HW 4 - Solutions
The solution for problem 2 is correct up to symmetry factors. See comments on returned homeworks
for information regarding these factors.
1
QFT Homework 4
ZJ Poh
Problem 1: Decay of a Scalar Field into Two Scalar Fields
Consider the following Lagrangian, involving two real scalar fields, Φ and φ:
L=
1
1
1
1
∂µ Φ2 − M 2 Φ2 +
∂µ φ2 − m2 φ2 − µΦφφ .
2
2
2
2
The last term is an interaction that allows a Φ particle to decay into two φ’s, provided that
M > 2m. Assuming that this condition is met, calculate the lifetime of the Φ to lowest order
in µ.
From the Lagrangian, we know the conjugate momenta for both fields:
∂L
= Φ̇
∂ Φ̇
∂L
= φ̇ .
πφ =
∂ φ̇
πΦ =
So, the Hamiltonian
Z
H=
d3 x
Z
=
d3 x
Z
=
d3 x
is
πΦ Φ̇ + πφ φ̇ − L
1
1
1
1
∂µ Φ2 + M 2 Φ2 −
∂µ φ2 + m2 φ2 + µΦφφ
2
2
2
2
1
1
1
1
1 2 1
Φ̇ +
∂i Φ2 + M 2 Φ2 + φ̇2 +
∂i φ2 + m2 φ2 + µΦφφ
2
2
2
2
2
2
≡ H0 + Hint ,
Φ̇2 + φ̇2 −
where
Hint =
Z
d3 x µΦφφ .
Using Peskin-Eq. (4.14) and Peskin-Eq. (4.19),
φI (x) = eiH0 (t−t0 ) φ(x)e−iH0 (t−t0 )
and
HI (x) = eiH0 (t−t0 ) Hint e−iH0 (t−t0 ) ,
the interaction Hamiltonian in the interacting pictures is
Z
HI (x) =
d3 x µeiH0 (t−t0 ) Φφφe−iH0 (t−t0 )
Z
=
d3 x µeiH0 (t−t0 ) Φe−iH0 (t−t0 ) eiH0 (t−t0 ) φe−iH0 (t−t0 ) eiH0 (t−t0 ) φe−iH0 (t−t0 )
Z
=
d3 x µΦI φI φI .
(1)
From here onwards, I will omit the subscript I for the interacting fields as all the fields that
we are dealing with are in the interacting picture.
1
Let the momentum of the decaying particle be k and that of the outgoing particle be p1 and
p2 . The decay rate can be calculated using Peskin-Eq. (4.86):
dΓ =
1
1
1 d3 p1 d3 p2
|M(k → p1 + p2 )|2 (2π)4 δ (4) (k − p1 − p2 ) ,
2M (2π)3 (2π)3 2Ep~1 2Ep~2
where M, the matrix element, is defined in Peskin-Eq. (4.73):
hp1 p2 |iT |ki = (2π)4 δ (4) (k − p1 − p2 )iM(k → p1 + p2 ) .
The matrix element can be calculated using Peskin-Eq. (4.90):
o
n RT
−i −T dt HI (t)
|ki
hp1 p2 |iT |ki =
lim
hp
p
|T
e
0
0 1 2
T →∞(1−i)
connected, amputated
Substituting Eq. (1), we have
n
o
R 4
hp1 p2 |iT |ki = hp1 p2 |T e−iµ d x Φφφ |ki
Z
= hp1 p2 |1 − iµ
d4 x T {Φφφ} + O(µ3 )|ki
Z
= −iµ
d4 x hp1 p2 |T {Φφφ} |ki + O(µ3 )
= −2iµ
Z
d4 x hp1 p2 |T {Φφφ}|ki + O(µ3 )
Note that terms with even powers of µ gives zero because there are odd number of fields
(including the incoming and outgoing fields). Also, I multiplied the matrix element by a
combinatorial factor of 2 because interchanging of the contractions on the φ’s produces the
same result.
Using Peskin-Eq. (4.94):
hp|φ(x) = h0|eip·x ,
the first term becomes
−2iµ
Z
d4 x ei(p1 +p2 −k)·x = −2iµ(2π)4 δ (4) (k − p1 − p2 ) ,
which implies
M = −2µ .
Now, substitute the matrix element into the decay rate and integrate the decay rate over all
outgoing momentum:
Z
1
1
1
d3 p1 d3 p2
|−2µ|2 (2π)4 δ (4) (k − p1 − p2 ) .
Γ=
3
3
2M
(2π) (2π) 2Ep~1 2Ep~2
2
In the rest frame of Φ, we have ~k = 0. So, we can split the delta function to obtain
Z
2µ2
d3 p1 d3 p2
1
1
Γ=
(2π)4 δ (3) (~p1 + p~2 )δ(E~k − Ep~1 − Ep~2 )
3
3
M
(2π) (2π) 2Ep~1 2Ep~2
Z
2µ2
d3 p
1
=
δ(M − 2Ep~ )
M
(2π)2 4Ep~2
Z
p
µ2 1 4π
2 1
2 + m2
δ
M
−
2
p
=
dp
p
2M (2π)2 2
Ep~2
where I evaluated the integral over the solid angle. Note that I divided the solid angle by 2
because the outgoing states are two identical scalar fields.
Using the following delta function identity
δ(x − x0 )
,
|f 0 (x0 )|
δ(f (x)) =
we can rewrite out delta function as
δ(p − p )
p
0
2
2
δ M −2 p +m =
2p0 /Ep~0
where
1√ 2
M − 4m2
p0 =
2
So, the decay rate becomes
and
µ2
Γ=
4πM
Z
Ep~0
dp p2
q
M
.
= p20 + m2 =
2
1 δ(p − p0 )
Ep~2 2p0 /Ep~0
µ2 p0
4πM 2Ep~0
µ2 √ 2
M − 4m2
=
8πM 2
=
So, the life-time of the decay is
τ=
1
8πM 2
= √
.
Γ
µ2 M 2 − 4m2
Feynman Rules for µΦφφ Theory:
x1
x2
= DF (x1 − x2 ) for φ field
x1
x2
= DF (x1 − x2 ) for Φ field
= −iµ
3
Z
d4 x
The Feynman diagram for this term is
x3 , p2
2 x1 , k
.
x2 , p1
4
Problem 2
The two point function for φ is
o
n
R 4
h0|T φ1 φ2 e−iµ d x Φφφ |0i
R
hΩ|T {φ1 φ2 } |Ωi =
lim
4
T →∞(1−i)
h0|T e−iµ d x Φφφ |0i
Consider
o
n
R
−iµ d4 x Φφφ
|0i
h0|T φ1 φ2 e
Z
(−i)2 2
4
4
4
= h0|T φ1 φ2 +
µ
d x3 d x4 φ1 φ2 Φ3 φ3 φ3 Φ4 φ4 φ4 + O(µ ) |0i
2!
Note that there are no term with odd power of µ because there will be odd number of fields
in between the time ordering symbol and by Wick’s theorem, we know that all those terms
are zero.
n
o
R 4
h0|T φ1 φ2 e−iµ d x Φφφ |0i
Z
(−i)2 2
µ
d4 x3 d4 x4 h0|T {φ1 φ2 Φ3 φ3 φ3 Φ4 φ4 φ4 } |0i + O(µ4 )
= h0|T {φ1 φ2 } |0i +
2!
= x1
+ 2 x1
x2 +
x1
x2 + 2 x1
x2
x2 + O(µ4 )
x2 + 4 x1
Explanation on the combinatorial factor: (Note: The factor of 1/2! from the Taylor series
will always be cancels out by exchanging the vertices.)
• Second Term: None because there is no other contraction that gives the same diagram.
Overall factor: 1
• Third Term: When contracting φ3 to φ4 to form the circle loop, there are two φ4 to
choose from.
Overall factor: 2
• Fourth Term: When contracting φ1 to φ3 , there are two φ3 to choose from.
Overall factor: 2
• Fifth Term: When contracting φ1 to φ3 , there are two ways φ3 to choose from. When
contracting φ3 to φ4 , there are two φ4 to choose from.
Overall factor: 4
5
So,
n
R
h0|T φ1 φ2 e−iµ
=
d4 x Φφφ
|0i
!
+ O(µ4 )
+2
x1
×
o
x2 + 2 x1
Consider the denominator,
n
R
h0|T e−iµ
=1+
(−i)2 2
µ
2!
=
d4 x Φφφ
Z
+2
o
!
x2 + O(µ4 )
x2 + 4 x1
|0i
d4 x1 d4 x2 h0|T {Φ1 φ1 φ1 Φ2 φ2 φ2 } |0i + O(µ4 )
+ O(µ4 )
So, the two point function for φ in Feynman diagram is
hΩ|T {φ1 φ2 } |Ωi
= x1
x2 + 2 x1
x2 + 4 x1
x2 + O(µ4 )
Note: The disconnected diagrams cancels.
The two point function for Φ is
Consider
o
n
R 4
h0|T Φ1 Φ2 e−iµ d x Φφφ |0i
R
hΩ|T {Φ1 Φ2 } |Ωi =
lim
4
T →∞(1−i)
h0|T e−iµ d x Φφφ |0i
n
o
R 4
h0|T Φ1 Φ2 e−iµ d x Φφφ |0i
Z
(−i)2 2
4
4
4
µ
d x3 d x4 Φ1 Φ2 Φ3 φ3 φ3 Φ4 φ4 φ4 + O(µ ) |0i
= h0|T Φ1 Φ2 +
2!
6
Note that there are no term with odd power of µ because of the same reasoning as above.
o
n
R
−iµ d4 x Φφφ
|0i
h0|T Φ1 Φ2 e
= x1
x2 +
x1
x2
x2 + O(µ4 )
+ 2 x1
=
x2 + 2 x1
+2
!
+ O(µ4 )
x1
x2 + 2 x1
!
x2 + O(µ4 )
The fourth term has a combinatorial factor of 2 because when contracting φ3 to φ4 , there
are two φ4 to choose from.
So, the two point function of Φ is
hΩ|T {Φ1 Φ2 } |Ωi = x1
x 2 + 2 x1
7
x2 + O(µ4 )
Problem 3
Consider the following scattering of two scalar particles to two scalar particles:
x1 , p
x3 , p0
x2 , q
x4 , q 0
The scattering matrix of this process is
n
o
R 4
Sp0 q0 ;pq = hp0 q 0 |T e−iµ d x Φφφ |pqi
Z
(−i)2 2
0 0
4
4
4
= hp q | 1 +
µ
d x5 d x6 T {Φ5 φ5 φ5 Φ6 φ6 φ6 } + O(µ ) |pqi
2!
Z
(−i)2 2
= δp0 q0 ,pq +
µ
d4 x5 d4 x6 hp0 q 0 |T {Φ5 φ5 φ5 Φ6 φ6 φ6 } |pqi + O(µ4 )
2!
Z
(−i)2 2
(S − 1)p0 q0 ;pq =
µ
d4 x5 d4 x6 hp0 q 0 |T {Φ5 φ5 φ5 Φ6 φ6 φ6 } |pqi + O(µ4 )
2!
Note: There are no terms with odd order of µ because the number of fields in the timeordering symbol is odd. By Wick’s theorem, all these terms are zero.
Applying Peskin-Eq. (4.94):
φ(x)|pi = e−ip·x |0i
hp|φ(x) = h0|eip·x
to the s-channel term, which is given by the following contraction,
2
(S − 1)p0 q0 ;pq, s-channel = (−iµ)
2
= (−iµ)
2
= (−iµ)
Z
Z
Z
Z
d4 x5 d4 x6 hp0 q 0 |T {φ6 φ6 Φ6 Φ5 φ5 φ5 } |pqi
0
0
d4 x5 d4 x6 h0|T {Φ6 Φ5 } |0ie−ip·x5 e−iq·x5 eip ·x6 eiq ·x6
0
0
d4 x5 d4 x6 DF0 (x5 − x6 )e−ip·x5 e−iq·x5 eip ·x6 eiq ·x6
i
d4 k
0
0
eik·(x5 −x6 ) e−ip·x5 e−iq·x5 eip ·x6 eiq ·x6
4
2
2
(2π) k − M + i
Z
d4 k
i
0
0
d4 x5 d4 x6
e−ix5 ·(p+q−k) eix6 ·(p +q −k)
= (−iµ)2
4
2
2
(2π) k − M + i
Z
4
dk
i
= (−iµ)2
(2π)4 δ (4) (p + q − k)(2π)4 δ (4) (p0 + q 0 − k)
(2π)4 k 2 − M 2 + i
i
= (−iµ)2
(2π)4 δ (4) (p + q − p0 − q 0 )
2
(p + q) − M 2 + i
2
= (−iµ)
d4 x5 d4 x6
Note: This expression is just for the diagram without the combinatorial factor. We need to
multiply it by the combinatorial factor.
8
To get t-channel, simply switch q and −p0 :
(S − 1)p0 q0 ;pq, t-channel = (−iµ)2
(p −
p0 ) 2
i
(2π)4 δ (4) (p + q − p0 − q 0 )
− M 2 + i
and to get u-channel, simply switch q and −q 0 .
(S − 1)p0 q0 ;pq, t-channel = (−iµ)2
(p −
q 0 )2
i
(2π)4 δ (4) (p + q − p0 − q 0 )
− M 2 + i
So, up to second order of µ,
i
(2π)4 δ (4) (p + q − p0 − q 0 )
(p +
− M 2 + i
i
+ 2(−iµ)2
(2π)4 δ (4) (p + q − p0 − q 0 )
0
2
(p − p ) − M 2 + i
i
+ 2(−iµ)2
(2π)4 δ (4) (p + q − p0 − q 0 )
0
2
(p − q ) − M 2 + i
(S − 1)p0 q0 ;pq = 2(−iµ)2
q)2
9
Instead of using Peskin-Eq. (4.94), we can also obtain this from the LSZ Formalism.
As a prerequisite, we need to know the following equations, which we derived when studying
scalar field theory (See Prof. Raby particle notes, pg. 204):
Z
d3 p
1 −i~p·~x †
e
a (p) + ei~p·~x a(p)
φ(x) =
3
(2π) 2Ep~
Z
←
→
†
a (p) =
d3 x φ† (x)i ∂t e−ip·x .
We will also be using a boundary condition statement:
Z
Z
Z
∂ 3
3
4
=
d
x
−
d
x
dx
∂x0
t=+∞
.
t=−∞
S-matrix element for the scattering is
in
Sα,pq = hψpout
0 q 0 |ψpq i
†
= hψpout
p )ψqin i
0 q 0 |ain (~
†
in
= lim hψpout
0 q 0 |a (p)ψq i
t→−∞
Z
←
→
out
= lim hψp0 q0 | d3 x φ† (x)i ∂0 e−ip·x ψqin i
t→−∞
Z
Z
←
→ −ip·x in
out
3
†
out
= lim hψp0 q0 | d x φ (x)i ∂0 e
ψq i − hψp0 q0 |
d4 x
t→+∞
←
→
∂ †
φ (x)i ∂0 e−ip·x ψqin i
0
∂x
The last line is obtained by using the boundary condition statement above.
Consider the first term,
Z
←
→
out
lim hψp0 q0 |
d3 x φ† (x)i ∂0 e−ip·x ψqin i
t→+∞
†
in
= lim hψpout
0 q 0 |a (p)ψq i
t→+∞
†
= hψpout
p)ψqin i
0 q 0 |aout (~
in
= haout (~p)ψpout
0 q 0 |ψq i
in
= δp0 ,p hψqout
0 |ψq i
So, we can rewrite the S-matrix as
Sα,pq = δp0 ,p Sq0 ,q + S p0 q0 ,qp ,
R 4
←
→
where S p0 q0 ,qp = hψpout
d x ∂x∂ 0 φ† (x)i ∂0 e−ip·x ψqin i. This term is the inseparable term be0 q0 |
cause we cannot separate it like the first term.
10
Instead of reducing only the particle with momentum p from the incoming state, we can
reduce both particles from the incoming state:
in
Sp0 q0 ,pq = hψpout
0 q 0 |ψpq i
†
p )a†in (~q )ψ0 i
= hψpout
0 q 0 |ain (~
†
†
= lim hψpout
0 q 0 |a (p)a (q)ψ0 i
t→−∞
Z
Z
←
→ −ip·x
←
→ −iq·y
out
3
†
3
†
= lim hψp0 q0 |
ψ0 i
d x φp (x)i ∂0 e
d y φq (y)i ∂0 e
t→−∞
Z
Z
←
→ −iq·y
←
→ −ip·x
3
†
3
†
out
ψ0 i
d y φq (y)i ∂0 e
d x φp (x)i ∂0 e
= lim hψp0 q0 |
t→+∞
Z
Z
←
→ −ip·x
←
→ −iq·y
∂ †
∂ †
out
4
4
− hψp0 q0 |
dx
φ (x)i ∂0 e
dy
φ (y)i ∂0 e
ψ0 i
∂x0 p
∂y 0 q
= δp0 q0 ,qp + S p0 q0 ,qp
So, we can rewrite the inseparable term as
(S − 1)p0 q0 ,qp = S p0 q0 ,qp
Calculating the inseparable term
Z
←
→ −ip·x in
∂ †
out
φ
(x)i
∂
ψq i
S p0 q0 ,qp = −hψp0 q0 |
d4 x
0e
∂x0
Z
→ −ip·x
∂
†
in ←
= −i
d4 x
hψpout
0 q 0 |φ (x)ψq i ∂0 e
0
∂x
Z
∂
out †
in
−ip·x
out
†
in −ip·x
= −i
d4 x
hψ
|φ
(x)ψ
i∂
e
−
hψ
|∂
φ
(x)ψ
ie
0
0
0
0
0
0
pq
q
pq
q
∂x0
Z
†
in
2 −ip·x
2 †
in −ip·x
= −i
d4 x hψpout
− hψpout
0 q 0 |φ (x)ψq i∂0 e
0 q 0 |∂0 φ (x)ψq ie
Z
−2 −ip·x
†
in
2 −ip·x
†
in ←
= −i
d4 x hψpout
− hψpout
0 q 0 |φ (x)ψq i∂0 e
0 q 0 |φ (x)ψq i ∂0 e
but
∂02 e−ip·x = −Ep~2 e−ip·x = (−~p 2 − m2 )e−ip·x = (∇2 − m2 )e−ip·x
So,
S p0 q0 ,qp = −i
Z
−2 −ip·x
†
in
2
2 −ip·x
†
in ←
− hψpout
d4 x hψpout
0 q 0 |φ (x)ψq i(∇ − m )e
0 q 0 |φ (x)ψq i ∂0 e
Performing integration by parts on ∇2 twice to move it to acting on φ† (x), we get (note
11
integration by parts twice so there is no change in sign)
Z
−2 −ip·x
−2
†
in ←
2 −ip·x
†
in ←
− hψpout
S p0 q0 ,qp = −i
d4 x hψpout
0 q 0 |φ (x)ψq i ∂0 e
0 q 0 |φ (x)ψq i( ∇ − m )e
Z
−2 −ip·x
←
−2
†
in
2 −ip·x
†
in ←
=i
d4 x hψpout
+ hψpout
0 q 0 |φ (x)ψq i(− ∇ + m )e
0 q 0 |φ (x)ψq i ∂0 e
Z
−
†
in ←
2 −ip·x
=i
d4 x hψpout
0 q 0 |φ (x)ψq i( x + m )e
Z
− −ip·x
†
in ←
S p0 q0 ,qp = i
d4 x hψpout
0 q 0 |φ (x)ψq iKx e
←
− ←
−
where Kx = x + m2 .
←
−
→
−
Note: 2 = 2 because the operator is squared, so the two minus signs cancel.
Now, we reduce the p0 particle from the outgoing state:
Z
−− −ip·x1
†
in ←
d4 x1 hψpout
(S − 1)p0 q0 ;pq = i
0 q 0 |φ (x1 )ψq iKx1 e
Z
−− −ip·x1
†
in ←
=i
d4 x1 ha†out (p0 )ψqout
0 |φ (x1 )ψq iKx1 e
Z
−− −ip·x1
0
†
in ←
=i
d4 x1 hψpout
0 |aout (p )φ (x1 )ψq iKx1 e
Consider the matrix element and substituting a(q 0 ):
0
†
in
out
0
†
in
hψqout
0 |aout (p )φ (x1 )ψq i = lim hψq 0 |a(p )φ (x1 )ψq i
t→+∞
Z
←
→
0
†
in
= 0lim i
d3 x3 eip ·x3 ∂x03 hψqout
0 |φ(x3 )φ (x1 )ψq i
x3 →+∞
Since x3 is at a later time than x1 , we can put in the time ordering symbol:
Z
←
→
0
out
0
†
in
†
in
hψq0 |aout (p )φ (x1 )ψq i = 0lim i
d3 x3 eip ·x3 ∂x03 hψqout
0 |T {φ(x3 )φ (x1 )}ψq i
x3 →+∞
Using the boundary condition identity:
0
†
in
hψqout
0 |aout (p )φ (x1 )ψq i
Z
←
→
0
†
in
= 0lim i
d3 x3 eip ·x3 ∂x03 hψqout
0 |T {φ(x3 )φ (x1 )}ψq i
x3 →−∞
Z
0 ←
†
in
4
ip ·x3 →
out
∂x03 hψq0 |T {φ(x3 )φ (x1 )}ψq i
+i
d x3 ∂x03 e
Writing the first term back in the previous notation and takes the derivatives of the second
term
0
†
in
0
†
in
out
hψqout
0 |aout (p )φ (x1 )ψq i = hψq 0 |ain (p )φ (x1 )ψq i
Z
0
4
†
in
+i
d x3 eip ·x3 ∂x20 hψqout
0 |T {φ(x3 )φ (x1 )}ψq i
3
2 ip0 ·x3
out
†
in
− ∂x0 e
hψq0 |T {φ(x3 )φ (x1 )}ψq i
3
12
(in)
The first term gives zero because ain (p0 ) annihilates ψq . There are no incoming states
with momentum p0 . For the second term, use the same argument as before that ∂02 e−ip·x =
(∇2 − m2 )e−ip·x :
Z
0
†
in
out
0
†
in
hψq0 |aout (p )φ (x1 )ψq i = i
d4 x3 eip ·x3 ∂x20 hψqout
0 |T {φ(x3 )φ (x1 )}ψq i
3
0
†
in
− ∇2x3 − m2 eip ·x3 hψqout
0 |T {φ(x3 )φ (x1 )}ψq i
Z
−−→
0
†
in
=i
d4 x3 eip ·x3 Kx3 hψqout
0 |T {φ(x3 )φ (x1 )}ψq i
Substituting this back into the S-Matrix element, we get
Z
−−→
−− −ip·x1
0
†
in ←
2
(S − 1)p0 q0 ;pq = i
d4 x1 d4 x3 eip ·x3 Kx3 hψqout
0 |T {φ(x3 )φ (x1 )}ψq iKx1 e
Repeat the same process to reduce q and q 0 , we get
4
(S − 1)p0 q0 ;pq = i
Z
−−→−−→
0
0
d4 x1 d4 x2 d4 x3 d4 x4 eiq ·x4 eip ·x3 Kx4 Kx3
←−−←−−
hψ0 |T {φ(x4 )φ(x3 )φ† (x2 )φ† (x1 )}|ψ0 iKx1 Kx2 e−ip·x1 e−iq·x2
Using Gell-Mann-Low Theorem, we know
hψ0 |T {φ(x4 )φ(x3 )φ† (x2 )φ† (x1 )}|ψ0 i
o
n
R
0
0
h0|T φ4 φ3 φ†2 φ†1 e−i dt H(t ) |0i
R 0 0
=
h0|T e−i dt H(t ) |0i
o
n
R
† † −iµ d4 x Φφφ
|0i
h0|T φ4 φ3 φ2 φ1 e
−iµ R d4 x Φφφ
=
h0|T e
|0i
where φ(xi ) = φi and all the fields and the Hamiltonian in RHS are in the interacting picture.
From class and argument similar to problem 2, we know that only connected and non-trivial
diagrams will survives. So,
hψ0 |T {φ(x4 )φ(x3 )φ† (x2 )φ† (x1 )}|ψ0 i
o
n
R
† † −iµ d4 x Φφφ
|0i
= h0|T φ4 φ3 φ2 φ1 e
Connected, non-trivial diagrams only
o
n
† †
= h0|T φ4 φ3 φ2 φ1 |0i
Z
n
o
(−iµ)2
+
d4 x5 d4 x6 h0|T φ4 φ3 φ†2 φ†1 Φ5 φ5 φ5 Φ6 φ6 φ6 |0i + O(µ4 )
2!
0
x1 , p
x3 , p
=4
0
x1 , p
x3 , p
+4
x2 , q
x4 , q 0
s-channel
Connected, non-trivial diagrams only
x1 , p
+ O(µ4 )
+4
x2 , q
x4 , q 0
t-channel
13
x3 , p0
x2 , q
x4 , q 0
u-channel
Note: There is a factor of 4 multiplying these diagrams because when contracting φ†1 to φ5 ,
there are two φ5 to choose from. In addition, when contracting φ6 to to the outgoing states,
there are two states to choose from.
Consider the s-channel diagram (excluding the combinatorial factor),
2
(−iµ)
= (−iµ)
2
Z
Z
Z
n
o
d4 x5 d4 x6 h0|T φ4 φ3 φ†2 φ†1 Φ5 φ5 φ5 Φ6 φ6 φ6 |0i
d4 x5 d4 x6 DF (x1 − x5 )DF (x2 − x5 )DF (x6 − x3 )DF (x6 − x4 )DF0 (x5 − x6 )
d4 p1 d4 p2 d4 p3 d4 p4 d4 k
(2π)4 (2π)4 (2π)4 (2π)4 (2π)4
i
i
i
i
i
2
2
2
2
2
2
2
2
2
p1 − m + i p2 − m + i p3 − m + i p4 − m + i k − M 2 + i
e−ip1 ·(x1 −x5 ) e−ip2 ·(x2 −x5 ) e−ip3 ·(x6 −x3 ) e−ip4 ·(x6 −x4 ) e−ik·(x5 −x6 )
Z
d4 p1 d4 p2 d4 p3 d4 p4 d4 k
2
= (−iµ)
d4 x5 d4 x6
(2π)4 (2π)4 (2π)4 (2π)4 (2π)4
i
i
i
i
i
2
2
2
2
2
2
2
2
2
p1 − m + i p2 − m + i p3 − m + i p4 − m + i k − M 2 + i
e−ip1 ·x1 e−ip2 ·x2 eip3 ·x3 eip4 ·x4 e−ix5 ·(−p1 −p2 +k) e−ix6 ·(p3 +p4 −k)
Z
d4 p1 d4 p2 d4 p3 d4 p4 d4 k
2
= (−iµ)
(2π)4 (2π)4 (2π)4 (2π)4 (2π)4
i
i
i
i
i
2
2
2
2
p1 − m2 + i p2 − m2 + i p3 − m2 + i p4 − m2 + i k 2 − M 2 + i
e−ip1 ·x1 e−ip2 ·x2 eip3 ·x3 eip4 ·x4
= (−iµ)
2
d4 x5 d4 x6
(2π)4 δ (4) (p1 + p2 − k)(2π)4 δ (4) (p3 + p4 − k)
Z
d4 p1 d4 p2 d4 p3 d4 p4
2
= (−iµ)
(2π)4 (2π)4 (2π)4 (2π)4
i
i
i
i
i
2
2
2
2
2
2
2
2
2
p1 − m + i p2 − m + i p3 − m + i p4 − m + i (p1 + p2 ) − M 2 + i
e−ip1 ·x1 e−ip2 ·x2 eip3 ·x3 eip4 ·x4
(2π)4 δ (4) (p1 + p2 − p3 − p4 )
14
Now, act the kinetic operators to the above expression:
Z
n
o ←−−←−−
−−→−−→
2
d4 x5 d4 x6 h0|T φ4 φ3 φ†2 φ†1 Φ5 φ5 φ5 Φ6 φ6 φ6 |0iKx1 Kx2
Kx4 Kx3 (−iµ)
Z
d4 p1 d4 p2 d4 p3 d4 p4
2
= (−iµ)
(2π)4 (2π)4 (2π)4 (2π)4
i
i
i
i
i
2
2
2
2
2
2
2
2
2
p1 − m + i p2 − m + i p3 − m + i p4 − m + i (p1 + p2 ) − M 2 + i
(−p4 + m2 )(−p3 + m2 )(−p2 + m2 )(−p1 + m2 )
e−ip1 ·x1 e−ip2 ·x2 eip3 ·x3 eip4 ·x4
(2π)4 δ (4) (p1 + p2 − p3 − p4 )
Z
d4 p1 d4 p2 d4 p3 d4 p4
i
2
= (−iµ)
e−ip1 ·x1 e−ip2 ·x2 eip3 ·x3 eip4 ·x4
4
4
4
4
2
(2π) (2π) (2π) (2π) (p1 + p2 ) − M 2 + i
(2π)4 δ (4) (p1 + p2 − p3 − p4 )
←−−
Note: Although Kx1 is acting to the left, but there is no negative sign multiplying it because
←−−
the derivative operator in Kx1 is second order. So, there are two minus signs and that gives
←−−
a plus sign. Similarly for Kx2 .
So, the matrix element for the s-channel is
Z
d4 p1 d4 p2 d4 p3 d4 p4
4
2
(S − 1)p0 q0 ;pq, s-channel = i (−iµ)
d4 x1 d4 x2 d4 x3 d4 x4
(2π)4 (2π)4 (2π)4 (2π)4
0
0
eiq ·x4 eip ·x3 e−ip·x1 e−iq·x2
i
e−ip1 ·x1 e−ip2 ·x2 eip3 ·x3 eip4 ·x4
2
2
(p1 + p2 ) − M + i
(2π)4 δ (4) (p1 + p2 − p3 − p4 )
Note: I moved the four exponentials into the momentum integrals because they are independent on p1 . . . p4 .
Do the x integrals to get four delta functions.
Z
i
d4 p1 d4 p2 d4 p3 d4 p4
2
(S − 1)p0 q0 ;pq, s-channel = (−iµ)
4
4
4
4
2
(2π) (2π) (2π) (2π) (p1 + p2 ) − M 2 + i
(2π)4 δ (4) (p1 + p)(2π)4 δ (4) (p2 + q)(2π)4 δ (4) (p3 + p0 )
(2π)4 δ (4) (p4 + q 0 )(2π)4 δ (4) (p1 + p2 − p3 − p4 )
i
= (−iµ)2
(2π)4 δ (4) (p + q − p0 − q 0 )
2
2
(p + q) − M + i
To get t-channel, simply switch q with −p0 :
(S − 1)p0 q0 ;pq, t-channel = (−iµ)2
(p −
p0 ) 2
15
i
(2π)4 δ (4) (p + q − p0 − q 0 )
− M 2 + i
and to get u-channel, simply switch q with −q 0 .
(S − 1)p0 q0 ;pq, t-channel = (−iµ)2
(p −
q 0 )2
i
(2π)4 δ (4) (p + q − p0 − q 0 )
− M 2 + i
So, up to second order of µ,
i
(2π)4 δ (4) (p + q − p0 − q 0 )
(p +
− M 2 + i
i
+ 4(−iµ)2
(2π)4 δ (4) (p + q − p0 − q 0 )
0
2
(p − p ) − M 2 + i
i
(2π)4 δ (4) (p + q − p0 − q 0 )
+ 4(−iµ)2
0
2
2
(p − q ) − M + i
(S − 1)p0 q0 ;pq = 4(−iµ)2
q)2
16
Problem 4: Particle Creation by a Classical Source
When there is a classical source in a free field, the Lagrangian is given by
1
1
L = (∂µ φ(x))2 − m2 φ(x)2 + j(x)φ(x) ,
2
2
(2)
where j(x) is a source. The conjugate momenta is
π=
∂L
= φ̇2
∂ φ̇
and the Hamiltonian is
Z
H=
d3 x π φ̇ − L
Z
1
1
1
=
d3 x φ̇2 − φ̇2 + (∂i φ)2 + m2 φ(x)2 − j(x)φ(x)
2
2
2
Z
1
1
1
=
d3 x φ̇2 + (∂i φ)2 + m2 φ(x)2 − j(x)φ(x)
2
2
2
Z
= H0 +
d3 x − j(x)φ(x) ,
(3)
where H0 is the Hamiltonian when the source is turned off. Also, the equation of motion is
∂L
∂L
∂µ
=0
−
∂(∂µ φ)
∂φ
∂µ ∂ µ φ − m2 φ2 − j(x) = 0
(∂ 2 + m2 )φ(x) = j(x)
(4)
Let φ0 (x) be the field when j(x) = 0, i.e. the source is turned off. From free field theory, we
know that
Z
1 −ip·x
d3 p
† ip·x
p
ap~ e
+ ap~ e
.
(5)
φ0 (x) =
(2π)3
2Ep~
When j(x) 6= 0, i.e. the source is turned on, we can write the field, i.e. the solution of Eq. (4)
as
Z
φ(x) = φ0 (x) + i
d4 y DR (x − y)j(y) .
(6)
Proof:
From Peskin-Eq. (2.56),
(∂ 2 + m2 )DR (x − y) = −iδ (4) (x − y)
This implies that DR (x − y) is a Green’s function of the Klein-Gordon Equation.
17
When we act ∂ 2 + m2 on Eq. (6), we get Eq. (4):
Z
2
2
2
2
4
(∂x + m )φ(x) = (∂x + m ) φ0 (x) + i
d y DR (x − y)j(y)
Z
2
2
d4 y (∂x2 + m2 )DR (x − y)j(y)
= (∂x + m )φ0 (x) + i
Z
=0+i
d4 y − iδ (4) (x − y)j(y)
= j(x)
So, Eq. (6) is the solution of Eq. (4).
Now, we want to combine the two terms in the RHS of Eq. (6). Using equation in the bottom
of page 29 of Peskin:
Z
d3 p
1
−ip·(x−y)
ip·(x−y)
e
−
e
h0|[φ(x), φ(y)]|0i 0= 0
x >y
(2π)3 2Ep~
and Peskin-Eq. (2.55):
DR (x − y) = θ(x0 − y 0 )h0|[φ(x), φ(y)]|0i ,
we can rewrite Eq. (6) as
0
0
φ(x) = φ0 (x) + θ(x − y )i
Z
d3 p
1
−ip·(x−y)
ip·(x−y)
dy
e
−
e
j(y) .
(2π)3 2Ep~
4
For x0 > y 0 , i.e. j(x) is turned off after it is turned on,θ(x0 − y 0 ) = 1. So,
Z
1
d3 p
−ip·(x−y)
ip·(x−y)
e
−
e
j(y)
φ(x) = φ0 (x) + i
d4 y
(2π)3 2Ep~
Z
Z
Z
1
d3 p
−ip·x
4
ip·y
ip·x
4
−ip·y
e
d y e j(y) − e
dy e
j(y) .
= φ0 (x) + i
(2π)3 2Ep~
Writing the source as its Fourier transform, we have
Z
Z
4
ip·y
∗
j(p) =
d y e j(y)
and
j (p) =
d4 y e−ip·y j ∗ (y) .
So,
φ(x) = φ0 (x) + i
Z
d3 p
1
e−ip·x j(p) − eip·x j ∗ (p) .
3
(2π) 2Ep~
Using Eq. (5),
Z
Z
d3 p
1 −ip·x
d3 p
1
† ip·x
p
φ(x) =
a
e
+
a
e
+
i
e−ip·x j(p) − eip·x j ∗ (p)
p
~
p
~
3
3
(2π)
(2π) 2Ep~
2Ep~
!
!
!
Z
d3 p
1
i
i
†
p
=
ap~ + p
j(p) e−ip·x + ap~ − p
j ∗ (p) eip·x
(2π)3
2Ep~
2Ep~
2Ep~
18
Using similar argument as the free scalar field, we can rewrite the Hamiltonian in terms of
the creation and annihilation operator:
!
!
Z
d3 p
i
i
H=
j ∗ (p)
ap~ + p
j(p)
Ep~ ap†~ − p
(2π)3
2Ep~
2Ep~
Z
d3 p
1
2
†
|j(p)|
=
Ep~ ap~ ap~ +
(2π)3
2Ep~
The energy of the system after the source has been turned off is then
Z
d3 p
1
2
†
|j(p)| |0i
Ep~ ap~ ap~ +
hHi = h0|
(2π)3
2Ep~
Z
d3 p
1
=
|j(p)|2
E
p
~
(2π)3
2Ep~
Z
d3 p 1
=
|j(p)|2 .
3
(2π) 2
The number of particles after the source has been turned off is
Z
d3 p
1
2
†
λ ≡ h0|N |0i = h0|
|j(p)| |0i
ap~ ap~ +
(2π)3
2Ep~
Z
d3 p
1
=
|j(p)|2 .
3
(2π) 2Ep~
(a) Show that the probability of the source creating no particles is
n R 4
o
2
P (0) = h0|T ei d x j(x)φI (x) |0i .
From Peskin-Eq. (4.90),
h~p1 · · · p~n |iT |~pA p~B i =
lim
T →∞(1+i)
n RT
h~
p
·
·
·
p
~
|T
e−i −T
0 1
n
dt HI (t)
o
|~pA p~B i0
From Eq. (3), we know that the interaction part of the Hamiltonian is
Z
Hint = − d3 x j(x)φ(x) .
From Peskin-Eq. (4.19), we know that the interaction Hamiltonian in the interaction
picture is
HI = eiH0 (t−t0 ) Hint e−iH0 (t−t0 )
Z
= − d3 x j(x)eiH0 (t−t0 ) φ(x)e−iH0 (t−t0 )
Z
=−
d3 x j(x)φI (x) ,
19
where to obtain the last line, we used Peskin-Eq. (4.14), the definition of the field in
the interaction picture.
Since we want to calculate the probability of the source creating no particle, the “in”
and “out” states should both be the Fock vacuum.
So, the probability becomes
o
n RT
lim h0|T e−i −T dt HI (t) |0i
T →∞(1+i)
n R 4
o
2
= h0|T ei d x j(x)φI (x) |0i
2
P (0) =
(7)
The exponent in the last line is positive because we can switch the integration limit of
t so that the time integral has a plus sign. Along with the minus sign of the spatial
integral of HI , we can a plus sign for the space-time integral.
(b) Evaluate P (0) of order j 2 and show that P (0) = 1 − λ + O(j 4 ).
Consider
n R 4
o
h0|T ei d x j(x)φI (x) |0i
Z
Z
i2
4
4
4
4
= h0| 1 + i
d x j(x)φ(x) +
d x1 d x2 T {j1 φ1 j2 φ2 } + O(j ) |0i ,
2
where I omitted the subscript
I
(8)
in all the fields, and defined ji ≡ j(xi ) and φi ≡ φ(xi ).
Notice that the even terms in Eq. (8) are zero because there are odd number of fields
in between the Fock vacuum and by Wick’s theorem, all these terms are zero.
So,
n R
h0|T ei
d4 x j(x)φI (x)
o
Z
i2
|0i = 1 +
2
d4 x1 d4 x2 j1 j2 h0|T {φ1 φ2 } |0i + O(j 4 ) .
Consider P (0) of order 2,
Z
Z
i2
i2
4
4
d x1 d x2 j1 j2 h0|T {φ1 φ2 } |0i =
d4 x1 d4 x2 j1 j2 DF (x1 − x2 ) .
2
2
Using equation Peskin-Eq. (2.60)
DF (x − y) = θ(x0 − y 0 )h0|φ(x)φ(y)|0i + θ(y 0 − x0 )h0|φ(y)φ(x)|0i
and Peskin-Eq. (2.50)
h0|φ(x)φ(y)|0i =
Z
20
d3 p
1 −ip·(x−y)
e
,
3
(2π) 2Ep~
(9)
we can rewrite Eq. (9) as
Z
i2
d4 x1 d4 x2 j1 j2 h0|T {φ1 φ2 } |0i
2
Z
i2
=
d4 x1 d4 x2 j1 j2 θ(x01 − x02 )h0|φ1 φ2 |0i + θ(x02 − x01 )h0|φ2 φ1 |0i
2
Z
1
i2
d3 p
0 −ip·(x2 −x1 )
0
0 −ip·(x1 −x2 )
0
)e
−
x
)e
+
θ(x
−
x
j
j
θ(x
=
d4 x1 d4 x2
1
2
1
2
2
1
2
(2π)3 2Ep~
If x01 > x02 ,
Z
Z
i2
i2
4
4
d x1 d x2 j1 j2 h0|T {φ1 φ2 } |0i =
2
2
Z
i2
=
2
Z
i2
=
2
Z
i2
=
2
i2
= λ.
2
If x02 > x01 ,
Z
Z
i2
i2
4
4
d x1 d x2 j1 j2 h0|T {φ1 φ2 } |0i =
2
2
Z
i2
=
2
Z
i2
=
2
Z
i2
=
2
i2
= λ.
2
1
d3 p
j1 j2 e−ip·(x1 −x2 )
3
(2π) 2Ep~
Z
Z
1
4
−ip·x1
d x1 j1 e
d4 x2 j2 eip·x2
2Ep~
1 ∗
j (p)j(p)
2Ep~
1
|j(p)|2
2Ep~
d4 x1 d4 x2
d3 p
(2π)3
d3 p
(2π)3
d3 p
(2π)3
d3 p
1
j1 j2 e−ip·(x2 −x1 )
3
(2π) 2Ep~
Z
Z
1
4
ip·x1
d x1 j1 e
d4 x2 j2 e−ip·x2
2Ep~
1
j(p)j ∗ (p)
2Ep~
1
|j(p)|2
2Ep~
d4 x1 d4 x2
d3 p
(2π)3
d3 p
(2π)3
d3 p
(2π)3
So, no matter regardless of the time ordering, P (0) up to order j 4 is
P (0) = 1 +
i2
λ + O(j 4 )
2
1
= 1 − λ + O(j 4 )
2
= 1 − λ2 + O(λ2 )
21
2
2
(c) Represent the term computed in part (b) and the whole perturbation series for P (0)
as a Feynman diagram. Show that P (0) = e−λ .
The Feynman diagram in position space of the P (0) is:
P (0) = 1 + x1
x2 + · · ·
where
x1
Each Propagator,
x2
x
Each External Point
= DF (x1 − x2 )
= (−i)
Z
d4 x j(x)
Intuitively, we can extrapolate the perturbative series as follows:
P (0) = 1+
+
1
2!
+
1
3!
+
1
4!
+· · ·
The combinatorial factor for i + 1 term is given by 1/i! because when contracting the
first incoming state with an outgoing states, there are i states to choose from. Note:
We need to consider i + 1 term because the first term in Taylor series for exponential
is n = 0. When contracting the second incoming states with an outgoing states, there
are i − 1 states to choose from and so on.
We can also calculate this perturbative series directly. Consider the 2n + 1 term,
n ∈ Z>0 , of Eq. (7):
2n Z
i
4
4
d x1 · · · d x2n j1 · · · j2n T {φ1 · · · φ2n } |0i
h0|
(2n)!
Z
i2n
=
d4 x1 · · · d4 x2n j1 · · · j2n h0|T {φ1 · · · φ2n } |0i
(2n)!
By Wick’s theorem, only terms will all fields contracted gives non-zero values. Also, notice that in order to obtain a non-zero value, there needs to be creation/annihilation operator pairs that have the same momentum. So, we can contract those creation/annihilation
operator pairs to get
Z
i2n
d4 x1 · · · d4 x2n j1 · · · j2n h0|T {φ1 · · · φ2n } |0i
(2n)!
Z
i2n
=
d4 x1 · · · d4 x2n j1 · · · j2n (DF (x1 − x2 ) · · · DF (x2n−1 − x2n ) + all other permutations)
(2n)!
There are (2n − 1)!! permutations in total. We can check that by writing out the first
22
few terms. Repeating similar analysis as part (b), we obtain
Z
i2n
d4 x1 · · · d4 x2n j1 · · · j2n h0|T {φ1 · · · φ2n } |0i
(2n)!
Z
i2n
=
(2n − 1)!! d4 x1 · · · d4 x2n j1 · · · j2n DF (x1 − x2 ) · · · DF (x2n−1 − x2n )
(2n)!
Z
i2n (2n)!
1
d3 p1
d3 pn 1
4
4
=
···
j1 · · · j2n eip1 ·(x1 −x2 ) · · · eipn ·(x2n−1 −x2n )
d
x
·
·
·
d
x
·
·
·
1
2n
n
3
3
(2n)! 2 n!
(2π)
(2π) 2Ep~1
2Ep~n
n Z
3
3
(−1)
d p1
1
d pn 1
= n
···
j(p1 )j ∗ (p1 ) · · · j(pn )j ∗ (pn )
···
3
3
2 n!
(2π)
(2π) 2Ep~1
2Ep~n
Z
(−1)n
1
d3 pn 1
d3 p1
= n
···
|j(p1 )|2 · · · |j(pn )|2
·
·
·
3
3
2 n!
(2π)
(2π) 2Ep~1
2Ep~n
(−1)n
= n λn
(10)
2 n!
So, we can write the perturbation series as
P (0) =
∞
X
(−λ/2)n
n=0
2
= e−λ/2 2 = e−λ
n!
(d) Compute the probability that the source creates one particle of momentum k. Perform
this computation first to O(j) and then to all others.
Similar to part (a), the probability of creating one particle with momentum k is
n R 4
o
2
i d x j(x)φI (x)
Pk (1) = hk|T e
|0i
Z
2
4
3
= hk| 1 + i
d x j(x)φ(x) + O(j ) |0i
= i
Z
2
d4 x j(x)hk|φ(x)|0i + O(j 3 )
Notice that all the terms with even number of φ gives zero by Wick’s Theorem.
Using Peskin-Eq. (4.94):
hp|φ(x) = h0|eip·x ,
the first term of the perturbation expansion becomes
Z
Z
4
i
d x j(x)hk|φ(x)|0i = i
d4 x j(x)hk|φ(x)|0i
Z
=i
d4 x j(x)eik·x
= ij(k)
So,
P (1) = ij(k) + O(j 3 )
23
2
= |j(k)|2 + O(j 3 )
To get a close expression for P (1) consider the m + 1 = 2n + 2 term, n ∈ Z>0 :
Z
im
d4 x1 · · · d4 xm j1 · · · jm hk|T {φ1 · · · φm } |0i
m!
One of the fields must contract with hk|. There are (2n + 1) permutations of this.
Without loss of generality, consider the case where the field with the largest time
component contracts with hk|:
Z
im
(2n + 1)
d4 x1 · · · d4 xm j1 · · · jm hk|φ1 T {φ2 · · · φm } |0i
m!
Using Peskin-Eq. (4.94), we have
Z
im
(2n + 1)
d4 x1 · · · d4 xm j1 · · · jm eik·x1 h0|T {φ2 · · · φm } |0i
m!
Z
i2n
= ij(k)
d4 x2 · · · d4 xm j2 · · · jm h0|T {φ2 · · · φm } |0i
(2n)!
Z
i2n
= ij(k)
d4 x1 · · · d4 x2n j1 · · · j2n h0|T {φ1 · · · φ2n } |0i
(2n)!
where I relabeled the fields in the last line. Using results from part (c), we have
Z
i2n
ij(k)
d4 x1 · · · d4 x2n j1 · · · j2n h0|T {φ1 · · · φ2n } |0i
(2n)!
(−λ/2)n
= ij(k)
n!
So, the probability of creating a particle with momentum k is
∞
X
(−λ/2)n
Pk (1) =
ij(k)
n!
n=0
2
= |j(k)|2 e−λ .
(e) Show that the probability of creating n particles is
P (n) =
Consider
n R
hk1 · · · kn |T ei
Consider the m + 1 term, m ≥ n.
1 n −λ
λ e .
n!
d4 x j(x)φ(x)
o
|0i
2
.
(11)
If m − n = 2l + 1, l ∈ Z>0 , then h | | i = 0 because there are odd number of fields in
the time-ordering symbol.
24
If m − n = 2l, then the mth term is
Z
im
d4 x1 · · · d4 xm j1 · · · jm hk1 · · · kn |T {φ1 · · · φm } |0i
m!
n of the fields must contract with the k outgoing states. There are m Pn permutations
of this. Note: The ordering does matter because contracting φi with φj is the same
as contracting φj with φi . Without loss of generality, consider the case where the n
latest fields contract with the outgoing states. Using Peskin-Eq. (4.94), the mth term
becomes
Z
in i2l
d4 x1 · · · d4 xm j1 · · · jm eik1 ·x1 · · · eikn ·xn h0|T {φn+1 · · · φm } |0i
(2l)!
Z
i2l
n
= i j(k1 ) · · · j(kn )
d4 xn+1 · · · d4 xm jn+1 · · · jm h0|T {φn+1 · · · φm } |0i
(2l)!
Z
i2l
n
d4 x1 · · · d4 x2l j1 · · · j2l h0|T {φ1 · · · φ2l } |0i
= i j(k1 ) · · · j(kn )
(2l)!
where in the last step, I relabeled the fields.
Using Eq. (10), we have
n
i j(k1 ) · · · j(kn )
Z
d4 x1 · · · d4 x2l j1 · · · j2l h0|T {φ1 · · · φ2l } |0i
(−λ/2)l
= i j(k1 ) · · · j(kn )
l!
n
So, Eq. (11) becomes
n
hk1 · · · kn |T ei
R
d4 x
j(x)φ(x)
o
|0i
2
= in j(k1 ) · · · j(kn )
∞
X
(−λ/2)l
l!
l=0
= in j(k1 ) · · · j(kn )e−λ/2
2
2
= |j(k1 )|2 · · · |j(kn )|2 e−λ
To find the probability of creation n particles with arbitrary momentum, we need to
integrate k1 · · · kn over all momentum space. Since we are integrating over all momentum space, the ordering of ki does not matter anymore. So, by integrating over all
momentum space, we overcounted by a factor of n!. So,
Z
1
d3 k1
d3 kn 1
1
·
·
·
·
·
·
|j(k1 )|2 · · · |j(kn )|2 e−λ
P (n) =
n!
(2π)3
(2π)3 Ep~1
Ep~n
1
= λn e−λ
n!
(f) Show
∞
X
P (n) = 1 ,
and
n=0
25
hN i =
∞
X
n=0
nP (n) = λ ,
and find h(N − hN i)2 i.
∞
X
P (n) =
n=0
hN i =
∞
X
nP (n) = e−λ
n=1
∞
∞
X
X
1 n
1 n −λ
λ e = e−λ
λ = e−λ eλ = 1
n!
n!
n=0
n=0
∞
∞
∞
X
X
X
n n
1
1 n
λ = λe−λ
λn−1 = λe−λ
λ = λe−λ eλ = λ
n!
(n
−
1)!
n!
n=1
n=1
n=0
(N − hN i)2 = N 2 + hN i2 − 2N hN i = N 2 + λ2 − 2N λ = N 2 + λ2 − 2 hN i λ
The first term is
N
2
=
∞
X
n2 P (n)
n=1
=e
−λ
= e−λ
∞
X
n2
n=1
∞
X
n!
λn
n
λn
(n − 1)!
n=1
∞
X
∞
1
n−1 n X
−λ
λ +
λn
=e
(n
−
1)!
(n
−
1)!
n=1
n=2
!
∞
X
1
= e−λ
λn + λeλ
(n
−
2)!
n=2
−λ
=e
λ2 eλ + λeλ
= λ2 + λ
So,
(N − hN i)2 = λ2 + λ + λ2 − 2λ2 = λ .
26
!