Home >Documents >Physics 772 Peskin and Schroeder Problem 3sandick/7640fall2013/Hmwk4Sols.pdf · Physics 772 Peskin...

# Physics 772 Peskin and Schroeder Problem 3sandick/7640fall2013/Hmwk4Sols.pdf · Physics 772 Peskin...

Date post:17-Feb-2018
Category:
View:284 times
Transcript:
• Physics 772

Peskin and Schroeder Problem 3.4

Problem 3.4a) We start with the equation @ m2 = 0. Define

RL

(! ,

! ) = 1

!

!2

!

!2

RR

(! ,

! ) = 1

!

!2 +

!

!2

Remember we showed in class (and it is shown in the text) that if L

transformsas a left-handed Weyl fermion, then 2

L

transforms as a right-handed fermion.Furthermore, remember that it was shown in the text and in the notes that11

2 1

2=

. Since is the lower-right block of , and since 12is

given in block-diagonal form by RL

in the upper-left block and RR

in the lowerright block, it follows that R1

R

RL

=

. Thus,

@ ! (1)

@

=

RR

R1L

(1)

@

= RR

R1L

@

We thus find

@ m2 ! (RR

R1L

@RL

mRR

2

= RR

( @ m2)

So this equation is Lorentz invariant.We have

@ = m2

@ = m2

We thus have

@

@

2

= m2(m2)= m2

We may write @2 = @

@

. This gives us

(@2 +m2) = 0

1

• b) We have

S =

Zd4x[ @+ m

2(T2 2)]

S =

Zd4x[(@

) m2(2 T2)]

Where we have used the fact that = . Partially integrating the first termgives

S =

Zd4x[@

m2(2 T2)]

= S

Varying with respect to a

gives us the Euler-Lagrange equation

(a

)ab

@

b

m2(

b

2ba

(a

) + (a

)2ab

b

) = 0

Moving the a

all the way to the left gives

@ m2 = 0

Similarly, varying with respect to a

yields

(@

b

)ba

(a

) =m

2(

b

2ba

(a

) + (a

)2ab

b

)

= mb

2ba

(a

)

We then get

= m(2)T

()@

= m2

which is the conjugate of the Majorana equation.c) We may write the Dirac Lagrangian as

L = (/@ m) =

L

@ L

+ R

@ R

m( L

R

+ R

L

)

Substituting in L

= 1, R = i22, we find

L = 1 @1 + (T2 2) @(i22)m(1(i

22) + (T2 2)1)= 1 @1 + T2 @2 m(

1

22 T2 21)= 1 @1 +

2 @2 + m(T2 21

1

22)

Note that if we take 2 = 1 we get (up to a factor of 2) the same Lagrangianwe had in part b).

2

• d) We showed in class that the Lagrangian was invariant under the symmetry ! e . Since this symmetry multiplies

L

and R

by the same phase, wefind the global symmetry

1 ! e12 ! e2

Plugging into the Lagrangian, it is clear that the phases cancel in each term.For the theory of part b), we find

@

J = @

()

= (@

)+ @= ( @) + @= m(2) +m2

= 2m

• but we do want to consider the possibility of an odd number of fields. So we cantake the theory from part b) as our starting point. Consider the Lagrangian

L =NX

i=1

i

@i

+m

2(T

i

2i

i

2i

)

Nothing in the Lagrangian acts on the i index, so its easy to see that, if M isa matrix in the fundamental representation of O(N) (so MTM = 1, and M isreal), then this Lagrangian is invariant under

i

! M ij

j

.e) So we quantize the Majorana theory of parts a) and b) by taking

{a

(x),b

(y)} = ab

(3)(x y)

Now for the Lagrangian in part b), the kinetic term (the term involving deriva-tives) is the same as for the Weyl theory of the left-handed spinor. So we maydefine

a

(x) = a

(x).

{a

(x),b

(y)} = ab

(3)(x y)

As we remember from computing the Hamiltonian for the Dirac theory (or anytheory where the kinetic term is linear), the time derivative term drops fromthe Hamiltonian, and we are only left with the other terms. So we get

H =

Zd3x[!

2(T2 2)]

=

Zd3x

2[!

!

[email protected] m(T2 2)]

=

Zd3x

2[(@0m2) T (@0 m2)m(T2 2)]

=

Zd3x

2[@0+

Zd3x[@0]

where in the second line we took the hermitian conjugate, and then interchangedthe two fields. Remember that satisfies the Klein-Gordon equation, we knowit satisfies the dispersion relation !2 = k2 +m2. So as usual, we expand insolutions of the wave equation (here, the Majorana equation), with annihilationoperators multiplying the solutions with space-time dependence epx. We thusget

H =

ZdN! a

N

aN

up to a normal-ordering constant. Equivalently, we can note from part c) thatthe Majorana Lagrangian is the same as the Dirac Lagrangian with the identi-fication

R

(x) = 2 L

(x). This is the same as saying C (x)C = (x). Thisidentification reduces implies the identification b!

p

= a!p

; in other words, iden-tifying a Majorana particle with its anti-particle. This means that the Hamil-tonian for the Majorana Lagrangian can be written in terms of the oscillatorswe had defined for the Dirac theory in the same way, but with the number ofoscillators reduced by a factor of two due to the identification.

4

• Physics 772

Peskin and Schroeder Problem 3.5

Problem 3.5

L = @@+ @+ F F

Under

= T2 = F + @2

F = @

we find

L = @@(T2) + @(T2)@+ @(F + @2)+(F + @2) @+ F ( @) + ( @)F

= @[email protected]+ (@)[email protected] @(F + @2)+(F + T2 @) @+ F ( @) + (@)F

= @[email protected]+ (@)[email protected] @( @2)+(T2 @) @

' T2(@2) 2(@2) + [email protected]@ [email protected]@

' 0

where the ' is used after integration by parts, where the total derivative termshave been dropped. So the change in the Lagrangian is indeed a total divergence.

b) We have

L =

mF +

1

2

mT2

+ (c.c.)

1

• Under the variation given above we have

(L) = m(T2)F +m( @) + 12

mT2(F + @2)

+

1

2

m(F + @2)T2+ (c.c.)

= mT2F m @+ 12

mT2F +1

2

mT @

+

1

2

m(TF 2T @)2+ (c.c.)

= mT2F m @+ 12

mT2F +1

2

mT @

+

1

2

mT2F 12

[email protected]+ (c.c.)

=

1

2

mT2F +1

2

mT @1

2

mT2F +1

2

[email protected]+ (c.c.)

=

1

2

ma2abbF +

1

2

[email protected]

1

2

ma2abbF +

1

2

[email protected]+ (c.c.)

= 0

We now have

L = @@+ @+ F F +mF +mF +1

2

mT2 12

m2

The Euler-Lagrange equations for F and F are then

F = mF = m

Substituting in gives us

L = @@m2+ @+m

2

(T2 2)

In this equation we see that the fermion has a mass term of the form given in

problem 3.4, and the mass is the same as that of the scalar.

c) We now have

L = @i @i + i @i + F

i Fi

+

Fi

@W []

@i+

2

@2W []

2j + c.c.

The top line of the Lagrangian is just the Lagrangian we considered in part a),

and we know that it is invariant under supersymmetry up to a total divergence.

2

• The variation of the second line is given by

L2 = [email protected] []

@i+ Fi

@2W []

2

@3W []

Ti

2j

+

2

@2W []

@[email protected]((i)

T2j + Ti

2j) + c.c.

= ( @i)@W []

@i+ Fi

@2W []

@[email protected](T2j) +

2

@3W []

@[email protected]@k(T2k)Ti 2j

+

2

@2W []

@[email protected]((Fi + @i2)T2j + Ti 2(Fj + @j2)) + c.c.

= @[email protected] []

@i+ Fi

@2W []

@[email protected](T2j) +

2

@3W []

@[email protected]@k(T2k)Ti 2j

T2j

+

2

@2W []

@[email protected]((2)[email protected] + Ti [email protected]) + c.c.

= @[email protected] []

@i+

1

2

@3W []

@[email protected]@k(a2ab

bk)(

ci

2cd

dj )

+

2

@2W []

@[email protected]([email protected] + Ti @j) + c.c.

= [email protected]@2W []

1

2

@3W []

@[email protected]@k(abk

ci

+

2

@2W []

@[email protected](aab

bj +

aj

ba

b)@i + c.c.

where in the last line we performed a partial integration of the first term, and

have explicitly expanded out the 2 Pauli matrices in the second term. Notethat the second term vanishes; one can see this by interchanging bk and

ci ,

while switching the b and c and k and i indices (and similarly for the second function pair). The remaining terms cancel, giving us

L2 = 0

If W = g3/3, then the Euler-Lagrange equations for F and F are

F = g2

F = g2

yielding

L = @@+ @ g2()2 + gT2 g2

The Euler-Lagrange equations for and are then given by

@2+ 2g22 + g2 = 0

@ 2g2 = 0

3

• Physics 772

Peskin and Schroeder Problem 3.6

Problem 3.6a) Peskin and Schroeder already give the 16 Lorentz structures, so we need tonormalize them properly. The list is: 1, 0, i, 5, 05, i5, 0i, ij . Onecan see that the trace of the product of any two dierent structures will vanishby explicit computation.

b) We may write

(u1Au2)(u3

Bu4) = ua1u

b2u

c3u

d4Mabcd

Mabcd = Aab

Bcd

Define the 16 matrices (Gcb) by (Gcb)ad = Gabcd, where a, b are the matrixindices and c and d are the labels of the 16 matrices. We then find

AabBcd =

X

r,s

sb

=X

rs

where the matrices Hrs are defined by (Hrs)cb = rcsb . Since the 16 Lorentz

structures we defined above define a complete set of 4x4 matrices (any 4x4 ma-trix is a linear combination of the matrices above), we can express the matricesHrs and Grs instead in that basis and write

AabBcd =

X

C,D

Dcb

(u1Au2)(u3

Bu4) =X

C,D

CABCD(u1Cu4)(u3

Du2)

for some coecients CCDAB .We may then write

(Aab)(Bcd) = C

ABCD(

Dcb)

(Aab)(Bcd)(

Cda)(

Dbc) = C

ABCD(

Dcb)(

Cda)(

Cbc)

(Aab)(Dbc)(

Bcd)(

Cda) = C

ABCD(

Cda)(

Dcb)(

Dbc)

Summing over indices a, b, c and d gives us

CABCD =1

1

• c) For (u1u2)(u3u4), we essentially have 1 = 2 = 1. So the completenessrelation tells us

(u1u2)(u3u4) =1

4

X

C

(u1Cu4)(u3

Cu2)

For (u1u2)(u3u4), note that the metric sign in the case of spatial indices( = 1, 2, 3) has the same eect as the factor in our basis of matrices. Since,up to this , we have A = B , we only have a non-zero CABCD if C = D (to seethis, just commute the A over to the B ... for all of our structures (A)2 = 1,so we are left with Tr[CD] ). Moreover, we are summing over the .

So for example, we see that the C = terms will cancel, since each onecommutes with for two choices of and anti-commutes for two choices. Thischoices which commute give an opposite sign in the trace from those whichanti-commute, so the sum over all terms vanishes.

We need only remember that 1 commutes with all and 5 anti-commuteswith all of them. anti-commutes with three of them and commutes with one,while 5 commutes with three and anti-commutes with one.

It follows that

(u1u2)(u3u4) =

X

C

(C)(u1Cu4)(u3

Cu2)

where (C) = 1 for C = 1, (C) = 1 for C = 5, (C) = 12 for C =

and (C) = 12 for C = 5.

2

of 9/9
Physics 772 Peskin and Schroeder Problem 3.4 Problem 3.4 a) We start with the equation ı ¯ σ · - ımσ 2 χ = 0. Deﬁne R L ( - ! , - ! β ) = 1 - ı - ! · - ! σ 2 - - ! β · - ! σ 2 R R ( - ! , - ! β ) = 1 - ı - ! · - ! σ 2 + - ! β · - ! σ 2 Remember we showed in class (and it is shown in the text) that if L transforms as a left-handed Weyl fermion, then σ 2 L transforms as a right-handed fermion. Furthermore, remember that it was shown in the text and in the notes that -1 1 2 γ μ 1 2 = μ γ . Since ¯ σ μ is the lower-right block of γ μ , and since 1 2 is given in block-diagonal form by R L in the upper-left block and R R in the lower right block, it follows that R -1 R ¯ σ μ R L = μ ¯ σ . Thus, ¯ σ · @ ! ¯ σ μ (-1 ) μ @ = μ R R ¯ σ R -1 L (-1 ) μ @ = R R ¯ σ R -1 L @ We thus ﬁnd ı ¯ σ · - ımσ 2 χ ! ı ¯ (R R σR -1 L · @ R L χ - ımR R σ 2 χ = R R (ı ¯ σ · - ımσ 2 χ ) So this equation is Lorentz invariant. We have ı ¯ σ · = ımσ 2 χ -ı ¯ σ · = ımσ 2 χ We thus have ıσ μ ¯ σ @ μ @ χ = ımσ μ @ μ σ 2 χ = ımσ 2 ¯ σ μ@ μ χ = ımσ 2 (-mσ 2 χ) = -ım 2 χ We may write @ 2 = σ μ ¯ σ @ μ @ . This gives us (@ 2 + m 2 )χ = 0 1
Embed Size (px)
Recommended

Documents

Documents

Documents

Documents

Documents

Documents

Documents

Documents

Documents

##### Study about Parent Essential Oil by Prof. Brian Peskin

Health & Medicine

Documents

Documents

Documents

Documents

Documents

Documents

Documents

Documents