2 Gauge Theories and the Standard Model
25
In a classical context this can be proven as follows. Consider a lagrangian
L =
1
2
|∂ μ φ|
2
− V (φ).
(2.45)
The potential V (φ) can be kept generic at this stage but, in the following, we will
be mostly interested in a renormalizable potential of the form (with no more than
quartic terms):
V (φ) = −
1
2
μ
2 φ
2
+
1
4
λ φ
4 .
(2.46)
Here by φ we mean a column vector with real components φ i (1=1,2. . . N) (complex
fields can always be decomposed into a pair of real fields), so that, for example,
φ 2 =
i φ 2
i . This particular potential is symmetric under a NxN orthogonal matrix
rotation φ = Oφ, where O is a SO(N) transformation. For simplicity, we have
omitted odd powers of φ, which means that we assumed an extra discrete symmetry
under φ ↔ −φ. Note that, for positive μ 2 , the mass term in the potential has the
“wrong” sign: according to the previous discussion this is the condition for the
existence of a non unique lowest energy state. More in general, we only assume
here that the potential is symmetric under the infinitesimal transformations
φ → φ
= φ + δφ, δφ i = iδθ
A t
A
ij φ j .
(2.47)
where δθ A are infinitesimal parameters and t A
ij are the matrices that represent the
symmetry group on the representation of the fields φ i (a sum over A is understood).
The minimum condition on V that identifies the equilibrium position (or the vacuum
state in quantum field theory language) is
(∂V /∂φ i )(φ i = φ
0
i ) = 0 .
(2.48)
The symmetry of V implies that
δV = (∂V /∂φ i )δφ i = iδθ
A (∂V /∂φ i )t
A
ij φ j = 0 .
(2.49)
By taking a second derivative at the minimum φ i = φ 0
i , given by the previous
equation, we obtain that, for each A:
∂ 2 V
∂φ k ∂φ i
(φ i = φ
0
i )t
A
ij φ
0
j +
∂V
∂φ i
(φ i = φ
0
i )t
A
ik = 0 .
(2.50)
The second term vanishes owing to the minimum condition, Eq. (2.48). We then find
∂ 2 V
∂φ k ∂φ i
(φ i = φ
0
i )t
A
ij φ
0
j = 0 .
(2.51)
25
In a classical context this can be proven as follows. Consider a lagrangian
L =
1
2
|∂ μ φ|
2
− V (φ).
(2.45)
The potential V (φ) can be kept generic at this stage but, in the following, we will
be mostly interested in a renormalizable potential of the form (with no more than
quartic terms):
V (φ) = −
1
2
μ
2 φ
2
+
1
4
λ φ
4 .
(2.46)
Here by φ we mean a column vector with real components φ i (1=1,2. . . N) (complex
fields can always be decomposed into a pair of real fields), so that, for example,
φ 2 =
i φ 2
i . This particular potential is symmetric under a NxN orthogonal matrix
rotation φ = Oφ, where O is a SO(N) transformation. For simplicity, we have
omitted odd powers of φ, which means that we assumed an extra discrete symmetry
under φ ↔ −φ. Note that, for positive μ 2 , the mass term in the potential has the
“wrong” sign: according to the previous discussion this is the condition for the
existence of a non unique lowest energy state. More in general, we only assume
here that the potential is symmetric under the infinitesimal transformations
φ → φ
= φ + δφ, δφ i = iδθ
A t
A
ij φ j .
(2.47)
where δθ A are infinitesimal parameters and t A
ij are the matrices that represent the
symmetry group on the representation of the fields φ i (a sum over A is understood).
The minimum condition on V that identifies the equilibrium position (or the vacuum
state in quantum field theory language) is
(∂V /∂φ i )(φ i = φ
0
i ) = 0 .
(2.48)
The symmetry of V implies that
δV = (∂V /∂φ i )δφ i = iδθ
A (∂V /∂φ i )t
A
ij φ j = 0 .
(2.49)
By taking a second derivative at the minimum φ i = φ 0
i , given by the previous
equation, we obtain that, for each A:
∂ 2 V
∂φ k ∂φ i
(φ i = φ
0
i )t
A
ij φ
0
j +
∂V
∂φ i
(φ i = φ
0
i )t
A
ik = 0 .
(2.50)
The second term vanishes owing to the minimum condition, Eq. (2.48). We then find
∂ 2 V
∂φ k ∂φ i
(φ i = φ
0
i )t
A
ij φ
0
j = 0 .
(2.51)