Physics 742, Standard Model: Homework #2 Solution

Physics 742, Standard Model: Homework #2 Solution

1 HIggs triplet model

(a) Problem 2.5 of the textbook [2.5.1] Since hypercharge, Y , is given to 0, the covariant derivative D? becomes,

D? ? - ig2W?ata - iY g1B?

= ? - ig2W?ata,

(1)

where

W3

W?ata

=

1

(W 1

2

+

W 2)

0

1 (W 1 - W 2)

2

0 1 (W 1 + W 2)

2

W3 W+

0

0

1 (W 1 - W 2)

2

W3

?

=

W

-

0

W

+

(2)

0 W- W3

?

The Lagrangian has a form of :

L

1 2

(D?)

(D?

)

(3)

If we expand non-derivative terms in the Lagrangian, we get

L (g2W?ata)(g2W a,?ta)

1[(W 3)2 + W +W -]1 + 2[W -W + + W +W -]2 + 3[w-W + + (W 3)2]3

+ 1W 3W +2 + 1(W +)23 + 2W +W 33

+ (c.c)

(4)

Using the charge conservation in each vertex, we can deduce the charge relations for 3 components of the higgs field. The off-diagonal terms gives :

q1 - q2 = 1

q1 - q3 = 2

q2 - q3 = 1

(5)

1

Since those equations produce two identities, we need the fact that the hypercharge is the average of electric charges;

Y = 0 = q1 + q2 + q3

(6)

From Eq.(5), (6), we can get q1 = 1, q2 = 0, q3 = -1.

[2.5.2] To see if the given VEV respect the Uem(1) gauge group, let us rotate infinitesimally the fields along the generator, Q. First, for the higgs doublet, , the hypercharge is 1/2, and the electric charge is given as,

1

1 1 0 1 10

10

Q = 2 t3 + Y = 2

0

-1

+ 2

0

1

=

0

0.

(7)

Under infinitesimal rotations,

ei3t3+Y

(1 + i3t3 + Y )

= + ,

(8)

where 3 = . The part becomes,

10 0

0

00

v 2

=

0

(9)

Thus, 0 is invariant under the Uem(1) rotation. Similarly, for higgs triplet, ,

ei3t3

(1 + i3t3)

= +

(10)

The part becomes,

10 0

0

0

0

0

0

1

(u + iw)

=

0

(11)

2

0 0 -1

0

0

Therefore, this VEV respect Uem(1) gauge group.

[2.5.3] When the triplet higgs field is minimum, = min, the only surviving one among the diagonal components in Eq.(4) is the second term,

L g22 1 (u + iw)[W -W + + W -W +] 1 (u + iw)

22

2

= g22 (u2 + w2)[W -W + + W -W +]

(12)

4

2

From the results, we get the mass of W ? contributed from the triplet higgs field, .

m2W ?

=

g22 (u2 2

+

w2),

(13)

while the masses of other bosons, Z, A are unchanged.

The mass expression for the four bosons are:

m2W ?

=

g22v2 4

+

g22 (u2 2

+

w2)

(14)

m2Z

=

v2 4

(g12

+ g22)

(15)

m2A = 0

(16)

Since cos isn't affected by the supplement of the higgs triplet field, , cos remains unchanged. However, the mass of mW has been changed, the mass relation MW = MZ cos doesn't hold.

(b) A Yukawa type coupling for neutrinos are given by,

yaL?( a) Lc .

(17)

Here, L denotes SU(2) doublet for electron field, E, and neutrino, . Let us explicitly expand this coupling in terms of each field. First, summation over a reads

y1L?(1) Lc + y2L?(2) Lc + y3L?(2) Lc .

(18)

Here, i, i are replaced by i, i to avoid confusion later. Now we expand the lepton doublet with each component of . For 1,

? E?

01 10

01 -1 0

c = -?c + E?Ec. Ec

(19)

For 2,

? E? 0 -i

0 1 c = i(?c + E?Ec).

(20)

i 0 -1 0 Ec

And lastly,

? E? 1 0

0 1 c = ?Ec + E?c.

(21)

0 -1 -1 0 Ec

Adding up all the components, we get the final result,

y1(-?c + E?Ec) + iy2(?c + E?Ec) + y3(?Ec + E?c).

(22)

3

The charges for each lepton field are shown below :

Field Q , ?, c 0

E -1 E?, Ec +1

As you can see, with the charge assignment of the higgs triplet, , that we have found previously, the charge conservation at some vertices doesn't work. Now let us fix the charge violation issue addressed above. We can rewrite the Eq.(22),

L = -y(1 - i2)?c + y(1 + i2)E?Ec + y3(?Ec + E?c)

=

- 2y0?c

+

2y-2

E?Ec

+

y-1(?Ec

+

E?c).

(23)

Here, we defined charge eigenstates for original components,

0 1 (1 - i2)

(24)

2

-1 3

(25)

-2 1 (1 + i2).

(26)

2

Then,

we

can

compute

-

?

- .

-

?

-

=

0

1 +

0 -i2 + 3

0

1 0

i2 0

0 -3

=

3

1 - i2

1 + i2 -3

-1

20

=

(27)

2-2 --1

To make charge at all vertices conserved, each charge eigenstate should be assigned charges as follows :

Q0 = 0

(28)

Q-1 = -1

(29)

Q-2 = -2.

(30)

By assuming that this interaction doesn't break electric charge, our new VEV should be invariant under Uem(1) rotations,

00

0 00

0

0

=

0

-1

0

0-1

=

-0-1

=

0

(31)

0 0 -2 0-2

-20-2

0

4

Thus we can see that only non-zero component of VEV is 00, for example,

v

2

0

=

0

(32)

0

Putting this new VEV into Eq.(23), the term with 0 only survives and we have

L -yv?c,

(33)

and the mass of the neutrino field reads,

m = 2yv.

(34)

From the argument that the hypercharge is identical to the average of charges, we can deduce that Y = -1. Now let us find the eigenvalue of T3 by taking commutator of T3 with . We first define corresponding to the charge eigenstate of ,

?

1

2

(1

i2)

(35)

so

that

-

?

-

for

original

components

and

for

charge

eigenstates

are

identical,

11 + 22 + 33 = -0 + 3-1 + +-2.

(36)

Then

the

eigenvalues

can

be

obtained

from

the

commutator

[T3,

i],

where

T3

=

1 2

3

,

[ 3 , -]0 2

=

[ 3 , 2

1 (1

2

+

i2)]0

=

1 (1

2

+

i2)0

=

-0

(37)

[

3 2

,

3]-1

=

0

(38)

[ 3 2

,

+]-2

=

[ 3 2

,

1

2

(1

-

i2)]-2

=

1

-

2

(1

-

i2)-2

=

- +-2.

(39)

From the results above, we get T3 for each component,

1

T3

=

0

(40)

-1

Therefore the charge expression Q = T3 + Y ,

1

-1

0

Q = T3 + Y

=

0

+

-1

=

-1

,

(41)

-1

-1

-2

5

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