7CCMMS43T/7CCMMS43U
Advanced QFT
Problem set E
This set of problems is about generating functionals. Here we work in Euclidean QFT. For
R
simplicity we use implicit integration over repeated variables: Ax Bx := d4 x A(x)B(x),
etc
Problem 1
Recall the generating functional of correlation functions
1 (3)
1 (2)
Jx Jy + Zx,y,z
Jx Jy Jz + . . .
Z[J] = 1 + Zx(1) Jx + Zx,y
2
3!
(n)
where Zx1 ,...,xn is the n-point correlation function (here we have normalized to Z[0] = 1:
the zero-point function is 1). Recall the generating functional of connected correlation
functions
W [J] = log(Z[J]),
expanded as
1 (2)
1 (3)
W [J] = Wx(1) Jx + Wx,y
Jx Jy + Wx,y,z
Jx Jy Jz + . . .
2
3!
(1)
(1)
Show that Wx = Zx (this is the one-point function, or vacuum expectation value).
Show that
(2)
(2)
Wx,y
= Zx,y
− Zx(1) Zy(1)
and argue that this only contains the connected diagrams contributing to the 2-point
function.
Problem 2
Now consider the effective action
1 (2)
Γ[ϕ] = Γ(0) + Γ(1)
x χx + Γx,y χx χy + . . .
2
(1)
where χx = ϕx − Wx . By using the Legendre transform
Γ[ϕ] + W [J] − Jx ϕx
= 0,
ϕ=ϕ[J]
ϕ[J]x =
δW [J]
δJx
(2)
show that Γ(0) = Γ(1) = 0 and that Γx,y = W (2)
as power series in J,
−1
x,y
. Hint: Start by writing ϕx and χx
1 (3)
(2)
Jy Jz + . . .
ϕx = Wx(1) + Wxy
Jy + Wxyz
2
Problem 3
Re-introduce ~ in the path integral formalism, and define Z[J] = eW [J]/~ . Show that the
classical limit of the effective action is the classical action,
lim Γ[ϕ] = S[ϕ].
~→0
Solutions
Problem 1
We just have to expand the log using log(1 + a) = a − a2 /2 + − . . .. We do:
1 (2)
(1)
log(Z[J]) = log 1 + Zx Jx + Zx,y Jx Jy + . . .
2
1 (2)
1 (1) 2
Jx Jy −
= Zx(1) Jx + Zx,y
Z Jx + . . .
2
2 x
2
(1)
(1) (1)
We write Zx Jx = Zx Zy Jx Jy . The results follow.
(1)
(2)
The argument is simple: in all diagrams contributing to the two-point function Zx,y ,
which are all two-leg diagrams, the disconnected ones will be diagrams composed of two
separate one-leg diagrams, one with the leg joining x the other with the leg joining y. So
the result is the sum of all connected two-leg diagrams, plus the product of the sum of all
one-leg diagrams at x, times the sum of all one-leg diagrams at y. The sum of all one-leg
(1)
(1) (1)
diagrams at x is the one-point function Zx . So subtracting the product Zx Zy , we take
away all disconnected diagrams.
Problem 2
First we expand ϕ in powers of J:
ϕx =
⇒ χx
δW
δJx
1 (3)
(2)
Jy Jz + . . .
= Wx(1) + Wxy
Jy + Wxyz
2
1 (3)
(2)
Jy + Wxyz
= Wxy
Jy Jz + . . .
2
(2)
Then we expand Γ:
1
1 (3)
(2)
(2)
(2)
(2)
Γ[ϕ] = Γ(0) + Γ(1)
W
J
+
W
J
J
y z + Γxy Wxz Wyv Jz Jv + . . .
x
xy y
2 xyz
2
1 (1) (3)
(2)
(2)
(2)
(2)
= Γ(0) + Γ(1)
W
J
+
Γ
W
+
Γ
W
W
x
xy y
xy
xz
yv Jz Jv + . . .
2 x xzv
(3)
Then we expand Jx ϕx :
Jx ϕx = Jx Wx(1) + Jx χx
1 (3)
(1)
(2)
= Jx Wx + Jx Wxy Jy + Wxyz Jy Jz
2
1 (3)
(1)
(2)
= Wx Jx + Wxy Jx Jy + Wxyz Jx Jy Jz + . . .
2
(4)
Putting all that together:
(0)
Γ
+
(2)
Γ(1)
x Wxy Jy
1 (1) (3)
(2)
(2)
(2)
Γ W + Γxy Wxz Wyv Jz Jv
+
2 x xzv
1 (2)
+Wx(1) Jx + Wxy
Jx Jy
2
1 (3)
(2)
Jx Jy Jz + . . .
−Wx(1) Jx − Wxy
Jx Jy − Wxyz
2
= 0.
(5)
Looking at order 0: Γ(0) = 0. Looking at order 1: Γ(1) = 0. Looking at order 2:
(2)
(2)
(2)
Γ(2)
xy Wxz Wyv = Wzv .
(2) −1
We then premultiply by (W (2) )−1
)vb to obtain the stated
ax and postmultiply by (W
relation.
Problem 3
Recall that (here S[ϕ] is the Euclidean action)
R
Dϕ e−S[ϕ]+Jx ϕx
Z[J] = R
Dϕ e−S[ϕ]
Let us put back the ~. By dimensional analysis, it divides the action:
R
Dϕ e−S[ϕ]/~+Jx ϕx /~
R
Z[J] =
Dϕ e−S[ϕ]/~
Therefore the limit ~ → 0 is obtained by a saddle point approximation,
c
c
e−S[ϕ [J]]/~+Jx ϕx /~
c
c
c
Z[J] ∼
= e−S[ϕ [J]]/~+Jx ϕx /~+S[ϕ [0]]/~
c [0]]/~
−S[ϕ
e
where ϕc [J] is the minimum of S[ϕ] − Jx ϕx , thus is the classical solution with a source
term:
δS[ϕ] = Jx .
δϕx ϕ=ϕc [J]
Taking the log,
lim W [J] = −S[ϕc [J]] + Jx ϕc [J]x + S[ϕc [0]].
~→0
Therefore, in the limit ~ → 0
δW [J]
ϕ[J]x =
=
δJx
δS[ϕ] −
+ Jy
δϕy ϕ=ϕc [J]
δϕc [J]y
+ ϕcx [J] = ϕcx [J]
δJx
and the effective action therefore satisfies, in the limit ~ → 0, the equation
Γ[ϕc [J]] + W [J] − Jx ϕc [J]x = 0 ⇒ Γ[ϕc [J]] = S[ϕc [J]].
Since this holds for all J, it holds for all functions ϕc [J] (functions of x) and the result
follows.