Masujima M. Applied Mathematical Methods in Theoretical Physics
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408 10 Calculus of Variations: Applications
Transformation law of the field operator is given by
ˆ
ψ(x) −→ U(, a)
ˆ
ψ(x)U
−1
(, a) = L
−1
()
ˆ
ψ(x
). (10.7.1)
Rational for the field operator transformation law is that the matrix element of the
field operator at the transformed point x
,
ˆ
ψ(x
), evaluated between the transformed
states should act like the wavefunction. So we demand that the following equation
is true:
φ
α
ˆ
ψ(x
)
φ
β
= L()
φ
α
ˆ
ψ(x)
φ
β
.
Thus we have
φ
α
U
−1
(, a)
ˆ
ψ(x
)U(, a)
φ
β
= L()
φ
α
ˆ
ψ(x)
φ
β
,
or, as the operator identity, we have
U
−1
(, a)
ˆ
ψ(x
)U(, a) = L()
ˆ
ψ(x),
which is inverted as
U(, a)
ˆ
ψ(x)U
−1
(, a) = L
−1
()
ˆ
ψ(x
).
Consider the infinitesimal Poincar´e transformation
x
µ
−→ x
µ = x
µ
+ δx
µ
; δx
µ
= δω
µν
x
ν
+δε
µ
.
For an infinitesimal Poincar
´
e transformation, we have the unitary operator
U(1 +δω, δε) ≡ 1 +iδε
µ
P
µ
−
i
2
δω
µν
J
µν
, (10.7.2)
where P
µ
(J
µν
) is the generator of space–time translation (rotation).
Composition law of two successive Poincar
´
e transformations parameterized by
(
1
, a
1
)and(
2
, a
2
)isgivenby
(
(
2
, a
2
)(
1
, a
1
) = (
2
1
,
2
a
1
+ a
2
),
U(
2
, a
2
)U(
1
, a
1
) = U(
2
1
,
2
a
1
+ a
2
).
(10.7.3)
Consider the following three successive Poincar´e transformations:
U
−1
(, a)U(1 + δω, δε)U(, a).
Sinceweknow
U
−1
(, a)U(, a) = U(1, 0),
10.7 Poincare Transformation and Spin 409
we have
U
−1
(, a) = U(
−1
, −
−1
a).
We can simplify the above Poincar
´
e transformations as
U
−1
(, a)U(1 + δω, δε)U(, a)
= U(
−1
, −
−1
a)U(1 +δω, δε)U(, a)
= U(
−1
, −
−1
a)U((1 +δω),(1+δω)a + δε)
= U(
−1
(1 +δω),
−1
[(1 +δω)a + δε] −
−1
a)
= U(1 +
−1
δω,
−1
δωa +
−1
δε).
Thus we have
U
−1
(, a)(1 +iδε
µ
P
µ
−
i
2
δω
µν
J
µν
)U(, a)
= 1 +i(
−1
δωa +
−1
δε)
λ
P
λ
−
i
2
(
−1
δω)
λκ
J
λκ
.
Recalling that (
−1
)
µ
ν
=
µ
ν
,wehave
(
−1
δω)
λκ
= (
−1
)
λ
µ
δω
µν
κ
ν
= δω
µν
λ
µ
κ
ν
,
(
−1
δωa +
−1
δε)
λ
= (
−1
)
λ
µ
(δω
µν
a
ν
+ δε
µ
)
=
λ
µ
δω
µν
a
ν
+
λ
µ
δε
µ
=
1
2
δω
µν
(
λ
µ
a
ν
−
λ
ν
a
µ
) +δε
µ
λ
µ
.
We thus have
U
−1
(, a)(1 +iδε
µ
P
µ
−
i
2
δω
µν
J
µν
)U(, a)
= 1 +iδε
µ
λ
µ
P
λ
−
i
2
δω
µν
{
λ
µ
κ
ν
J
λκ
+ (a
µ
λ
ν
− a
ν
λ
µ
)P
λ
}.
Hence, for arbitrary δε and δω,wehave
(
U
−1
(, a)P
µ
U(, a) =
λ
µ
P
λ
,
U
−1
(, a)J
µν
U(, a) =
λ
µ
κ
ν
J
λκ
+(a
µ
λ
ν
− a
ν
λ
µ
)P
λ
.
We further specialize to the following case:
µ
ν
= η
µ
ν
+ δω
µ
ν
, a
µ
= δε
µ
.
P
µ
transformation:
U
−1
(1 +δω, δε)P
µ
U(1 +δω, δε) = P
µ
+ δω
µν
P
ν
.
410 10 Calculus of Variations: Applications
We recall
(
U(1 + δω, δε) = 1 +iδε
ν
P
ν
−
i
2
δω
λκ
J
λκ
,
U
−1
(1 +δω, δε) = 1 − iδε
ν
P
ν
+
i
2
δω
λκ
J
λκ
.
To the first order in the infinitesimals, δε and δω,wehave
P
µ
− iδε
ν
[P
ν
, P
µ
] +
i
2
δω
λκ
[J
λκ
, P
µ
] = P
µ
+ δω
µν
P
ν
.
From δε term, we obtain
[P
ν
, P
µ
] = 0. (10.7.4)
From δω term, we observe
δω
µν
P
ν
= δω
λκ
η
λµ
P
κ
=
1
2
δω
λκ
(η
λµ
P
κ
− η
κµ
P
λ
),
and obtain
i[P
µ
, J
λκ
] = η
κµ
P
λ
− η
λµ
P
κ
. (10.7.5)
J
µν
transformation:
U
−1
(, a)J
µν
U(, a) = J
µν
−δω
λ
µ
J
λν
−δω
κ
ν
J
µκ
+ δε
µ
P
ν
− δε
ν
P
µ
.
Thus we have
J
µν
− iδε
σ
[P
σ
, J
µν
] +
i
2
δω
λκ
[J
λκ
, J
µν
] = J
µν
+δε
µ
P
ν
−δε
ν
P
µ
−δω
λ
µ
J
λν
−δω
κ
ν
J
µκ
.
We observe that
δω
λ
µ
J
λν
+ δω
κ
ν
J
µκ
= η
µκ
δω
κλ
J
λν
+ η
νλ
δω
λκ
J
µκ
=
1
2
δω
λκ
(η
µκ
J
νλ
− η
µλ
J
νκ
+η
νλ
J
µκ
− η
νκ
J
µλ
).
We thus obtain
i[J
µν
, J
λκ
] = η
µλ
J
νκ
− η
µκ
J
νλ
+η
νκ
J
µλ
− η
νλ
J
µκ
. (10.7.6)
We finally consider the transformation property of the field operator with the
allowanceforthecaseofthemulticomponent field operator. Transformation law of
the field operator is given by
ˆ
ψ
ρ
(x) −→ U(, a)
ˆ
ψ
ρ
(x)U
−1
(, a) = (L
−1
())
ρσ
ˆ
ψ
σ
(x
).
10.8 Conservation Laws and Noether’s Theorem 411
Infinitesimal translation:
(
U(1, δε) = 1 +iδε
µ
P
µ
,
x
µ
= x
µ
+ δε
µ
,
(L( = 1))
ρσ
= δ
ρσ
.
We observe that
ˆ
ψ
ρ
(x) + iδε
µ
[P
µ
,
ˆ
ψ
ρ
(x)] =
ˆ
ψ
ρ
(x + δε) =
ˆ
ψ
ρ
(x) + δε
µ
∂
µ
ˆ
ψ
ρ
(x).
We thus obtain
i[P
µ
,
ˆ
ψ
ρ
(x)] = ∂
µ
ˆ
ψ
ρ
(x). (10.7.7)
Infinitesimal rotation:
(
U(1 +δω,0) = 1 −
i
2
δω
µν
J
µν
,
x
µ
= x
µ
+ δω
µν
x
ν
,
(L(1 + δω))
ρσ
= (1 +
i
2
δω
µν
S
µν
)
ρσ
.
We observe that
ˆ
ψ
ρ
(x) −
i
2
δω
µν
[J
µν
,
ˆ
ψ
ρ
(x)] = (1 −
i
2
δω
µν
S
µν
)
ρσ
ˆ
ψ
σ
(x + δω ·x)
= (1 −
i
2
δω
µν
S
µν
)
ρσ
(
ˆ
ψ
σ
(x) + δω
µν
x
ν
∂
µ
ˆ
ψ
σ
(x))
=
ˆ
ψ
ρ
(x) −
i
2
δω
µν
(S
µν
)
ρσ
ˆ
ψ
σ
(x) + δω
µν
x
ν
∂
µ
ˆ
ψ
ρ
(x)
=
ˆ
ψ
ρ
(x) −
i
2
δω
µν
(S
µν
)
ρσ
ˆ
ψ
σ
(x) +
1
2
δω
µν
(x
ν
∂
µ
− x
µ
∂
ν
)
ˆ
ψ
ρ
(x).
We thus obtain
[J
µν
,
ˆ
ψ
ρ
(x)] = (S
µν
)
ρσ
ˆ
ψ
σ
(x) + i(x
ν
∂
µ
− x
µ
∂
ν
)
ˆ
ψ
ρ
(x)
= (S
µν
)
ρσ
ˆ
ψ
σ
(x) + (x
µ
1
i
∂
ν
− x
ν
1
i
∂
µ
)
ˆ
ψ
ρ
(x). (10.7.8)
The mixing matrix S
µν
of the multicomponent field is thus identified as the spin
of the field operator.
10.8
Conservation Laws and Noether’s Theorem
There exists the close relationship between the continuous symmetries of the
Lagrangian L(q
r
(t),
˙
q
r
(t), t) (the Lagrangian density L(ψ
a
(x), ∂
µ
ψ
a
(x))) and the con-
servation laws, commonly known as Noether’s theorem.
412 10 Calculus of Variations: Applications
We shall first consider classical mechanics, described by the Lagrangian
L(q
r
(t),
˙
q
r
(t), t). Suppose that when we perform the variation,
q
r
(t) −→ q
r
(t) +δq
r
(t),
we have
δL(q
r
(t),
˙
q
r
(t), t) =
d
dt
δ, (10.8.1)
where δ is the function of q
r
(t)’s and
˙
q
r
(t)’s. Then, without the use of Lagrange
equation of motion, we have
f
r=1
∂L
∂q
r
δq
r
+
∂L
∂
˙
q
r
δ
˙
q
r
−
d
dt
δ = 0.
Now, with the use of Lagrange equation of motion, we have
f
r=1
d
dt
∂L
∂
˙
q
r
δq
r
+
∂L
∂
˙
q
r
d
dt
δq
r
−
d
dt
δ =
d
dt
f
r=1
∂L
∂
˙
q
r
δq
r
− δ
= 0.
(10.8.2)
We have the conserved Noether charge δQ as
δQ =
f
r=1
∂L
∂
˙
q
r
δq
r
− δ, (10.8.3)
which is independent of t.
For the ease of presentation, we consider the n particle system with the three
spatial dimension,
x
a
= (q
3(a−1)+1
, q
3(a−1)+2
, q
3(a−1)+3
), a = 1, ..., n, f = 3n.
(1) δx
a
= δ,afixeddisplacementwithL(q
r
,
˙
q
r
, t)invariant,
δQ =
n
a=1
p
a
·δ = δ ·
n
a=1
p
a
= δ ·
n
a=1
∇
˙x
a
L,
∇
˙x
a
=
ˆ
e
x
∂
∂
˙
x
a
+
ˆ
e
y
∂
∂
˙
y
a
+
ˆ
e
z
∂
∂
˙
z
a
.
The following quantity is conserved:
P =
n
a=1
∇
˙x
a
L. (10.8.4)
10.8 Conservation Laws and Noether’s Theorem 413
Total momentum conservation results from the translational invariance of L.
(2) δx
a
= δθ × x
a
, a fixed rotation about the origin with L(q
r
,
˙
q
r
, t) invariant,
δQ =
n
a=1
∇
˙x
a
L · (δθ × x
a
) = δθ ·
n
a=1
x
a
× ∇
˙x
a
L.
The following quantity is conserved:
L =
n
a=1
x
a
× ∇
˙x
a
L. (10.8.5)
Total angular momentum conservation results from the rotational invariance of L.
(3) The Lagrangian L(q
r
,
˙
q
r
, t) does not depend on t explicitly. Then we have
δq
r
=
˙
q
r
δt and δ
˙
q
r
(t) =
¨
q
r
δt under t −→ t + δt,
so that, with the use of Lagrange equation of motion, we have
δL =
f
r=1
∂L
∂q
r
˙
q
r
+
∂L
∂
˙
q
r
¨
q
r
δt =
f
r=1
d
dt
∂L
∂
˙
q
r
˙
q
r
+
∂L
∂
˙
q
r
d
dt
˙
q
r
δt
=
d
dt
f
r=1
∂L
∂
˙
q
r
˙
q
r
δt =
dL
dt
δt.
The following quantity is conserved:
E =
f
r=1
∂L
∂
˙
q
r
˙
q
r
−L. (10.8.6)
Total energy conservation results from the explicit t independence of L.
We shall now consider classical field theory, described by the Lagrangian density
L(ψ
a
(x), ∂
µ
ψ
a
(x)). Suppose that when we perform the variation,
ψ
a
(x) → ψ
a
(x) + δψ
a
(x),
we have δL(ψ
a
(x), ∂
µ
ψ
a
(x)) as
δL(ψ
a
(x), ∂
µ
ψ
a
(x)) = ∂
µ
δ
µ
, (10.8.7)
where δ
µ
is the function of ψ
a
(x)’s and ∂
µ
ψ
a
(x)’s. Then, without the use of
Euler–Lagrange equation of motion, we have
a
∂L
∂ψ
a
δψ
a
+
∂L
∂(∂
µ
ψ
a
)
δ(∂
µ
ψ
a
)
− ∂
µ
δ
µ
= 0.
414 10 Calculus of Variations: Applications
Now, with the use of Euler–Lagrange equation of motion, we have
a
∂
µ
∂L
∂(∂
µ
ψ
a
)
δψ
a
+
∂L
∂(∂
µ
ψ
a
)
∂
µ
(δψ
a
)
− ∂
µ
δ
µ
= ∂
µ
a
∂L
∂(∂
µ
ψ
a
)
δψ
a
− δ
µ
= 0. (10.8.8)
We have the conserved Noether current δC
µ
as
δC
µ
=
a
π
µ
a
δψ
a
−δ
µ
,withπ
µ
a
≡
∂L
∂(∂
µ
ψ
a
)
. (10.8.9)
(4.) δx
λ
= ε
λ
, a space–time translation with ε
λ
constant
δψ
a
= ε
λ
∂
λ
ψ
a
,andδ
µ
= ε
µ
L,
where we assume that the Lagrangian density L has no explicit space–time
dependence.
The conserved ‘‘current’’ δC
µ
is given by
δC
µ
=
a
π
µ
a
ε
λ
∂
λ
ψ
a
− ε
µ
L = ε
λ
a
π
µ
a
∂
λ
ψ
a
− η
µλ
L
.
We call the object in the bracket as the canonical energy–momentum tensor
µλ
,
µλ
=
a
π
µ
a
∂
λ
ψ
a
−η
µλ
L,with∂
µ
µλ
= 0. (10.8.10)
We can define the conserved four-vector P
λ
, from the canonical energy–
momentum tensor
µλ
given above, by
P
λ
=
d
3
x
0λ
such that
d
dx
0
P
λ
= 0, (10.8.11)
which is the energy–momentum conservation law. The space–time translation
charge P
λ
generates the space–time translation as it should.
(5) δx
µ
= ε
αβ
(η
αµ
x
β
− η
βµ
x
α
), Lorentz transformation with ε
αβ
infinitesimal and
antisymmetric with respect to α and β,
δψ
a
= ε
αβ
{δ
ab
(x
α
∂
β
− x
β
∂
α
) + S
αβ
(a,b)
}ψ
b
.
Here, S
µλ
(a,b)
is the spin matrix of the multicomponent field ψ
a
,givenby
S
µλ
(a,b)
=
0 for scalar field,
(i2)[γ
µ
, γ
λ
]
ab
for spinor field,
δ
µ
a
δ
λ
b
− δ
λ
a
δ
µ
b
for vector field.
(10.8.12)
10.8 Conservation Laws and Noether’s Theorem 415
Since the Lorentz transformation law is coordinate dependent, the derivative of the
field variable, ∂
µ
ψ
a
, transforms as
δ∂
µ
ψ
a
= ε
αβ
%/
δ
ab
(x
α
∂
β
− x
β
∂
α
) + S
αβ
(a,b)
0
∂
µ
+ δ
ab
(η
α
µ
∂
β
− η
β
µ
∂
α
)
'
ψ
b
,
resulting in the Lorentz covariance condition
π
µa
S
αβ
(a,b)
∂
µ
ψ
b
+
∂L
∂ψ
a
S
αβ
(a,b)
ψ
b
= π
β
a
∂
α
ψ
a
−π
α
a
∂
β
ψ
a
,
δ
µ
= ε
αβ
(η
µβ
x
α
− η
µα
x
β
)L.
The conserved ‘‘current’’ δC
µ
is given by
δC
µ
= (π
µ
a
δψ
a
− δ
µ
) = ε
αβ
x
α
µβ
− x
β
µα
+ π
µ
a
S
αβ
(a,b)
ψ
b
.
We write the third-rank tensor inside the bracket as
M
µαβ
= x
α
µβ
− x
β
µα
+ π
µ
a
S
αβ
(a,b)
ψ
b
,with∂
µ
M
µαβ
= 0. (10.8.13)
We can define the conserved second-rank tensor L
αβ
, from the third-rank tensor
M
µαβ
given above, by
L
αβ
=
d
3
xM
0αβ
,suchthat
d
dx
0
L
αβ
= 0, (10.8.14)
which is the angular momentum conservation law. To be more precise, the spatial
(i, j)-components provide the angular momentum conservation law and the time-
spatial (0, k)-components provide the conservation law of the Lorentz boost charge.
The conserved Lorentz charge L
αβ
generates the Lorentz transformation as it
should.
The energy–momentum tensor
µλ
is not symmetric with respect to µ and λ
in general. We can construct the symmetric energy–momentum tensor T
µλ
from
the explicit inclusion of the spin degrees of freedom. We define the symmetric
energy–momentum tensor T
µλ
by
T
µλ
=
µλ
+ ∂
σ
σµλ
, (10.8.15)
with
σµλ
=
1
2
a,b
(π
σ
a
S
µλ
(a,b)
−π
µ
a
S
σλ
(a,b)
− π
λ
a
S
σµ
(a,b)
)ψ
b
. (10.8.16)
The new tensor T
µλ
differs from the old tensor
µλ
bythedivergenceof
σµλ
with
respect to x
σ
. The construction of the symmetric energy–momentum tensor T
µλ
is properly accomplished with the inclusion of the spin degrees of freedom. We
416 10 Calculus of Variations: Applications
can define the conserved four-vector
˜
P
λ
, from the symmetric energy–momentum
tensor T
µλ
given above, by
˜
P
λ
=
d
3
xT
0λ
such that
d
dx
0
˜
P
λ
= 0, (10.8.17)
with the requisite identity,
˜
P
λ
= P
λ
.
We can also define the third-rank tensor
˜
M
µαβ
in terms of the symmetric
energy–momentum tensor T
µβ
given above by
˜
M
µαβ
= x
α
T
µβ
− x
β
T
µα
,with∂
µ
˜
M
µαβ
= 0, (10.8.18)
also leading to the angular momentum conservation law
˜
L
αβ
=
d
3
x
˜
M
0αβ
such that
d
dx
0
˜
L
αβ
= 0, (10.8.19)
with the requisite identity
˜
L
αβ
= L
αβ
.
The space–time translation and the Lorentz transformation are unified to
the 10-parameter Poincar
´
e transformation. Beyond the 10-parameter Poincar
´
e
transformation, we have the 15-parameter conformal group, which constitutes the
space–time translation, the Lorentz transformation, the scale transformation (the
dilatation) and the conformal transformation, where the last one consists of the
inversion, the translation followed by the inversion.
We consider the scale transformation and the conformal transformation.
(6.) δx
µ
= x
µ
δρ, the scale transformation with δρ infinitesimal,
δψ
a
= δρ(d + x · ∂)ψ
a
,
and
δ
µ
= δρx
µ
L,
with the scale invariance condition
−4L + π
µa
(d + 1)∂
µ
ψ
a
+
∂L
∂ψ
a
·d · ψ
a
= 0,
where d is the scale dimension of the field ψ
a
,givenby
d =
(
1, for boson field,
32, for fermion field.
(10.8.20)
The above-stated scale invariance condition demands that the scale dimension of
the Lagrangian density L must be 4. Namely, any dimensional parameter in the
Lagrangian density L is forbidden for the scale invariance . The mass term in the
Lagrangian density violates the scale invariance condition.
10.8 Conservation Laws and Noether’s Theorem 417
The conserved ‘‘current’’ δC
µ
is given by
δC
µ
= π
µ
a
δψ
a
− δ
µ
= δρ(x
α
µα
+ π
µ
a
· d · ψ
a
).
We write the four-vector inside the bracket as
D
µ
= x
α
µα
+ π
µ
a
·d · ψ
a
,with∂
µ
D
µ
= 0. (10.8.21)
The dilatation charge D is given by
D =
d
3
xD
0
such that
d
dx
0
D = 0. (10.8.22)
The dilatation charge D generates the scale transformation as it should.
(7.) δx
µ
= δc
α
(2x
α
x
µ
− η
αµ
x
2
), the conformal transformation with the four-vector
δc
α
infinitesimal,
δψ
a
= δc
α
%
(2x
α
x
λ
− η
αλ
x
2
)∂
λ
ψ
a
+2x
λ
(δ
ab
η
λα
d − S
λα
(a,b)
)ψ
b
'
,
and
δ
µ
= δc
α
(2x
α
x
µ
− η
αµ
x
2
)L,
with the scale invariance condition
−4L + π
µa
(d + 1)∂
µ
ψ
a
+
∂L
∂ψ
a
· d · ψ
a
= 0,
plus the second condition that the field virial V
µ
defined below must be a total
divergence,
V
µ
≡ π
αa
(δ
ab
η
αµ
d − S
αµ
(a,b)
)ψ
b
= ∂
λ
σ
λµ
for some σ
λµ
.
The conserved ‘‘current’’ δC
µ
is given by
δC
µ
= δc
α
((2x
α
x
λ
− η
α
λ
x
2
)
µλ
+ 2x
λ
π
µ
a
(δ
ab
η
λα
d − S
λα
(a,b)
)ψ
b
− 2σ
αµ
).
We write the second-rank tensor inside the bracket as
K
αµ
= (2x
α
x
λ
−η
α
λ
x
2
)
µλ
+ 2x
λ
π
µ
a
(δ
ab
η
λα
d − S
λα
(a,b)
)ψ
b
− 2σ
αµ
,
with ∂
µ
K
αµ
= 0. (10.8.23)
The conformal charge K
α
is given by
K
α
=
d
3
xK
α0
such that
d
dx
0
K
α
= 0. (10.8.24)