Nakahara M. Geometry, Topology and Physics

Подождите немного. Документ загружается.

Here is the Lagrangian density. The Euler–Lagrange equation now takes the

form

∂

∂x



∂

∂(∂

φ)



−

∂

∂φ

= 0. (1.208)

The Lagrangian density of a free scalar ﬁeld is

(φ, ∂

φ) =−

(∂

φ∂

φ + m

). (1.209)

The Euler–Lagrange equation derived from this Lagrangian density is the Klein–

Gordon equation

(

− m

)φ = 0, (1.210)

where

= ∂

∂

=−∂

+∇

The vacuum-to-vacuum amplitude in the presence of a source J has the path

integral representation 0, ∞|0, −∞

∝ Z

[J ],where

[J ]=



φ exp







+ J φ +

εφ



(1.211)

where the iε term has been added to regularize the path integral.

Integration by

parts yields

[J ]=



φ exp





dx (

{φ( −m

)φ + iεφ

}+J φ)



. (1.212)

Let φ

be the classical solution to the Klein–Gordon equation in the presence

of the source,

(

− m

+ iε)φ

=−J. (1.213)

The solution is easily found to be

(x ) =−



dy (x − y)J (y) (1.214)

where (x − y) is the Feynman propagator

(x − y) =

−1

(2π)



ik(x −y)

+ m

− iε

. (1.215)

Here d denotes the spacetime dimension. Note that (x − y) satisﬁes

(

− m

+ iε)(x − y) = δ

(x − y).

It is easy to show that (exercise) the functional Z

[J ] is now written as

[J ]=Z

[0]exp



−



dx dyJ(x)(x − y)J (y)



. (1.216)

Alternatively, we can introduce the imaginary time τ = ix

to Wick rotate the time axis.

It is instructive to note that the propagator is conversely obtained by the functional

derivative of Z

[J ],

(x − y) =

[0]

[J ]

δ J (x)δ J (y)



J =0

. (1.217)

The amplitude Z

[0] is the vacuum-to-vacuum amplitude in the absence of

the source and may be evaluated as follows. Let us introduce the imaginary time

= τ = ix

. Then, we obtain

[0]=



φ exp





dx φ(

− m

)φ



=[Det(

− m

)]

−1/2

, (1.218)

where

= ∂

+∇

and the deteminant is understood in the sense of section 1.4,

namely it is the product of eigenvalues with a relevant boundary condition.

A free complex scalar ﬁeld theory has a Lagrangian density

=−∂

∗

∂

φ −m

|φ|

+ J φ

∗

+ J

∗

φ (1.219)

where the source terms have been included. The generating functional is now

given by

[J, J

∗



φ φ

∗

exp





dx (

− iε|φ|

)





φ φ

∗

exp





dx {φ

∗

( − m

+ iε)φ + J

∗

φ + J φ

∗

}



(1.220)

The propagator is now given by

(x − y) =

[0, 0]

[J, J

∗

]

δ J

∗

(x )δ J(y)



J =J

∗

. (1.221)

By substituting the Klein–Gordon equations

(

− m

)φ

=−J ( − m

)φ

∗

=−J

∗

(1.222)

we separate the generating functional as

[J, J

∗

]=Z

[0, 0]exp



− i



dx dyJ

∗

(x )(x − y)J (y)



(1.223)

where

[0, 0]=



φ φ

∗

exp



− i



dxφ

∗

( − m

− iε)φ



=[Det(

− m

)]

−1

. (1.224)

Wick rotation has been made to occur at the last line.

1.6.2 Interacting scalar ﬁeld

It is possible to add interaction terms to the free ﬁeld Lagrangian (1.209),

(φ, ∂

φ) =

(φ, ∂

φ) − V (φ). (1.225)

The possible form of V (φ) is restricted by the symmetry and renormalizability of

the theory. A typical form of V is a polynomial

V (φ) =

(n ≥ 3, n ∈ )

where the constant g ∈

controls the strength of the interaction. The generating

functional is deﬁned similarly to the free theory as

Z[J ]=



φ exp





dx {

φ( − m

)φ − V (φ) + J φ}



. (1.226)

The presence of V (φ) makes things slightly more complicated. It can be handled

at least perturbatively as

Z[J ]=



φ exp



−i



dxV(φ)



exp





{

+ J φ

}



= exp



−i



dxV



δ J (x)





φ exp





{

+ J φ

}



= exp



−i



dxV



δ J (x)



[J ]

∞



k=0



...



(−i)

× V



δ J (x

)



...V



δ J (x

)



[J ]. (1.227)

The generating functional Z [J ] generates the vacuum expectation value

of the T -product of ﬁeld operators, also known as the Green function

,...,x

),as

,...,x

) ≡0|T [φ(x

)...φ(x

)]|0

(−i)

δ J (x

)...δJ (x

)

Z[J ]



J =0

. (1.228)

Since this is the nth functional derivative of Z[J ] around J = 0, we obtain the

functional Taylor expansion of Z[J ] as

Z[J ]=

∞



n=1





i=1



J (x

)



0|T [φ(x

)...φ(x

)]|0

=0|T e



dxJ(x )φ(x )

|0. (1.229)

The connected n-point functions are generated by W [J ] deﬁned by

Z[J ]=e

−W [J ]

. (1.230)

The effective action [φ

] is deﬁned by the Legendre transformation

[φ

]≡W [J ]−



dτ dxJ φ

(1.231)

where

≡φ

δW [J ]

δ J

. (1.232)

The functional [φ

] generates one-particle irreducible diagrams.

1.7 Quantization of a Dirac ﬁeld

The Lagrangian of the free Dirac ﬁeld ψ is

ψ(i

∂ − m)ψ, (1.233)

where

∂ = γ

∂

. In general

A ≡ γ

. Variation with respect to

ψ yields the

Dirac equation

∂ − m)ψ = 0. (1.234)

The Dirac ﬁeld, in canonical quantization, satisifes the anti-commutation

relation

{

ψ(x

, x), ψ(x

, y)}=δ(x − y). (1.235)

Accordingly, it is expressed as a Grassmann number function in path integrals.

The generating functional is

[¯η, η]=



ψ exp







ψ(i

∂ − m)ψ +

ψη +¯ηψ



(1.236)

where η, ¯η are Grassmannian sources.

The propagator is given by the functional derivative with respect to the

sources,

S(x − y) =−

[¯η, η]

δ ¯η(x)δη(y)

(2π)



ikx

k − m − iε

= (i

∂ + m + iε)(x − y)

(1.237)

where (x − y) is the scalar ﬁeld propagator.

By making use of the Dirac equations

∂ − m)ψ =−η

ψ(i

←−

∂ + m) =¯η (1.238)

the generating functional is cast into the form

[¯η, η]=Z

[0, 0]exp



−i



dx dy ¯η(x)S(x − y)η(y)



. (1.239)

After Wick rotation τ = ix

, the normalization factor is obtained as

[0, 0]=Det(i

∂ − m) =



(1.240)

where λ

is the ith eigenvalue of the Dirac operator i

∂ − m.

1.8 Gauge theories

At present, physically sensible theories of fundamental interactions are based

on gauge theories. The gauge principle—physics should not depend on how we

describe it—is in harmony with the principle of general relativity. Here we give

a brief summary of classical aspects of gauge theories. For further references, the

reader should consult those books listed at the beginning of this chapter.

1.8.1 Abelian gauge theories

The reader should be familiar with Maxwell’s equations:

div B = 0 (1.241a)

∂ B

∂t

+ curl E = 0 (1.241b)

div E = ρ (1.241c)

∂ E

∂t

− curl E =−j. (1.241d)

The magnetic ﬁeld B and the electric ﬁeld E are expressed in terms of the vector

potential A

= (φ, A) as

B = curl AE=

∂ A

∂t

− grad φ. (1.242)

Maxwell’s equations are invariant under the gauge transformation

→ A

+ ∂

χ (1.243)

where χ is a scalar function. This invariance is manifest if we deﬁne the

electromagnetic ﬁeld tensor F

µν

≡ ∂

− ∂







0 −E

−E

0 B

−B

0 B

−B







. (1.244)

From the construction, F is invariant under (1.243). The Lagrangian of the

electromagnetic ﬁelds is given by

=−

µν

+ A

(1.245)

where j

= (ρ, j).

Exercise 1.11. Show that (1.241a) and (1.241b) are written as

∂

µν

+ ∂

νξ

+ ∂

ξµ

= 0 (1.246a)

while (1.241c) and (1.241d) are

∂

µν

= j

(1.246b)

where the raising and lowering of spacetime indices are carried out with the

Minkowski metric η = diag(−1, 1, 1, 1). Verify that (1.246b) is the Euler–

Lagrange equation derived from (1.245).

Let ψ be a Dirac ﬁeld with electric charge e. The free Dirac Lagrangian

ψ(iγ

∂

+ m)ψ (1.247)

is clearly invariant under the global gauge transformation

ψ → e

−ieα

ψ →

ψe

ieα

(1.248)

where α ∈

is a constant. We elevate this symmetry to invariance under the local

gauge transformation,

ψ → e

−ieα(x)

ψ →

ψe

ieα(x)

. (1.249)

The Lagrangian transforms under (1.249) as

ψ(iγ

∂

+ m)ψ →

ψ(iγ

∂

+ eγ

∂

α + m)ψ. (1.250)

Since the extra term e∂

α looks like a gauge transformation of the vector

potential, we couple the gauge ﬁeld A

with ψ so that the Lagrangian has a local

gauge symmetry. We ﬁnd that

ψ[iγ

(∂

− ieA

) + m]ψ (1.251)

is invariant under the combined gauge transformation,

ψ → ψ



= e

−ieα(x)

ψ →



ψe

ieα(x)

→ A



= A

− ∂

α(x).

(1.252)

Let us introduce the covariant derivatives,

∇

≡ ∂

− ieA

∇



≡ ∂

− ieA



. (1.253)

The reader should verify that ∇

ψ transforms in a nice way,

∇



= e

−ieα(x)

∇

ψ. (1.254)

The total quantum electrodynamic (QED) Lagrangian is

QED

=−

µν

ψ(iγ

∇

+ m)ψ. (1.255)

Exercise 1.12. Let φ = (φ

+ iφ

√

2 be a complex scalar ﬁeld with electric

charge e. Show that the Lagrangian

= η

µν

(∇

φ)

†

(∇

φ) + m

†

φ (1.256)

is invariant under the gauge transformation

φ → e

−ieα(x)

φφ

†

→ φ

†

ieα(x)

→ A

− ∂

α(x). (1.257)

1.8.2 Non-Abelian gauge theories

The gauge transformation just described is a member of a U(1) group, that

is a complex number of modulus 1, which happens to be an Abelian group.

A few decades ago, Yang and Mills (1954) introduced non-Abelian gauge

transformations. At that time, non-Abelian gauge theories were studied from

curiosity. Nowadays, they play a central role in elementary particle physics.

Let G be a compact semi-simple Lie group such as SO(N) or SU(N).The

anti-Hermitian generators {T

} satisfy the commutation relations

, T

]= f

αβ

(1.258)

where the numbers f

αβ

are called the structure constants of G.AnelementU

of G near the unit element can be expressed as

U = exp(−θ

). (1.259)

We suppose a Dirac ﬁeld ψ transforms under U ∈ G as

ψ → U ψ

ψ →

ψU

†

. (1.260)

[Remark: Strictly speaking, we have to specify the representation of G to which

ψ belongs. If readers feel uneasy about (1.260), they may consider ψ is in the

fundamental representation, for example.]

Consider the Lagrangian

ψ[iγ

(∂

+ g

) + m]ψ (1.261)

where the Yang–Mills gauge ﬁeld

takes its values in the Lie algebra of G,that

is,

can be expanded in terms of T

= A

. (Script ﬁelds are anti-

Hermitian.) The constant g is the coupling constant which controls the strength

of the coupling between the Dirac ﬁeld and the gauge ﬁeld. It is easily veriﬁed

that

is invariant under

ψ → ψ



= Uψ

ψ →



ψU

†

→



= U

†

+ g

−1

U∂

†

(1.262)

The covariant derivative is deﬁned by ∇

= ∂

+ g

as before. The covariant

derivative ∇

ψ transforms covariantly under the gauge transformation

∇



= U∇

ψ. (1.263)

The Yang–Mills ﬁeld tensor is

µν

≡ ∂

µ ν

− ∂

ν µ

+ g[

]. (1.264)

The component F

µν

= ∂

− ∂

+ gf

βγ

. (1.265)

If we deﬁne the dual ﬁeld tensor ∗

µν

≡

µνκλ

κλ

, it satisﬁes the Bianchi

identity,

∗

µν

≡ ∂

∗

µν

+ g[

, ∗

µν

]=0. (1.266)

Exercise 1.13. Show that

µν

transforms under (1.262) as

µν

→ U

µν

†

. (1.267)

From this exercise, we ﬁnd a gauge-invariant action

=−

tr(

µν

) (1.268a)

where the trace is over the group matrix. The component form is

=−

µνα

µν

tr(T

) =

µνα

(1.268b)

where we have normalized {T

} so that tr(T

) =−

αβ

. The ﬁeld equation

derived from (1.268) is

µ µν

= ∂

µ µν

+ g[

µν

]=0. (1.269)

1.8.3 Higgs ﬁelds

If the gauge symmetry is manifest in our world, there would be many observable

massless vector ﬁelds. The absence of such ﬁelds, except for the electromagnetic

ﬁeld, forces us to break the gauge symmetry. The theory is left renormalizable if

the symmetry is broken spontaneously.

Let us consider a U(1) gauge ﬁeld coupled to a complex scalar ﬁeld φ, whose

Lagrangian is given by

=−

µν

+ (∇

φ)

†

(∇

φ) − λ(φ

†

φ − v

)

. (1.270)

The potential V (φ) = λ(φ

†

φ − v

)

has minima V = 0at|φ|=v.The

Lagrangian (1.270) is invariant under the local gauge transformation

→ A

− ∂

αφ→ e

−ieα

φφ

†

→ e

ieα

†

. (1.271)

This symmetry is spontaneously broken due to the vacuum expectation value

(VEV) φ of the Higgs ﬁeld φ. We expand φ as

φ =

√

[v +ρ(x)]e

iα(x)/v

∼

√

[v +ρ(x) + iα(x)]

assuming v = 0. If v = 0, we may take the unitary gauge in which the phase of

φ is ‘gauged away’ so that φ has only the real part,

φ(x) =

√

(v + ρ(x)). (1.272)

If we substitute (1.272) into (1.270) and expand in ρ,wehave

=−

µν

∂

ρ∂

ρ +

+ 2vρ + ρ

)

−

λ(4v

+ 4vρ

+ ρ

). (1.273)

The equations of motion for A

and ρ derived from the free parts are

∂

νµ

+ 2e

= 0 ∂

∂

ρ + 2λv

ρ = 0. (1.274)

From the ﬁrst equation, we ﬁnd A

must satisfy the Lorentz condition ∂

= 0.

The apparent degrees of freedom of (1.270) are 2(photon) +2(complex scalar) =

4. If VEV = 0, we have 3(massive vector) +1(real scalar) = 4. The ﬁeld A

has

a mass term with the wrong sign and so cannot be a physical degree of freedom.

The creation of massive ﬁelds out of a gauge ﬁeld is called the Higgs mechanism..

1.9 Magnetic monopoles

Maxwell’s equations unify electricity and magnetism. In the history of physics

they should be recognized as the ﬁrst attempt to unify forces in Nature. In spite

of their great success, Dirac (1931) noticed that there existed an asymmetry in

Maxwell’s equations: the equation div B = 0 denies the existence of magnetic

charges. He introduced the magnetic monopole, a point magnetic charge, to make

the theory symmetric.

1.9.1 Dirac monopole

Consider a monopole of strength g sitting at r = 0,

div B = 4πgδ

(r). (1.275)

It follows from (1/r) =−4πδ

(r) and ∇(1/r ) =−r/r

that the solution of

this equation is

B = gr/r

. (1.276)

The magnetic ﬂux  is obtained by integrating B over a sphere S of radius R so

that

 =

B · dS = 4πg. (1.277)

What about the vector potential which gives the monopole ﬁeld (1.276)? If

we deﬁne the vector potential A

−gy

r(r + z)

= 0 (1.278a)

we easily verify that

curl A

= gr/r

+ 4πgδ(x)δ(y)θ (−z). (1.279)

We have curl A

= B except along the negative z-axis (θ = π ). The singularity

along the z-axis is called the Dirac string and reﬂects the poor choice of the

coordinate system. If, instead, we deﬁne another vector potential

r(r − z)

−gx

r(r − z)

= 0 (1.278b)

we have curl A

= B except along the positive z-axis (θ = 0) this time. The

existence of a singularity is a natural consequence of (1.277). If there were a

vector A such that B = curl A with no singularity, we would have, from Gauss’

law,

 =

B · dS =

curl A · dS =



div(curl A) dV = 0

where V is the volume inside the surface S. This problem is avoided only when

we abandon the use of a single vector potential.

Exercise 1.14. Let us introduce the polar coordinates (r,θ,φ). Show that the

vector potentials A

and A

are expressed as

(r) =

g(1 −cos θ)

r sin θ

ˆe

(1.280a)

(r) =−

g(1 +cos θ)

r sin θ

ˆe

(1.280b)

where ˆe

=−sin φ ˆe

+ cos φ ˆe