lanxicy.com

第一范文网 文档专家

第一范文网 文档专家

Particle Physics 2010

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 、Z0、γ、膠子 g）

(a) s green (b) s red

ν

e-

ν

c green (c)

es green

d red

d green

d red

u 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

mc > 2m A, B . This is similar to the ABC model. But we want to add two more interactions. The total interaction Lagrangian is:

LI =

1 1 g 1 AAC + g 2 BBC + g 3 ABC . 2 2

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 i g1 A A B C i g2 B A C i g3 B

a. Draw the lowest order Feynman diagrams for the scattering process:

A( p1 ) + 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 g2 is 0. That is, the CBB vertex disappears. Now draw the lowest order Feynman diagram (just one) for the decay C ( p1 ) → B( p 2 ) + B( p3 ) . 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 g3 is 0 (g1,2 are not 0). Calculate, in the CM frame, the dependence of A( p1 ) + B( p 2 ) → A( p 3 ) + B( p 4 ) scattering cross section

dσ on the d?

r r scattering angle θ and initial momentum magnitude p ≡ p1 = 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 g1 and g2 are both 0 (g3 is not 0). So we come back to the original ABC model. Calculate again, in the CM frame, the dependence of A( p1 ) + B ( p 2 ) → A( p 3 ) + B ( p 4 ) scattering cross section

dσ on the scattering angle θ d?

r r and initial momentum magnitude p ≡ p1 = 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 = mC = m D . In this model, we assume only the interaction:

LI = 1 g ? ABCD . 2

a. Write down the Feynman rule for the interaction vertex. Consider the scattering process A( p1 ) + B ( p 2 ) → C ( p3 ) + D( p 4 ) . Calculate, in the CM frame, the dependence of A( p1 ) + B ( p 2 ) → C ( p3 ) + D( p 4 ) scattering cross section dσ on the scattering angle d?

r r θ and initial momentum magnitude p ≡ p1 = 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( p1 ) + 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 ? ( p1 ) → e ? ( p 2 ) + γ ( p 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 ? ( p1 ) + ? ? ( p 2 ) → e ? ( p 3 ) + ? ? ( p 4 ) + γ 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.

? φ1 ? ?φ ? r ? x? ? ? Consider a set of 3 real scalar fields Φ ≡ ? φ 2 ? . Think of them as Φ ≡ ? φ y ? , components of an ?φ ? ?φ ? ? 3? ? z?

imaginary 3 vector and then O(3) transformation as the imaginary 3 dimensional rotation. Assume its Lagrangian is:

L=

r r 2 1 r ?r 1 2r r ??Φ ?? Φ + ? Φ ?Φ ? λ Φ ?Φ . 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 r 1 r r r r 2 1 r r 1 r H = π ? π + ?Φ ? ?Φ ? ? 2 Φ ? Φ + λ Φ ? Φ 2 2 2

(

)

And for constant Φ, the potential is r r r r 2 1 V (Φ ) = ? ? 2 Φ ? Φ + λ Φ ? Φ . 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 r r π = 0, ?Φ = 0, Φ = constant . Due to presence of negative mass square, the minimum of the energy does not occur at Φ = 0 . And you will see very soon a minimum at nonzero Φ break

the symmetry O(3).

(

)

r r r r 2 r 1 a. The value of the potential V (Φ) = ? ? 2 Φ ? Φ + λ Φ ? Φ depend only on the length of Φ : 2 r r r r Φ ? Φ = φ12 + φ 22 + φ32 . Find the value v 2 of Φ ? Φ that gives a minimum of the potential. b. You can now see that the minimum occurs at a sphere (in the (φ1 , φ 2 , φ3 , ) space) satisfying r r the condition Φ ? Φ = φ12 + φ 22 + φ32 = 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

(

)

?0? ? ? direction Φ 0 ≡ ? 0 ? . Now we can study the effects of O(3) on this minimum. O(3) is just ?v? ? ?

3D rotation. The generators of O(3) are just the angular momenta. So a small “rotation”

? 0 0 0 ? ? φ1` ? ? ? ? ? around x axis will change the field by ?Φ = L x Φ = ? 0 0 ? i ? × ? φ 2 ? , a small “rotation” ?0 i 0 ? ?φ ? ? ? ? 3? ? 0 0 i ? ? φ1 ? ? ? ? ? around y axis will change the field by ?Φ = L y Φ = ? 0 0 0 ? × ? φ 2 ? and a small ? ? i 0 0 ? ?φ ? ? ? ? 3? ? 0 ? i 0 ? ? φ1` ? ? ? ? ? “rotation” around y axis will change the field by ?Φ = L z Φ = ? i 0 0 ? × ? φ 2 ? . Which ?0 0 0? ?φ ? ? ? ? 3?

rotation will keep the minimum we choose invariant, ie. ?Φ 0 ≡ L x , y , z Φ 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 Φ 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.

相关文章:

更多相关标签:

- 媒体称经调查山西人大代表4妻10子情况属实
- (1)按照现代的粒子理论,可以将粒子分为三大类：媒介子、轻子和强子,质
- 图19-2-8中互相垂直的两条虚线MN及M′N′表示两种介质的分界面及
- 如图所示,同种介质中两列简谐机械波相对传播,实线表示的波向右传播,虚线
- 如图所示,虚线a、b、c代表电场中的三个等势面,实线为一带正电的粒子仅
- 在图中虚线表示两种介质的界面及其法线,实线表示一条光线斜射向界面后发生
- 媒体素养教育是什麼东东ppt
- 如图所示,虚线...表示匀强电场中的4个等势面。两个带电粒子M.N(重
- 图中的虚线a、b、c、d表示匀强电场中的4个等势面。两个带电粒子M、N
- 如图所示,虚线a、b、c、d表示匀强电场中的4个等势面.两个带电粒子M
- 实例教你说课技巧
- FBI阅人术—用最短的时间了解一个人
- 电大复习
- 北师大英语词根记忆法
- 营养师教材(第二篇 食物营养与食品卫生)