Nuclear Effective Field Theory
Paulo Bedaque
Lawrence-Berkeley Laboratory
Extracting low energy information from QCD in a model
independent way:
No nucleons a chiral perturbation theory
One nucleon a heavy baryon chiral perturbation theory
Two or more nucleons a
Nuclear effective theory
Hierarchy of scales:
8 MeV, 45 MeV
NN scale (spin singlet),
momentum in the
deuteron (spin triplet)
m , 270 MeV
Fermi momentum
in nuclei
m
QCD scale
Two consequences:
Bound states within the EFT range of validity
Nuclear EFT is non-perturbative
Two possible EFT’s
“pionfull” Q
mr
~
mp <<
“pionless” Q ~ 1/a << mp
Pionless theory: two-nucleons
2
L   (i 0 
)  C0 () |  |4 C2 () |  |2  2 |  |2 
2M
†
C0 ()

4π
M
C0 () 2 (  ik )
C0 ( )3 (  ik ) 2
1
4π
+Λ+…-ik
MC0 (Λ)
(1/a 1S = 8 MeV
0
1 r
k cotδ  - + 0 k 2 +…
a 2
4π
1
+ Λ MC0 (Λ)
a
fine tuned
cancellation
1/ a 3 S = 45 MeV)
1
That’s why nuclear physics
exists !
another way of looking at the fine tuning:
 d c0  c0 (1  c0 )
d
c0 ()  4 M C0 ()

trivial fixed
point
non-trivial
fixed point
Assuming this is the only fine tuning:
• Expansion in powers of Q/m, keep Qa to all orders
T
1
1 r0 2
  k 
a 2
r0 k 2

(1 

1  ik
2 1  ik
a
a
1
 ik
)
C2 is NLO, not NNLO
• Naïve dimensional analysis fails
• C0 is the only non-perturbative operator
A good example: neutrino-deuteron collisions (Butler, Chen)
Haxton et al. : no
exchange currents
Kubodera et al. : a model of
meson exchange currents
5% difference
Both calculations are reproduced by EFT with two different values of
L1A (Nσ 2σi τ 2 N)† Nτ 2 τ - σ 2 N,
L1A
5 fm3
The same constant appears on pp fusion,
m capture on deuterium, triton beta decay
For the three-body (“pionless”) :
†
3
How large is D 0 (ψ ψ) ?
1
MΛ 4
naïve dimensional
analysis
would appear
only at
NNNLO
ultraviolet finite
k
p
t(k,p)
p 1/a, k
For
1
,k
a
p t(k, p) =
1
p2
D0 would not not run
and would not needed
at leading order
:
p
4
3
1
Λ2

dq  p + pq+ q
0 q ln  p2 - pq+ q2
2
2

 q t(k,q)

t(k, p)
 s)
sin(
8
6
ps-1, 1 =
3 s cos(  s )
2
L=0, S=1/2: triton, helium 3, bosons
Two kinds of
channels:
All others: Pauli principle, centrifugal barrier
All others:
1)
1
2
=- ,
t(k, p)
1
p
3.17…
'
2)
 dq ( (q, p) -  K(q, p) ) t
0
~ 1/Q2
'
'
(q) = -  dq (
)

~ 1/L2
Three-body force no
needed until very high
orders, a lot of
predictive power
Neutron-deuteron elastic phase shifts
L=0, S=3/2
L=2, S=1/2
+ = AV18 + UX (Kievski et al.)
= LO,
m=Schmelzbach et al.
= NLO,
= NNLO
L=0, S=3/2 scattering length: a(EFT)=5.09 + 0.89 + 0.35 + …=6.33m0.05 fm
a(Exp)=6.35m0.02 fm
3H, 3He
(and bosons):
1)   1: t(k, p)
1
is0 +1
p
2)
, s0
'
 dq ( (q, p) -  K(q, p) ) t
harder in the UV
'
'
(q) = -  dq (
0
~1/Q2 or ~1/QL (zero mode)
change in
on-shell
amplitude
 1.006…

~1/QL
)
t  ' (p) = t  (p) + Csin(s0log(p))
Adjust H(L) so:
'
2
Q
 dq ( + H()) t  (q) = 0 + O(


2
)
2
2
three-body D ()  a H()  a sin( s0 log( / )  arctg s0 )
0
2
 2 sin( s0 log( / )  arctg s0 )
force:
limit cycle: L
g e p/s0 L
At higher orders:
SUW(4) invariant three-body force
terms are enhanced
Neutron-deuteron elastic phase shifts: L=0, S=1/2
x = AV18 + UX (Kievski et al.)
= LO,
i= Schmelzbach et al.
= NLO,
= NNLO
blue band describes the variation between =200 g 600 MeV
Phillips line:
one 3-body free parameter
one line
“Pionfull” EFT
(expansion on Q/ and m/)
2
f2
i †
L   (i 0 
 g A  i ) 
tr ( m † m )  Btr†  C0 () | |4 C2 () |  |2 2 |  |2 
2M
4
†
  e
2
i 2
f
 0 2
, 
 



0


2 
Restrictions from c symmetry
Potential:
Amplitude:
 dependence ?
Perturbatively this is inconsistent, but we now know better
m ln 
2
perturbative: destroys chiral expansion
non-perturbative:
still inconsistent
lattice extrapolations,
isospin breaking,
cosmology
k 2 ln 
destroys the momentum expansion
momentum expansion is
consistent
Some NN phase shifts (Epelbaum et al.):
500<<600
3S
=LO
e1
1
=NLO
=NNLO*
N couplings fit
=Nijmegen PWA
Neutral pion photoproduction (Beane, Lee, van Kolck)
2
3k d
|q 0  Ed  e L Ld
8q d 
2
10-3
E d (0)  (1.79  0.2) 
m
 m 
  


3
E
EXP
d
10-3
(0)  (1.45  0.09) 
m