PHYSICS 513: QUANTUM FIELD THEORY HOMEWORK 4 1

Physics 513, Quantum Field Theory

Homework 4
Due Tuesday, 30th September 2003
Jacob Lewis Bourjaily

1. We have defined the coherent state by the relation

(Z )
d3 k ηk a†k
|{ηk }i ≡ N exp √ |0i.
(2π)3 2Ek
For my own personal convenience throughout this solution, I will let
Z
d3 k ηk a†k
A≡ √ .
(2π)3 2Ek
£ ¤ η
a) Lemma: ap , eA = √ p eA .
2Ep
proof: First we note that from simple Taylor expansion (which is justified here),
A2 A3
eA = 1 + A + + + ...
2 3!
Clearly ap commutes with 1 so we may write,
£ ¤ 1 1
ap , eA = [ap , A] + [ap , A2 ] + [ap , A3 ] + . . . ,
2 3!
1 1 ¡ ¢
= [ap , A] + ([ap , A]A + A[ap , A]) + [ap , A]A2 + A[ap , A]A + A[ap , A]A + . . . ,
µ 2 2 3 4
3! ¶
∗ A A A
= [ap , A] 1 + A + + + + ... ,
2 3! 4!
= [ap , A]eA .
Note that the step labelled ‘*’ is unjustified. To allow the use of ‘*’ we must show that
[ap , A] is an invariant scalar and therefore commutes with all the A’s. This is shown by
direct calculation.
Z
d3 k ηk
[ap , A] = √ [ap , a†k ],
(2π)3 2Ek
Z
d3 k ηk
= √ p − ~k),
(2π)3 δ (3) (~
(2π)3 2Ek
ηp
=p .
2Ep
η
This proves what was required for ‘*.’ √ p is clearly a scalar because η and Ep are real
2Ep
numbers only. But by demonstrating the value of [ap , A] we can complete the proof of the
required lemma. Clearly,
£ ¤ ηp
ap , eA = [ap , A]eA = p eA .
2Ep
‘ ’
£ ¤ óπ²ρ ²́δ²ι δ²Äιξαι
A A A
It is clear from the definition of the commutator that ap e = ap , e + e ap . Therefore it
is intuitively obvious, and also proven that
ap |{ηk }i = N ap eA |0i,
¡£ ¤ ¢
= N ap , eA + eA ap |0i,
ηp
=Np |0i + N eA ap |0i,
2Ep
ηp
∴ ap |{ηk }i = p ap |{ηk }i. (1.1)
2Ep
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óπ²ρ ²́δ²ι δ²Äιξαι
2 JACOB LEWIS BOURJAILY

b) We are to compute the normalization constant N so that h{ηk }|{ηk }i = 1. I will proceed
by direct calculation.
1 = h{ηk }|{ηk }i,
R d3 k √
ηk ak

= N ∗ h0|e (2π)3 2Ek

|{ηk }i,
R d3 k ηk
√
= N ∗ h0|e (2π)3 2Ek
|{ηk }i
because we know that ak |{ηk }i = √ηk |{ηk }i. So clearly
2Ek
R d3 k
2
ηk

1 = |N |2 e (2π)3 2Ek ,
R d3 k
2
ηk
− 12
∴N =e (2π)3 2Ek .
c) We will find the expectation value of the field φ(x) by direct calculation as before.
Z
d3 p 1 ¡ i~p·~x ¢
φ(x) = h{ηk }|φ(x)|{ηk }i = h{ηk }| 3
p ap e + a†p e−i~p·~x |{ηk }i,
(2π) 2Ep
 
Z
d3 p 1  
 h{ηk }|ap ei~p·~x |{ηk }i + h{ηk }|a†p e−i~p·~x |{ηk }i,
= p 
(2π) 3
2Ep | {z } | {z }
act with ap to the right act with a†p to the left
Z Ã !
d3 p 1 ηp i~
p·~
x ηp −i~
p·~
x
= p p e +p e ,
(2π)3 2Ep 2Ep 2Ep
Z
d3 p ηp
= cos(~
p · ~x).
(2π)3 Ep
d) We will compute the expected particle number directly.
Z
d3 p †
N = h{ηk }|N |{ηk }i = h{ηk }| a ap |{ηk }i,
(2π)3 p
Z µ ¶
d3 p †
= h{η k }|a a
p p |{η k }i ,
(2π)3 ←−−−− −−−−→
Z
d3 p ηp2
= .
(2π)3 2Ep
e) To compute the mean square dispersion, let us recall the theorem of elementary probability
theory that
2
h(∆N )2 i = N 2 − N .
We have already calculated N so it is trivial to note that
Z 3 3
2 d kd p ηk2 ηp2
N = .
(2π)6 4Ek Ep
Let us then calculate N 2 .
Z 3 3
d kd p †
N 2 = h{ηk }|N 2 |{ηk }i = h{ηk }| a ak a†p ap |{ηk }i,
(2π)6 k
Z 3 3
d kd p ηk ηp
= p h{ηk }|ak a†p |{ηk }i,
(2π)6 2 Ek Ep
Z 3 3
d kd p ηk ηp ³ 3 (3) ~ †
´
= p (2π) δ (k − p
~) + h{η k }|ap a k |{ηk }i ,
(2π)6 2 Ek Ep
Z Z 3 3
d3 k ηk2 d kd p ηk2 ηp2
= + .
(2π)3 2Ek (2π)6 4Ek Ep
It is therefore quite easy to see that
Z
2 2 d3 k ηk2
h(∆N ) i = N2 −N = .
(2π)3 2Ek
 PHYSICS 513: QUANTUM FIELD THEORY HOMEWORK 4 3

2. We are given the Lorentz commutation relations,

[J µν , J ρσ ] = i(g νρ J µσ − g µρ J νσ − g νσ J µρ + g µσ J νρ ).
a) Given the generators of rotations and boosts defined by,
1
Li = ²ijk J jk K i = J 0i ,
2
we are to explicitly calculate all the commutation relations. We are given trivially that
[Li , Lj ] = i²ijk Lk .
Let us begin with the K’s. By direct calculation,
[K i , K j ] = [J 0i , J 0j ] = i(g 0i J 0j − g 00 J ij − g ij J 00 + g 0j J i0 ),
= −iJ ij ;
= −2i²ijk Lk .
Likewise, we can directly compute the commutator between the L and K’s. This also will
follow by direct calculation.
1
[Li , K j ] = ²lk [J ilk , J 0j ],
2
1
= ²ilk i(g l0 J ij − g i0 J lj − g lj J i0 + g ij J l0 ),
2
= i²ijk J 0k ;
= i²ijk K k .
We were also to show that the operators
i 1 1
J+ = (Li + iK i ) J−i
= (Li − iK i ),
2 2
could be seen to satisfy the commutation relations of angular momentum. First let us
compute,
1£ i ¤
[J+ , J− ] = (L + iK i ), (Lj − iK i ) ,
4
1¡ i j ¢
= [L , L ] + i[K i , Lj ] − i[Li , K j ] + [K i , K j ] ,
4
= 0.
In the last line it was clear that I used the commutator [Li , K j ] derived above. The next
two calculations are very similar and there is a lot of ‘justification’ algebra in each step.
There is essentially no way for me to include all of the details of every step, but each can be
verified (e.g. i[K i , Lj ] = −i[Lj , K i ] = (−i)i²jik K k = −²ijk K k ...etc). They are as follows:
i j 1£ i ¤
[J+ , J+ ]= (L + iK i ), (Lj + iK j ) ,
4
1¡ i j ¢
= [L , L ] + i[K i , Lj ] + i[Li , K j ] + i[Li , K i ] − [K i , K j ] ,
4
1 ¡ ijk k ¢
= i² L − ²ijk K k − ²ijk K k + i²ijk Lk ,
4
1
= i²ijk (Lk + iK k ) = i²ijk J+ k
.
2
Likewise,
i j 1£ i ¤
[J− , J− ]= (L − iK i ), (Lj − iK j ) ,
4
1¡ i j ¢
= [L , L ] − i[K i , Lj ] − i[Li , K j ] + i[Li , K i ] − [K i , K j ] ,
4
1 ¡ ijk k ¢
= i² L + ²ijk K k + ²ijk K k + i²ijk Lk ,
4
1
= i²ijk (Lk − iK k ) = i²ijk J− k
.
2
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4 JACOB LEWIS BOURJAILY

b) Let us consider first the (0, 12 ) representation. For this representation we will need to satisfy
i 1 i 1 i σi
J+ = (L + iK i ) = 0 i
J− = (L − iK k ) = .
2 2 2
σi iσ i
This is obtained by taking Li = 2 and K i = 2 . The transformation law then of the (0, 12 )
representation is
µν
Φ(0, 12 ) −→ e−iωµν J Φ(0, 12 ) ,
i
Li +β j K j )
= e−i(θ Φ(0, 21 ) ,
i i j j
− iθ 2σ + β 2K
=e Φ(0, 12 ) .
σi i
The calculation for the ( 12 , 0) representation is very similar. Taking Li = 2 and K i = − σ2 ,
we get
i 1 σi 1
J+ = (Li + iK i ) = J−i
= (Li − iK k ) = 0.
2 2 2
Then the transformation law of the representation is
µν
Φ( 12 ,0) −→ e−iωµν J Φ( 12 ,0) ,
i
Li +β j K j )
= e−i(θ Φ( 21 ,0) ,
i i j j
− iθ 2σ − β 2K
=e Φ( 12 ,0) .
Comparing these transformation laws with Peskin and Schroeder’s (3.37), we see that
ψL = Φ( 12 ,0) ψR = Φ(0, 12 ) .
3. a) We are given that Ta is a representation of some Lie group. This means that
[Ta , Tb ] = if abc Tc
by definition. Allow me to take the complex conjugate of both sides. Note that [Ta , Tb ] =
[(−Ta ), (−Tb )] in general and recall that f abc are real.
[Ta , Tb ]∗ = (if abc Tc )∗ ,
[Ta∗ , Tb∗ ] = −if abc Tc∗ ,
∴ [(−Ta∗ ), (−Tb∗ )] = if abc (−Tc∗ ).
So by the definition of a representation, it is clear that (−Ta∗ ) is also a representation of the
algebra.
b) As before, we are given that Ta is a representation of some Lie group. We will take the
Hermitian adjoint of both sides.
[Ta , Tb ]† = (if abc Tc )† ,
(Ta Tb )† − (Tb Ta )† = −if abc Tc† ,
Tb† Ta† − Ta† Tb† = −if abc Tc† ,
[Tb† , Ta† ] = −if abc Tc† ,
∴ [Ta† , Tb† ] = if abc Tc† .
So by the definition of a representation, it is clear that Ta† is a representation of the algebra.
a
c) We define the spinor representation of SU (2) by Ta = σ2 so that
µ ¶ µ ¶ µ ¶
1 0 1 1 0 −i 1 1 0
T1 ≡ T2 ≡ T3 ≡ .
2 1 0 2 i 0 2 0 −1
We will consider the matrix S = iσ 2 . Clearly S is unitary because (iσ 2 )(iσ 2 )† = 1. Now,
one could proceed by direct calculation to demonstrate that
µ ¶ µ ¶ µ ¶
1 0 −1 1 0 −i 1 −1 0
ST1 S † = = −T1∗ ST2 S † = = −T2∗ ST3 S † = = −T3∗ .
2 −1 0 2 i 0 2 0 1
This clearly demonstrates that the representation −Ta∗ is equivalent to that of Ta .
 PHYSICS 513: QUANTUM FIELD THEORY HOMEWORK 4 5

i i
d) From our definitions of our representation of SO(3, 1) using J+ and J− , it is clear that
i † i
(J+ ) = J− .
This could be expressed as if ( 21 , 0)† = (0, 12 ), or, rather L† = R. So what we must ask
ourselves is, does there exist a unitary matrix S such that
SLS † = L but SKS † = −K ?
If there did exist such a unitary transformation, then we could conclude that L and R
are equivalent representations. However, this is not possible in our SO(3, 1) representation
because both L and K are represented strictly by the Pauli spin matrices so that iK = L =
σ
2 . It is therefore clear that there cannot exist a transformation that will change the sign
of K yet leave L alone. So the representations are inequivalent.
‘ ’
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