É { ú g¶ É { ú g¶ æ# - 大阪大学osksn2.hep.sci.osaka-u.ac.jp/~kazu/class_2010/par_phys...今回の目次ヒッグス機構 GWS模型弱い相互作用におけるV-A型結合

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素粒子物理学2素粒子物理学序論B

2010年度講義第7回

今回の目次

ヒッグス機構GWS模型弱い相互作用におけるV-A型結合フェルミオンの質量SU(2)⊗U(1)ワインバーグ角ゲージボソンの質量

GWS模型の実験的検証中性カレントWとZの質量

2

ヒッグス機構

ゲージ対称性の破れ

ゲージボソンの質量項を手で入れると

となりゲージ対称性を保てないゲージボソンの質量はゼロでないとならない実際に光子とグルーオンの質量はゼロしかし 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

2つのフェルミオンの質量が等しければゲージ対称

4

m2AµAµ → m2(Aµ + ∂µΛ)(Aµ + ∂µΛ) �= m2AµAµ

ψ ≡�

pn

�ψ ≡

�pn

�L = ψ(i �∂ −m)ψ

L� ≡ ψ�(i �∂ −m)ψ� = ψU †(i �∂ −m)Uψ = L

ヒッグス場の自発的対称性の破れスピン０のスカラー場　　（＝ヒッグス場）を導入

真空が　　　　　　　　　　　　　に固定される（＝対称性が自発的に破れる）と

5

φ(x)

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)






L = "µ!!"µ! ! V (|!|2) , (9.121)

with

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


L =1

2("µ!1"

µ!1 + "µ!2"µ!2) ! V

!!21 + !2

2

2

". (9.123)


!21 + !2

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


"µ"L

"("µ!i)=

"L"!i

$ "2!i = !"V

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


H "#

d3x($

i

#i!i ! L) =#

d3x!1

2

$

i

[!2i + ($%!i)

2] + V". (9.126)

V = µ2φ∗φ + λ|φ∗φ|2

宇宙初期（高温） μ2>0

現在　　（低温） μ2<0相転移

η

ξ

∴ L =12(∂µξ)2 +

12(∂µη)2 + µ2η2 + (const) + (higher term)

φ(x) =�

1/2{v + η(x) + iξ(x)}

ゴールドストンボソンスカラー場の質量項

φ1 = v, φ2 = 0 (v =�−µ2/λ)

ヒッグス機構　U(1)対称性

ヒッグス場とゲージ場を考える

ゲージ変換は　　　　　　　　　　　　とparametrizeする(ρとθ実数)ゲージ変換の自由度を利用してというゲージを選ぶ（＝ユニタリーゲージ）　　が実数となるようなゲージを選んだ (　　　　 )このゲージで対称性が自発的に破れると

6

496 CHAPTER 9. THE STANDARD MODEL

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)




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)

φ�(x)

φ1 =�

1/2{v + χ(x)} φ2 = 0

L =12(∂µχ∂µχ− µ2χ2)− 1

4FµνFµν +

e2v2

2AµAµ +

e2

2AµAµ(2vχ + χ2)

ゲージボソンの質量項

Λ(x) =θ(x)

e

ヒッグス機構　SU(2)対称性

ラグランジアンの形はU(1)のときと同じ

ヒッグス場は二重項

自発的対称性の破れ

対称性が破れた後のラグランジアン

7


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)

φ(x) =�

01√2v

�+

�0

1√2χ(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(∂µχ∂µχ− µ2χ2)− 1

4F i

µνF iµν +g2v2

8Ai

µAiµ +g2

8(2vχ + χ2)Ai

µAiµ

ヒッグス機構　まとめ

　　　　　　　　　　という形のポテンシャルの存在実はポテンシャルを作るのがヒッグス（というスカラー粒子）でなくてもよい

対称性が破れる前の世界では、ヒッグスは2つの自由度を持った（粒子・反粒子）荷電粒子のように振る舞う

対称性が破れる（真空がゼロでない期待値を持つ）と

8

V = µ2φ∗φ + λ|φ∗φ|2

(Dµφ)∗(Dµφ) = [(∂µ + ieAµ)φ]∗[(∂µ + ieAµ)φ]

�φ� → v√2






L = "µ!!"µ! ! V (|!|2) , (9.121)

with

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


L =1

2("µ!1"

µ!1 + "µ!2"µ!2) ! V

!!21 + !2

2

2

". (9.123)


!21 + !2

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


"µ"L

"("µ!i)=

"L"!i

$ "2!i = !"V

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


H "#

d3x($

i

#i!i ! L) =#

d3x!1

2

$

i

[!2i + ($%!i)

2] + V". (9.126)

e2AµAµφ∗φ ⇒ e2v2

2AµAµ

ゲージ場の質量は既知のゲージ場の結合定数と既知の真空期待値に依存

⇔ ヒッグス場の質量は未知のパラメータμに依存

Glshow-Weinberg-Salam （GWS）模型

模型のスタート

左巻きフェルミオンをSU(2)doublet, 右巻きをsinglet

ハイパーチャージ（Y）の導入SU(2)⊗U(1)変換（後で導入）での保存量

10

9.4. ELECTRO-WEAK UNIFICATION AND THE STANDARD MODEL 507

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)


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)



SU(2)とU(1)対称性

SU(2)とU(1)ゲージ場の導入簡単のためにレプトンの質量を生み出す項のみ

11


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)

V-A 結合

SU(2)不変なラグランジアン

12




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)






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)



⇐ 望む形が得られた


of a scalar field. Now we would like to make the Lagrangian invariant under thespace-time dependent isospin transformation given by

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

!Ti !

"i

2

", (9.78)

where #i(x) (i = 1, 2, 3) are arbitrary real functions of x = (t, $x) and the real ‘couplingconstant’ g is separated out.

As in the case of QED, we will introduce vector fields and assign certain transfor-mation rule for such fields. Our goal is to define the covariant derivative Dµ and thetransformation of $Aµ such that Dµ transforms the same way as ! itself:

(Dµ!)! = U(x)Dµ! . (9.79)

Then, the Lagrangian !iD/ ! would be invariant

L! = !!i(D/ !)! = !U †(x)iU(x)D/ ! = !iD/ ! = L . (9.80)

When the derivative is taken of !! = U(x)!, there will be extra factors that containthree independent functions %µ#i(x) that somehow need to be cancelled by the changein the vector fields. Thus, to accomplish the task, we need at least three real vectorfields $Aµ(x) = (A1

µ, A2µ, A

3µ) that can be changed independently under the yet-to-be-

defined local transformation. Since index i of Aiµ is not a Lorentz index, we will not

distinguish superscript and subscript. We will try the covariant derivative of the form

Dµdef! %µ + ig $Aµ · $T ! %µ + igAµ , (9.81)

where we have defined the ‘inner product’ of a 3-component quantity $X with $T :

Xdef! $X · $T , (9.82)

which is a 2 " 2 matrix in the isospin space. Explicitly, Aµ ! $Aµ · $T with Ti = "i/2gives

Aµ = Aiµ

"i

2

=1

2

#A1

µ

!0 11 0

"+ A2

µ

!0 #ii 0

"+ A3

µ

!1 00 #1

"$

=1

2

%A3

µ

$2A+

µ$2A"

µ #A3µ

&

(9.83)

where

A±µ

def! 1$2(A1

µ % iA2µ) . (9.84)

カイラルフェルミオン

左巻きと右巻きフェルミオンは別粒子として振る舞う＝カイラル対称

質量は、左巻き粒子と右巻き粒子の結合の強さ（カイラリティの破れの度合い）を表す

13


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)

ψi �∂ψ = (ψR + ψL)i(ψR + ψL)= ψRi �∂ψR + ψLi �∂ψL + ψRi �∂ψL + ψLi �∂ψR

= ψRi �∂ψR + ψLi �∂ψL




# = #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)

ψL → e−iαψL, ψR → e+iαψR質量が無ければカイラル変換（まとめて　　　　　　）に対して不変ψ → eiαγ5

ψ

⇐ カイラル対称

レプトンの質量

湯川結合　　　　　を導入する

ヒッグス場が自発的に対称性を破ると

14


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

�νL, lL

� �φ+

φ0

�lR + hν

�νL, lL

� �φ∗

0

−φ∗+

�νR + 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

φ+ → 0 , φ0 →v√2

vhl√2[(lLlR) + c.c.] +

vhν√2

[(νLνR) + c.c.] =vhl√

2ll +

vhν√2

νν

ml =vhl√

2, mν =

vhν√2

hl, hν(= Yl, Yν)

ゲージボソンに質量を与えるヒッグス場によってフェルミオンも質量を獲得（省エネ？）

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最重要フェルミオン質量もゲージボソンに質量を与えるヒッグス機構によって生成されているのか？実験的に ”質量∝結合定数” を確認する必要がある

GWS模型ではフリーパラメータ　　　　　　　　　　　　　　　（全てのフェルミオンに対してそれぞれ湯川結合が必要）が激増⇐ 理論的に美しくない

15

SU(2) ⊗ U(1) 対称性の導入

SU(2)⊗U(1)変換

ゲージ不変になっている例として1つの項だけチェック

　　　　　　　　　となるcovariant derivativeを探す⇒（興味のある人は自分で確認してみて下さい）一般に

16


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




% &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



= 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

QEDの確認

真空はU(1)em変換で対称性を保存

付随するゲージボソン（光子）の質量はゼロのままヒッグス（複素スカラー場）二重項には4つの自由度3つのゲージボソン（W±とZ0）に質量⇒ 4(自由度)-3(質量を持つゲージボソン)　=1個の実在粒子（＝ヒッグス）

相互作用項の所在

17


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




= 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)


なので　　　　　　の中に






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

φ二重項空間

ワインバーグ角

U(1)ゲージ場とSU(2)ゲージ場の中性成分は混合しているワインバーグ角　　がその混合角

18






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








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)

θW

ゲージボソンの質量

Covariant derivative を整理

ヒッグス場の対称性が自発的に破れると

19



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)

(Dµφ)†(Dµφ) =12∂µχ∂µχ +

v2g2

4W ∗

+µWµ+ +

v2g2

8ZµZµ

+2vχ + χ2

4(g2W ∗

+µWµ+ +

g2

2ZµZµ)

GWS模型のラグランジアン

実験事実を高い精度で再現しますが、美しくないですね

20


where sR,L are the values of S = T3 ! sin2 !W Q for the corresponding right-handedand left-handed fields. Namely, Z0 couples to the right-handed as well as left-handedcomponents of ". Since s!R = 0, however, the coupling of Z0 to # is always purelyleft-handed. This is true regardless of the neutrino mass (massive or not).

Let’s move on to the Higgs-fermion coupling. As we have seen in (9.182), it resultsin the fermion masses. If we repeat the derivation without ignoring $, we obtain thesame result with the replacement a " a + $. Namely,

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

"= !(a + $)h"#

2"" ! (a + $)h!#

2## . (9.230)

The fermion masses are identified as

m" =h"a#

2, m! =

h!a#2

, (9.231)

and in terms of the masses, the Higgs-fermion terms are now

!m""" ! m! ## ! m"

a$("") ! m!

a$(##) . (9.232)

Note that the coupling of the Higgs ($) to fermion is proportional to the mass of thefermion.

Putting everything together, the SU(2)$U(1) Lagrangian (9.190) in the unitarygauge with the scalar field expanded around the vacuum %V AC = (0, a/

#2) is now

L = #(i&/ ! m!)# + "(i&/ ! m")" +1

2(&µ$&µ$ ! µ2$2)

! 1

4F i

µ!Fiµ! + m2

W W !+µ

W µ+ ! 1

4Gµ!G

µ! +m2

Z

2ZµZ

µ

+ eAµ("'µ") ! g#2

!W µ

+ (#'µPL") + c.c."

! g Zµ

!#'µ(s!LPL + s!RPR)# + "'µ(s"LPL + s"RPR)"

"

+2a$ + $2

4

#g2W !

+µW µ

+ +g2

2ZµZ

µ$

! m"

a$("") ! m!

a$(##) , (9.233)

with

e = g sin !W , g =%

g2 + g"2 , sin !W =g

g, cos !W =

g"

g, (9.234)

and s!R,L , s"R,L are the values of S = T3 ! sin2 !W for the corresponding fields. Whenwe extend this Lagrangian to include three doublets of leptons and three doublets of


where sR,L are the values of S = T3 ! sin2 !W Q for the corresponding right-handedand left-handed fields. Namely, Z0 couples to the right-handed as well as left-handedcomponents of ". Since s!R = 0, however, the coupling of Z0 to # is always purelyleft-handed. This is true regardless of the neutrino mass (massive or not).

Let’s move on to the Higgs-fermion coupling. As we have seen in (9.182), it resultsin the fermion masses. If we repeat the derivation without ignoring $, we obtain thesame result with the replacement a " a + $. Namely,

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

"= !(a + $)h"#

2"" ! (a + $)h!#

2## . (9.230)

The fermion masses are identified as

m" =h"a#

2, m! =

h!a#2

, (9.231)

and in terms of the masses, the Higgs-fermion terms are now

!m""" ! m! ## ! m"

a$("") ! m!

a$(##) . (9.232)

Note that the coupling of the Higgs ($) to fermion is proportional to the mass of thefermion.

Putting everything together, the SU(2)$U(1) Lagrangian (9.190) in the unitarygauge with the scalar field expanded around the vacuum %V AC = (0, a/

#2) is now

L = #(i&/ ! m!)# + "(i&/ ! m")" +1

2(&µ$&µ$ ! µ2$2)

! 1

4F i

µ!Fiµ! + m2

W W !+µ

W µ+ ! 1

4Gµ!G

µ! +m2

Z

2ZµZ

µ

+ eAµ("'µ") ! g#2

!W µ

+ (#'µPL") + c.c."

! g Zµ

!#'µ(s!LPL + s!RPR)# + "'µ(s"LPL + s"RPR)"

"

+2a$ + $2

4

#g2W !

+µW µ

+ +g2

2ZµZ

µ$

! m"

a$("") ! m!

a$(##) , (9.233)

with

e = g sin !W , g =%

g2 + g"2 , sin !W =g

g, cos !W =

g"

g, (9.234)

and s!R,L , s"R,L are the values of S = T3 ! sin2 !W for the corresponding fields. Whenwe extend this Lagrangian to include three doublets of leptons and three doublets of

S = T3 − sin2 θW QsR,Lは、右巻き、左巻き粒子の　　　　　　　　　の値

+quark mass term

模型から理論へ- GWS模型の実験的検証 -

中性カレント

中性カレント反応はワインバーグ角依存性があるの測定からワインバーグ角を測定できる

歴史的には偏極電子ビームを使ったeD散乱における非対称（ビームの偏極度による違い）測定が先行した電磁相互作用があるとろこには中性カレントによる弱い相互作用も存在する

22

e−e−

νµ νµ

ZS = T3 − sin2 θW QsR,L


Using (9.217) for Dµ, we have

LiD/ L = Li!/L ! g!2L(W/ +T+ + W/ "T")L !e LQA/ L !g LSZ/ L

"RiD/ "R = "Ri!/"R (Ti = 0) (Q = 0) !g "RSZ/ "R

#RiD/ #R = #Ri!/#R (Ti = 0) !e #RQA/ #R !g #RSZ/ #R

sum (a) (b) (c) (d)

(9.222)

We kept the last term of "RiD/ "R for a bookkeeping purpose even though S = Y +T3 =0 for "R. The sum (a) gives just the free field kinetic terms for the fermions as wehave seen in (9.175):

(a) = "Li!/"L + #Li!/#L + "Ri!/"R + #Ri!/#R = "i!/" + #i!/# , (9.223)

where we have used (9.169). The sum (b) gives the V ! A coupling of the fermionsto W±:

(b) = ! g!2("L, #L)

!0 W/ +

W/ " 0

" !"L

#L

"= ! g"

2

#"LW/ +#L + #LW/ ""L

$

= ! g"2

#W µ

+ ("$µPL#) + c.c.$. (9.224)

We have already seen in (9.205) that

(c) = eAµ(#$µ#) . (9.225)

The last part (d) is the coupling of the massive neutral vector Zµ to the fermions.Noting that S = T3 ! sin2 %W Q, its value for each multiplet is

S =!

s!L 00 s"L

"(L) , S = s!R ("R) , S = s"R (#R) , (9.226)

with %s!L = 1

2

s"L = !12 + sin2 %W

, s!R = 0 , s"R = sin2 %W , (9.227)

the sum (d) can be written as

(d) = !g(s!L "LZ/ "L + s"L #LZ/ #L + s!R "RZ/ "R + s"R #RZ/ #R)

= !g#"Z/ (s!LPL + s!RPR)" + #Z/ (s"LPL + s"RPR)#

$

= !g Zµ

#"$µ(s!LPL + s!RPR)" + #$µ(s"LPL + s"RPR)#

$. (9.228)

Thus, the coupling of Zµ to fermions has the form

!g Zµ &$µ(sLPL + sRPR)& (& = ", #) , (9.229)

σ(νµe)sin2 θW (νµe) = 0.2324± 0.0083

WとZの質量

Wの質量の予測が以下の測定から可能中性カレントの実験電荷

ミューオンの寿命測定

GWS模型の関係を使えば

Zの質量も予測可能1983年に予測通りの質量を持ったWとZが発見された

23

sin2 θW = 0.23


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)

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)

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,mW =

vg

2, mZ =

vg

2→ mW

mZ=

g

g= cos θW

コメント（1）

ワインバーグ角、WとZの質量、その他が非常に高い精度で測定され、GWS模型の正しさが証明された⇒ 模型から理論へ「GWS理論＋QCD」を一般に標準理論（SM）と呼ぶSMは真の理論の低エネルギー極限と考えられている予言不能なフリーパラメータが多過ぎる全ての力が統一されていないなどなどGWS模型の提唱から40年、WとZの発見から25年経った現在、SMを超える新しい物理現象（beyond the SM, New Physics）の発見が待たれている

24

コメント（2）

（前のページと矛盾しているが）GWS模型は確立したと思われているが、ヒッグスセクターについては何一つ実証されていない測定されているのは全てゲージセクター

25


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)

V = µ2φ∗φ + λ|φ∗φ|2

v =�−µ2/λ

v =2mW

g→ v = 246 GeV

ヒッグス場のポテンシャルの形はわからない⇒ ヒッグス質量も第一原理からは予測不能

今回のまとめ

ヒッグス機構ヒッグス場の自発的対称性の破れにより、ゲージボソンとフェルミオンの双方に質量を与えるゲージ対称性によりゲージボソンの質量は本来ゼロカイラル対称性によりフェルミオンの質量は本来ゼロ

GWS模型SU(2)⊗U(1)とヒッグス機構

GWS模型の実験的検証ゲージセクターは高い精度で合格

26

Documents

É { ú g¶ É { ú g¶ æ# - 大阪大学osksn2.hep.sci.osaka-u.ac.jp/~kazu/class_2010/par_phys...今回の目次 ヒッグス機構 GWS模型 弱い相互作用におけるV-A型結合

É { ú g¶ É { ú g¶ æ# - 大阪大学osksn2.hep.sci.osaka-u.ac.jp/~kazu/class_2010/par_phys...今回の目次ヒッグス機構 GWS模型弱い相互作用におけるV-A型結合