Physics 205a, Problem Set 5 Solutions

Physics 205a, Problem Set 5 Solutions
Du Pei
Nov 6, 2012
8.1
From Eq. (8.11), we have:
∫
(−∂x 2 + m 2 )∆(x − x ′ ) =
which is just Eq.(8.12).
8.2
∫
=
d 4 k (−∂x 2 + m 2 )e ik(x−x′ )
∫
(2π) 4 k 2 + m 2 =
− iϵ
d 4 k
(2π) 4 eik(x−x′) = δ 4 (x − x ′ ),
d 4 k (k 2 + m 2 )e ik(x−x′ )
(2π) 4 k 2 + m 2 − iϵ
Notice:
∫
∆(x−x ′ ) =
d 4 k e ik(x−x′ )
∫
(2π) 4 k 2 + m 2 − iϵ =
d 4 k e i⃗ k(⃗x−⃗x ′ )−ik 0 (t−t ′ ) (
(2π) 4 2ω
1
k 0 + ω − iϵ − 1
k 0 − ω − iϵ
)
.
Now we do the integral over k 0 . If t − t ′ < 0 we can close the contour in the
upper half plane, while if t − t ′ > 0 we can close the contour in the lower half
plane. So we have:
which is just Eq.(8.13).
∫
∆(x − x ′ ) =θ(t ′ d 3⃗ k
− t)
(2π) 4 ei⃗ k(⃗x−⃗x ′ )+iω(t−t ′ ) 2πi
2ω
∫
− θ(t − t ′ d 3⃗ k
)
(2π) 4 ei⃗ k(⃗x−⃗x ′ )−iω(t−t ′ ) −2πi
2ω
∫
=i ˜dk e i⃗ k(⃗x−⃗x ′ )−iω|t−t ′| ,
8.3
[ ∫
(∂t 2 − ∇ 2 + m 2 ) iθ(t − t ′ ) ˜dk ]
e ik(x−x′ )
∫
∫
∫
= iθ(t − t ′ ) ˜dk ( ⃗ k 2 + m 2 )e ik(x−x′) + ∂ 0
[iδ(t − t ′ ) ˜dk e ik(x−x′) + iθ(t − t ′ )
∫
= ∂ 0
[iδ(t − t ′ ) ˜dk ] ∫
e i⃗ k(⃗x−⃗x ′ )
+ iδ(t − t ′ ) ˜dk (−iω)e ik(x−x′ )
∫
∫
= δ(t − t ′ ) ˜dk ωe i⃗ k(⃗x−⃗x ′) + iδ ′ (t − t ′ ) ˜dk e i⃗ k(⃗x−⃗x ′) .
˜dk e ik(x−x′ ) ]
1

Similarly,
[ ∫
(∂t 2 − ∇ 2 + m 2 ) iθ( ′ t − t) ˜dk ]
e −ik(x−x′ )
∫
∫
= δ(t − t ′ ) ˜dk ωe i⃗ k(⃗x−⃗x ′) − iδ ′ (t − t ′ ) ˜dk e i⃗ k(⃗x−⃗x ′) .
Combining those results we have:
∫
(∂x+m 2 2 )∆(x−x ′ ) = δ(t−t ′ )
8.4
∫
˜dk 2ωe i⃗ k(⃗x−⃗x ′) = δ(t−t ′ )
Bringing Eq. (3.19) into the time-ordered product gives:
d 3 k
(2π) 3 ei⃗ k(⃗x−⃗x ′) = δ 4 (x−x ′ ).
⟨0|Tϕ(x 1 )ϕ(x 2 )|0⟩
∫
)
= ˜dk 1˜dk2 ⟨0|T
(a ⃗k1 e i⃗ k 1 ⃗x 1 −iω 1 t 1
+ a † ⃗
e −i⃗ k 1 ⃗x 1 +iω 1 t 1
)(a ⃗k2 e i⃗ k 2 ⃗x 2 −iω 2 t 2
+ a † k1 ⃗
e −i⃗ k 2 ⃗x 2 +iω 2 t 2
|0⟩
k2
∫
]
= ˜dk
[θ(t 1 − t 2 )e −i⃗ k 1 ⃗x 1 +iω 1 t 1
e −i⃗ k 2 ⃗x 2 +iω 2 t 2
+ θ(t 2 − t 1 )e −i⃗ k 1 ⃗x 1 +iω 1 t 1
e i⃗ k 2 ⃗x 2 −iω 2 t 2
∫
∫
= θ(t 2 − t 1 ) ˜dke ik(x 2−x 1 ) + θ(t 1 − t 2 ) ˜dke −ik(x 2−x 1 ) = 1 i ∆(x 2 − x 1 ).
8.5
For x 0 > y 0 , we can close the contour in the lower k 0 plane. The integral will
only vanish if both poles are above the real axis. So in the denominator, we
should have −(k 0 − iϵ) 2 + m 2 + ⃗ k 2 = k 2 + m 2 + iϵk 0 . Then we have for the
retarded potential:
∫
∆ ret (x − y) =
d 4 k e ik(x−y)
(2π) 4 k 2 + m 2 + iϵk 0 .
And similarly, for x 0 < y 0 we can close the contour in the upper k 0 plane, so
the pole prescription that will work for the advanced potential is:
∫
∆ adv (x − y) =
d 4 k e ik(x−y)
(2π) 4 k 2 + m 2 − iϵk 0 .
2