k * x
aˆ k , aˆ k 2 2k k k
3
3
d k
k k
k k
2 2k
3
k x x k 2k exp i k x
Hˆ
d 3k
L * m 2 *
2k 2
d 3k
2k 2
3
3
I
H 0 *0* L
ˆ ˆ
3
2 ˆ ˆ
ˆ
ˆ
ˆ
H d x
m
t
t
a k exp ik x c* k exp ik x 0
k k
k aˆ k aˆ k cˆ k cˆ k 2 2k 0
k : aˆ k aˆ k cˆ k cˆ k :
3
7.3.
The Dirac Equation and spin-1/2 Particles
K-G eq. is 2nd order in t partials → E < 0 solutions.
Remedy: find eq. that’s 1st order in t partials.
Dirac’s choice:
p m x i m x 0
Correct 4-momentum → must satisfy the K-G eq.
2
2
0 i m x 2im m x m x
2
→
→
1
1
2
2
, , 2
Set of all linear combinations of is called a Dirac algebra.
It is a special case C(V 4(1) ) of the Clifford algebra C(V n(s) ) [see Choquet].
E.g., C(V 1(0) ) = C, C(V 2(0) ) = quarternions , C(V 3(3) ) = Pauli algebra
Standard representation:
0 1
1
0
1
Pauli matrices:
For j = ½ ,
1
sj j
2
0 i
i
0
2
1 0
0
1
3
Spin ½
i , j 2 i i j k k
Angular momenta commutator
, 2
Pauli algebra
i
Dirac eq.:
j
0
j
j
0
I 0
0
0 I
j
ij
iσ
I it m
0
I i t m
iσ
Lorentz Covariance and Spin
Transformation of the Matrices :
Dirac eq.
i
m x 0
is Lorentz covariant if is a scalar.
are 44 matrices → must be a 4 1 matrix
→ covariance not automatic.
Components of γ matrices: ( γμ)αβ with μ = 0,1,2,3 and α, β = 1,2,3,4.
Lorentz transformation:
x x ' x
Covariance →
i
'
i
'
x ' x S x
' m ' x 0
' m S x 0
S 1 ' S '
spinor
→
→
S i S 1 ' S ' m x 0
' S 1 ' S
→
' ' S S 1
The Matrix S
Infinitesimal transformation:
' ' ' O 2
1
S I i O 2
4
' ' 0
1
S 1 I i O 2
4
( No need to distinguish primed and unprimed indices in
since such information does not appear explicitly in S. )
' ' S S 1
1
1
I
i
I
i
4
4
→
' '
i ' '
1
4
1
i
4
( μ → λ )
1
1
1
i
4
2
2
→
, 2i
1
1
i →
4
2
→
a, b, c a, b , c b, a, c
, C
a , b , c b, a , c
, , , ,
4C
→
C ,
Simplest ansatz:
4C 2i
, const 2 const
2C , ,
i
,
2
Generators
Infinitesimal (inverse) Lorentz transformation:
x ' x ' x ' x O 2
x x O 2
Scalar :
' x x
x
x x x
1 x
x
x
1
x
x
x 2
x
x
1
i x p x p
2
1
i x p x p
2
' x 1 i x p x p x
2
1
' x 1 i x p x p
1
2
x
Spinor :
1
' x i x
4
1
1
i 1 i x p x p
4
2
1
1
i x p x p
2
2
1
2
1
2
M M
x
1
x
exp
i
M
x
2
1
x p x p
M x p x p I
2
' x I i M
M
x
→ ½ (42 4) = 6 independent components
Divided into 2 groups :
K M
j
0j
1 i jk jk
J M
2
i
Spin
1
2
x exp i M x
ω0 j ~ boost
→
K j = generator of boost along j-axis.
ωi j ~ rotation
→
J i = rotation generator in the j-k plane ( i j k cyclic )
M
→
1
x p x p I
2
1
1
1
J i i j k M j k i j k j k x j p k x k p j
2
2
2
i j k k , j i j k 2 i k j m m 2 i i j j m i m j j m
2i i i
jj
4i
i
i j k k , j 4 i i
1 i jk jk i i jk j k
,
4
8
i
i jk
8
0
j
j 0 k 0 k 0 j
k
k
j
0
0
0
0
i i j k j k k j
8
0
1 i 0
j k
k
j
2 0 i
0
i
1 i jk j k
1
i
i
x p x k p j x p p x x p
2
2
→
1
J i i jk
2
i
1
1 jk
j k
k
j
x
p
x
p
2 0
2
J ΣL
1 i 0
2 0 i
i
0
i
r
p
I
i
Li r p I
i
J = total angular momentum, L = orbital angular momentum → = spin ( ½)
Pauli-Lubanski 4-Vector
A simple 4-vector description of the spin is the Pauli-Lubanski 4-vector
1
1
1
W M p x p x p p
2
2
2
1
p
p p p p 0
since
4
Wμ is a pseudovector since εμνλσ is a tensor density of weight +1 .
1
1
W0 0 p i j k i j p k k p k Σ p W 0
4
4
1 i jk
4
jk
i
1
1
Wi i p i 0 j k 0 j p k i j 0 k j 0 p k i j k 0 jk p 0
4
4
1
1
i j k 0 j j 0 p k i j k 0 jk p 0 4 i j k K j p k i j k j k p 0
4
4
K p i p 0
i
W i K p i p 0
i
1
K j M 0 j 0 j x0 p j x j p0 I
2
Caution: T iμaμ is the ith component of a 4-vector, but T i jaj is not.
→
i j k K j pk K p
i j k i j k
→
i
has no tensorial meaning in Minkowski space.
i jk K p
j
k
Let aμ be a 4-vector with spatial part
Then
i jk
K p K p
j
k
ai K p
i
i
ai a i K p i jk K j pk
i
W2 = WμWμ is invariant → it can be evaluated in any convenient coordinate system.
Consider W2 acting on the rest frame of a plane wave with kμ = ( m, 0 ).
p k 0
→
W 2 k W 0W 0 W iW i k
→
W 2 m2 Σ2
m2 Σ2 k
is the spin operator in the rest frame of the particle with eigenvalue s(s + 1).
Some Properties of the γ Matrices
,
2
0 2
→
j 2
I
I
0 i m 0 i 0 i 0 j j m 0
→
i 0 i 0 j j m 0
→
H i 0 j j m 0 0 j p j m 0
H
0
→
0
j
→
j
0
p j m
j
0
0
0
0 2
H 0 j p j m 0
j
0 0
j j
0 0
H
j
5 i 0 1 2 3
5 2
1
i
4!
0 1 2
3 0 1 2
3
0 2
where
1 2
3 1 2
3
for
0 2
1 2
2 2
3 2
5 0 1 2 3 5
Define
a a
Dirac equation:
( not a Lorentz invariant since γ is not a 4-vector )
i m 0 p m
Pauli-Lubanski 4-vector:
1
W , p 5
4
Ex 7.6
I
Conjugate Wavefunction and the Dirac Action
i
m x 0
0 0
→
→
0 i m 0
i
m 0 i m
0 0
i m
4
0
i m
S d x i m d x i m
4
a a
Conjugate
wavefunction
Conjugate eq.
Dirac action
Probability Current
i
m x 0
→
i m 0
0 i i
→
j 0
γμ is not a 4-vector → need to show jμ is.
'
j ' x ' x 0
' x x S 0 ' S x
1
S exp i
4
1
1
S exp i exp i 0 0
4
4
1
0 exp i 0 0S 1 0
4
,
i
2
i
0 , 0
2
0 0
S 0 S 1 0
→
j ' x x S 0 ' S x
x 0 S 1 ' S x
' x x
' j x
j 0 0 0
4-vector.
~ probability density
Bilinear Covariants ( Tensors )
Dirac algebra is spanned by 16 basis “numbers”, e.g., I, γ5, γμ, γμγ5, σμν.
Tensor Type
Bilinears
Transformation
Scalar
S x x x
S x S x
Pseudoscalar
P x x 5 x
P x det ' P x
x x x
V ' x ' V x
Vector
Pseudovector
Tensor
V
A x x 5 x A ' x det ' ' A x
T x x x
T ' ' x ' ' T x
Covariant Spin Polarization
Spin 1/2 particle in its rest frame:
k m, 0
n 0, n
n = unit vector along axis of spin polarization.
→
k 2 m2
n2 1
n2 1
k n 0
In a frame in which the particle is moving with momentum p = k ,
k k 0, k
Since
we have
n n0 , n
k 2 k 2
k
0 2
k n k n
n 2 n2
k m
2
mn0 k n
n
n 1
n0
k n
m
0 2
2
→
2
k n k n
k 0n 0 k n 0
By symmetry, n can involve only n and k
n a k b n
→
n a k 2abk n b n a k 2abmn b
2
2
2
2
2
2
2
0
2
n
0 2
1
k n ak 2 bk n ak 2 bmn0 k 0n 0
a k 2abmn k b m n
2
Square:
→
0
2
2
2
k n
0 2
0 0 2
2
0 2
2
0 2
0 2
b m n k n k k 2 k 2
→
4
2
k
a
m n
→
b2 1
b = +1 since n = n
when k = 0.
k n
n
0
2
m k 0 m
k m
k
k n
k n
k n
n
,
k n
n
k n
0
0
m
m k m
m k m
Rest frame:
0
0
0
W 0 , mΣ
→
W n m Σ n
Plane Wave Solutions
k x eikx uk
Free particle (plane wave) solution:
i
m x 0
a
uk k
bk
→
ak
→
k 0 m I
0
k σ
1
k σ bk
k0 m
a k 0 m ak k σbk
k
0
0
k m I bk k σ ak k m bk
bk
, 2
i
a σb σ aib j i j
k σ
2
k I
2
→
j
m uk 0
k σ
1
1
ak 0
k σ 0
k σ ak
k m
k m
i , j 2 i i j k k
k
→
spinor
1
k σ ak
k0 m
k 0 2 m2 I k σ 2
→
ij
i j i i j k k i j
→
a i b j i i j k k i j i a b σ a b I
k 0 2 m2 k 2
→
k 0 k
k m 2 k 2
ak
exp i t k x e i k xu
k x 1
k
k
k
σ
a
k
m
k
1
k
σ
b
k
k x k m
exp i k t k x
bk
k x k x k k
k
Dirac equation :
k 0 k
1
k
σ
b
k
ik x
k m
exp
i
t
k
x
e
vk
k
bk
m uk 0
k m k m k k m
k 0 k
k
2
m vk 0
1
k k m 2
2
k k m2 k 2 m2 0
Interpretation
k0
Rest frame:
0
vk
bk
ak
uk
0
Choose uks and vks to be the eigenstates of the spin Σn :
1
s
Σ nuk s s uk s
Σ nvk s s vk s
2
W n uks msuks
W n vks msvks
→
uk
0
0
vk
where
1
0
0
1
Frame pμ = kμ:
ei k x
k x Ck 1
m k σ
k
ks ks ks ks 1
→
1
k
σ
i k x
k x Dk k m
e
Ck Dk
k m
2m
Charge Conjugation
Charge conjugation : particle ↔ antiparticle
Principle of minimal coupling :
i eA m 0
Charge conjugation: → C
i eA m C 0
Conjugate equation 0 i eA m → T i eA m T 0
Set
T C 1 C
C 1 i eA m C T 0
→
C C T C 0T *
0
2 0
C i
2
i
Standard representations →
Exercise:
k x
C
k x
i 2
0
Massless Spin 1/2 Particles
Dirac equation for massless particle:
i p 0
No rest frame → spin polarization specified by helicity
h Σ kˆ
1
1
1
W , p 5 5 , p 5 p
4
4
4
1
1
1
5 2 p 5 p 5 p p
4
2
2
Massless particle:
1
W 5 p
2
→ Plane wave is eigenfunction of W if it is an eigenfunction of γ5.
R L
→
R
5 R
1
I 5
2
1 5
I R
2
L
1
I 5
2
5 L
1 5
I L
2
1 5
W p
2
5 R R
5 L L
If is a plane wave with wavevector kμ ,
then R and L are eigenfunctions of W with eigenvalue ½ kμ and ½ kμ, resp.
m=0 →
k0 k
W0 Σp
→
h R Σ pˆ R
h L Σ pˆ L
1 0
1
W R R
k
2
chiral projections
1 0
1
W L L
k
2
γ5 = chirality operator
Only for massless particles do the chiral projections have definite helicities.
7.4.
Dirac eq: ρ 0
Spinor Field Theory
but k0 = ωk
→ 2nd quantization needed for proper interpretation.
Dirac action:
S d 4 x i i ij m i j j
Momentum conjugate to i :
Hamiltonian density :
Hamiltonian :
i
S
i j 0ji i 0 0 i i
i
0 i
H i0 i L i i 0 i i i i j m i j j
H d 3 x i 0 0
2nd Quantization
ˆ x
ˆ
d 3k
2 2k
x
3
bˆ
ks
e ik x uk s dˆks e i k x vk s
s
d 3k
2 2k
3
k 0 k
bˆks eik x uk s 0 dˆk s e i k x vk s 0
s
k 0 k
uk s uk s vk s vk s 2k s s
Normalization:
k k 0 , k
uk s 0 vk s vk s 0 uk s 0
3
i k x
0
ˆ e ik t
ˆ
d
x
e
u
x
b
ks
ks
→
3
i k x
0
ˆ eik t
ˆ
d
x
e
v
x
d
ks
ks
i 0ˆ x
→
Hˆ
d 3k
2 2k
3
d 3k
2 2k
3
→
→
Ex 7.4
bˆk s d 3 x eik x uk s 0ˆ x
k 0 k
dˆks d 3 x eik x vk s 0ˆ x
k 0 k
ˆ e ik x u dˆ e i k x v
b
k ks
ks
ks
ks
s
k
s
bˆks bˆk s dˆk s dˆks
k 0 k
Non-negativeness of H requires anticommutation relations:
bˆk s , bˆk s dˆk s , dˆk s
bˆk s , bˆk s dˆk s , dˆk s
bˆk s , bˆk ' s '
dˆ k s , dˆ k ' s ' 0
2 2 k s s k k
3
Using commutation relations leads to causality violation
(operators with space-like separation would not commute)
Hˆ
with
d 3k
2 2k
3
s
d 3k k 0
s
ˆ bˆ dˆ dˆ
b
k ks ks ks ks
removed by normal ordering.
Field Operator Version
ˆ x, t , ˆ x, t ˆ x, t , ˆ x, t 0
i
j
i
ˆ x, t , ˆ x, t i
i
j
ij
j
x x
Hˆ d 3 x :ˆ i 0 0ˆ :
Nˆ d 3 x : ˆj 0 : d 3 x :ˆ ˆ : d 3 x :ˆ 0 ˆ :
d 3k
2 2k
3
s
bˆks bˆk s dˆks dˆk s
( Not positive-definite )
Spatial parts of the Dirac wavefunctions should be anticommutating.
7.5.
Weyl and Marjorana Spinors
Weyl (Chiral) Representation ( for Massless Particles ) :
0
I
0
→
I
0
I 0
5
0 I
0 1
1 0
i 2
i
0
i
i
0
C
0
0
2 I
0
C C 0 T
i * i
0
Chiral Solutions
Weyl representation:
m=0 →
k = ( 0,0,k )
E.g.,
k
k0 k
→
1
0
uR
0
0
m k
k uk s k vk s 0
( u and v are linearly dependent )
0 0
0 0
0
kI 3k
k
3
0 0
0
kI k
0 2k
0
0
0
uL
0
0
1
I 0
uR
uR
0 I 0 0
5
mI
k 0 I k
m 0
k
I
k
mI
2k
0
0
0
0
0
0
0
I 0 0 0
uL
uL
0 I
5
General k
0
k
k I σk
→
k
uR k
0
Writing
k I σk
0
0
uL k
k
k aI σ b
k I σ k k 0
we have
0 k I σ k aI σ b k a I σ ak k b σ k σ b
k a k b I σ ak k b i k b
Simplest solution:
→
→
b ck
k a k c 0
2
a kc
ak k kc 0
k c k I σ k
c = normalization constant.
Normalization
k c k I σ k
k k3
c 1
2
k
i
k
k k3
k c k I σ k c 1
k i k 2
For c real,
k k3
k1 i k 2 1
c k1 i k 2
3
k k 0
k 1 i k 2
k 1 i k 2 0
c
3
3
k k 1
k
k
k 1 i k 2
0 1 k k3
k c
c
k
1
3
2
1
0
k
k
k i k
*
uR
0
0
I
A11
0 a A
21
a
A11
0
A21
I
0
0
0
uL 0
I
I
0
A12 a
A a
11
0
a
a
A21a
A22 0
A21a
A12 0
a
A22 a
A12a
0
a
A12a
A22a
σ aσ b a bI i a b σ
→
uR k 0 uR k c 2 T k I σ k k I σ k
2c2 T k I k σ k 2 k c 2 k I σ k 11
2
2 k c2 k k 3
0
uR k 0 u R k 2 k c 2 k k 3
σ k i σ k σ ei
k i I i k ei σ
i σ k σ ei σ k k i I i k ei σ
σ k i σ k σ k k i I i k ei σ
σ k k i i k ei kI k k ei σ
σ k k i kk i ei k
→
2
σ
2 σ k k i k i
2
uR k i uR k i c 2 T k I σ k i k I σ k
2k i c 2 T k I σ k 2k i c 2 k k 3
uR k uR k 2k c 2 k k 3
uR k uR k 2k c 2 k k 3
Setting
c
1
k k
3
→
k
σk
I
3
k
k k
k
uR k uR k uL k uL k 2k
uR k uL k uL k uR k 0
orthonormality
Charge Conjugation
The chiral solution are also related by charge conjugation:
uRC k C 0T uR* k
0
0 0
0 I
I * k
0 0
0
0 0
0
* k * k k uL k
uRC k uL k
0 0
u k C u k
I
0
C
L
0T *
L
0
I 0
0 * k
0 * k * k k
u k
0 0 0 R
uLC k uR k
2nd Quantization
ˆ x
d 3k
2
3
bˆR k e i k xuR k dˆR k ei k xu L k bˆL k e i k xu L k dˆL k ei k xu R k
2k
ˆ x ˆ R x ˆ L x
ˆ R x
ˆ L x
d 3k
ˆ k e i k xu k dˆ k ei k xu k
b
R
R
L
R
3
2 2 k
d 3k
ˆ k e i k xu k dˆ k ei k xu k
b
L
L
R
L
3
2 2 k
Matrix form:
Weyl representation:
R 0
R L
0 L
Dirac equation:
0
i 0 i σ
i i
i
i
σ
0
0
m
i 0 i σ R
i m i i σ
m
0
L
m
i 0 i σ R i 0 i σ L mR
0
m
i 0 i σ
L i 0 i σ R mL
→
i0 i σ L mR
i0 i σ R mL
0
i 0 i σ R
i R i R
i
i
σ
0
0
0
0
0
m L
i
i
σ
R mL
0
i R m L
0
i 0 i σ 0
i L i L
i
i
σ
0
0
L
i 0 i σ R mR
0 m R
0
i L m R
Weyl Spinors
From Ex.7.9:
Action:
L R R L
L L R R
S d 4 x i m
d 4 x Li L R i R m L R R L
m=0 →
S d 4 x Li L Ri R
( R and L decoupled )
→ Either R or L alone describes a self-consistent theory.
Spinors in this reduced theory are called Weyl spinors.
E.g., in a theory involving only L, the operators are
bˆL k , dˆR k , bˆL k , dˆR k
→ there are only left-handed particles and right-handed antiparticles.
For a theory involving only R, the opposite is true.
Both theories are entirely equivalent, physically as well as mathematically.
Majorana Spinors
A spin ½ particle which is its own anti-particle can access only 2 of the 4 spinor
states allowed by the Dirac equation.
The field operator M for such a particle is called a Majorana spinor.
By definition
MC x M x
If the particle is also massless, we have
ˆ M x
→
d 3k
ˆ k e i k x u k ˆ k ei k xu k ˆ k e i k xu k ˆ k ei k xu k
R
R
R
L
L
L
L
R
3
2 2 k
1
S d x i M M
2
4
→
→
1
S
i M 0
M
0 M 2
3
ˆA k , ˆB k 2 2 k AB k k
ˆ x, t , ˆ x, t i x x I
M
M
A, B R, L
Proof ?
For non-interacting massless particles, Weyl & Majorana spinors are equivalent.
E.g.
ˆM x ˆ L x ˆ LC x
d 3k
ˆ k e i k x u k dˆ k ei k x u k
b
L
L
R
L
3
2
2
k
bˆL k ei k x uR k dˆR k e i k xu R k
Corresponding action:
S d 4x
1
4
i M M d x i L L
2
This equivalence is broken in the presence of interactions.
7.6.
Particles of Spin 1 and 2
Maxwell equations in a source-free region:
0 F A A WA A
Proca equation:
WA m2 A 0
Consider the gauge transformation:
Setting
In this context,
WA A 0
WA m2 A A 0
WA m2 A W A m2 A 0
→
→
W A
A 0
→
A 0
( Klein-Gordon eq for Aμ )
A A A
gives
A A 0
is called the Lorentz condition.
Aμ is said to be in the Lorentz gauge.
Spin
Lorentz transformation:
A x ' A x
or
x ' x S x
Comparing with
A x A x
gives
S
Hence, the “spin” part of the infinitesimal generators M are simply the
generators of the Lorentz transformation on x.
The spin part of Ki ( i ) denotes a boost (rotation):
0
1
iK 1
0
0
0
0
1
i
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0 0
0 0
0 1
1 0
0
0
iK 2
1
0
0
0
0
0
0 0
0 0
2
i
0 0
0 1
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
3
iK
0
1
0
0
0
0
0
0
i 3
0
0
0 0
0 1
1 0
0 0
0
0
0
0
1
0
0
0
0
0
0
0
Plane wave solution:
Lorentz condition →
Ak x eikx
polarization vector εμ
is a column matrix
Ak i keikx 0
→
k 0
For a massive particle, k μ = ( m, 0 ) in the rest frame.
The Pauli- Lubanski vector becomes
W 0, Σm
→
0
0
0
0
Σ Σ i2
2
0
0 2 I
0
W W W W m2 Σ Σ
0
1
0
0
0
0
1
0
0
0
0
1
→
s s 1 2
→
s 1
Secular equations for all three matrices i are the same, i.e.,
2 2 1 0
0,0,1, 1.
so that
The corresponding (orthonormalized) eigenvectors for 3 are
0
1 1
1
2 i
0
0
0
0
0
1
0
1 1
1
2 i
0
1
0
0
0
0
ε0 is not an acceptable solution since
0
k m 0
; 1,0, 1
( Lorentz condition violated )
form a basis for the polarization of the spin 1 particle.
Wrt this basis, ΣΣ is effectively a unit matrix.
k k ,0,0, k
For massless particles
A 0
W0
h 0 3
k
k k 0 3 0
→
→ 0 3
The last condition is satisfied by
0
1 1
1
2 i
0
k
L
k
0
1 1
1
2 i
0
→
→
→
→
→
1
0
L 0 0
0
1
polarization εL is longitudinal
corresponding plane wave is pure gauge ( ~ Aμ = 0 )
photon has only 2 polarizations h = 1
complications in quantization scheme
path integral
Gravitons
B
g x g
x h x
Source free regions :
To 1st order in h(x) :
Field eq.
→
B
g
x
T x 0
R 0
Field eq. →
1
h , h , h ,
2
R R , ,
, ,
1
1
h , h , h , h , h , h ,
2
2
h , h , h , h , 0
h,, h,, h, h,, 0
h Wh h h 0
W h h h h 0
0
0
Gauge Invariance
g symmetric → at most 10 independent components in h .
2 different h are equivalent if they lead to the same g in different coordinates.
i.e., (gauge) transformations on g that leave all physical properties unchanged.
Gauge invariance further reduces number of independent components of h.
Infinitesimal coordinate transformation:
x x x x
x x
g x g x g x
x x
x x
g x g x x x g x
x x
g x g x x g x x O 2
g x g x g x x g x x O 2
Set
g x h x
g x h x
To 1st order terms in both h and ,
h x h x x x
h x h x x x
h h
gauge transformation
Harmonic Gauge Condition
Field eq.
→
W h h h h 0
1
1
Wh h h h h 0
2
2
1
Wh q q 0
q h h
where
2
Gauge transformation:
h x h x x x
1
q x h x h x
2
1
h h 2
2
1
h h
2
q
q x W x
1
q h h
2
W q
g
q x q x W x
→
q 0
harmonic gauge condition
1
g g g , g , g ,
2
1
1
g g g g g g g g g
2
2
1
g h g h
2
g h g h h g h h h
→
g
Field eq. →
g
1
1
h
h
h
h
~0
2
2
Wh 0
to 1st order in h in
harmonic gauge
( massless K-G eq ) → graviton is massless
Spin
W h h h h 0
Field eq.
Plane wave solution:
hk x eikx
is a symmetric polarization tensor (at most 10 independent components)
In the harmonic gauge,
Gauge condition
→
→
k k 0
1
1
h h i k i k e ik x 0
2
2
1
k k
4 constraints on
2
q x q x W x
E.g.
Wh 0
→
W x 0
x i eikx with k k 0
for transformation
between harmonic gauges
x i eikx
h x h x x x
→
h x k k e i k x
Writing
→
hk x eikx
hk x eikx
k k
further reduce the number of independent components of to 2.
~ helicity states of h = 2.
→ Gravition is massless, spin 2.
Helicity states of h = 0, 1 correspond to purely gauge
degrees of freedom and have no physical significance.
Experimental proof of gravitational waves : binary pulsar
No graviton has yet been detected .
7.7.
Scalar Field
Wave Equations in Curved Spacetime
S d 4 x g x Lmatter L field
Lmatter g x * m 2 *
g x * m 2 *
L field R x *
= dimensionless coupling constant, R(x) = Ricci curvature scalar, Λ = 0.
field is the only choice that allows a dimensionless coupling constant.
Note:
but
Euler-Lagrange equations
S d 4 x g x L ,
For the degrees of freedom:
→
L
L
S d x g x
0
4
ab a b a b
L
L
d x g x d x g x
4
L
4
Covariant Gauss’ theorem [see eq(A.23), appendix A.4] →
δ = 0 on S
L
L
4
d
x
g
x
dS
g
x
0
L
L
0
Hence, the covariant Euler-Lagrange equations are
L g x * m 2 * R x *
L
L
0
*
*
g 0
→
→
m 2 g 0
g m 2 R 0 m 2 R
There is no known physical principle that can be used to determine .
Effects of spacetime curvature are too small for measurement of .
The arbitrary case ξ = 0 is called minimal coupling.
Conformal transformation :
Conformal coupling:
g x x g x
2
(x) = arbitrary
real function
For ξ= 1/6 and m = 0,
m 2 R 0
is invariant under a conformal transformation if
x 1 x x
Vierbeins
Let ya be local coordinates at point X with large-scale coordinates xμ = Xμ.
Spacetime is locally flat → g(y) = η.
Transformation matrices :
x
e a X a
y
x X
a
y
ea X
x
x X
x y a x
e a x e x a
x
y x
Orthogonality:
a
x y b
y b
e a x e x a a ab
y
y x
b
Local inertial frames should vary smoothly from point to point.
→ For a fixed, eμa (x) is a vector field specifying the ya axis at every point x.
e x , a 0,1, 2,3 is called a vierbein, a tetrad, or a frame field.
e x , 0,1, 2,3
Likewise
a
a
Verbein e is a 2nd rank tensor whose and a indices are associated with g(x) & ab .
g x ea x eb x ab
ab ea x eb x g x
The 16 components of e carry 2 kinds of information.
• 10 components specifying g .
• 6 components specifying boost & rotation relating each local frame to a
fixed reference Minkowkian frame.
V x ea x V a x
coordinate vector
V a x ea x V x
Lorentz vector
Spin Connection
Parallel transport of a Lorentz vector such as
V a x x dx V a x ba x V b x dx
V
→
x e a x V x
a
V a x x a x
ba x = spin connection
e a x dx e a x e a , x dx
V x x dx e a x dx V a x x dx
e a x e a , x dx V a x ba x V b x dx
V x e a , x V a x e a x ba x V b x dx
V x e a , x e a x V x e a x ba x eb x V x dx
V x e a , x e a x e a x ba x eb x V x dx
V x x dx V x e a , x e a x e a x ba x eb x V x dx
V x
xV x dx
x e a , x ea x e a x ba x eb x
→
(a)
ec e d
ec e d e a , ea ec e d e a ba eb
ec a d e a , c a bd ba ec e d , cd
→
e c
e c e a , ea e c e a ba eb
ba ea e b , ea e b
V V V
V V
V
e c a :
→
→
V a V a ba V b
→
Va Va ab Vb
e c
x e c e a , ea e c e a ba eb e c , e a ca
e c e c ,
e c ca e a 0
e a 0
Vierbein, like g, is invariant under parallel transport
Reminder: e serves the role of g that converts between a and types of indices.
→
e a 0
is the compatibility condition between ω & η.
→ magnitudes and angles be invariant under parallel transport:
abV a x x dx V b x x dx abV a x V b x
V a x x dx V a x ba x V b x dx
→
L.H.S. =
To 1st order in dx:
→
ab V a x ca x V c x dx V b x bd x V d x dx
ab ca x V c x V b x ab bd x V a x V d x 0
ad ca x cb bd x V c x V d x 0
This can be shown to imply
g 0
→ d c x cd x 0
i.e., Γ = metric connection.
Spinors
Parallel transport of a spinor :
→
x x dx x x x dx
x x
Ω = 44 spin matrix
S x x x
Scalar field:
x x dx x x dx 0 x x x 0dx
To 1st order in dx :
S x x dx x x dx x x dx
x x x 0dx x x x dx
S x x x x dx x x 0 x dx
S x x x 0 x 0 x dx
S is invariant →
x 0 x 0 0
Lorentz vector :
V a x x a x
V a x x dx x x dx a x x dx
V a x x a x 0 x 0 a x dx
V a x ba x V b x dx
→
x a x 0 x 0 a x ba x x b x
a x 0 x 0 a ba x b
x 0 x 0 0
→
a x x a ba x b
a , x ba x b
x ab x ab
Consider ansatz
, 2i
0 0
→
x 0 x 0 0
for
i
,
2
ab , 0 0
→
i
2
*
0 0 ab 0 0
*
ab
0 x 0 ab
x
x
ab
x ab x ab
→
*
ab
x ab x
a , x bc x a , bc 2i bc x ab c a c b
a
b
x
b
cb x
ac
b
1
bc x a c b a b c
2
1 ac
cb x b abbc x c
2
→
i
4
bc x bc x
i
4
bc x bc x
→
→
*
ab
x ab x
x ab x
is automatically satisfied since ω is real
ab
i
1
ab
bc x
bc x ,
8
4
Dirac Equation
μ are defined only for inertial frames → must be mediated by e.
Covariant Dirac equation is
Set
x ea x a
→
i e a x a m x 0
i x m x 0
ab
a
b
2
e
x
e
x
2g
x
,
x
e
x
e
x
,
x
a b
a b
( Covariant Clifford / Dirac algebra )
Covariant action :
S d 4 x g x i x m
g det e a
There’s no field term because the coupling constant cannot be dimensionless.
E.g., a term like R
requires to have the dimension of length.
Vacuum State Problem
Concept of the vacuum is problematic even in Minkowkian spacetime:
The vacuum state, which by definition contains no particle to an inertial
observer, will appear to an accelerating observer as a thermal bath of
particles with temperature proportional to the acceleration.
A complete proof of this statement is rather involved.
We’ll demonstrate it for the special case of a massless 2-D Hermitian scalar field.
Rindler Coordinates
Massless spin 0 particles in a 2-D Minkowskian spacetime.
Rindler coordinates: (η,ξ)
Inertial frame: (t, x)
t
1
e sinh
x
coth
t
x
1
x
1
t
→
e cosh
→
x t
α>0
for x > 0
Rindler wedge
Inverse:
x2 t 2
1
2
e
2
→
1
ln x 2 t 2
1
1
t
1
tanh 1
ln
x
2 1
t
x 1 ln x t
t
xt
x
t
→
1
e sinh
x
1
e cosh
dt e sinh d cosh d
dx e
d 2 dt 2 dx 2
→
cosh d sinh d
e 2 d 2 d 2
For an observer at fixed . The eq of his worldline in an inertial frame is
x t
2
2
1
2
e
2
a
2
p
const
His velocity in the inertial frame is
→
x u x a
His acceleration:
2
2 2
a
2
p
dx
2x
2t 0
dt
→
u
→
dx t
dt x
x
1
a p 1 u2
du
1 t
2 u 1 u2
dt
x x
3/ 2
ap
Let inertial frame S be moving with velocity v with respect to S. Then
1 v
a
2 3/ 2
1 uv
3
( see Ex.2.2 )
a
If S coincides instantaneously with the rest frame of the particle, then v = u ,
a
a 1 u
2 3/ 2
ap
1
1 u
2 3/ 2
→
a
ap is the proper acceleration of the observer.
Proper time of the observer is obtained by setting dξ= 0.
d 2 e 2 d 2 d 2
→
d 2 e2 d 2
With a proper choice of coordinate origins, we have
→
d e d
e
2D Massless Hermitian Scalar Field (x, t)
ˆ x, t
dq
aˆ q ei q t iq x aˆq e i q t i q x
2 2 q
q q
Klein-Gordon equation inside the Rindler wedge :
2
2
2 2
2
0 2 2 g e
2 2
t x
Plane wave solution :
→
1
ln
k , ei k
1
xt 1
ln x 2 t 2 ln x t
xt
k
ik /
ln x t x t
k , exp i
Expanding in terms of k (ξ,η),
ˆR ,
→
dk
bˆk ei k i k bˆk ei k i k
2 2 k
ˆ
i k i k
bˆk d e
k ˆR , i
R ,
dq k q aˆq k q aˆq
Caution: bk and bk+ are confined in the Rindler wedge but not aq and aq+.
In general, transformation that relates different sets of solutions to a
wave equation is called a Bogoliubov transformation.
Average Particle Numbers
N k , k 0 bˆk bˆk 0
| 0 = vacumm state in Minkowskian spacetime →
0 aˆq 0
aˆq 0 0
dk
N k , k = number of particles with momentum between k
2 2 k
and k+dk as seen by a Rindler observer.
Problem: If volume → , then N(k,k) → even if the density of particles is finite.
Remedy: Use N(k,k) then take limit k → k.
N k , k dq dq 0 aˆq k* q aˆq k* q aˆq k q aˆq k q 0
dq dq k* q k q 0 aˆq aˆq 0
dq dq k* q k q q q
dq dq k* q k q 2 2 q q q
Involved manipulation →
4 dq k* q k q q
N k , k 2 2 k k k e
2 k
1
1
N k , k 2 2 k k k e
e
2 k
1
1
2 k
1
1
= Bose-Einstein occupation number with
Consider the observer at fixed = ξ0 .
A positive energy plane wave is
2 k
2
k e
0
2 k
exp i k e 0 ik
→
k
k BT
e
His proper time is
which represents a particle with energy
→
N k , k 0
0
k k e
0
k BT
ap
e
2
2
0
QED