# QFT II Homework 1 ```Homework #1
Rafael Matheus Gonçalves Ribeiro
PGF5828 - Quantum Field Theory II
University de São Paulo
September 14, 2020
1.β-Functions of Yukawa Theory
Consider the Lagrangian given by
1
1
L = ∂&micro; φ∂ &micro; φ − m2 φ2 + ψ̄(i∂/ − M )ψ − ig ψ̄γ5 ψφ
(1)
2
2
In the homework 7 we derived the counterterms δφ , δψ , δλ and δg and with this we will
compute the β-functions for λ and g .
a) The counterterm were show to be
−g 2
Λ2
log
2
8π
M2
2
4
−3λ
3g
Λ2
δλ =
−
log
32π 2
2π 4
M2
δφ =
−g 2
Λ2
log
32π 2
M2
2
g
Λ2
δg =
log
16π 2
M2
δψ =
(2)
(3)
Using the definition given by Peskin and Schoreder’s book
g2
Aφ = −γφ = − 2
8π
2
3g
3λ
Bλ =
−
2
2π
32π 2
g2
Aψ = −γφ = −
32π 2
−g
Bg =
16π 2
(4)
(5)
(6)
(7)
Therefore we see that,
β(g) = −2gBg − 2gAψ − gAφ
g2
g2
g2
5g 3
= 2g
+
2g
+
g
=
16π 2
32π 2
8π 2
16π 2
(8)
(9)
and
=2
β(λ) = −2Bλ − 4λAφ
3λ2
3g 4
g2
−
+
4λ
32π 2 2π 2
8π 2
3λ2 + 8λg 2 − 48g 4
=
16π 2
1
(10)
(11)
(12)
b) To find the flow, we have that
(
ẋ = x(3 − x − 2y)
ẏ = y(2 − x − y)
(13)
Then find the fixed
ẋ
0
=
ẏ
0
Which is (0, 0), (0, 2), (3, 0). Now we gonna linearize the system near the fixed point
3 − 2x − 2y
−2x
W =
−y
2 − x − 2y
(14)
2.RG Description of the Ferromagnetic Transition
The Landau-Ginzburg description of a ferromagnet corresponds to d dimensional Euclidean
φ4 theory, with the action
Z
1
1
λ
SE [φ] = dd x{ (∂&micro; φ)2 + m2 φ2 + φ4 }
(15)
2
2
4!
Then, to begin we will split the low and high energy mode as
φ(x) = φl (x) + φh (x)
(16)
Here, we will follow the notation of the lecture.
a) So we have the Sef f , that can be written as
Z
−Sef f [φl ]
−S[φl ]
e
=e
Dφ e−S[φl ,φh ] e−S[φh ]
2
(17)
Figure 1: Integrating out the high energy modes. The double lines correspond to φh , whereas
the external single lines are φl , the low energy modes. (a): first order contribution to the
φl two-point function. (b): lowest order contribution to the four-point function. These
diagrams result in shifts in λ and m2
Performing the integration in the high energy modes
e−Sef f [φl ] = e−S[φl ] &lt; e−S[φl ,φh ] &gt;h
(18)
making the expansion
1
e−Sef f [φl ] = e−S[φl ] exp{− &lt; S[φl , φh ] &gt; + &lt; S[φl , φh ]2 &gt;}
Z
Z2
2 Z
1
λ
λ
−S[φl ]
2
2 2
d
2 2
d
2 2
=e
d xφl φh +
d xφl φh d yφl φh + . . .
1−
4!
2 16
(19)
(20)
The second term in the expansion, refers tho the diagram (b) in the figure 1. So now, let’s
make us re-scaling of the momentum according to
k → k = bk
x
x→x=
b
This change in the scale is to obtain the contribution m2 , then the diagram (a) is
λ
4
Z
λ
λ
b
dd k
1
d
2
(2π) k + m2
Z
dd q
φ(−q)φ(q)
(2π)d
(21)
Here we allow nonzero mass but since m2 is mass term in the IR, it is only a small perturbation inside the integral over the high energy interval. Thus the shift in δm2 is
δm2 ≡
λ
I1 − m2 I2
2
(22)
Where Iα is define as
Z
λ
Iα =
λ
b
dd k 1
(2π)d k 2α
, α = 1, 2
To obtain this definition, we consider the summation over the high energy modes in the expression 21. Using the fact that we are near the vicinity of the critical point and anticipating
3
that we are interested in no more than the expansion of the β-function for small values of
the coupling, we now expand the integrand to first order.
From the change the scale, we have
d d x = bd d d x 0
1
∂&micro; = ∂&micro;0
b
and the action
Z
Sef f [φl ] =
d 0 d
d xb
1 −2 0 0&micro;
1
1
b ∂&micro; φ∂ φ + (m2 + δm2 )φ2 + (λ + δλ)φ4
2
2
4!
(23)
Now we recover the original form of the original Lagrangian now with all terms with primed.
Shift in the φ
Z
d−2
1 −2 0 d−2
bd 2
bd
d 0
0&micro;
2
2
4
Sef f [φl ] = d x
b ∂&micro; (b 2 φ)∂ (b 2 φ) + (m + δm )φ + (λ + δλ)φ
(24)
2
2
4!
So
φ0 = b
d−2
2
φ
From equation (24), we can see that
(
m0 = b2 (m2 + δm2 )
λ0 = b4−d (λ + δλ)
Now to find the δm2 and δλ we need to compute from the equation (24). Let’s begin with
the (21).
These integrals are computed by switching to polar coordinates
Z 1
Ωd
Iα = Ωd
dkk d−2α−1 =
(1 − b2α−d )
(25)
d − 2α
b−1
where
Ωd =
1 2π d/2
(2π)2 Γ(d/2)
Also we need to eliminate the dependence of the cutoff Λ. To do this we rescale the momenta
to momenta in units of the cutoff Λ. Therefore
k→
k
λ
in all momenta. After this the length scale is measure in units of the inverse cutoff, x → xΛ.
We need to change the parameters then
m2 → m2 Λ
(26)
4−d
(27)
λ → λΛ
4
We need to solving the diagram (b) in the figure (1). The presence of the four external legs,
it’s contribution will be proportional is to φ4l . Moreover, momentum conservation implies that
the momenta carried by the internal lines of the diagram will depend on both fast internal
momenta and the external momenta carried by the fields φl . However, we can simplify this
analysis by neglecting the dependence on the latter from the outset, this can be done cause
the integration over the internal momentum followed by a Taylor expansion in the slow
momenta would generate expressions of the structure F (k1 , k2 , k3 )φ(k1 )φ(k2 )φ(k3 )φ(−k1 −
k2 − k3 ), where k1,2,3 represent slow momenta and F is some polynomial. Taking account of
the small momenta would thus generate derivatives action on an operator of fourth order in
φ, a combinations that we saw above is irrelevant.
Neglecting the external momenta, diagram (b) lead to the result.
Z
Z
Z
λ2
1
dd k
λ2 I2
1
2
d
4
&lt; S[φl , φh ] &gt;} '
d xφl
=
dd xφ4l + O(λ2 m2 ) (28)
2
16
(2π)d (m2 + k 2 )2
16
Evaluating the integral and rescaling, we find
Z
λ2 Ωd 1 − b4−d
λ
(4)
4−d
−
dd xφ4l
S [φ] = b
4!
16 d − 4
(29)
b)
Combining the results,
m = b m2 +
2
2
m2 λΩd
Ωd
2−d
4−d
(1 − b ) −
(1 − b )
2(d − 2)
2(d − 4)
3 λ2 Ωd (1 − b4−d )
4−d
λ=b
λ−
2
d−4
Now we define = d − 4, and making the expansion for 1 and Ω4−d ≈ Ω4 =
obtain the final result as
λ
2
2
−2
2 ln b
2
m =b m +
(1 − b ) − m λ
32π 2
16π 2
3λ2
ln b
λ = (1 + ln b) λ −
16π 2
(30)
(31)
1
.
8π 2
We
(32)
(33)
Make the derivatives to find the β-functions, we find that
dm2
λ
m2 λ
= 2m2 +
−
d ln b
16π 2 16π 2
dλ
3 2
= λ −
λ
d ln b
16π 2
(34)
(35)
With we equating the right-hand sides of eqs. (34) and (35) to zero. We find the Gaussian
1
16π 2
∗
2∗
∗
fixed point (m2∗
1 , λ1 ) = (0, 0) and non-trivial point (m2 , λ2 ) = (− 6 , 3 ).
To see the behaviour near the fixed point we employ a first-order expansion to come up with
matrix equations predicting the flow.
5
For our ferromagnet, the matrices for the two sets of fixed points are
1
1
2 − 3 16π
2 16π
2
2
, W2 =
W1 =
0 0
−
(36)
For the Guassian fixed point at (0, 0) (corresponding to W1 ) we have λ0 = λ. If &gt; 0 (that
is, we are examining the physics in less than four dimensions since d = 4−) then λ increases
as we look at larger scales. In that case λ is a relevant variable and will be important to the
physics.
c)
Figure 2: Renormalization group flow for a ferromagnet. FM labels ‘ferromagnet’, PM
labels ‘paramagnet’.
6
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