QFT I Problem Homework

Homework 16, QFT I Professor: Sergey Kovalenko Assistant Professor: Marcela Gonzalez Student: Daniel Salinas-Arizmendi Universidad Técnica Federico Santa María

July 5, 2016

Problem 1. Prove: 0 (a) y = 0 : 0 0 y = : (b) Solution 1. (a) The chiral representation of the gamma matrix is given by: 0

= where

= (1; ~ ) and 0

0

(1)

0

= (1; ~ ) : So 0

0

=

0

0 1

=

0

0

0

0

1 0

0

0 1

0 0

=

0

0 1

0

0

0

(2)

1 0

1 0

0

=

0

Moreover y

= =

y

0

(3)

0 0 (

)

1

y

( y

) 0

!

but, (

y

) = 1y ; ~ y

where we used

k y

= (1; ~ ) = =

y

) = 1y ; ~ y

, and (

= (1; ~ ) =

;

(Pauli matrix are hermitian) y

0

=

(4)

0

Therefore 0

(b) De…ning

=

i 2

0

0

y

=

(5)

; 0

i 2 i 2 i 2 i 2 i 2 i 2

= = = = = =

0

0

0

0

0

0 0

y

y y

( $ ) 0

( $ )

( $ ) y

y

y y y

i 2

= =

i 2

=

y

where we used the previous result we used following notation

(6)

y

0

0

=

y

and

0 0

Xa Yb + Xb Ya = Xa Yb + (a $ b) :

(7)

Problem 2. Show that: 1 (a) y5 = 5 = 5 = ( 5 ) : i (b) 5 = 4! : 2 (c) ( 5 ) = 1 y 0 (d) 5 = 0 5 : 1 Demostration,

0:

used chiral representation of 0 0

=

0 1

1 0

0 1

1 0

2

=

1 0

0 1

= 11 : Furthermore

=1

Solution 2. (a) We used di…nition y 5

3 y

= i

5

2 y

=

i 1 y

0 1 2 3:

So

0 y

(8)

0 3 0 0 2 0 0 1 0 0 0 0

= i = i 0 = i = +i = i = 5 where we used 0 f ; g = 2g :

0

=(

5

= = =

3 2 1 0 2 3 1 0 2 1 3 0 1 2 3

y

0 0

) ,

i 0 ig00 i 0

= i 0 5 =

= 1 and the anticommutation relations

(9)

1 2 3 0 g11 1 g22 2 g33 3 1 2 3

1 2 3

where we used for up and down indices the metric tensor de…ned by g (b) We …nd

= = = = = = =

= = =

i 4! i 0 k + k 0 4! i 0 i j k k 0 k i + k0 + ki 0ijk 4! i 0 i j k + k0ij k 0 i j + ki0 k i 0 + 0ijk 4! i 0 i j k + k0ij k 0 i j + ki0j k i 0 j + 0ijk 4! i 4 0ijk 0 i j k 4! i 0 1 j k + 02jk 0 2 j k + 03jk 0 3 j k 01jk 3! i 0 1 2 3 + 0132 0 1 3 2 + 0213 0 2 1 3 0123 3! + 0231 0 2 3 1 + 0312 0 3 1 2 + 0321 0 3 2 1 i 6 0 1 2 3 3! i 0 1 2 3 5

3

=diag(1; 1; 1; 1) :

(10)

kij

kij0

k i j

k i j 0

where we used the de…nition fo the symbol (c) Also 5 = i 0 1 2 3 = i 0 1 2 3 2 5)

(

=

0123

1:

0 1 2 3

= = = = = = = = = =

(11)

0 1 2 3 1 0 3 2

3

3 3 3

0 1 1 3 0 2 0 1 0 2 0 1 3 1 2 0 0 2 1 0 0 1 2 1 1 2 3 2 2 3

2 3 1 2 3 2 3 1 2 3 2 3

3 3

1

where we used the anticommutation relation of the gamma matrices and that 0 k k 0 = 1 and k = 1: Demotration of k =1 k k

k k

= 1 0

(d) We used

0

0

=(

5

0

0

k

(12)

0

k

0

k

k

0

= =

k

0

=

k

0 1

1 y

) and y

y 5

=

5;

y 5

=

y

=

0

0

=

0

0

= i

we have (13)

5

0

=

0

=

0

i

0 1 2 3

0 1 2 3 0 0 5 0 5

Problem 3. Show that: (a) f 5 ; g = 0: (b) [ 5 ; ] = 0: Solution 3. (a) Using the de…nition commutator 0 1 2 + 5 5 =i 4

5 3

=i

0 1 2 3

+i

; we have to the anti-

0 1 2 3

(14)

For

=0 5;

0

0 1 2 3 0

= i

= i = i 0 = i = 0 For

1

1 0 2 3 0 0 1 2

0 1 2 3 1

= i

= i = i 0 = 0

+i +i 0 3 +i

(15)

0 0 1 2 3 0 1 2 3 0 0 1 2 3

1 0 1 2 3

+i

0 1 2 1 3

(16)

0 1 1 2 3

i

1 1 2 3

i

0 1 1 2 3

0 1 2 3 2

+i

2 0 1 2 3

=2 5;

2

= i = = =

For

0 1 2 0 3

=1 5;

For

0 0 1 2 3

+i

i i

0 1 2 2 3 0 1 2 2 3

(17)

0 2 1 2 3

i +i

0 1 2 2 3

0

=3 5;

3

We conclude that f 5 ; (b) Using the fact that [A; BC]

5;

]

= = = =

0 1 2 3 3

+i

3 0 1 2 3

0 1 2 3 3 0 1 2 3 3

0 1 3 2 3

0 1 2 3

i +i 3 i

0 3 1 2 3

= i = i = i = 0

0 1 2 3 3

(18)

g = 0: = ABC BCA = (AB + BA) C B (AC + CA) = fA; Bg C B fA; Cg

And using the de…nition [

= i

=

i [ ; 2 5 i [ ; 2 5 i (f 5 ; 2 0

i 2

[

;

] ; we have ]

]

(20)

( $ )

g

f

5

(19)

5;

g)

( $ )

we used that f

5;

g = 0: 2

Problem 4. Prove a/ = a2 Solution 4. We have a/a/ =

a a 1 a a ( 2 g a a a a a2

= = = =

(21) +

)

Problem 5. Derive the following identities tith contractions of the gamma matrices: (a) = 4: (b) = 2 : (c) = 4g : (d) = 2 : (e) = 12: 5 (f ) = 4: 5 (g) = 0: (h) = 4 : 5 (i) = 0: 5 (j) = 12 5 : Solution 5. (a) We have = =

0 0

1

+

1

+

2 2

+

3 3

(22)

4

(b) Using the previous result we …nd = = = =

(

+ 2g + 2g

4 2

)

(23)

+2

(c) Using the previous results we have =

+ 2g

=

+ 2g

= 2 = 2 =

+2 ;

4g 6

(24)

(d) Using the previous results we …nd =

(

+ 2g

=

)

(25)

+ 2g

= =

4g 2

=

2

+2 +g

(e) Using the previous results we have = = = = =

i2 4

( 1 4 1 4

)

(26) +

2

4

4

1 (2 4 + 4 4 + 4 4 4 16

2

2 4)

(f ) We have 5 5

5

= =

4

where we used the anticonmutation relation f (g) Using the previous results we …nd 1 4 1 g 4 1 g 4 1 g 4 1 g 4

= = = = = =

(27)

5

5;

g = 0, and

5

5

= 1:

(28) + 4g

2

4g 4

4 4

4 4

+4

0

7

2 +

4g

+

4g

(h) Using the previous results we have = = =

i 8 i 8 i 8

(

) + +

+ =

= = = (i) Using that [

i ( 2 8 4g i 16 ( 8 i 4 [ ; 2 4 5;

5;

+ 4g

+2

2

+ 4g

+ 4g +2

) ]

] = 0 and previous resulst we …nd 5

(j) Using that [

(29)

5

= = =

5

(30) 0

0

] = 0 and

= 12

5

5

= =

12

(31) 5

Problem 6. Prove the following identities with traces of gamma matrices: (a) Tr = 0: (b) Tr = 4g : (c) Tr = 4 (g g g g + g g ): (d) Tr( 5 ) = 0: (e) Tr 5 = 0: (f ) Tr a/1 a/2n+1 = 0: (g) Tr 5 = 0: Solution 6. (a) As (

2 5)

Tr

= 1; we can say = Tr =

Tr

8

(32)

5 5 5

5

where used that

;

5

= 0: Now we remember the traza properties2 T r (ABC) = T r (BCA)

(33)

We have Tr

So the only option is Tr (b) We have

=

Tr

=

Tr

(34)

5 5

=0

Tr

1 Tr + Tr 2 1 Tr ; 2 1 T r (2g 1) 2 g T r (1) 4g

= = = = =

(35)

(c) Using the previous results we …nd Tr

=

Tr

+ T r 2g

(36)

= Tr

8g g

=

+ 8g g

8g g

Tr

+ 8g g + 8g g

or 2T r

= 8g g

8g g

+ 8g g

(37)

Tr

= 4 (g g

g g

+g g )

(38)

or (d) Analogous to part (a) Tr (

So the only option is T r (

5)

5)

= T r ( 5 0 0) = T r ( 0 0 5) = T r ( 5)

(39)

= 0:

2 Demostration:

T r (ABC)

= =

X

X

n

hnjABCjni =

m

hmjBCAjmi = T r (BCA)

X

n;m

hnjAjmi hmjBCjni =

9

X

n;m

hmjBCjni hnjAjmi

(e) Using the previuos results we have Tr

= Tr

5

= Tr = = =

(40)

5 5

T r 5 4g 4g T r ( 5 ) 0

(f ) Using the previuos results we …nd T r a/1

a/2n+1

= Tr

/1 5 5a

a/2n+1

=

Tr

/1 5 5a

=

T r a/1

a/2n+1

=

T r a/1

a/2n+1

(41)

a/2n+1

5

5 5

So the only option is T r a/1 a/2n+1 = 0: (f ) Using the previuos results we have Tr

= =

5

where we used Tr(odd number of

iT r

0

Problem 7. Calculate Tr (p/

(42)

0 1 2 3

0 s) = 0 (part (e)).

m)

(1

5 ) (q/ +

m)

:

Solution 7. We have T r (p/

m)

= T r p/ q/ mT r = T r p/ q/ = T r (p/1/q/1) /

(1

5 ) (q/ +

+ mT r p/

T r p/

2

q/

m Tr T r p/

m) + mT r

mT r p/ 5 q/

5

4m2 g

5q

where we used Tr[a/b/c/d] / = 4 [(a b) (c d) + (a d) (b c) previuos results (1

+ m Tr

m Tr

= 4 [(p 1) (q 1) + (p 1) (1 q) (p q) (1 1)] + 4i = 4 p q +p q (p k) g + i p q 4m2 g

Problem 8. Calculate

5

2

2

5 q/

Tr p

5 q/

5 ) (p/

10

m)

:

p q

4m2 g

(a c) (b d)] and the

Solution 8. Using the previous results (prob.5 part (a) ; (b))

= = = =

(1 p/

m) 5 ) (p/ m +m 5 p/ 5 p 4m p m 5 5 2 p 4m 2 5 p 4m 5 2p/ 4m 2 5 p/ 4m 5

11

(43)