Examination 1 1

8/13/2019 Examination 1 1

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Physics 513, Quantum Field Theory

Examination 1Due Tuesday, 28th October 2003

Jacob Lewis BourjailyUniversity of Michigan, Department of Physics, Ann Arbor, MI 48109-1120

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2 JACOB LEWIS BOURJAILY

1. a) We are to verify that in the Schodinger picture we may write the total momentum operator,

P=

d3x (x)(x),

in terms of ladder operators as

P= d3p

(2)3 papap.

Recall that in the Schrodinger picture, we have the following expansions for the fields and in terms of the bosonic ladder operators

(x) =

d3p

(2)312Ep

eipx

ap+ap

; (1.1)

(x) =

d3p

(2)3 (i)

Ep2

eipx

ap ap

. (1.2)

To begin our derivation, let us compute (x).

(x) = d3p

(2)3

12Ep apeipx +apeipx,=

d3p

(2)312Ep

ipape

ipx ipapeipx

,

=

d3p

(2)312Ep

ipeipx

ap ap

.

Using this and (1.2) we may write the expression for P directly.

P=

d3x (x)(x),

=

d3x

d3kd3p

(2)61

2

EkEp

pei(p+k)x

ak a

k

ap ap

,

=

d3kd3p

(2)61

2

EkEp

p(2)3(3)(p + k)

ak ak

ap a

p

,

=

d3p

(2)31

2p

ap ap

ap ap

.

Using symmetry we may show that apap= apa

p. With this, our total momentum becomes,

P=

d3p

(2)31

2p

apap+apap

.

By adding and then subtracting apap inside the parenthesis, one sees that

P= d3p

(2)3

1

2p 2apap+ [ap, ap],

=

d3p

(2)3p

apap+ [ap, ap]

.

Unfortunately, we have precisely the same problem that we had with the Hamiltonian: thereis an infinite baseline momentum. Of course, our justification here will be identical to the oneoffered in that case and so

P =

d3p

(2)3 p apap. (1.3)

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PHYSICS 513: QUANTUM FIELD THEORY EXAMINATION 1 3

b) We are to verify that the Dirac charge operator,

Q=

d3x (x)(x),

may be written in terms of ladder operators as

Q= d3p

(2)3s

aspa

sp b

spb

sp

.

Recall that we can expand our Dirac s in terms of fermionic ladder operators.

a(x) =

d3p

(2)312Ep

eipxs

aspu

sa(p) +b

spv

sa(p)

; (1.4)

b(x) =

d3p

(2)312Ep

eipxr

arpu

rb (p) +b

rpv

rb (p)

. (1.5)

Therefore, we can compute Q by writing out its terms explicitly.

Q=

d3x (x)(x),

=

d3xd3kd3p

(2)61

2

EkEpei(pk)x

r,s

ark u

rb (k) +b

rkv

rb (k)

aspu

sa(p) +b

spv

sa(p)

,

=

d3kd3p

(2)61

2

EkEp(2)3(3)(p k)

r,s

ark u

rb (k) +b

rkv

rb (k)

aspu

sa(p) +b

spv

sa(p)

,

=

d3p

(2)31

2Ep

r,s

arpu

rb (p) +b

rpv

rb (p)

aspu

sa(p) +b

spv

sa(p)

,

=

d3p

(2)31

2Ep

r,s

arpa

spu

rb (p)u

sa(p) +a

rpb

spu

rb (p)v

sa(p)

+brpaspv

rb (p)u

sa(p) +b

rpb

spv

rb (p)v

sa(p) ,

=

d3p

(2)31

2Ep

r,s

arpa

spu

rb (p)u

sa(p) +b

rpb

spv

rb (p)v

sa(p)

,

=

d3p

(2)31

2Ep2Ep

rsr,s

arpa

sp+b

rpb

sp

,

=

d3p

(2)3

s

aspa

sp+b

spb

sp

,

We note that by symmetry bspbsp = b

spb

sp . By using its anticommutation relation to rewrite

bspbsp and then dropping the infinite baseline energy as we did in part (a), we see that

Q = d3p(2)3 s

aspasp bspbsp. (1.6)o


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2. a) We are to show that the matrices

(J) =i

,

generate the Lorentz algebra,

[J

, J

] =i (g

J

g

J

g

J

+g

J

) .We are reminded that matrix multiplication is given by (AB) = AB

. Recall that in

homework 5.1, we showed that

(J) =i

g g

.

Let us proceed directly to demonstrate the Lorentz algebra.

[J, J] =

g g

+

g g

,

= g

1

+ g

2

+ g

3

g

4

+g

5

g

6

g

7+

g

8,

=(g

g

1&8

)g + (g

g

2&6

)g

+ (g

g

3&7

)g (g

g

4&5

)g,

[J, J] =i (gJ gJ gJ +gJ) . (2.1)

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b) Like part (a) above, we are to show that the matrices

S

=

i

4 [

,

],generate the Lorentz algebra,

[S, S] =i (gS gS gS +gS) .

As Pascal wrote, I apologize for the length of this [proof], for I did not have time to make itshort. Before we proceed directly, lets outline the derivation so that the algebra is clear. First,we will fully expand the commutator ofS withS. We will have 8 terms. For each of thoseterms, we will use the anticommutation identity = 2g to rewrite the middle ofeach term. By repeated use of the anticommutation relations, it can be shown that

= + 2(g g +g g +g g), (2.2)

This will be used to cancel many terms and multiply the whole expression by 2 before we contractback to terms involving Ss. Let us begin.

[S, S ] = 1

16([ , ]) ,

=1

16([, ] [, ] [, ] + [, ]) ,

=1

16( +

+ + ) ,

=1

16

2g 2g +

2g + + 2g

2g + + 2g

+ 2g 2g +.


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Now, the rest of the derivation is a consequence of (2.2). Because each term isequal to its complete antisymmetrization together with six g -like terms, all termsnot involving the metric tensor will cancel each other. When we add all of the contributionsfrom all of the cancellings, sixteen of the added twenty-four terms will cancel each other andthe eight remaining will have the effect of multiplying each of the g -like terms by two. Soafter this is done in a couple of pages of algebra that I am not courageous enough to type, thecommutator is reduced to

[S, S] = 1

4(g( ) g( ) +g ( ) +g( )) .

[S, S] =i (gS gS gS +gS) . (2.3)

o

c) We are to show the explicit formulations of the Lorentz boost matrices () along thex3 directionin both vector and spinor representations. These are generically given by

() =ei2J

,

whereJ are the representation matrices of the algebra and parameterize the transforma-tion group element.

In the vector representation, this matrix is,

() =

cosh() 0 0 sinh()0 1 0 00 0 1 0

sinh() 0 0 cosh()

. (2.4)

In the spinor representation, this matrix is

() =

cosh(/2) sinh(/2) 0 0 0

0 cosh(/2) + sinh(/2) 0 00 0 cosh(/2) + sinh(/2) 0

0 0 0 cosh(/2) sinh(/2)

So,

() =

e/2 0 0 0

0 e/2 0 0

0 0 e/2 0

0 0 0 e/2

. (2.5)

d) No components of the Dirac spinor are invariant under a nontrivial boost.

e) Like part (c) above, we are to explicitly write out the rotation matrices ( ) corresponding to arotation about the x3 axis.

In the vector representation, this matrix is given by

() =

1 0 0 00 cos sin 00 sin cos 00 0 0 1

. (2.6)In the spinor representation, this matrix is given by

() =

ei/2 0 0 0

0 ei/2 0 0

0 0 ei/2 0

0 0 0 ei/2

. (2.7)

f) The vectors are symmetric under 2 rotations and so are unchanged under a complete rotation.

Spinors, however, are symmetric under 4 rotations are therefore only half-way back under a2 rotation.


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3. a) Let us define the chiral transformation to be given by ei5

. How does the conjugatespinor transform?

We may begin to compute this transformation directly.

= 0,

= (ei5

)0,

=ei5

0.

When we expand ei5

in its Taylor series, we see that because 0 anticommutes with each ofthe5 terms, we may bring the0 to the left of the exponential with the cost of a change in thesign of the exponent. Therefore

ei5

. (3.1)

b) We are to show the transformation properties of the vector V = .We can compute this transformation directly. Note that 5 anticommutes with all .

V = ei5

ei5

,

=

ei5

ei5

,= = V .

Therefore,

V V. (3.2)

c) We must show that the Dirac Lagrangian L = (i m) is invariant under chiral trans-formations in the the massless case but is not so when m = 0.

Note that because the vectors are invariant, . Therefore, we may directly computethe transformation in each case. Let us say that m = 0.

L= i L = iei

5

ei5

,

= iei5

ei5

,

= i= L.Therefore the Lagrangian is invariant ifm = 0. On the other hand, ifm= 0,

L= i m L = i e

i5mei5

,

= i me2i5 =L.

It is clear that the Lagrangian is not invariant under the chiral transformation generally.

d) The most general Noether current is

j = L

()(x)

L

()(x) L

x,

where is the total variation of the field and x is the coordinate variation. In the chiral

transformation,x = 0 and is the Dirac spinor field. So the Noether current in our case isgiven by,

j5 = L

() +

L

().

Now, first we note that

L

()= i and

L

()= 0.

To compute the conserved current, we must find . We know =ei5

(1 +i5),so i5. Therefore, our conserved current is

j5 =5. (3.3)

Note that Peskin and Schroeder write the conserved current as j5 = 5. This is essen-tially equivalent to the current above and is likewise conserved.


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e) We are to compute the divergence of the Noether current generally (i.e. when there is a possiblynon-zero mass). We note that the Dirac equation implies that = im and

=im. Therefore, we may compute the divergence directly.

j5 =(

)5 5,

=()5+ 5,

=im5 im5,

j5 =i2m

5. (3.4)

Again, this is consistent with the sign convention we derived for j5 but differs from Peskin andSchroeder.

4. a) We are to find unitary operatorsCandPand an anti-unitary operator Tthat give the standardtransformations of the complex Klein-Gordon field.

Recall that the complex Klein-Gordon field may be written

(x) = d3p

(2)3

12Ep apeipx +bpeipx;(x)

d3p

(2)312Ep

ape

ipx +bpeipx

.

We will proceed by ansatz and propose each operators transformation on the ladder operatorsand then verify the transformation properties of the field itself.

ParityWe must to define an operator P such that P(t, x)P = (t, x). Let the parity transfor-

mations of the ladder operators to be given by

PapP =aap and PbpP

=bbp.

We claim that the desired transformation will occur (with a condition on ). Clearly, thesetransformations imply that

P(t, x)P =

d3p

(2)312Ep

aape

ipx +b bpe

ipx

(t, x).

If we want P(t, x)P = (t, x) up to a phase a, then it is clear that a must equal b in

general. More so, however, if we want true equality we demand that a= b = 1.

Charge ConjugationWe must to define an operator Csuch that C(t, x)C =(t, x). Let the charge conjugation

transformations of the ladder operators be given by

CapC =bp and CbpC

=ap.

These transformations clearly show that

C(t, x)C =

d3p

(2)312Ep

bpe

ipx +apeipx

=(t, x).

Time ReversalWe must to define an operator T such that T(t, x)T =(t, x). Let the anti-unitary time

reversal transformations of the ladder operators be given by

TapT =ap and TbpT

=bp.

Note that when we act with Ton the field, because it is anti-unitary, we must take the complexconjugate of each of the exponential terms as we bring Tin. This yields the transformation,

T(t, x)T

= d3p

(2)3

12Ep apeipx +bpeipx =(t, x).


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b) We are to check the transformation properties of the current

J =i[() ()],

underC, P, and T. Let us do each in turn.

ParityNote that under parity, .

PJP =Pi[() ()]P,

=i[PPPP PPPP],

=i[(t, x)((t, x)) ((t, x))(t, x)],

PJP =J. (4.1)

Charge Conjugation

CJC =Ci[() ()]C,

=i[CCCC CCCC],

=i[() ()],

CJC =J. (4.2)

Time ReversalNote that under time reversal, and that T is anti-unitary.

TJT =Ti[() ()]T,

=i[TTTT TTTT],

=i[TT(TT) + (T

T)TT],

=i[(t, x)((t, x)) ((t, x))(t, x)],

TJT =J. (4.3)

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