ゲージ対称性とヒッグス機構

ゲージ原理　ー例としてQEDー

QEDラグランジアン

U(1)変換を考える ---(1)

明らかに変換を考える ---(2)

2

L = !(i !" "m)! " 14Fµ!Fµ! " eAµ(!#µ!)

# Le + L" + Lint

!(x)! ei!!(x)

Le ! i(e!i!!)"µ#µ(ei!!)"m!!

= i!#µ! " (!"µ!)#µ$ "m!!

= Le + %Le , %Le # (!"µ!)#µ$

!L! = 0 !Lint = 0Aµ(x)! Aµ(x) + !µ!(x)

!Le = 0 , !L = 0

!Lint = !e("µ!)#$µ#

Fµ! ! !!Aµ " !µA!

ゲージ原理

θ= -eΛとおいて変換(1),(2)を同時に行うと…

逆の論理も真からスタートしてU(1)変換に対して不変という要請を課すと、ゲージ場Aμ(x)を導入して、が必要になる。言い換えると相互作用の形が決まる。

U(1)群による変換：QEDSU(2)群：弱い相互作用SU(3)群：強い相互作用

3

!Le = !!Lint ! !L = 0

QEDラグランジアンがU(1)変換に対して不変になった

Le

Lint

ゲージ原理：ゲージ対称性を課すことによって相互作用の形が規定される

結合の強さは各群で共通

Covariant Derivative

位相変換すると　　の項から　　が生成されてしまうゲージ場はこれをキャンセルするために必要（＝変換性も決まってしまう）になる

　　　　　　　を導入すると

　がと同じ変換をする

4

なので

!µ" !µ!

Dµ = !µ + ieAµ

(Dµ!)! = e"ie!(x)(Dµ!)

!Dµ!

!! = e"ie!(x)!, Aµ(x)! Aµ(x) + "µ!(x)

ゲージボソンの質量項

ゲージボソンの質量項を手でいれると

　となりゲージ対称性を破ってしまう実際に光子とグルーオンは質量ゼロmW ≠ 0, mZ ≠ 0 ⇐ ゲージ対称性が破れている　　（かのように見える）なんらかのカラクリが必要（＝ヒッグス機構。　後で説明）

フェルミオンの質量も手で入れると上手くいかない

5

m2AµAµ ! m2(Aµ + !µ!)(Aµ + !µ!) "= m2AµAµ

SU(2)対称性

陽子と中性子からなるアイソスピン二重項を真似て

をgeneratorとするSU(2)変換を考える

QEDと同様に、ゲージ変換でラグランジアンが不変になることを要求すると、ゲージ場Aμ(x)が必要ただしQEDと違ってゲージ場が３つ

を満たすcovariant derivative Dμを探せばよい。すると…

6

! !!

ud

"L = !i !"!

Ti !!i

2! ! !! = U(x)! U(x) = e!ig!i(x)Ti

!!!"Aµ(x) = (A1

µ, A2µ, A3

µ)

L! = !!i(!D!)! = !U†(x)iU(x) !D! = !i !D! = L

(Dµ!)! = U(x)Dµ!

SU(2)変換の特徴

　　　　　　　　　　　　　　　を試す

整理すると

同様にゲージ場のkinematic termについてゲージ不変性を要求

を定義すれば

7

Dµ ! !µ + ig"#Aµ ·"#T ! !µ + ig !Aµ

9.2. NON-ABELIAN GAUGE SYMMETRY 485

To make sure that we know what we are dealing with, let us write down Dµ! explicitly:

Dµ! = ("µ + igAµ)!

ud

"=

!"µu"µd

"+ i

g

2

#A3

µu +!

2A+µ d!

2A!µ u " A3

µd

$

. (9.85)

It is straightforward to find the transformation of #Aµ that realizes (Dµ!)" =U(x)Dµ!. Substituting Dµ = "µ " igAµ in this,

(Dµ!)" = U(x)Dµ!

# ("µ + igA"µ)(U!) = U("µ + igAµ)!

#%("µU)! + !

!!"""U("µ!)

&+ igA"

µU! = !!!"""U("µ!) + igUAµ!

# ("µU + igA"µU)! = igUAµ! , (9.86)

which should hold for any !. Thus,

"µU + igA"µU = igUAµ , (9.87)

or right-multiplying by U †

A"µ = UAµU

† +i

g("µU)U † . (9.88)

At this point, the transformation is quite obscure. To see it more clearly, let usexamine a small transformation. Noting that U and U † can be written as

U(x) = e!ig! ($ $ $iTi) # U †(x) = eig!†= eig! , (9.89)

and using eBAe!B = A + [B, A] + · · ·, the transformation (9.88) becomes

A"µ = e!ig!Aµe

ig!

' () *Aµ " ig[$, Aµ]

+i

g("µe

!ig!)' () *

("ig"µ$)e!ig!

eig! = Aµ " ig[$, Aµ] + "µ$

# %Aµ = "µ$ " ig[$, Aµ] . (9.90)

Or, writing out the T -matrices and using [Ti, Tj] = i&ijkTk,

%AkµTk = ("µ$k)Tk " ig [$iTi, A

jµTj]

' () *$iA

jµ [Ti, Tj]' () *i&ijkTk

= ("µ$k + g&ijk$iAjµ)Tk . (9.91)


To make sure that we know what we are dealing with, let us write down Dµ! explicitly:

Dµ! = ("µ + igAµ)!

ud

"=

!"µu"µd

"+ i

g

2

#A3

µu +!

2A+µ d!

2A!µ u " A3

µd

$

. (9.85)

It is straightforward to find the transformation of #Aµ that realizes (Dµ!)" =U(x)Dµ!. Substituting Dµ = "µ " igAµ in this,

(Dµ!)" = U(x)Dµ!

# ("µ + igA"µ)(U!) = U("µ + igAµ)!

#%("µU)! + !

!!"""U("µ!)

&+ igA"

µU! = !!!"""U("µ!) + igUAµ!

# ("µU + igA"µU)! = igUAµ! , (9.86)

which should hold for any !. Thus,

"µU + igA"µU = igUAµ , (9.87)

or right-multiplying by U †

A"µ = UAµU

† +i

g("µU)U † . (9.88)

At this point, the transformation is quite obscure. To see it more clearly, let usexamine a small transformation. Noting that U and U † can be written as

U(x) = e!ig! ($ $ $iTi) # U †(x) = eig!†= eig! , (9.89)

and using eBAe!B = A + [B, A] + · · ·, the transformation (9.88) becomes

A"µ = e!ig!Aµe

ig!

' () *Aµ " ig[$, Aµ]

+i

g("µe

!ig!)' () *

("ig"µ$)e!ig!

eig! = Aµ " ig[$, Aµ] + "µ$

# %Aµ = "µ$ " ig[$, Aµ] . (9.90)

Or, writing out the T -matrices and using [Ti, Tj] = i&ijkTk,

%AkµTk = ("µ$k)Tk " ig [$iTi, A

jµTj]

' () *$iA

jµ [Ti, Tj]' () *i&ijkTk

= ("µ$k + g&ijk$iAjµ)Tk . (9.91)

!Akµ = "µ#k + g$ijk#iA

jµ

⇐非可換な(non-abelian)SU(2)変換の特徴

F kµ! ! !!Ak

µ " !µAk! + g"ijkAi

µAj!

L = !i !D! " 14F i

µ!F iµ! がゲージ不変になる

SU(2)対称性における相互作用

の各項を書き下す

8

A±µ =1!2(A1

µ "A2µ)

L = !i !D! " 14F i

µ!F iµ!

488 CHAPTER 9. THE STANDARD MODEL

To summarize what we have accomplished, the Lagrangian is now

L = !iD/ ! ! 14F

iµ!F

iµ!

Dµ " "µ + igAiµTi

F kµ! " "!Ak

µ ! "µAk! + g#ijkAi

µAj!

! "!

ud

", Ti "

$i

2

, (9.103)

where Ai(x) (i = 1, 2, 3) are real vector fields. Then, this Lagrangian is invariantunder the local SU(2) transformation given by

!! = U(x)! , U(x) = e"ig"i(x)Ti

%Akµ = "µ&k + g#ijk&iAj

µ (for small &i), (9.104)

where &i(x) (i = 1, 2, 3) are arbitrary di!erentiable real functions of x = (t, 'x). This isa remarkable result: it means that we can mix up u and d fields and do so di!erentlyat di!erent space-time, yet the law of physics remains invariant as long as the vectorfields simultaneously transform in a specific way. In the process, we have also specifiedthe interactions between the vector fields and the fermions as well as those amongthe vector fields.

The interaction terms between the fermions and the vectors are found in !iD/ !.Using the explicit form (9.85), we have

!iD/ ! = (u, d)i

#!"/u"/d

"+ i

g

2

$A/3u +

#2A/+d#

2A/"u ! A/3d

%&

= i(u"/u + d"/d) ! g

2

'#2(uA/ +d + dA/ "u) + uA/ 3u ! dA/ 3d

(

) *+ ,g#2

'A+

µ (u(µd) + A"µ (d(µu)

(+

g

2

'A3

µ(u(µu) ! A3µ(d(µd)

(

. (9.105)

The first term is the free field kinetic terms for u and d fields and the rest givesthe couplings of A±,3 to the fermions. The term A"

µ (d(µu) is simply the complexconjugate of A+

µ (u(µd) and we see that the ‘charged’ spin-1 particles A± couple tou and d just as the W± bosons couple to the actual u and d quarks exept that thequark-W coupling is V ! A instead of V . There are also interactions between the‘neutral’ spin-1 particle A3 and the fermions. These may be graphically represented


as

A

u

d

+ A

u,d

3

. (9.106)

On the other hand, the self-interactions of gauge bosons are embedded in !F iµ!F

iµ!/4:

!1

4F k

µ!Fkµ! = !1

4(!!A

kµ ! !µA

k! + g"ijkA

iµA

j!)(!

!Akµ ! !µAk! + g"lmkAlµAm!)

= !1

4

!(!!A

kµ ! !µA

k!)(!

!Akµ ! !µAk!)

+2g"ijkAiµA

j!(!

!Akµ ! !µAk!) + 2g2"ijk"lmkAiµA

j!A

lµAm!". (9.107)

The first terms is the standard kinetic terms of free spin-1 fields. The second and thethird terms are the self-interactions among the spin-1 fields:

. (9.108)

These self-couplings are required by the non-Abelian gauge invariance. When thegauge group is Abelian, then the structure constants i"ijk are zero and thus there isno self-interaction of gauge bosons. Thus, it is of great importance to check if suchself-couplings of gauge bosons, which hold also for the standard model, are indeedoperating in nature. One of the most direct signature is the interaction

e+e! " Z0 " W+W!

+

e!

et

Z0

W!

W+

. (9.109)

Such interaction was observed in 1997 at the e+e! collider at CERN.Well, we have jumped ahead a little too far. The gauge bosons we have introduced

are massless since there is no mass term, m2AiµA

iµ in the gauge invariant Lagrangian


as

A

u

d

+ A

u,d

3

. (9.106)


iµ!/4:

!1

4F k

µ!Fkµ! = !1

4(!!A

kµ ! !µA

k! + g"ijkA

iµA

j!)(!


= !1

4

!(!!A

kµ ! !µA

k!)(!

!Akµ ! !µAk!)

+2g"ijkAiµA

j!(!


j!A

lµAm!". (9.107)


. (9.108)


e+e! " Z0 " W+W!

+

e!

et

Z0

W!

W+

. (9.109)





as

A

u

d

+ A

u,d

3

. (9.106)


iµ!/4:

!1

4F k

µ!Fkµ! = !1

4(!!A

kµ ! !µA

k! + g"ijkA

iµA

j!)(!


= !1

4

!(!!A

kµ ! !µA

k!)(!

!Akµ ! !µAk!)

+2g"ijkAiµA

j!(!


j!A

lµAm!". (9.107)


. (9.108)


e+e! " Z0 " W+W!

+

e!

et

Z0

W!

W+

. (9.109)




SU(2)の実験的検証

ゲージボソンの自己結合の存在 ⇔ U(1)対称性からはゲージボソンの自己結合は生じない

LEPの実験結果によって　　　　　　　　　　　　　　　Z(γ)WW結合の存在を　　　　　　　　　　　　　　　確認gauge cancellation: それ　　　　　　　　　　　　ぞれのダイアグラムで相殺　　　　　　　　　　　することによって無限大の　　　　　　　　　　　発散を打ち消す

9

Theoretical Framework 1.4. W-pair production at LEP

e!

e+

!e

W!

W+

e!

e+

"

W!

W+

e!

e+

Z

W!

W+

Figure 1.1: The three Feynman diagrams, referred to as CC03, which contribute at treelevel to the process e+e!!W+W!. Two diagrams contain a triple gauge-boson vertex ofthe type VWW, indicated by the shaded circles.

contribute at tree level to the process e+e!!W+W!, are referred to as CC03 diagrams.They are shown in Fig 1.1. The matrix element for W-pair production at tree level isthe sum of the matrix elements for these three diagrams separately. Actually, a fourthdiagram exists at tree level in the SM where a Higgs boson is exchanged through the s-channel, but its amplitude is proportional to the electron mass and can thus be neglected.However, at very high energies this diagram needs to be taken into account to ensure aproper behaviour of the cross section.

1.4.1 Helicity Amplitudes

To study the e!ect of anomalous couplings on the W-pair production process, it is instruc-tive to express the matrix elements in terms of the helicity states of the two W bosons,M(#, $, $"). The helicities of the W! and W+ are given by $ and $", incoming e! and e+

helicities are #/2 and "#/2, with the assumption that the electrons are massless.It is convenient to define reduced matrix elements by extracting some common factors:

M(#, $, $"; ") =#

2e2#M!,","!(")dJ0!,!"("). (1.23)

The angle " is the production angle of the W! with respect to the incoming e!. Theleading angular dependence is given in terms of the d-functions dJ0

!,!" [79], where J0 =max(|#|, |#$|) gives the lowest angular momentum contributing to a given helicity com-bination. Two out of the nine possible helicity combinations give J0 = 2, with both W’soppositely, transversely polarised (±,$) thus |#$| = 2. The other seven possible helicityconfigurations all have J0 = 1. The explicit form of the d functions for all possible helicitycombinations is given in the last column of Table 1.2.

The reduced matrix elements are not partial wave amplitudes since they can still havea " dependence due to partial waves with J > J0. The two s-channel diagrams onlycontribute to the seven helicity contributions that have J0 = 1, since angular momentumconservation in the decay of a spin-1 particle dictates that J = 1. The t-channel diagramon the other hand, can form all nine possible helicity combinations and contributions from

15

0

10

20

30

160 180 200!s (GeV)

"W

W (p

b)YFSWW/RacoonWWno ZWW vertex (Gentle)only #e exchange (Gentle)

LEPPRELIMINARY

17/02/2005

自発的対称性の破れスピン０のスカラー場Φ(x)を導入

真空が Φ1=v, Φ2=0 ( ) に固定（＝自発的対称性の破れ）されると

10

9.3. SPONTANEOUS BREAKDOWN OF SYMMETRIES 493

given by the curvature of the groove at the bottom. The potential V is no longersymmetric with respect to the vacuum. This is a simple example of spontaneousbreakdown of symmetry.

As the universe cools down it may so happen that one part of the string settlesdown in one groove and other part in another groove. Then, there will be a transitionpoint from one groove to the other as shown in Figure 9.1(b). This is an example oftopological solitons and called kink. A kink will move around carrying some energy,and will not be annihilated unless it meets another kink of the opposite sign oftransition. This particular soliton is possible only for one-dimensional field; there are,however, varieties of topological solitons that are predicted to exist in our universesuch as monopoles and instantons. These topological objects should exist if we takethe posturate of spontaneous symmetry breaking seriously.

9.3.2 Breakdown of global U(1) symmetry

Consider a charged spin-0 field !(x) with some potential V which depends only on|!|2 = !!!:

L = "µ!!"µ! ! V (|!|2) , (9.121)

with

! " 1#2(!1 + i!2) (!1,2 : real) , (9.122)

or, in terms of the two real fields !1 and !2,

L =1

2("µ!1"

µ!1 + "µ!2"µ!2) ! V

!!21 + !2

2

2

". (9.123)

This Lagrangian is invariant under the global phase rotation !" = ei!!; namely, ithas a global U(1) symmetry. In terms of !1,2, the Lagrangian and in particular thepotential V are invariant under rotation in the !1-!2 plane. Now, suppose that thepotential has a minimum away from ! = 0 for whatever the reason. Since V isrotationally invariant, it would have a shape looking like the bottom of a wine bottleas shown in Figure 9.2 with the minimum of the potential forming a circle

!21 + !2

2 = a2 (a > 0 : real) . (9.124)

The vacuum is by definition the solution with the lowest energy. The Euler-Lagrangeequations for !1,2 using the Lagrangian (9.123) are

"µ"L

"("µ!i)=

"L"!i

$ "2!i = !"V

"!i(i = 1, 2) , (9.125)

and the total energy is given by (with #i = !i)

H "#

d3x($

i

#i!i ! L) =#

d3x!1

2

$

i

[!2i + ($%!i)

2] + V". (9.126)

9.3. SPONTANEOUS BREAKDOWN OF SYMMETRIES 493

given by the curvature of the groove at the bottom. The potential V is no longersymmetric with respect to the vacuum. This is a simple example of spontaneousbreakdown of symmetry.

As the universe cools down it may so happen that one part of the string settlesdown in one groove and other part in another groove. Then, there will be a transitionpoint from one groove to the other as shown in Figure 9.1(b). This is an example oftopological solitons and called kink. A kink will move around carrying some energy,and will not be annihilated unless it meets another kink of the opposite sign oftransition. This particular soliton is possible only for one-dimensional field; there are,however, varieties of topological solitons that are predicted to exist in our universesuch as monopoles and instantons. These topological objects should exist if we takethe posturate of spontaneous symmetry breaking seriously.

9.3.2 Breakdown of global U(1) symmetry

Consider a charged spin-0 field !(x) with some potential V which depends only on|!|2 = !!!:

L = "µ!!"µ! ! V (|!|2) , (9.121)

with

! " 1#2(!1 + i!2) (!1,2 : real) , (9.122)

or, in terms of the two real fields !1 and !2,

L =1

2("µ!1"

µ!1 + "µ!2"µ!2) ! V

!!21 + !2

2

2

". (9.123)

This Lagrangian is invariant under the global phase rotation !" = ei!!; namely, ithas a global U(1) symmetry. In terms of !1,2, the Lagrangian and in particular thepotential V are invariant under rotation in the !1-!2 plane. Now, suppose that thepotential has a minimum away from ! = 0 for whatever the reason. Since V isrotationally invariant, it would have a shape looking like the bottom of a wine bottleas shown in Figure 9.2 with the minimum of the potential forming a circle

!21 + !2

2 = a2 (a > 0 : real) . (9.124)

The vacuum is by definition the solution with the lowest energy. The Euler-Lagrangeequations for !1,2 using the Lagrangian (9.123) are

"µ"L

"("µ!i)=

"L"!i

$ "2!i = !"V

"!i(i = 1, 2) , (9.125)

and the total energy is given by (with #i = !i)

H "#

d3x($

i

#i!i ! L) =#

d3x!1

2

$

i

[!2i + ($%!i)

2] + V". (9.126)

V = µ2!!! + "|!!!|2

宇宙初期（高温） μ2>0

現在　　（低温） μ2<0相転移

v =!!µ2/!

! L =12(!µ")2 +

12(!µ#)2 + µ2#2 + (const) + (higher term)

!(x) =!

1/2{v + "(x) + i#(x)}

η

ξ

ゴールドストンボソンスカラー場の質量項

ヒッグス機構　例としてU(1)対称性の破れ

次のラグランジアンからスタート

ゲージ変換は　　　　　　　　　　　　とparametrizeして

Φ’(x)が実数となるようなϑ(x)を選んでいる ⇐ ゲージ変換の自由度（ユニタリーゲージ）このゲージで自発的対称性の破れが起きると

11


away’. This degree of freedom, as we will see, is not lost, but it will give mass to thegauge boson increasing the number of degree of freedom of the gauge boson from two(helicity ±1) to three (helicity 0,±1).

We start from the charged spin-0 Lagrangian (9.121), and make it gauge invariant.Thus, we introduce a real spin-1 field Aµ, and replace !µ by Dµ:

L = (Dµ")!(Dµ") ! V (|"|2) ! 1

4Fµ!F

µ!

Dµ " !µ + ieAµ , Fµ! " !!Aµ ! !µA!

, (9.133)

which is invariant under the gauge transformation given by

""(x) = e#ie!(x)" , A"µ(x) = Aµ(x) + !µ!(x) , (9.134)

where !(x) is an arbitrary real function of x. Namely, if "(x) and Aµ(x) are solutionsof the equations of motion, then the gauge-transformed fields ""(x) and A"(x) givenabove for any !(x) are also solutions representing the same physical states and inter-action. The is the standard Lagrangian where a charged spin-0 particle is interactingwith photon.

Let us parametrize the complex field "(x) as

"(x) " 1#2("1 + i"2) " #(x)ei"(x) , (#(x) > 0, $(x) : real) . (9.135)

This form is motivated by the ease of gauging away one of the two degrees of free-dom. In fact, for a given physical state represented by "(x) we can use the gaugetransformation

""(x) = e#i"(x)"(x) (9.136)

to make $(x) = 0 for all x. The resulting wave function ""(x) is real, and stillrepresents the same physical state as long as Aµ(x) is transformed at the same timeaccording to (9.134). The condition $(x) = 0 is a gauge condition just like the gaugeconditions for QED listed in (6.46) where the conditions were imposed on the vectorfield Aµ(x). Here the condition is imposed on the scalar field rather than on thevector field. The particular gauge we have chosen is called the unitary gauge, or aU -gauge, which has the advantage that the remaining fields are physical. Preciselyspeaking, a unitary gauge is the gauge in which all remaining scalar fields correspondto oscillations perpendicular to the oscillations of massless Goldstone bosons as shownin Figure 9.2. In the following, we will stay in the unitary gauge, and thus we have"(x) real or equivalently "2 = 0.

Now assume that the potential V has the same wine-bottle form of (9.128). With"2 = 0, we have

V = b("21 ! a2)2 . (9.137)




L = (Dµ")!(Dµ") ! V (|"|2) ! 1

4Fµ!F

µ!

Dµ " !µ + ieAµ , Fµ! " !!Aµ ! !µA!

, (9.133)


""(x) = e#ie!(x)" , A"µ(x) = Aµ(x) + !µ!(x) , (9.134)



"(x) " 1#2("1 + i"2) " #(x)ei"(x) , (#(x) > 0, $(x) : real) . (9.135)


""(x) = e#i"(x)"(x) (9.136)



V = b("21 ! a2)2 . (9.137)

というゲージ変換を考える




L = (Dµ")!(Dµ") ! V (|"|2) ! 1

4Fµ!F

µ!

Dµ " !µ + ieAµ , Fµ! " !!Aµ ! !µA!

, (9.133)


""(x) = e#ie!(x)" , A"µ(x) = Aµ(x) + !µ!(x) , (9.134)



"(x) " 1#2("1 + i"2) " #(x)ei"(x) , (#(x) > 0, $(x) : real) . (9.135)


""(x) = e#i"(x)"(x) (9.136)



V = b("21 ! a2)2 . (9.137)




L = (Dµ")!(Dµ") ! V (|"|2) ! 1

4Fµ!F

µ!

Dµ " !µ + ieAµ , Fµ! " !!Aµ ! !µA!

, (9.133)


""(x) = e#ie!(x)" , A"µ(x) = Aµ(x) + !µ!(x) , (9.134)



"(x) " 1#2("1 + i"2) " #(x)ei"(x) , (#(x) > 0, $(x) : real) . (9.135)


""(x) = e#i"(x)"(x) (9.136)



V = b("21 ! a2)2 . (9.137)




L = (Dµ")!(Dµ") ! V (|"|2) ! 1

4Fµ!F

µ!

Dµ " !µ + ieAµ , Fµ! " !!Aµ ! !µA!

, (9.133)


""(x) = e#ie!(x)" , A"µ(x) = Aµ(x) + !µ!(x) , (9.134)



"(x) " 1#2("1 + i"2) " #(x)ei"(x) , (#(x) > 0, $(x) : real) . (9.135)


""(x) = e#i"(x)"(x) (9.136)



V = b("21 ! a2)2 . (9.137)

!2 = 0

L =12(!µ"!µ"! µ2"2)! 1

4Fµ!Fµ! +

e2v2

2AµAµ +

e2

2AµAµ(2v" + "2)

!1 =!

1/2{v + "(x)}

ゲージボソンの質量項

ヒッグス機構　SU(2)の場合

ラグランジアン

ヒッグス場を二重項にする

自発的対称性の破れ

ゲージボソンの質量項の由来

12


freedom (two helicities). When the energy scale is small enough, or the universe coolsdown, the scalar field would settle around a minimum of the potential and oscillateabout the vacuum. Then, one of the two degrees of the freedom of the scalar, namelywhat would have been the massless Goldstone boson in the case of global symmetrybreaking, can be gauged away and the gauge boson acquires a mass resulting in threehelicities. Thus, total number of degree of freedom does not change when the localsymmetry is broken. This mechanism of breakdown of a local symmetry leading to amassive gauge boson is called the Higgs mechanism and the scalar field that breaksthe symmetry is often called the Higgs field.

9.3.4 Breakdown of local SU(2) symmetry

As before, we will use a spin-0 field to break the symmetry, but one complex field willnot do. In order to break the symmetry, it should be some non-trivial representationof SU(2); namely, the vacuum state should change under a SU(2) transformation:

U!V AC != !V AC (U : a SU(2) transformation) . (9.143)

Let us take it to be a SU(2) doublet of two complex fields !+ and !0:

! "!

!+

!0

"=

# 1!2(!+1 + i!+2)

1!2(!01 + i!02)

$

, (9.144)

where !+i, !0i (i = 1, 2) are real. The subscripts ‘+’ and ‘0’ are used simply to distin-guish the two scalar fields at this point. Then, the local SU(2)-invariant Lagrangianis given by (9.111):

L = (Dµ!)†(Dµ!) # V (!†!) # 1

4F i

µ!Fiµ!

Dµ " "µ + igAiµTi , F k

µ! " "!Akµ # "µAk

! + g#ijkAiµA

j!

(9.145)

where we have added the potential V (!†!), which is also gauge invariant because

(!†!)" = !†U †(x)U(x)! = !†! . (9.146)

We assume that the potential has the form

V = b(2!†! # a2)2 (9.147)

whose minimum is at

2!†! = !2+1 + !2

+2 + !021 + !0

22 = a2 . (9.148)

!(x) =!

01!2v

"+

!0

1!2"(x)

"







! "!

!+

!0

"=

# 1!2(!+1 + i!+2)

1!2(!01 + i!02)

$

, (9.144)


L = (Dµ!)†(Dµ!) # V (!†!) # 1

4F i

µ!Fiµ!



! + g#ijkAiµA

j!

(9.145)


(!†!)" = !†U †(x)U(x)! = !†! . (9.146)


V = b(2!†! # a2)2 (9.147)

whose minimum is at

2!†! = !2+1 + !2

+2 + !021 + !0

22 = a2 . (9.148)







! "!

!+

!0

"=

# 1!2(!+1 + i!+2)

1!2(!01 + i!02)

$

, (9.144)


L = (Dµ!)†(Dµ!) # V (!†!) # 1

4F i

µ!Fiµ!



! + g#ijkAiµA

j!

(9.145)


(!†!)" = !†U †(x)U(x)! = !†! . (9.146)


V = b(2!†! # a2)2 (9.147)

whose minimum is at

2!†! = !2+1 + !2

+2 + !021 + !0

22 = a2 . (9.148)

L =12(!µ"!µ"! µ2"2)! 1

4F i

µ!F iµ! +g2v2

8Ai

µAiµ +g2

8(2v" + "2)Ai

µAiµ

!!" # v$2g2

4!†!Ai

µAiµ ! g2v2

8Ai

µAiµ

GWS模型

カイラルフェルミオン

左巻きと右巻き粒子は別粒子として振る舞う

質量は、左巻きと右巻きとの結合の強さ（カイラリティの破れ）を表す

14

9.4. ELECTRO-WEAK UNIFICATION AND THE STANDARD MODEL 503

the Lagrangian will still be gauge invariant, but the lepton and neutrino will havethe same mass. We might introduce Yukawa-type couplings between the scalar andthe fermions, but this has to be done in a SU(2)-invariant way, and as long as !and " form a SU(2) doublet, one cannot write down a coupling that gives di!erentmasses to them. The solution to the fermion masses as well as the V ! A coupling,as we will see, is to form a SU(2) doublet by the left-handed parts of ! and " whilekeeping their right-handed parts as SU(2) singlets. Additional problems are thatW±(80.2 GeV) and Z0(91.2 GeV) actually have di!erent masses and that we wouldlike the electromagnetic interaction to be included in the theory. We will see thatthese two problems are solved by introducing a local U(1) symmetry that is di!erentfrom the U(1) symmetry of of QED, and let the corresponding gauge boson ‘mix’with the neutral component of the three gauge bosons of SU(2). This then leads tothe ‘prediction’ that the ratio of the masses of W and Z is related to the ratio of theelectric charge e to the weak coupling constant g.

First, we will see how V ! A interaction can be introduced and how " and ! canhave di!erent masses. A fermion field # (an ordinary 4-component spinor, not adoublet) can be written as a sum of the right-handed and left-handed parts as

# = #R + #L , #(RL ) " P(R

L )# , P(RL ) "

1 ± $5

2. (9.168)

As we have seen in (3.353), the right-handed and left-handed components decouplein the massless limit, or more precisely, they satisfy separate equations of motion. Interms of Lagrangian, the massless fermion Lagrangian can be separated into right-handed and left-handed parts as

#i%/# = (#R + #L)i%/(#R + #L)

= #Ri%/#R + #Li%/#L + #Ri%/#L + #Li%/#R! "# $0, as shown below

= #Ri%/#R + #Li%/#L , (9.169)

where we have used

#Ri%/#L = PR#! "# $#PL

i$µ%µPL# = #i$µ%

µ PRPL! "# $0

# = 0 , (9.170)

and similarly #Li%/#R = 0. On the other hand, the mass term involves both #R and#L:

m## = m(#R + #L)(#R + #L) = m(#R#L + #L#R) = m#R#L + c.c. , (9.171)

where we have used#R#R = PR#! "# $

#PL

PR# = # PLPR! "# $0

# = 0 , (9.172)




# = #R + #L , #(RL ) " P(R

L )# , P(RL ) "

1 ± $5

2. (9.168)


#i%/# = (#R + #L)i%/(#R + #L)


= #Ri%/#R + #Li%/#L , (9.169)

where we have used

#Ri%/#L = PR#! "# $#PL

i$µ%µPL# = #i$µ%

µ PRPL! "# $0

# = 0 , (9.170)




#PL

PR# = # PLPR! "# $0

# = 0 , (9.172)

なので

!i !"! = (!R + !L)i(!R + !L)= !Ri !"!R + !Li !"!L + !Ri !"!L + !Li !"!R

= !Ri !"!R + !Li !"!L




# = #R + #L , #(RL ) " P(R

L )# , P(RL ) "

1 ± $5

2. (9.168)


#i%/# = (#R + #L)i%/(#R + #L)


= #Ri%/#R + #Li%/#L , (9.169)

where we have used

#Ri%/#L = PR#! "# $#PL

i$µ%µPL# = #i$µ%

µ PRPL! "# $0

# = 0 , (9.170)




#PL

PR# = # PLPR! "# $0

# = 0 , (9.172)




# = #R + #L , #(RL ) " P(R

L )# , P(RL ) "

1 ± $5

2. (9.168)


#i%/# = (#R + #L)i%/(#R + #L)


= #Ri%/#R + #Li%/#L , (9.169)

where we have used

#Ri%/#L = PR#! "# $#PL

i$µ%µPL# = #i$µ%

µ PRPL! "# $0

# = 0 , (9.170)




#PL

PR# = # PLPR! "# $0

# = 0 , (9.172)

V-A結合

左巻きをSU(2)doublet, 右巻きをsinglet にする

SU(2)不変なラグラジアンは

15


and similarly !L!L = 0.In the massless limit, we treat the right-handed and left-handed parts as if they

are di!erent particles, and assume that the left-handed parts of " and # transformunder SU(2) as a doublet while the right-handed parts are invariant under SU(2)(i.e. singlets):

Ldef!

!"L

#L

", "R , #R , (9.173)

L! = ei!iTiL#Ti =

$i

2

$, " !

R = "R , #!R = #R (Ti = 0) . (9.174)

Then, using (9.169), the free Lagrangian of massless " and # can be written as

L = "i%/" + #i%/#

= ("Ri%/"R + "Li%/"L) + (#Ri%/#R + #Li%/#L)

= Li%/L + "Ri%/"R + #Ri%/#R , (9.175)

which is invariant under the (global) SU(2) transformation given by (9.174). Whenthis Lagrangian is made locally SU(2) invariant (or ‘gauged’), the Lagrangian be-comes

L = LiD/ L + "RiD/ "R + #RiD/ #R " 1

4F i

µ"Fiµ" (9.176)

where Dµ = %µ + igAiµTi with Ti = $i/2 for the doublet L and Ti = 0 for "R and

#R. Since Ti = 0 for right-handed fermions, there is no term that couples them to thegauge bosons. The coupling of the gauge bosons to the fermions is then found in inLiD/ L which is given by (9.105) with the simple replacement (u, d) # ("L, #L):

LiD/ L = i("L%/"L + #L%/#L)

+g$2

%A+

µ ("L&µ#L) + A"µ (#L&µ"L)

&+

g

2

%A3

µ("L&µ"L) " A3µ(#L&µ#L)

&. (9.177)

The desired V " A coupling is found in the second term:

g$2A+

µ ("L&µ#L) =g$2A+

µ (PL"'()*"PR

&µPL#) =g$2A+

µ (" PR&µPL' () *&µPL

#)

=g

2$

2A+

µ ["&µ(1 " &5)#] , (9.178)

and A"µ (#L&µ"L) is simply the conjugate of A+

µ ("L&µ#L). This has exactly the sameform as the W -e" coupling (5.240) we have been using and it will survive withoutmodification through the spontaneous symmetry breaking which will give mass to thegauge bosons.




Ldef!

!"L

#L

", "R , #R , (9.173)

L! = ei!iTiL#Ti =

$i

2

$, " !

R = "R , #!R = #R (Ti = 0) . (9.174)


L = "i%/" + #i%/#

= ("Ri%/"R + "Li%/"L) + (#Ri%/#R + #Li%/#L)

= Li%/L + "Ri%/"R + #Ri%/#R , (9.175)


L = LiD/ L + "RiD/ "R + #RiD/ #R " 1

4F i

µ"Fiµ" (9.176)



LiD/ L = i("L%/"L + #L%/#L)

+g$2

%A+

µ ("L&µ#L) + A"µ (#L&µ"L)

&+

g

2

%A3

µ("L&µ"L) " A3µ(#L&µ#L)

&. (9.177)


g$2A+

µ ("L&µ#L) =g$2A+

µ (PL"'()*"PR

&µPL#) =g$2A+


#)

=g

2$

2A+

µ ["&µ(1 " &5)#] , (9.178)






Ldef!

!"L

#L

", "R , #R , (9.173)

L! = ei!iTiL#Ti =

$i

2

$, " !

R = "R , #!R = #R (Ti = 0) . (9.174)


L = "i%/" + #i%/#

= ("Ri%/"R + "Li%/"L) + (#Ri%/#R + #Li%/#L)

= Li%/L + "Ri%/"R + #Ri%/#R , (9.175)


L = LiD/ L + "RiD/ "R + #RiD/ #R " 1

4F i

µ"Fiµ" (9.176)



LiD/ L = i("L%/"L + #L%/#L)

+g$2

%A+

µ ("L&µ#L) + A"µ (#L&µ"L)

&+

g

2

%A3

µ("L&µ"L) " A3µ(#L&µ#L)

&. (9.177)


g$2A+

µ ("L&µ#L) =g$2A+

µ (PL"'()*"PR

&µPL#) =g$2A+


#)

=g

2$

2A+

µ ["&µ(1 " &5)#] , (9.178)






Ldef!

!"L

#L

", "R , #R , (9.173)

L! = ei!iTiL#Ti =

$i

2

$, " !

R = "R , #!R = #R (Ti = 0) . (9.174)


L = "i%/" + #i%/#

= ("Ri%/"R + "Li%/"L) + (#Ri%/#R + #Li%/#L)

= Li%/L + "Ri%/"R + #Ri%/#R , (9.175)


L = LiD/ L + "RiD/ "R + #RiD/ #R " 1

4F i

µ"Fiµ" (9.176)



LiD/ L = i("L%/"L + #L%/#L)

+g$2

%A+

µ ("L&µ#L) + A"µ (#L&µ"L)

&+

g

2

%A3

µ("L&µ"L) " A3µ(#L&µ#L)

&. (9.177)


g$2A+

µ ("L&µ#L) =g$2A+

µ (PL"'()*"PR

&µPL#) =g$2A+


#)

=g

2$

2A+

µ ["&µ(1 " &5)#] , (9.178)



SU(2) singlet なので　ゲージボソンと結合しない




Ldef!

!"L

#L

", "R , #R , (9.173)

L! = ei!iTiL#Ti =

$i

2

$, " !

R = "R , #!R = #R (Ti = 0) . (9.174)


L = "i%/" + #i%/#

= ("Ri%/"R + "Li%/"L) + (#Ri%/#R + #Li%/#L)

= Li%/L + "Ri%/"R + #Ri%/#R , (9.175)


L = LiD/ L + "RiD/ "R + #RiD/ #R " 1

4F i

µ"Fiµ" (9.176)



LiD/ L = i("L%/"L + #L%/#L)

+g$2

%A+

µ ("L&µ#L) + A"µ (#L&µ"L)

&+

g

2

%A3

µ("L&µ"L) " A3µ(#L&µ#L)

&. (9.177)


g$2A+

µ ("L&µ#L) =g$2A+

µ (PL"'()*"PR

&µPL#) =g$2A+


#)

=g

2$

2A+

µ ["&µ(1 " &5)#] , (9.178)






Ldef!

!"L

#L

", "R , #R , (9.173)

L! = ei!iTiL#Ti =

$i

2

$, " !

R = "R , #!R = #R (Ti = 0) . (9.174)


L = "i%/" + #i%/#

= ("Ri%/"R + "Li%/"L) + (#Ri%/#R + #Li%/#L)

= Li%/L + "Ri%/"R + #Ri%/#R , (9.175)


L = LiD/ L + "RiD/ "R + #RiD/ #R " 1

4F i

µ"Fiµ" (9.176)



LiD/ L = i("L%/"L + #L%/#L)

+g$2

%A+

µ ("L&µ#L) + A"µ (#L&µ"L)

&+

g

2

%A3

µ("L&µ"L) " A3µ(#L&µ#L)

&. (9.177)


g$2A+

µ ("L&µ#L) =g$2A+

µ (PL"'()*"PR

&µPL#) =g$2A+


#)

=g

2$

2A+

µ ["&µ(1 " &5)#] , (9.178)



フェルミオンの質量　例としてレプトン

湯川結合 hl, hν を導入する（⇐恣意的）

ここでも対称性が自発的に破れたとする

16


Separating the right-handed parts as singlets allows us to assign Yukawa couplingssuch that ! and " acquire di!erent masses as we will see now. We take the SU(2)doublet of scalar fields and its conjugate doublet [see (9.57)]

# !!

#+

#0

"" #c !

!#!

0

##!+

"(9.179)

together with the fermion multiplets as in (9.173) and form Yukawa couplings thatare invariant under SU(2):

h!(L#)"R + h"(L#c)!R + c.c.

= h!(!L, "L)

!#+

#0

""R + h"

(!L, "L)!

#!0

##!+

"!R + c.c.

= h![(!L"R)#+ + ("L"R)#0] + h" [(!L!R)#!0 # ("L!R)#!

+] + c.c. , (9.180)

where h! and h" are two real coupling constants which can be di!erent. We did notinclude (L#)!R and (L#c)"R since later on we will assign electrical charge +1 to #+

and 0 to #0, and such terms will violate the conservation of electrical charge (theelectrical charge of " is #1 and that of ! is zero). We could also form the conjugatedoublet of L and combine it with the two scalar doublets; this, however, results inthe same Lagrangian. Now, when the symmetry is broken by

#+ " 0 , #0 "a$2, (9.181)

the above Yukawa couplings will result in the mass terms of the fermions: using(9.171),

ah!$2[("L"R) + c.c] +

ah"$2

[(!L!R) + c.c] =ah!$

2"" +

ah"$2

!! . (9.182)

Clearly, " and ! can have di!erent masses when h! %= h" which is possible since "R and!R belong to di!erent ‘multiplets’ (actually, singlets) of SU(2) and thus could formseparate interaction terms h!(L#)"R and h"(L#c)!R with di!erent coupling constantsin (9.180).

Including the terms for the scalar field, the Lagrangian at this point is

L = LiD/ L + !RiD/ !R + "RiD/ "R + (Dµ#)†(Dµ#) # V (#†#)

##h!(L#)"R + h"(L#c)!R + c.c.

$# 1

4F i

µ"Fiµ" . (9.183)

The next step is to incorporate the electro-magnetic interaction in such a way that itwill create mass di!erence between W± and Z0. Since in the end we need an unbrokenU(1) gauge symmetry of QED, we will search a U(1) symmetry which is left in this

= hl

!!L, lL

" #"+

"0

$lR + h!

!!L, lL

" #"!

0

!"!+

$!R + c.c.


Separating the right-handed parts as singlets allows us to assign Yukawa couplingssuch that ! and " acquire di!erent masses as we will see now. We take the SU(2)doublet of scalar fields and its conjugate doublet [see (9.57)]

# !!

#+

#0

"" #c !

!#!

0

##!+

"(9.179)

together with the fermion multiplets as in (9.173) and form Yukawa couplings thatare invariant under SU(2):

h!(L#)"R + h"(L#c)!R + c.c.

= h!(!L, "L)

!#+

#0

""R + h"

(!L, "L)!

#!0

##!+

"!R + c.c.

= h![(!L"R)#+ + ("L"R)#0] + h" [(!L!R)#!0 # ("L!R)#!

+] + c.c. , (9.180)

where h! and h" are two real coupling constants which can be di!erent. We did notinclude (L#)!R and (L#c)"R since later on we will assign electrical charge +1 to #+

and 0 to #0, and such terms will violate the conservation of electrical charge (theelectrical charge of " is #1 and that of ! is zero). We could also form the conjugatedoublet of L and combine it with the two scalar doublets; this, however, results inthe same Lagrangian. Now, when the symmetry is broken by

#+ " 0 , #0 "a$2, (9.181)

the above Yukawa couplings will result in the mass terms of the fermions: using(9.171),

ah!$2[("L"R) + c.c] +

ah"$2

[(!L!R) + c.c] =ah!$

2"" +

ah"$2

!! . (9.182)

Clearly, " and ! can have di!erent masses when h! %= h" which is possible since "R and!R belong to di!erent ‘multiplets’ (actually, singlets) of SU(2) and thus could formseparate interaction terms h!(L#)"R and h"(L#c)!R with di!erent coupling constantsin (9.180).

Including the terms for the scalar field, the Lagrangian at this point is

L = LiD/ L + !RiD/ !R + "RiD/ "R + (Dµ#)†(Dµ#) # V (#†#)

##h!(L#)"R + h"(L#c)!R + c.c.

$# 1

4F i

µ"Fiµ" . (9.183)

The next step is to incorporate the electro-magnetic interaction in such a way that itwill create mass di!erence between W± and Z0. Since in the end we need an unbrokenU(1) gauge symmetry of QED, we will search a U(1) symmetry which is left in this

!+ ! 0 , !0 !v"2

vhl!2[(lLlR) + c.c.] +

vh!!2

[(!L!R) + c.c.] =vhl!

2ll +

vh!!2

!!

ml =vhl!

2, m! =

vh!!2

湯川結合定数の測定

もしヒッグス（らしい粒子）が発見されたら湯川結合定数の測定は急務ヒッグス機構が正しいかどうかGWS模型が正しいかどうか

17

ハイパーチャージ

ハイパーチャージを導入

ラグランジアンのまとめ

18


L

(!L , "L)!R "R

#

(#+ , #0)

lepton number N 1 1 1 0

hyper charge Y !12 0 !1 1

2

T3 (+12 ,!1

2) 0 0 (+12 ,!1

2)

Q (0,!1) 0 !1 (+1, 0)

Table 9.1: The assignment of quantum numbers for the SU(2) " U(1) gauge theorythat leads to the standard model.

!R, "R, and # can be written as

U = ei!N!i"Y . (9.188)

For each particle, the electric charge Q can be expressed in terms of Y and T3 as

Q = Y + T3 , (9.189)

as can be easily verified in Table 9.1. Since the average value of T3 for any multipletis zero, the hyper charge Y assigned as above is the average electrical charge of eachSU(2) multiplet. It can be shown that, once the electrical charges of the particles aregiven, the requirement that Q be a linear combination of T3 and Y determines theassignments of Y up to an overall multiplicative constant.

Thus, our gauge group is the direct product of the isospin SU(2) and the hypercharge U(1) corresponding to the phase $. In order to make the Lagrangian invariantunder the local SU(2)"U(1) transformation, we introduce three gauge fields %Wµ forthe three generators %T of SU(2) and one gauge field Bµ for the generator Y of U(1).The Lagrangian is then

L = LiD/ L + !RiD/ !R + "RiD/ "R + (Dµ#)†(Dµ#) ! V (#†#)

!!h#(L#)"R + h$(L#c)!R + c.c.

"! 1

4Fiµ$F

iµ$ ! 14Gµ$Gµ$

, (9.190)

with

L =#

!L

"L

$, # =

##+

#0

$

%Fµ$ = &$%Wµ ! &µ

%W$ + g %Wµ " %W$ , Gµ$ = &$Bµ ! &µB$

. (9.191)


L

(!L , "L)!R "R

#

(#+ , #0)



2

T3 (+12 ,!1

2) 0 0 (+12 ,!1

2)

Q (0,!1) 0 !1 (+1, 0)



U = ei!N!i"Y . (9.188)


Q = Y + T3 , (9.189)




!!h#(L#)"R + h$(L#c)!R + c.c.

"! 1

4Fiµ$F

iµ$ ! 14Gµ$Gµ$

, (9.190)

with

L =#

!L

"L

$, # =

##+

#0

$

%Fµ$ = &$%Wµ ! &µ

%W$ + g %Wµ " %W$ , Gµ$ = &$Bµ ! &µB$

. (9.191)


L

(!L , "L)!R "R

#

(#+ , #0)



2

T3 (+12 ,!1

2) 0 0 (+12 ,!1

2)

Q (0,!1) 0 !1 (+1, 0)



U = ei!N!i"Y . (9.188)


Q = Y + T3 , (9.189)




!!h#(L#)"R + h$(L#c)!R + c.c.

"! 1

4Fiµ$F

iµ$ ! 14Gµ$Gµ$

, (9.190)

with

L =#

!L

"L

$, # =

##+

#0

$

%Fµ$ = &$%Wµ ! &µ

%W$ + g %Wµ " %W$ , Gµ$ = &$Bµ ! &µB$

. (9.191)


L

(!L , "L)!R "R

#

(#+ , #0)



2

T3 (+12 ,!1

2) 0 0 (+12 ,!1

2)

Q (0,!1) 0 !1 (+1, 0)



U = ei!N!i"Y . (9.188)


Q = Y + T3 , (9.189)




!!h#(L#)"R + h$(L#c)!R + c.c.

"! 1

4Fiµ$F

iµ$ ! 14Gµ$Gµ$

, (9.190)

with

L =#

!L

"L

$, # =

##+

#0

$

%Fµ$ = &$%Wµ ! &µ

%W$ + g %Wµ " %W$ , Gµ$ = &$Bµ ! &µB$

. (9.191)

SU(2) ⊗ U(1) 対称

SU(2) ⊗ U(1) 変換として

本当にゲージ不変になっているか一つだけチェック

となるcovariant derivativeを探す ➜ 本当になってるか各自確認してみて下さい一般に

19


The SU(2) ! U(1) gauge transformation is

!! = U(x)! , U(x) = e"ig!"(x)·!T"ig!#(x)Y (! = L, "R, #R, and $)

% &Wµ = 'µ&( + g&( ! &Wµ , %Bµ = 'µ)

. (9.192)

where g and g! are real constants and &T and Y vary appropriately depending onthe field they are acting on; namely, they are the representations of the generatorsof SU(2) ! U(1) in the space of the fields they are attached to. We have alreadyproven the invariance of the gauge field terms F i

µ$Fiµ$ and Gµ$Gµ$ under the gauge

transformation defined as above. The interaction terms are invariant by construction.For example, using the values of Y as defined in Table 9.1,

Y =

(L!

12

$!

12

) #!R"1

=!L (e"ig!"·!T+ig! !

2 ))†e"ig!"·!T"ig! !2 )

" #$ %eig!"·!T e"ig! !

2 e"ig!"·!T e"ig! !2

$&eig!##R = (L$)#R . (9.193)

How should the covariant derivative be defined? What we want is that D!µ(U(x)!) =

U(x)(Dµ!) holds for any representation of SU(2), since it will then assure the invari-ance of the kinetic terms LiD/ L, "RiD/ "R, #RiD/ #R, and (Dµ$)†(Dµ$). The answeris

Dµ = 'µ + ig &Wµ · &T + ig!BµY , (9.194)

as we will demonstrate below. What we would like to prove is that D!µU(x) = U(x)Dµ

holds in any representation. What we know is that such relation holds separately forSU(2) and U(1):

('µ + ig &W !µ · &T )e"ig!"·!T = e"ig!"·!T ('µ + ig &Wµ · &T ) ,

('µ + ig!B!µY )e"ig!#Y = e"ig!#Y ('µ + ig!BµY ) , (9.195)

where the first is identical to the definition in (9.103), and the second is obtainedfrom (9.68) by the replacement e " g!Y and ! " ). Now, since Y is just a numberfor any multiplet of SU(2), it should commute with &T in any representation:

[&T , Y ] = 0 , (9.196)

and[B!

µY, e"ig!"·!T ] = 0 , [ &Wµ · &T , e"ig!#Y ] = 0 . (9.197)

Then, together with (9.195), we have

D!µU(x) =

#$$$$$$$$$$$$$$$$$$('µ + ig &W !

µ · &T" #$ %

&W !µ " &Wµ

+ig! B!µY" #$ %

commute

) e"ig!"·!T e"ig!#Y




% &Wµ = 'µ&( + g&( ! &Wµ , %Bµ = 'µ)

. (9.192)




Y =

(L!

12

$!

12

) #!R"1

=!L (e"ig!"·!T+ig! !

2 ))†e"ig!"·!T"ig! !2 )

" #$ %eig!"·!T e"ig! !

2 e"ig!"·!T e"ig! !2

$&eig!##R = (L$)#R . (9.193)



Dµ = 'µ + ig &Wµ · &T + ig!BµY , (9.194)






[&T , Y ] = 0 , (9.196)

and[B!

µY, e"ig!"·!T ] = 0 , [ &Wµ · &T , e"ig!#Y ] = 0 . (9.197)


D!µU(x) =

#$$$$$$$$$$$$$$$$$$('µ + ig &W !

µ · &T" #$ %

&W !µ " &Wµ

+ig! B!µY" #$ %

commute


D!µU(x) = U(x)Dµ


= e!ig!"·!T!""""""""""""""""""(!µ + ig"B"

µY! "# $B"

µ # Bµ

+ig "Wµ · "T! "# $

commute

) e!ig!#Y

= e!ig!"·!T e!ig!#Y (!µ + ig "Wµ · "T + ig"BµY )

= U(x)Dµ , (9.198)

which holds for any representation of "T and Y . Clearly, if there are more symmetriesto be gauged, one can simply add the corresponding terms of the form (couplingconstant) $ (vector field) $ (generator) to the covariant derivative:

Dµ = !µ + i%

i

giAiµGi (Gi : generators) . (9.199)

Next, we will see how the gauge bosons acquire di!erent masses and how thephoton-lepton coupling emerges. We assume again that the scalar potential has theform V = b(2#†# " a2)2, and the choose the unitary gauge that leaves only the realpart of #0 to be non-zero. When the energy scale is low enough, the scalar field isthen written as a small oscillation around the vacuum as in (9.150). Since we ‘lost’three degrees of freedom of the scalar field, we expect that there will be three massivegauge bosons leaving one gauge boson massless. Which one stays massless? Beforejumping into the calculation, we can guess what may happen. Note that in the spaceof the # doublet, Q = Y + T3 is

Q = Y + T3 =1

2

&1 00 1

'+

1

2

&1 00 "1

'=

&1 00 0

'(for #) , (9.200)

which is equibvalent to say that #+ has Q = 1 and #0 has Q = 0. Then, the vacuumis invariant under the phase rotation proportional to Q:

ei$Q#V AC = #V AC , (9.201)

which means that the vacuum has zero electrical charge. Thus, the vacuum doesnot break the U(1) symmetry generated by Q, and we expect that the gauge bosonassociated with it should remain massless. Also, such gauge boson should couple tothe electrical charge and is in fact identified as the photon.

If we had chosen the vacuum expectation value (or the gauge) to be

#V AC =& a#

20

', (9.202)

then we would still define the photon as the massless gauge boson that corresponds tothe symmetry of the vacuum. Namely, the generator Q of such symmetry is redefinedsuch that ei$Q#V AC = #V AC or equivalently Q#V AC = 0. In other words, #+ is redefined




% &Wµ = 'µ&( + g&( ! &Wµ , %Bµ = 'µ)

. (9.192)




Y =

(L!

12

$!

12

) #!R"1

=!L (e"ig!"·!T+ig! !

2 ))†e"ig!"·!T"ig! !2 )

" #$ %eig!"·!T e"ig! !

2 e"ig!"·!T e"ig! !2

$&eig!##R = (L$)#R . (9.193)



Dµ = 'µ + ig &Wµ · &T + ig!BµY , (9.194)






[&T , Y ] = 0 , (9.196)

and[B!

µY, e"ig!"·!T ] = 0 , [ &Wµ · &T , e"ig!#Y ] = 0 . (9.197)


D!µU(x) =

#$$$$$$$$$$$$$$$$$$('µ + ig &W !

µ · &T" #$ %

&W !µ " &Wµ

+ig! B!µY" #$ %

commute


実在のゲージボソンとの対応真空は電荷を持たない：

QをgeneratorとするU(1)対称性を破らない付随するゲージボソン（光子）が質量ゼロのまま

という形をしていれば

20


= e!ig!"·!T!""""""""""""""""""(!µ + ig"B"

µY! "# $B"

µ # Bµ

+ig "Wµ · "T! "# $

commute

) e!ig!#Y


= U(x)Dµ , (9.198)


Dµ = !µ + i%

i



Q = Y + T3 =1

2

&1 00 1

'+

1

2

&1 00 "1

'=

&1 00 0

'(for #) , (9.200)


ei$Q#V AC = #V AC , (9.201)



#V AC =& a#

20

', (9.202)



= e!ig!"·!T!""""""""""""""""""(!µ + ig"B"

µY! "# $B"

µ # Bµ

+ig "Wµ · "T! "# $

commute

) e!ig!#Y


= U(x)Dµ , (9.198)


Dµ = !µ + i%

i



Q = Y + T3 =1

2

&1 00 1

'+

1

2

&1 00 "1

'=

&1 00 0

'(for #) , (9.200)


ei$Q#V AC = #V AC , (9.201)



#V AC =& a#

20

', (9.202)



as electrically charge-neutral, which could be accomplished by assigning Y = !1/2for the scalar doublet. With appropriate rewriting of the Yukawa terms to conserveQ, one would then recreate the physical system identical to the one obtained by thechoice (9.150) for !V AC. The Lagrangian (9.190) is in fact symmetric under SU(2)transformation; namely, we could swap "L and #L keeping their right-handed partnersthe same and the physics it describes does not change. In particular, the electricalcharges at this point has no familiar physical significance. Only after the symmetryis broken, and only after the gauge field that corresponds to the symmetry of thevacuum is identified as the photon, the electrical charge as we know emerges as thequantum number the photon field couples to.

What is the gauge field Aµ associated with Q = Y +T3? Could it be Aµ = W 3µ+Bµ?

Not exactly, the criteria is that it should appear in the covariant derivative as ieQAµ

coupling to the electric charge

Dµ = $µ + · · · + ieQAµ + · · · , (9.203)

where e is the absolute value of the electron charge. For the rest of this chapter, it isunderstood that e is a positive quantity. Then, it would lead to the desired photon-lepton coupling even though we have separated the right-handed and left-handedparts: noting that for the lepton doublet L, Q is

Q = Y + T3 = !1

2

!1 00 1

"+

1

2

!1 00 !1

"=

!0 00 !1

"(for L) (9.204)

and that Q = !1 for "R, we see that LiD/ L+ "RiD/ "R with Dµ = $µ + · · ·+ieQAµ + · · ·contains

! L(eQA/ )L# $% &!e "LA/ "L

! "R(e Q#$%&!1

A/ )"R = e( "RA/ "R + "LA/ "L# $% &"A/ " similarly to (9.169)

) = eAµ("%µ") . (9.205)

Writing down Dµ of (9.194) explicitly,

Dµ = $µ + ig(W 1µT1 + W 2

µT2) + i(gW 3µT3 + g!BµY ) . (9.206)

The idea is to mix W 3µ and Bµ by an orthogonal matrix as

!Bµ

W 3µ

"=

!cos &W ! sin &W

sin &W cos &W

" !Aµ

Zµ

", (9.207)

such that either Aµ or Zµ couples to Q = Y +T3. Substituting this in (gW 3µT3+g!BµY ),

gW 3µT3 + g!BµY

= g(sin &W Aµ + cos &W Zµ)T3 + g!(cos &W Aµ ! sin &W Zµ)Y

= (g sin &W T3 + g! cos &W Y )Aµ + (g cos &W T3 ! g! sin &W Y )Zµ . (9.208)






Dµ = $µ + · · · + ieQAµ + · · · , (9.203)


Q = Y + T3 = !1

2

!1 00 1

"+

1

2

!1 00 !1

"=

!0 00 !1

"(for L) (9.204)


! L(eQA/ )L# $% &!e "LA/ "L

! "R(e Q#$%&!1


) = eAµ("%µ") . (9.205)


Dµ = $µ + ig(W 1µT1 + W 2

µT2) + i(gW 3µT3 + g!BµY ) . (9.206)


!Bµ

W 3µ

"=

!cos &W ! sin &W

sin &W cos &W

" !Aµ

Zµ

", (9.207)


gW 3µT3 + g!BµY








Dµ = $µ + · · · + ieQAµ + · · · , (9.203)


Q = Y + T3 = !1

2

!1 00 1

"+

1

2

!1 00 !1

"=

!0 00 !1

"(for L) (9.204)


! L(eQA/ )L# $% &!e "LA/ "L

! "R(e Q#$%&!1


) = eAµ("%µ") . (9.205)


Dµ = $µ + ig(W 1µT1 + W 2

µT2) + i(gW 3µT3 + g!BµY ) . (9.206)


!Bµ

W 3µ

"=

!cos &W ! sin &W

sin &W cos &W

" !Aµ

Zµ

", (9.207)


gW 3µT3 + g!BµY



Q = !1 for R






Dµ = $µ + · · · + ieQAµ + · · · , (9.203)


Q = Y + T3 = !1

2

!1 00 1

"+

1

2

!1 00 !1

"=

!0 00 !1

"(for L) (9.204)


! L(eQA/ )L# $% &!e "LA/ "L

! "R(e Q#$%&!1


) = eAµ("%µ") . (9.205)


Dµ = $µ + ig(W 1µT1 + W 2

µT2) + i(gW 3µT3 + g!BµY ) . (9.206)


!Bµ

W 3µ

"=

!cos &W ! sin &W

sin &W cos &W

" !Aµ

Zµ

", (9.207)


gW 3µT3 + g!BµY



なので　　　　　　　　の中に

という項を含む

ワインバーグ角

ワインバーグ角を導入すると

21






Dµ = $µ + · · · + ieQAµ + · · · , (9.203)


Q = Y + T3 = !1

2

!1 00 1

"+

1

2

!1 00 !1

"=

!0 00 !1

"(for L) (9.204)


! L(eQA/ )L# $% &!e "LA/ "L

! "R(e Q#$%&!1


) = eAµ("%µ") . (9.205)


Dµ = $µ + ig(W 1µT1 + W 2

µT2) + i(gW 3µT3 + g!BµY ) . (9.206)


!Bµ

W 3µ

"=

!cos &W ! sin &W

sin &W cos &W

" !Aµ

Zµ

", (9.207)


gW 3µT3 + g!BµY








Dµ = $µ + · · · + ieQAµ + · · · , (9.203)


Q = Y + T3 = !1

2

!1 00 1

"+

1

2

!1 00 !1

"=

!0 00 !1

"(for L) (9.204)


! L(eQA/ )L# $% &!e "LA/ "L

! "R(e Q#$%&!1


) = eAµ("%µ") . (9.205)


Dµ = $µ + ig(W 1µT1 + W 2

µT2) + i(gW 3µT3 + g!BµY ) . (9.206)


!Bµ

W 3µ

"=

!cos &W ! sin &W

sin &W cos &W

" !Aµ

Zµ

", (9.207)


gW 3µT3 + g!BµY



!W :






Dµ = $µ + · · · + ieQAµ + · · · , (9.203)


Q = Y + T3 = !1

2

!1 00 1

"+

1

2

!1 00 !1

"=

!0 00 !1

"(for L) (9.204)


! L(eQA/ )L# $% &!e "LA/ "L

! "R(e Q#$%&!1


) = eAµ("%µ") . (9.205)


Dµ = $µ + ig(W 1µT1 + W 2

µT2) + i(gW 3µT3 + g!BµY ) . (9.206)


!Bµ

W 3µ

"=

!cos &W ! sin &W

sin &W cos &W

" !Aµ

Zµ

", (9.207)


gW 3µT3 + g!BµY




It does not matter which of Aµ or Zµ we pick as the photon candidate; let’s pick Aµ.Then, in order for Aµ to couple to Q = Y + T3 we should have

g sin !W = g! cos !W , (9.209)

or

sin !W =g!

g, cos !W =

g

gwith g

def!!

g2 + g!2 . (9.210)

The mixing angle !W is called the Weinberg angle. Then, we have

gW 3µT3 + g!BµY = g sin !W" #$ %

! e

(T3 + Y" #$ %Q

)Aµ + (g cos !W T3 " g! sin !W Y"#$%Q " T3

)Zµ

= e QAµ +&(g cos !W + g! sin !W" #$ %

g2+g!2

g = g

)T3 " g! sin !W" #$ %g sin2 !W

Q'Zµ

= e QAµ + g(T3 " sin2 !W Q)Zµ , (9.211)

where we have identified the electric charge as

e = g sin !W . (9.212)

The W 1,2µ terms can be rewritten as

W 1µT1 + W 2

µT2 =1#2(W+µT+ + W"µT") . (9.213)

where we have defined

W±µ

def! 1#2(W 1

µ $ iW 2µ) (W #

+µ! W"µ) , (9.214)

andT±

def! T1 ± iT2 . (9.215)

For an isospin doublet, T± are explicitly,

T+ =(

0 10 0

), T" =

(0 01 0

)(doublet) . (9.216)

The covariant derivative can now be written as

Dµ = "µ + ig#2

(W+µT+ + W"µT") + ie QAµ + ig SZµ

Sdef! T3 " sin2 !W Q

, (9.217)



g sin !W = g! cos !W , (9.209)

or

sin !W =g!

g, cos !W =

g

gwith g

def!!

g2 + g!2 . (9.210)



! e

(T3 + Y" #$ %Q


)Zµ

= e QAµ +&(g cos !W + g! sin !W" #$ %

g2+g!2

g = g

)T3 " g! sin !W" #$ %g sin2 !W

Q'Zµ

= e QAµ + g(T3 " sin2 !W Q)Zµ , (9.211)


e = g sin !W . (9.212)


W 1µT1 + W 2

µT2 =1#2(W+µT+ + W"µT") . (9.213)


W±µ

def! 1#2(W 1

µ $ iW 2µ) (W #

+µ! W"µ) , (9.214)

andT±

def! T1 ± iT2 . (9.215)


T+ =(

0 10 0

), T" =

(0 01 0

)(doublet) . (9.216)


Dµ = "µ + ig#2



, (9.217)



g sin !W = g! cos !W , (9.209)

or

sin !W =g!

g, cos !W =

g

gwith g

def!!

g2 + g!2 . (9.210)



! e

(T3 + Y" #$ %Q


)Zµ

= e QAµ +&(g cos !W + g! sin !W" #$ %

g2+g!2

g = g

)T3 " g! sin !W" #$ %g sin2 !W

Q'Zµ

= e QAµ + g(T3 " sin2 !W Q)Zµ , (9.211)


e = g sin !W . (9.212)


W 1µT1 + W 2

µT2 =1#2(W+µT+ + W"µT") . (9.213)


W±µ

def! 1#2(W 1

µ $ iW 2µ) (W #

+µ! W"µ) , (9.214)

andT±

def! T1 ± iT2 . (9.215)


T+ =(

0 10 0

), T" =

(0 01 0

)(doublet) . (9.216)


Dµ = "µ + ig#2



, (9.217)



g sin !W = g! cos !W , (9.209)

or

sin !W =g!

g, cos !W =

g

gwith g

def!!

g2 + g!2 . (9.210)



! e

(T3 + Y" #$ %Q


)Zµ

= e QAµ +&(g cos !W + g! sin !W" #$ %

g2+g!2

g = g

)T3 " g! sin !W" #$ %g sin2 !W

Q'Zµ

= e QAµ + g(T3 " sin2 !W Q)Zµ , (9.211)


e = g sin !W . (9.212)


W 1µT1 + W 2

µT2 =1#2(W+µT+ + W"µT") . (9.213)


W±µ

def! 1#2(W 1

µ $ iW 2µ) (W #

+µ! W"µ) , (9.214)

andT±

def! T1 ± iT2 . (9.215)


T+ =(

0 10 0

), T" =

(0 01 0

)(doublet) . (9.216)


Dµ = "µ + ig#2



, (9.217)

ゲージボソンの質量

自発的に対称性が破れると

22



g sin !W = g! cos !W , (9.209)

or

sin !W =g!

g, cos !W =

g

gwith g

def!!

g2 + g!2 . (9.210)



! e

(T3 + Y" #$ %Q


)Zµ

= e QAµ +&(g cos !W + g! sin !W" #$ %

g2+g!2

g = g

)T3 " g! sin !W" #$ %g sin2 !W

Q'Zµ

= e QAµ + g(T3 " sin2 !W Q)Zµ , (9.211)


e = g sin !W . (9.212)


W 1µT1 + W 2

µT2 =1#2(W+µT+ + W"µT") . (9.213)


W±µ

def! 1#2(W 1

µ $ iW 2µ) (W #

+µ! W"µ) , (9.214)

andT±

def! T1 ± iT2 . (9.215)


T+ =(

0 10 0

), T" =

(0 01 0

)(doublet) . (9.216)


Dµ = "µ + ig#2



, (9.217)



g sin !W = g! cos !W , (9.209)

or

sin !W =g!

g, cos !W =

g

gwith g

def!!

g2 + g!2 . (9.210)



! e

(T3 + Y" #$ %Q


)Zµ

= e QAµ +&(g cos !W + g! sin !W" #$ %

g2+g!2

g = g

)T3 " g! sin !W" #$ %g sin2 !W

Q'Zµ

= e QAµ + g(T3 " sin2 !W Q)Zµ , (9.211)


e = g sin !W . (9.212)


W 1µT1 + W 2

µT2 =1#2(W+µT+ + W"µT") . (9.213)


W±µ

def! 1#2(W 1

µ $ iW 2µ) (W #

+µ! W"µ) , (9.214)

andT±

def! T1 ± iT2 . (9.215)


T+ =(

0 10 0

), T" =

(0 01 0

)(doublet) . (9.216)


Dµ = "µ + ig#2



, (9.217)



g sin !W = g! cos !W , (9.209)

or

sin !W =g!

g, cos !W =

g

gwith g

def!!

g2 + g!2 . (9.210)



! e

(T3 + Y" #$ %Q


)Zµ

= e QAµ +&(g cos !W + g! sin !W" #$ %

g2+g!2

g = g

)T3 " g! sin !W" #$ %g sin2 !W

Q'Zµ

= e QAµ + g(T3 " sin2 !W Q)Zµ , (9.211)


e = g sin !W . (9.212)


W 1µT1 + W 2

µT2 =1#2(W+µT+ + W"µT") . (9.213)


W±µ

def! 1#2(W 1

µ $ iW 2µ) (W #

+µ! W"µ) , (9.214)

andT±

def! T1 ± iT2 . (9.215)


T+ =(

0 10 0

), T" =

(0 01 0

)(doublet) . (9.216)


Dµ = "µ + ig#2



, (9.217)


which is valid for any field as long as appropriate representations are used for Ti andQ.

The masses of gauge bosons arises from the term (Dµ!)†(Dµ!) as before. In theunitary gauge we are using, the only nonzero component of ! is !0, and thus we canset Q = 0 in Dµ! [see (9.204)]. This immediately tells us that Aµ drops out of Dµ!and thus there will be no mass term for photon. Note that this is directly relatedto the fact that the vacuum is invariant under the transformation generated by Q asseen in (9.201). Then, Dµ! can be written as

Dµ! =!"µ + i

g!2

(W+µT+ + W!µT!) + ig T3Zµ

"!

=#"µ +

i

2

$gZµ

!2gW+µ!

2gW!µ "gZµ

%& '0

1"2!01

(

=1!2

'i2g!

2W+µ!01

"µ!01 " i2 gZµ!01

(

. (9.218)

Noting that !01 and Zµ are real, we then have

(Dµ!)†(Dµ!) =g2

4!0

21W+

#µW+

µ +1

2("µ!01 "

i

2gZµ!01)

#("µ!01 "i

2gZµ!01)

=1

2"µ!01"

µ!01 +g2

4!0

21W+

#µW+

µ +g2

8!0

21ZµZ

µ

(!01 = a + #)

=1

2"µ#"µ# +

(a + #)2

4(g2W+

#µW+

µ +g2

2ZµZ

µ)

=1

2"µ#"µ# +

a2g2

4W+

#µW+

µ +a2g2

8ZµZ

µ

+2a# + #2

4(g2W+

#µW+

µ +g2

2ZµZ

µ) (9.219)

Now we can identify the mass terms of the gauge fields (the form is m2A#µA

µ for acharged vector and (m2/2)AµAµ for a neutral vector):

mW =ag

2, mZ =

ag

2# mW

mZ=

g

g= cos $W . (9.220)

The interaction between the fermions and gauge bosons are embedded in

LiD/ L + %RiD/ %R + &RiD/ &R . (9.221)

(Dµ!)†(Dµ!) =12"µ#"µ# +

v2g2

4W !

+µWµ+ +

v2g2

8ZµZµ

+2v# + #2

4(g2W !

+µWµ+ +

g2

2ZµZµ)

WとZの質量

e-d散乱実験から電荷から

ミューオンの寿命の測定

以上からWの質量の予言可能さらに　　　　　　　　　　　　　　　　　　　の関係を使えば Z の質量も予言可能1983年：予言通りの質量で W, Z が発見される

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quarks, and add the strong interaction, which is a unbroken SU(3) gauge theory ofquarks and gluons called the quantum chromo-dynamics (QCD), then the resultingtheory is called the standard model. Now, what do we know about the parametersappearing in the Lagrangian above?

First, we know that the electron neutrino is very light. The current best upperlimit on the mass of !e is

m!e < 3eV (9.235)

obtained from the "-decay of H3 (tritium, made of one proton and two neutrons)

H3(pnn) ! He3(ppn) + e! + !e (9.236)

where the energy spectrum of e! at the high-end point is studied. If the neutrinomass is precisely zero, then the relation m! = h!a/

"2 indicates that h! = 0, namely

Higgs does not couple to the neutrino. Since T3 = 0 and Q = 0 for right-handedneutrino, it does not couple to photn nor to Z (because S = T3 + sin2 #W Q = 0).Since W couples to leptons through Ti, W does not couple to right-handed neutrinowhich is SU(2) singlet. Thus, if the neutrino is massless, the right-handed neutrinocannot be created by any interaction, and thus may be altogether ignored. In fact, ifthe neutrino mass is exactly zero, we could have started by omitting !R in (9.173).The original electro-weak unification was proposed in such form without right-handedneutrino.

Next, let’s check the coupling constants e and g. We know the value of e:

$ # e2

4%$ 1

137! e = 0.303 , (9.237)

and the masses and W± and Z0 are measured to be!

mW = 80.22 ± 0.26 (GeV)

mZ = 91.187 ± 0.007 (GeV)! cos #W =

mW

mZ= 0.880 . (9.238)

Then, the weak coupling constant g is ‘predicted’ to be

g =e

sin #W=

e"1 % cos2 #W

= 0.64 . (9.239)

On the other hand, the value of g can be independently extracted from the muondecay. In (5.341), we found

GF =g2

4"

2 m2W

= 1.166 & 10!5 GeV!2 ! g = 0.65 , (9.240)

which is a remarkable agreement! Historically, e and GF were known first and thensin2 #W was measured by the parity violation in the e%d(electron-deuteron) scattering,




m!e < 3eV (9.235)


H3(pnn) ! He3(ppn) + e! + !e (9.236)





$ # e2

4%$ 1

137! e = 0.303 , (9.237)


mW = 80.22 ± 0.26 (GeV)

mZ = 91.187 ± 0.007 (GeV)! cos #W =

mW

mZ= 0.880 . (9.238)


g =e

sin #W=

e"1 % cos2 #W

= 0.64 . (9.239)


GF =g2

4"

2 m2W

= 1.166 & 10!5 GeV!2 ! g = 0.65 , (9.240)





m!e < 3eV (9.235)


H3(pnn) ! He3(ppn) + e! + !e (9.236)





$ # e2

4%$ 1

137! e = 0.303 , (9.237)


mW = 80.22 ± 0.26 (GeV)

mZ = 91.187 ± 0.007 (GeV)! cos #W =

mW

mZ= 0.880 . (9.238)


g =e

sin #W=

e"1 % cos2 #W

= 0.64 . (9.239)


GF =g2

4"

2 m2W

= 1.166 & 10!5 GeV!2 ! g = 0.65 , (9.240)


v =2mW

g! v = 246 GeV

mW =vg

2, mZ =

vg

2! mW

mZ=

g

g= cos !W

sin2 !W = 0.23

ここまでのまとめ

ゲージ対称性ラグランジアンがゲージ変換に対して不変、という要求を課すことによって相互作用が決まる変換を規定する群SU(3): 強い力SU(2): 弱い力U(1): 電磁気力

GWS模型電磁気力と弱い相互作用は SU(2)⊗U(1) ゲージ対称性により統一的に扱うことができるゲージ対称性からゼロであるべき素粒子の質量はヒッグス機構によって生成されるヒッグスが未発見であることを除き高い精度で合格

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ゲージ対称性とヒッグス機構 - Osaka...

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