Physics 513, Quantum Field Theory

Examination 1
Due Tuesday, 28
th
October 2003
Jacob Lewis Bourjaily
University of Michigan, Department of Physics, Ann Arbor, MI 48109-1120
1
2 JACOB LEWIS BOURJAILY
1. a) We are to verify that in the Schodinger picture we may write the total momentum operator,
P =
_
d
3
x (x)(x),
in terms of ladder operators as
P =
_
d
3
p
(2)
3
p a

p
a
p
.
Recall that in the Schrodinger picture, we have the following expansions for the elds and
in terms of the bosonic ladder operators
(x) =
_
d
3
p
(2)
3
1
_
2E
p
e
ipx
_
a
p
+ a

p
_
; (1.1)
(x) =
_
d
3
p
(2)
3
(i)
_
E
p
2
e
ipx
_
a
p
a

p
_
. (1.2)
To begin our derivation, let us compute

(x).
(x) =
_
d
3
p
(2)
3
1
_
2E
p
_
a
p
e
ipx
+ a

p
e
ipx
_
,
=
_
d
3
p
(2)
3
1
_
2E
p
_
ipa
p
e
ipx
ipa

p
e
ipx
_
,
=
_
d
3
p
(2)
3
1
_
2E
p
ipe
ipx
_
a
p
a

p
_
.
Using this and (1.2) we may write the expression for P directly.
P =
_
d
3
x (x)(x),
=
_
d
3
x
d
3
kd
3
p
(2)
6
1
2

E
k
E
p
pe
i(p+k)x
_
a
k
a

k
__
a
p
a

p
_
,
=
_
d
3
kd
3
p
(2)
6
1
2

E
k
E
p
p(2)
3

(3)
(p +k)
_
a
k
a

k
__
a
p
a

p
_
,
=
_
d
3
p
(2)
3
1
2
p
_
a

p
a
p
_
_
a
p
a

p
_
.
Using symmetry we may show that a
p
a

p
= a
p
a

p
. With this, our total momentum becomes,
P =
_
d
3
p
(2)
3
1
2
p
_
a

p
a
p
+ a
p
a

p
_
.
By adding and then subtracting a

p
a
p
inside the parenthesis, one sees that
P =
_
d
3
p
(2)
3
1
2
p
_
2a

p
a
p
+ [a
p
, a

p
]
_
,
=
_
d
3
p
(2)
3
p
_
a

p
a
p
+ [a
p
, a

p
]
_
.
Unfortunately, we have precisely the same problem that we had with the Hamiltonian: there
is an innite baseline momentum. Of course, our justication here will be identical to the one
oered in that case and so
P =
_
d
3
p
(2)
3
p a

p
a
p
. (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 =
_
d
3
x

(x)(x),
may be written in terms of ladder operators as
Q =
_
d
3
p
(2)
3

s
_
a
s
p
a
s
p
b
s
p
b
s
p
_
.
Recall that we can expand our Dirac s in terms of fermionic ladder operators.

a
(x) =
_
d
3
p
(2)
3
1
_
2E
p
e
ipx

s
_
a
s
p
u
s
a
(p) + b
s
p
v
s
a
(p)
_
; (1.4)

b
(x)

=
_
d
3
p
(2)
3
1
_
2E
p
e
ipx

r
_
a
r
p
u
r
b
(p) + b
r
p
v
r
b
(p)
_
. (1.5)
Therefore, we can compute Q by writing out its terms explicitly.
Q =
_
d
3
x

(x)(x),
=
_
d
3
x
d
3
kd
3
p
(2)
6
1
2
_
E
k
E
p
e
i(pk)x

r,s
__
a
r
k
u
r
b
(k) + b
r
k
v
r
b
(k)
__
a
s
p
u
s
a
(p) + b
s
p
v
s
a
(p)
__
,
=
_
d
3
kd
3
p
(2)
6
1
2
_
E
k
E
p
(2)
3

(3)
(p k)

r,s
__
a
r
k
u
r
b
(k) + b
r
k
v
r
b
(k)
__
a
s
p
u
s
a
(p) + b
s
p
v
s
a
(p)
__
,
=
_
d
3
p
(2)
3
1
2E
p

r,s
__
a
r
p
u
r
b
(p) + b
r
p
v
r
b
(p)
__
a
s
p
u
s
a
(p) + b
s
p
v
s
a
(p)
__
,
=
_
d
3
p
(2)
3
1
2E
p

r,s
_
a
r
p
a
s
p
u
r
b
(p)u
s
a
(p) + a
r
p
b
s
p
u
r
b
(p)v
s
a
(p)
+b
r
p
a
s
p
v
r
b
(p)u
s
a
(p) + b
r
p
b
s
p
v
r
b
(p)v
s
a
(p)
_
,
=
_
d
3
p
(2)
3
1
2E
p

r,s
_
a
r
p
a
s
p
u
r
b
(p)u
s
a
(p) + b
r
p
b
s
p
v
r
b
(p)v
s
a
(p)
_
,
=
_
d
3
p
(2)
3
1
2E
p
2E
p

rs

r,s
_
a
r
p
a
s
p
+ b
r
p
b
s
p
_
,
=
_
d
3
p
(2)
3

s
_
a
s
p
a
s
p
+ b
s
p
b
s
p
_
,
We note that by symmetry b
s
p
b
s
p
= b
s
p
b
s
p
. By using its anticommutation relation to rewrite
b
s
p
b
s
p
and then dropping the innite baseline energy as we did in part (a), we see that
Q =
_
d
3
p
(2)
3

s
_
a
s
p
a
s
p
b
s
p
b
s
p
_
. (1.6)

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4 JACOB LEWIS BOURJAILY
2. a) We are to show that the matrices
(J

= i
_

_
,
generate the Lorentz algebra,
[J

, J

] = i (g

+ g

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

= A

. Recall that in
homework 5.1, we showed that
(J

= i
_
g

_
.
Let us proceed directly to demonstrate the Lorentz algebra.
[J

, J

] =
_

_
_
g

_
+
_

_
_
g

_
,
=

. .
1
+

. .
2
+

. .
3

. .
4
+

. .
5

. .
6

. .
7
+

. .
8
,
= (

. .
1&8
)g

+ (

. .
2&6
)g

+ (

. .
3&7
)g

. .
4&5
)g

,
[J

, J

] = i (g

+ g

) . (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 (g

+ g

) .
As Pascal wrote, I apologize for the length of this [proof], for I did not have time to make it
short. Before we proceed directly, lets outline the derivation so that the algebra is clear. First,
we will fully expand the commutator of S

with S

. We will have 8 terms. For each of those

terms, we will use the anticommutation identity

= 2g

to rewrite the middle of

each term. By repeated use of the anticommutation relations, it can be shown that

+ 2(g

+ g

+ g

), (2.2)
This will be used to cancel many terms and multiply the whole expression by 2 before we contract
back to terms involving S

s. Let us begin.
[S

, S

] =
1
16
([

]) ,
=
1
16
([

] [

] [

] + [

]) ,
=
1
16
(

) ,
=
1
16
_
2g

2g

2g

+ 2g

2g

+ 2g

+ 2g

2g

_
.
PHYSICS 513: QUANTUM FIELD THEORY EXAMINATION 1 5
Now, the rest of the derivation is a consequence of (2.2). Because each

term is
equal to its complete antisymmetrization

together with six g

-like terms, all terms

not involving the metric tensor will cancel each other. When we add all of the contributions
from all of the cancellings, sixteen of the added twenty-four terms will cancel each other and
the eight remaining will have the eect of multiplying each of the g

-like terms by two. So

after this is done in a couple of pages of algebra that I am not courageous enough to type, the
commutator is reduced to
[S

, S

] =
1
4
(g

) g

) + g

) + g

)) .
[S

, S

] = i (g

+ g

) . (2.3)

o


c) We are to show the explicit formulations of the Lorentz boost matrices () along the x
3
direction
in both vector and spinor representations. These are generically given by
() = e

i
2

,
where J

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 0
0 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 0
0 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 a
rotation about the x
3
axis.
In the vector representation, this matrix is given by
() =
_
_
_
_
1 0 0 0
0 cos sin 0
0 sin cos 0
0 0 0 1
_
_
_
_
. (2.6)
In the spinor representation, this matrix is given by
() =
_
_
_
_
e
i/2
0 0 0
0 e
i/2
0 0
0 0 e
i/2
0
0 0 0 e
i/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 a
2 rotation.
6 JACOB LEWIS BOURJAILY
3. a) Let us dene the chiral transformation to be given by e
i
5
. How does the conjugate
spinor

transform?
We may begin to compute this transformation directly.

0
,
= (e
i
5
)

0
,
=

e
i
5

0
.
When we expand e
i
5
in its Taylor series, we see that because
0
anticommutes with each of
the
5
terms, we may bring the
0
to the left of the exponential with the cost of a change in the
sign of the exponent. Therefore

e
i
5
. (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

=

e
i
5

e
i
5
,
=

e
i
5
e
i
5
,
=

= 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 compute

the transformation in each case. Let us say that m = 0.
L =

i

=

ie
i
5

e
i
5

,
=

i

e
i
5
e
i
5

,
=

i

= L.
Therefore the Lagrangian is invariant if m = 0. On the other hand, if m = 0,
L =

i

m L

=

i

e
i
5
me
i
5
,
=

i

me
2i
5
= 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 eld and x

is the coordinate variation. In the chiral

transformation, x

= 0 and is the Dirac spinor eld. So the Noether current in our case is
given by,
j

5
=
L
(

)
+
L
(

.
Now, rst we note that
L
(

)
=

i

and
L
(

)
= 0.
To compute the conserved current, we must nd . We know

= e
i
5
(1 + i
5
),
so i
5
. Therefore, our conserved current is
j

5
=

5
. (3.3)
Note that Peskin and Schroeder write the conserved current as j

5
=

5
. This is essen-
tially equivalent to the current above and is likewise conserved.
PHYSICS 513: QUANTUM FIELD THEORY EXAMINATION 1 7
e) We are to compute the divergence of the Noether current generally (i.e. when there is a possibly
non-zero mass). We note that the Dirac equation implies that

= im and

=
im

. Therefore, we may compute the divergence directly.

5
= (

,
= (

5
+

,
= im

5
im

5
,

5
= i2m

5
. (3.4)
Again, this is consistent with the sign convention we derived for j

5
but diers from Peskin and
Schroeder.
4. a) We are to nd unitary operators C and P and an anti-unitary operator T that give the standard
transformations of the complex Klein-Gordon eld.
Recall that the complex Klein-Gordon eld may be written
(x) =
_
d
3
p
(2)
3
1
_
2E
p
_
a
p
e
ipx
+ b

p
e
ipx
_
;

(x)
_
d
3
p
(2)
3
1
_
2E
p
_
a

p
e
ipx
+ b
p
e
ipx
_
.
We will proceed by ansatz and propose each operators transformation on the ladder operators
and then verify the transformation properties of the eld itself.
Parity
We must to dene an operator P such that P(t, x)P

= (t, x). Let the parity transfor-

mations of the ladder operators to be given by
Pa
p
P

=
a
a
p
and Pb
p
P

=
b
b
p
.
We claim that the desired transformation will occur (with a condition on ). Clearly, these
transformations imply that
P(t, x)P

=
_
d
3
p
(2)
3
1
_
2E
p
_

a
a
p
e
ipx
+

b
b

p
e
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 Conjugation
We must to dene an operator C such that C(t, x)C

(t, x). Let the charge conjugation

transformations of the ladder operators be given by
Ca
p
C

= b
p
and Cb
p
C

= a
p
.
These transformations clearly show that
C(t, x)C

=
_
d
3
p
(2)
3
1
_
2E
p
_
b
p
e
ipx
+ a

p
e
ipx
_
=

(t, x).
Time Reversal
We must to dene 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
T a
p
T

= a
p
and T b
p
T

= b
p
.
Note that when we act with T on the eld , because it is anti-unitary, we must take the complex
conjugate of each of the exponential terms as we bring T in. This yields the transformation,
T (t, x)T

=
_
d
3
p
(2)
3
1
_
2E
p
_
a
p
e
ipx
+ b

p
e
ipx
_
= (t, x).
8 JACOB LEWIS BOURJAILY
b) We are to check the transformation properties of the current
J

= i[

) (

)],
under C, P, and T . Let us do each in turn.
Parity
Note that under parity,

.
PJ

= Pi[

) (

)]P

,
= i[P

PP

],
= i[

(t, x)(

(t, x)) (

(t, x))(t, x)],

PJ

= J

. (4.1)
Charge Conjugation
CJ

= Ci[

) (

)]C

,
= i[C

CC

],
= i[(

) (

],
CJ

= J

. (4.2)
Time Reversal
Note that under time reversal,

and that T is anti-unitary.

T J

= T i[

) (

)]T

,
= i[T

T T

],
= i[T

T T

) + (

)T T

],
= i[

(t, x)(

(t, x)) (

(t, x))(t, x)],

T J

= J

. (4.3)