Seminar Report On LFD PDF
Seminar Report On LFD PDF
Seminar Report On LFD PDF
Submitted by:
Sudeep Saha (Roll:194120007)
Under the supervision of Prof. Asmita Mukherjee
Department of Physics
November 16, 2019
i
Contents
1 Introduction 1
2 Light-Front Dynamics 2
2.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
2.2 Dispersion Relation . . . . . . . . . . . . . . . . . . . . . . . . 2
8 Conclusion 22
ii
1 Introduction
1
quantization surface. In field theory, the Poincare generators are constructed
from the symmetric energy-momentum tensor T µν
Z
P = d3 xT 0µ ,
µ
Z
µν
M = d3 x[xν T 0µ − xµ T 0ν ].
2 Light-Front Dynamics
2.1 Definition
2
we know for a free massive particle
k 2 = m2
or (k 0 )2 − (k 3 )2 − (k 1 )2 − (k 2 )2 = m2
or (k 0 + k 3 )(k 0 − k 3 ) − (k ⊥ )2 = m2
or k + k − − k ⊥ = m2
− (k ⊥ )2 + m2
or k = (1)
k+
From the above dispersion relation we can infer the following conclusions:
(1) Unlike the relativistic dispersion relation, there is no square root.
(2) Dependence of the energy k − on the transverse momentum k ⊥ is like
non-relativistic dispersion relation.
(3) For k + positive (negative) , k − is positive (negative).
(4) The dependence of energy on k ⊥ and k + is multiplicative. Large energy
can result from large k ⊥ or small k + .
We will consider the generators of boost and rotation in equal time dynamics.
Let us consider a boost along negative 3-axis. The relationship between
the space-time coordinates of the frame S̃, which is moving along the negative
3-axis with a speed v relative to another frame S, is given by,
x̃0 = γ(x0 + βx3 ) (2)
x̃3 = γ(x3 + βx0 ) (3)
where
v 1
β= and γ = p (4)
c 1 − β2
In terms of rapidity parameter φ, we can write,
3
Then we have
x˜0
coshφ 0 0 sinhφ
˜1
x 0 1 0 0
˜2 = B 3
0
(6)
x 0 1 0
x˜3 sinhφ 0 0 coshφ
4
Next, we consider the rotations of a vector V~ about the 3-axis through an
angle θ in a clockwise manner. So we must have
V˜1
1
V
˜2 3 2
V = R V (11)
V˜3
3
V
Where the rotation matrix
cosθ sinθ 0
R3 = −sinθ cosθ 0 (12)
0 0 1
The generator J 3 of this rotation is given by,
0 1 0
1 ∂R3
J3 = |θ = 0 = −i −1 0 0 (13)
i ∂θ
0 0 0
In the 4-dimension we can define the rotation generators as,
0 0 0 0
0 0 1 0
J 3 = −i
0 −1 0
(14)
0
0 0 0 0
Similarly,
0 0 0 0
0 0 0 −1
J 2 = −i
0
(15)
0 0 0
0 1 0 0
and
0 0 0 0
0 0 0 0
J 1 = −i
0
(16)
0 0 1
0 0 −1 0
i
We note that that J i matrices are hermitian so the operators eiθJ are unitary.
Also note that the generators of rotations J i leave x0 = 0 invariant, whereas
the boost operators K i do not leave the x= 0 invariant.
5
3.1 Group Algebra
Now we will verify that our matrix representations of the generators of the
homogeneous Lorentz group satisfies the group algebra. To do that we will
compute the following commutation relations:
(i)
[J 1 , J 2 ] = J 1 J 2 − J 2 J 1
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 −1 0 0 0 0
J 1 J 2 = −1
0 0 0 1 0 0 0 0 = 0
−1 0 0
0 0 −1 0 0 1 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 −1 0 0 0 0 0 0 −1 0
J 2 J 1 = −1
0 0 0 0 0 0 0 1 = 0
0 0 0
0 1 0 0 0 0 −1 0 0 0 0 0
0 0 0 0
0 0 1 0
J 1J 2 − J 2J 1 = 0 −1 0 0 = iJ
3
0 0 0 0
Similarly taking cyclic permutation of the other generators we can prove the
following commutation relations,
[J 2 , J 3 ] = iJ 1 and [J 3 , J 1 ] = iJ 2
So, in general we can write,
[J i , J j ] = iijk J k (17)
(ii)
[J 1 , K 2 ] = J 1 K 2 − K 2 J 1
0 0 0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
J 1 K 2 = −1
0 0 0 1 1 0 0 0 = 0
0 0 0
0 0 −1 0 0 0 0 0 1 0 0 0
6
0 0 1 0 0 0 0 0 0 0 0 −1
0 0 0 0 0 0 0 0 0 0 0 0
K 2 J 1 = −1
1 =
0 0 0 0 0 0 1 0 0 0 0
0 0 0 0 0 0 −1 0 0 0 0 0
0 0 0 1
0 0 0 0
J 1J 2 − J 2J 1 =
0 0
= iK 3
0 0
1 0 0 0
Similarly taking cyclic permutation of the other generators we can prove the
following commutation relations,
[J 2 , K 3 ] = iK 1 and [J 3 , K 1 ] = iK 2
So, in general we can write,
[J i , K j ] = iijk K k (18)
(iii)
[K 1 , K 2 ] = K 1 K 2 − K 2 K 1
0 1 0 0 0 0 1 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0 −1 0
K 1 K 2 = −1
0 0 0 0 1 0 0 0 = 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
0 0 1 0 0 1 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0 0 0
K 2 K 1 = −1
1 0 0 0 0 0 0 0 = 0 −1 0 0
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0
1 2 2 1
0 0 1 0 = −iJ 3
K K − K K = −1
0 −1 0 0
0 0 0 0
Similarly taking cyclic permutation of the other generators we can prove the
following commutation relations,
[K 2 , K 3 ] = −iJ 1 and [K 3 , K 1 ] = −iJ 2
7
So, in general we can write,
[K i , K j ] = −iijk J k (19)
Now, we define
E 1 = −K 1 + J 2
Explicitly
0 −1 0 0
−1 0 0 −1
E 1 = −i
0 0 0 0
(24)
0 1 0 0
8
For infinitesimal φ,
1 −φ 0 0
1 −φ 1 0 −φ
eiφE = 1 + iφE 1 =
0
(25)
0 1 0
0 φ 0 1
Under this transformation,
x̃0 = x0 − φx1 , (26)
x̃1 = −φx0 + x1 − φx3 = x1 − φx+
x̃2 = x2
x̃3 = φx1 + x3
thus,
x̃+ = x̃0 + x̃3 = x0 − φx1 + φx1 + x3 = x0 + x3 = x+ (27)
The x̃+ variable remains invariant under E 1 . The action of E 1 is like a
Galilean boost in non-relativistic mechanics.
9
Define, Define E 2 = −K 1 − J 1 . Explicitly we can write,
0 0 −1 0
0 0 0 0
E 2 = −i −1 0 0 −1
(32)
0 0 1 0
For infinitesinal φ,
1 0 −φ 0
2 0 1 0 0
eiφE = 1 + iφE 2 =
−φ
(33)
0 1 −φ
0 0 1 0
Under this transformation,
x̃0 = x0 − φx2 ,
x̃1 = x1 ,
x̃2 = −φx0 + x2 − φx3 = x2 − φx+ ,
x̃3 = φx2 + x3 . (34)
Thus,
x̃+ = x+ (35)
Thus the action of E 2 is also just like a Galilean boost in non-relativistic
mechanics.
10
Under this transformation,
x̃0 = x0 − φx2 ,
x̃1 = x1 ,
x̃2 = x2 − φx0 + φx3 ,
x̃3 = −φx2 + x3 . (38)
Thus,
x̃+ = x+ − 2φx2 (39)
For rotation about 3-axis, we have for infinitesimal φ,
1 0 0 0
3 0 1 −φ 0
eiφJ = 1 + iφJ 3 =
0 φ 1
(40)
0
0 0 0 1
Under this transformation,
x̃0 = x0 ,
x̃1 = x1 + φx2 ,
x̃2 = x2 − φx1 +,
x̃3 = x3 . (41)
Thus,
x̃+ = x+ (42)
Thus, among the six generators of the homogeneous Lorentz group the four
generators K 3 , J 3 , E 1 , and E 2 leaves x+ = 0 invariant while the two other
generators F 1 and F 2 do not leave x+ = 0 invariant.
We can explicitly verify,
[F 1 , F 2 ] = 0, [J 3 , F i ] = iij F j i, j = 1, 2 (43)
with 12 = −21 = 1, 11 = 22 = 0.
Thus, J 3 , F 1 and F 2 form a close algebra.
Also,
[E 1 , E 2 ] = 0, [K 3 , E i ] = iE i (44)
Thus, K 3 , E 1 , and E 2 form a closed algebra.
11
In the next section, we will derive the transformation properties of the
~ and the magnetic field B
electric field E ~ under Lorentz transformation using
Maxwell’s equations and Light-Front variables
12
we can get
∂B 1 ∂B 2 ∂B 3
+ + =0 (53)
∂x ∂y ∂z
∂E 3 ∂E 2 ∂B 1
− + =0 (54)
∂y ∂z ∂t
∂E 1 ∂E 3 ∂B 2
− + =0 (55)
∂z ∂x ∂t
∂E 2 ∂E 1 ∂B 3
− + =0 (56)
∂x ∂y ∂t
From equation-54 we can get
∂E 3 ∂(E 2 − B 1 ) ∂(E 2 + B 1 )
− + =0 (57)
∂y ∂x+ ∂x−
Similarly, from equation-55 we can get
∂E 3 ∂(E 1 + B 2 ) ∂(E − B 2 )
− + =0 (58)
∂x ∂x+ ∂x−
From equation-53 and 56 we can get
∂(E 1 − B 2 ) ∂(E 2 + B 1 ) ∂B 3
− −2 + =0 (59)
∂y ∂x ∂x
Now, under the boost along negative 3-axis the variables x+ and x− changes
by a scale factor as derived in equation-22 and equation-23. So using those
transformations we will have
∂ −φ ∂ ∂ φ ∂ ∂ ∂ ∂ ∂
= e , = e , = , = (60)
∂ x̃+ ∂x+ ∂ x̃− ∂x− ∂ x̃ ∂x ∂ ỹ ∂y
So, we must have
∂E 3 2
φ ∂(E − B )
1 2
−φ ∂(E + B )
1
−e +e =0 (61)
∂ ỹ ∂ x̃+ ∂ x̃−
∂E 3 1
φ ∂(E + B )
2 1
−φ ∂(E − B )
2
−e +e =0 (62)
∂ x̃ ∂ x̃+ ∂ x̃−
∂(E 1 − B 2 ) ∂(E 2 + B 1 ) φ ∂B
3
− − 2e =0 (63)
∂ ỹ ∂ x̃ ∂ x̃+
13
Now, if we require to form an invariance between the equations-(57,58,59)
and equations-(61,62,63) then we would have
Ẽ 3 = E 3 (64)
(Ẽ 2 − B̃ 1 ) = eφ (E 2 − B 1 )
(Ẽ 2 + B̃ 1 ) = e−φ (E 2 + B 1 )
(Ẽ 2 + B̃ 2 ) = eφ (E 1 + B 2 )
(Ẽ 1 + B̃ 2 ) = e−φ (E 1 − B 2 )
From the above set of equations it follows that
Ẽ 3 = E 3
B̃ 3 = B 3
Ẽ 1 = E 1 coshφ + B 2 sinhφ (65)
Ẽ 2 = E 2 coshφ − B 1 sinhφ
B̃ 1 = B 1 coshφ − E 2 sinhφ
B̃ 2 = B 2 coshφ + E 1 sinhφ
In the following section we will consider the free scalar field theory in the
light-front formulation. We will also discuss the the equal-x+ commutation
relations and propagator for the scalar field theory. Before get into that
discussion we will discuss some notation, conventions and useful relations
which will be needed in our further discussions.
14
Scalar product
x · y = (x0 )2 − (x3 )2 − (x⊥ )2 (70)
1 1
= x+ y − + x− y + − x⊥ y ⊥
2 2
The metric tensor is
1
0 0 0
2 0 2 0 0
1 0 0 0 µν 2 0 0 0
gµν = 2
0 0 −1 0 g = 0 0 −1 0 (71)
0 0 0 −1 0 0 0 −1
Thus
1 1
x− = x+ , x+ = x− (72)
2 2
Partial derivatives
∂ + = 2∂− , ∂ − = 2∂+ (73)
Four dimensional volume element
1
d4 x = dx0 dx1 dx2 dx3 = dx+ dx− d2 x⊥ (74)
2
Three dimensional volume element
1
[d3 x] = dx− d2 x⊥ (75)
2
Lorentz invariant volume element in momentum space
3 dk + d2 k ⊥
[d k] = (76)
2(2π)3 k +
The step function
θ(x) = 0, x<0 (77)
= 1, x>0 (78)
The anti-symmetric step function
(x) = θ(x) − θ(−x) (79)
∂
= 2δ(x) (80)
∂x
Where the delta function
|x| = x(x) (81)
15
6 Scalar Field Theory In Terms Of Light-Front Variables
16
6.1 Equal x+ commutation relation
For free field theory, the equal time commutator of φ(x) and φ(y) for arbitrary
x and y is given by
Z
1
[φ(x), φ(y)]x0 =y0 = 3
d4 k δ(k 4 − m2 )θ(k 0 )[e−ik·(x−y) − eik·(x−y) ] (91)
(2π)
We know k + = k 0 + k 3 and k − = k 0 − k 3 . Thus,
k+ + k−
0
k =
2
Now, let’s calculate this integral in light-front variables.
[φ(x), φ(y)]x+ =y+ (92)
Z 2 ⊥ Z +∞ Z +∞ + −
1 dk k + k
= 3
dk + dk − δ(k + k − − (k ⊥ )2 − m2 )θ
2 (2π) −∞ −∞ 2
1 − − − ⊥ ⊥ ⊥ 1 − − − ⊥ ⊥
+ + 1 + + + 1 +
−y ⊥ ))
[e−i( 2 k (x −y )+ 2 k (x −y )−k ·(x −y )) − ei( 2 k (x −y )+ 2 k (x −y )−k ·(x ]x+ =y+
Z 2 ⊥ Z +∞ + Z +∞ ⊥ 2 2
+ −
1 dk dk (k ) + m k + k
= dk − δ(k − − θ (93)
2 (2π)3 −∞ k+ −∞ k+ 2
1 − + + 1 + − − ⊥ ⊥ ⊥ 1 − + + 1 + − − ⊥ ⊥ ⊥
[e−i( 2 k (x −y )+ 2 k (x −y )−k ·(x −y )) − ei( 2 k (x −y )+ 2 k (x −y )−k ·(x −y )) ]x+ =y+
Z 2 ⊥ Z +∞ +
1 dk dk −i( 1 k+ (x− −y− )−k⊥ ·(x⊥ −y⊥ )) i( 21 k + (x− −y − )−k ⊥ ·(x⊥ −y ⊥ ))
= [e 2 − e ]
2 (2π)3 −∞ k+
Changing k ⊥ → k −⊥ in the 2nd integral
d k k⊥ ·(x⊥ −y⊥ ) +∞ dk + −i k+ (x− −y− )
Z 2 ⊥ Z
1 +
i k2 (x− −y − )
= e [e 2 −e ]
2(2π) (2π)2 −∞ k
+
Z +∞ +
i 2 ⊥ ⊥ 1 dk −i k+ (x− −y− ) +
i k2 (x− −y − )
= δ (x − y ) [e 2 − e ]
2 2πi −∞ k +
Z +∞ + Z +∞
i 2 ⊥ 1 dk k+ − − 1 dk + −i k+ (y− −x− )
= lim δ (x − y ⊥ )[ e −i 2 (x −y )
− e 2 ]
→0 2 2πi −∞ k + + i 2πi −∞ k + + i
i
= δ 2 (x⊥ − y ⊥ )[θ(x− − y − ) − θ(y − − x− )]
2
i
= (x− − y − )δ 2 (x⊥ − y ⊥ )
2
So
i
[φ(x), φ(y)]x+ =y+ = (x− − y − )δ 2 (x⊥ − y ⊥ ) (94)
2
17
Here, we have used the definition
+∞
e−itx
Z
1
θ(x) = lim dt
→0 2πi −∞ t + i
If we compare the commutator relation with the equal-time commutator,
namely,
[φ(x), φ(y)]x0 =y0 = 0,
for x0 = y 0 , the two fields are separated by a space-like interval. So, to
respect microscopic causality the commutator has to be zero. But, this is not
the case for equal x+ commutator. for x+ = y + , if x⊥ 6= y ⊥ , the two fields
are separated by a space-like distance and so the commutator has to vanish.
On the other hand, for x+ = y + and x⊥ = y ⊥ , the two fields are separated
by a light like distance and so the commutator need not vanish.
6.2 Propagator
We consider now the scalar field propagator. Let S¯B denote the scalar field
propagator in light-front theory.
Equation of motion
[∂ + ∂ − − (∂ ⊥ )2 + m2 ]φ(x) = 0
The Green’s function is defined as
[∂ + ∂ − − (∂ ⊥ )2 + m2 ]G(x − y) = −δ 4 (x − y) (95)
Where
d4 k −ik·(x−y)
Z
G(x − y) = e G(k)
(2π)4
So,
d4 k −ik·(x−y)
Z
+ − ⊥ 2 2
[∂ ∂ − (∂ ) + m ] e G(k)
(2π)4
d4 k −ik·(x−y)
Z
⊥ 2 2
= [4∂+ ∂− − (∂ ) + m ] e G(k)
(2π)4
d4 k
Z
= 4
[−k + k − + (k ⊥ )2 + m2 ]e−ik·(x−y) G(k)
(2π)
d4 k
Z
= 4
[−k 2 + m2 ]e−ik·(x−y) G(k)
(2π)
18
We know
d4 k −ik·(x−y)
Z
4
δ (x − y) = e
(2π)4
So, to satisfy equation-95, we must have
1
G(k) = (96)
k 2 − m2
So, the Green’s function is given by
d4 k e−ik·(x−y)
Z
G(x − y) = (97)
(2π)4 k 2 − m2
So, the propagator is given by,
d4 k e−ik·(x−y)
Z
S¯B = lim (98)
→0 (2π)4 k 2 − m2 + i
Now
k 2 − m2 + i = 0
or, k − k + − (k ( ⊥))2 − m2 + i = 0
− (k ⊥ )2 + m2
or, k ≈ − i (99)
k+
So, the k − integration has a pole.
S¯B (x − y) (100)
dk − dk + d2 k ⊥ e−ik·(x−y)
Z
= lim
2k + (2π)4 k − − (k⊥ )2++m2 − i
→0
k
1 − + +
+ 2 ⊥
e−i[ 2 k (x −y )]
Z Z
dk d k −i[ 12 k + (x− −y − )−k ⊥ ·(x⊥ −y ⊥ )] −
= lim e dk ⊥ 2 2
2k + (2π)4 k − − (k )k++m − i
→0
dk + d2 k ⊥
Z ⊥ 2 2
−i[ 12 k + (x− −y − )−k ⊥ ·(x⊥ −y ⊥ )] −i[( 21 (k k) ++m −i)(x+ −y + )]
= lim e (−2πi)e
→0 2k + (2π)4
dk + d2 k ⊥ −ik·(x−y)
Z
=−i e
2k + (2π)3
So
dk + d2 k ⊥
Z
iS¯B = e−ik·(x−y) (101)
2k + (2π)3
19
Now
dk + d2 k ⊥
Z
φ(y)|0i = |kieik·y
2k + (2π)3
dk + d2 k ⊥
Z
h0|φ(x) = + 3
e−ik·x hk|
2k (2π)
So, it is straight forward to show that
dk + d2 k ⊥
Z
h0|φ(x)φ(y)|0i = + 3
e−ik·(x−y) (102)
2k (2π)
Thus we can write
iS¯B (x − y) =h0|T + φ(x)φ(y)|0i (103)
=θ(x+ − y + )h0|φ(x)φ(y)|0i + θ(y + − x+ )h0|φ(y)φ(x)|0i
=iSBF (x − y)
Where SBF is the Feynman propagator for the scalar field. Thus, for a scalar
field, light-front propagator is the same as Feynman propagator.
From the Lagrangian density, we can find the stress tensor T µν and from that,
we can construct the four-momentum P µ and generalised angular momentum
M µν . They are defined as follows
Z
1
µ
p = dx− d2 x⊥ T +µ , (104)
Z 2
1
M =µν
dx− d2 x⊥ [xν T +µ − xµ T +ν ] (105)
2
M µν is anti-symmetric and has six independent components. Poincare alge-
bra in terms of P µ and M µν is given by
[P µ , P ν ] = 0, (106)
[P µ , M ρσ ] = i[g µρ P σ − g µσ P ρ ], (107)
[M µν , M ρσ ] = i[−g µρ M νσ + g µσ M νρ − g νσ M µρ + g νρ M µσ ] (108)
In light-front dynamics P − is the Hamiltonian and P + and P i (i = 1, 2) are
the momenta. M +− = 2K 3 and M +i = E i are the boosts. M 12 = J 3 and
20
M −i = F i are rotations.
Now
T µν = ∂ µ ∂ ν − g µν L. (109)
with
1 1
L = ∂µ ∂ µ − m2 φ2 (110)
2 2
The momentum operators are given by
Z
1
P+ = dx− d2 x⊥ ∂ + φ∂ + φ, (111)
2 Z
1
Pi = dx− d2 x⊥ ∂ + φ∂ i φ. (112)
2
The Hamiltonian operator is given by
Z
− 1
P = dx− d2 x⊥ [∂ i φ∂ i φ + m2 φ2 ]. (113)
2
The boost generators are(at x+ = 0)
Z
1
K3 = dx− d2 x⊥ x− ∂ + φ∂ + φ, (114)
4 Z
1
Ei = dx− d2 x⊥ xi ∂ + φ∂ + φ. (115)
2
The rotation generators are
Z
1
3
J =− dx− d2 x⊥ ∂ + φ[x1 ∂ 2 φ − x2 ∂ 1 φ], (116)
Z 2
1
Fi = − dx− d2 x⊥ [x− ∂ + φ∂ i φ − xi (∂ ⊥ φ · ∂ ⊥ φ + m2 φ2 )]. (117)
2
In this section, we will construct the Poincare generators of free scalar field
theory in Fock representation.
Now Z
1
+
P = dx− d2 x⊥ (∂ + φ)(∂ + φ) (118)
2
∂ + φ(x) = 2∂− φ
dk + d2 k ⊥ +
Z
=i + 3
k [−a(k)e−ik·x + a† (k)eik·x ]
2k (2π)
21
Substituting this in equation-118 we will get after performing the integration
dk + d2 k ⊥ +
Z
1
+
P =− + 3
k [a(k)a(−k) − a(k)a† (k) − a† (k)a(k) + a† (k)a† (−k)]
2 2k (2π)
(119)
Now, ~k = (k , k , k ), and −~k means a component with negative k . But
+ 1 2 +
8 Conclusion
22
kinematical subgroup of generators in the Poincare group increased. This
~ and B
formulation also produced the right transformation equations for E ~ of
Maxwell’s equations. In the context of scalar field theory in terms of light-
front variables the equal x+ commutator of two fields φ(x) and φ(y) is not
in general zero, but it respects the microscopic causality. The scalar field
light-front propagator, on the other hand, remains the same as the Feynman
propagator of equal-time theory. Finally, the Poincare generators of scalar
field theory can be expressed in Fock representation.
References
[1] A. Lahiri and Palash B. Pal, A First Book of Quantum Field Theory
[2] James D. Bjorken and Sidney D. Drell, Relativistic Quantum Fields
[3] A. Das, Lectures on Quantum Field Theory
[4] David tong, Lectures on Quantum Field Theory, University of Cam-
bridge
[5] Avaroth Harindranath, An Introduction to Light-Front Dynamics for
Pedestrians, arXiv:hep-ph/9612244v2
23