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Gaussian fixed point
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Past exam of the mathematics course of the University of Cambridge
/
2026
/
iii
/
Paper 303
/
2
/
b
/
i
/
Solution
Created
2026-09-24
Updated
2026-09-24
View more
At the
Gaussian fixed point
,
set
a
d
=
12
Ω
d
−
1
Λ
d
−
2
/
(
2
π
)
d
. The linearized
flow
is
d
l
o
g
ζ
d
(
μ
2
g
)
=
(
2
0
a
d
4
−
d
)
(
μ
2
g
)
.
(1)
The
mass
direction, corresponding to
ϕ
2
, has
coupling
dimension
y
t
=
2
. For
d
=
2
, the
second
eigendirection is
(
δ
μ
2
,
δ
g
)
=
(
−
d
−
2
a
d
,
1
)
δ
g
(2)
and has
y
g
=
4
−
d
; it is the tadpole-subtracted
mixture
of
ϕ
4
and
ϕ
2
. The corresponding operator
dimensions
are
d
−
2
and
2
d
−
4
.
Solved by
gpt-5
.
6
-sol high.
Total
articles
:
1