Please work alone on these problems. They are easier than they look like. You can call me at
0922492150 if you have question. Good luck and have a good summer!
1.
請問下圖中虛線所代表媒介子分別是什麼粒子？（ W
± 、 Z
0 、γ、膠子 g ）
ν
(a) e
- s green
(b) s red
ν c green
(c) e - s green d red d green d u red red
2.
既然電磁交互作用與弱交互作用都是同一個 SU(2) × U(1) 規範場論的一部分，兩者作用強度
應該差不多，事實上兩者強度差距極大，請簡單解釋原因。
3.
ABC model:
Consider a theory with three spin-0 particles without charge: A, B, C . They are described by three real scalar fields: A ( x ) , B ( x ) , C ( x ). The masses of A and B are identical: m
A
 m
B while m c

2 m
A , B
. This is similar to the ABC model. But we want to add two more interactions. The total interaction Lagrangian is:
L
I

1
2 g
1
AAC

1
2 g
2
BBC
 g
3
ABC .
The one halves in the first two terms are to compensate for the fact that there are two identical fields in the terms. Anyway, the vertices in the theory are ：
C C C i g
1
Dra i g
2
Dra i g
3
Dra
A A B B A w st w st w the lowe the a.
Draw the lowest order Feynman diagrams for the scattering process: lowe the lowe st
B order
Feyn man order
Feyn man order
Feyn man

A ( p
1
)

B ( p
2
)

A ( p
3
)

B ( p
4
) . (There are two!) The letters in the parentheses are the corresponding momenta. Write down the corresponding Feynman Amplitudes. b.
Continue from (a), but assume that g
2
is 0. That is, the CBB vertex disappears. Now draw the lowest order Feynman diagram (just one) for the decay C ( p
1
)

B
   
2 3
. Write down the Feynman amplitude.
Hint: It is a diagram with a loop (or a triangle) composed of lines of different particles.
This is an example where the tree contribution vanishes but the loop correction gives the leading contribution. c.
Continue from (a), but assume that g
3
is 0 ( g
1,2
are not 0). Calculate, in the CM frame, the dependence of A ( p
1
)

B ( p
2
)

A ( p
3
)

B ( p
4
) scattering cross section d
 d

on the scattering angle
θ and initial momentum magnitude p

 p
1

 p
2
. Only its relations with
θ and p are needed. You can drop all the multiplicative constants. d.
Continue from (a), but this time assume that g
1 and g
2
are both 0 ( g
3
is not 0). So we come back to the original ABC model. Calculate again, in the CM frame, the dependence of A ( p
1
)

B ( p
2
)

A ( p
3
)

B ( p
4
) scattering cross section d
 d

on the scattering angle
θ and initial momentum magnitude p

 p
1

 p
2
.
4.
ABCD model
Consider a theory with four spin-0 particles without charge: A, B, C, D . They are described by four scalar real fields: A ( x ) , B ( x ) , C ( x ), D ( x ). Their masses are identical: m
A
 m
B
 m
C
 m
D
. In this model, we assume only the interaction:
L
I

1
2 g

ABCD . a.
Write down the Feynman rule for the interaction vertex. Consider the scattering process A ( p
1
)

B ( p
2
)

C ( p
3
)

D ( p
4
) . Calculate, in the CM frame, the dependence of A ( p
1
)

B ( p
2
)

C ( p
3
)

D ( p
4
) scattering cross section d
 d

on the scattering angle
θ and initial momentum magnitude p

 p
1

 p
2
. Only its relations with
θ and p are needed. You can drop all the multiplicative constants.
Comment: This is the result for a structure-less scattering. Compare it to the answers in 3c,
3d where there is a propagating mediating particle. From the experimental data, we can tell which the case is. b.
Consider the process A ( p
1
)

B ( p
2
)

A ( p
3
)

B ( p
4
) . Draw the lowest order Feynman diagrams (I think there are two) for the scattering. Write down the amplitude (you don’t need to calculate) for any one of the diagrams (one is enough!).

5.
This is a problem set about QED. a.
Prove that an electron in isolation cannot emit one photon because of momentum conservation. That is, the process e

 
1
 e

( p
2
)
   
3
(the letters in the parentheses are the corresponding momenta) is prohibited. b.
However, if electron gets a scattering by some heavy particle, it could emit a photon in the process. For example, the electron-muon scattering process e

 
1
  
( p
2
)
 e

( p
3
)
  
( p
4
)
 
is possible. Draw the lowest order Feynman diagrams for the process. Write down the amplitude for any one of the diagrams (one is enough).
6.
O(3) symmetry breaking
Don’t be frightened. O(3) is actually just SU(2). And it’s even simpler since you can imagine it to be just like a rotation in an imaginary 3D space.
Consider a set of 3 real scalar fields
 

1




2
3


. Think of them as

 x



 y z


, components of an imaginary 3 vector and then O(3) transformation as the imaginary 3 dimensional rotation.
Assume its Lagrangian is:
L 
1
2



   

 
1
2

2

 

  
 
 



2
.
Note that the mass square term has a wrong sign. The first term is just the normal kinetic energy of a Klein-Gordon Field. The second and third term could be considered as the negative of the potential. Since the Lagrangian is composed only of the inner products of 3 vectors, it is invariant under O(3), ie. the 3D rotation. The Hamiltonian can be written as
H 
1
2


 


1
2

  

  
1
2

2

 

  
 
 



2
And for constant Φ, the potential is
V (

)
 
1
2

2

 

  
 
 



2
.
The phenomenon of Spontaneous Symmetry Breaking can be studied even in classical field theory. It (SSB) is the situation when the minimum of the Hamiltonian is not invariant under the symmetry of the Hamiltonian itself, in this case O(3). Here the minimum of H occurs at



0 ,



0 ,
  constant . Due to presence of negative mass square, the minimum of the energy does not occur at
 
0 . And you will see very soon a minimum at nonzero

break the symmetry O(3).

a.
The value of the potential V (

)
 
1
2

2

 

  
 
 



2
depend only on the length of

:
  
1
2  
2
2  
3
2 . Find the value v
2 of
 
that gives a minimum of the potential. b.
You can now see that the minimum occurs at a sphere (in the
 
1
,

2
,

3
,

space) satisfying the condition
  
1
2  
2
2  
3
2  v
2 . We can choose any point on the sphere as the minimum (Physical results are independent of the choice. After all we can use a rotation to rotate your choice minimum to any direction). Let’s choose the nonzero to lie in the z direction

0
0
 

0 v


. Now we can study the effects of O(3) on this minimum. O(3) is just
3D rotation. The generators of O(3) are just the angular momenta. So a small “rotation” around x axis will change the field by
 
L x

0
 

0
0
0
0 i
0

0 i 

 


1

2

3
`


, a small “rotation” around y axis will change the field by
 
0
L y
  

0
 i
0
0
0 i
0
0




1




3
2


and a small
“rotation” around y axis will change the field by
 
L z
  

 i
0

0
0 0 i 0
0
0


 


1 `

2

3


. Which rotation will keep the minimum we choose invariant, ie.

0

L x , y , z

0

0 . Those rotations that change the minimum are not symmetries of the minimum.
Comments: The minimum of Hamiltonian we calculate here will translate directly into the vacuum expectation value of the field
 in quantum field theory. Nonzero VEV breaks part of the O(3) symmetry just like you have demonstrated in (b) into a smaller subgroup which remain a symmetry of the vacuum.