Hyperon sigma terms for 2+1 quark flavours

Abstract

QCD lattice simulations determine hadron masses as functions of the quark
masses. From the gradients of these masses and using the Feynman-Hellmann
theorem the hadron sigma terms can then be determined. We use here a novel
approach of keeping the singlet quark mass constant in our simulations which
upon using an SU(3) flavour symmetry breaking expansion gives highly
constrained (i.e. few parameter) fits for hadron masses in a multiplet. This is
a highly advantageous procedure for determining the hadron mass gradient as it
avoids the use of delicate chiral perturbation theory. We illustrate the
procedure here by estimating the light and strange sigma terms for the baryon
octet.

Full text

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arXiv:1110.4971v1 [hep-lat] 22 Oct 2011
DESY 11-168
Edinburgh 2011/27
Liverpool LTH 922
October 22, 2011
Hyperon sigma terms for 2 + 1 quark flavours
R. Horsleya, Y. Nakamurab, H. Perltc,
D. Pleiterde, P. E. L. Rakowf, G. Schierholzeh,
A. Schillerc, H. St¨ ubeni, F. Wintera
and J. M. Zanottia
– QCDSF-UKQCD Collaboration –
aSchool of Physics and Astronomy, University of Edinburgh,
Edinburgh EH9 3JZ, UK
bRIKEN Advanced Institute for Computational Science,
Kobe, Hyogo 650-0047, Japan
cInstitut f¨ ur Theoretische Physik, Universit¨ at Leipzig,
04109 Leipzig, Germany
dJ¨ ulich Research Centre, JSC, Room 324,
52425 J¨ ulich, Germany
eInstitut f¨ ur Theoretische Physik, Universit¨ at Regensburg,
93040 Regensburg, Germany
fTheoretical Physics Division, Department of Mathematical Sciences,
University of Liverpool, Liverpool L69 3BX, UK
hDeutsches Elektronen-Synchrotron DESY,
22603 Hamburg, Germany
iKonrad-Zuse-Zentrum f¨ ur Informationstechnik Berlin,
14195 Berlin, Germany
Abstract
QCD lattice simulations determine hadron masses as functions of the
quark masses. From the gradients of these masses and using the Feynman–
Hellmann theorem the hadron sigma terms can then be determined. We
use here a novel approach of keeping the singlet quark mass constant in our
simulations which upon using an SU(3) flavour symmetry breaking expan-
sion gives highly constrained (i.e. few parameter) fits for hadron masses
in a multiplet. This is a highly advantageous procedure for determining
the hadron mass gradient as it avoids the use of delicate chiral perturba-
tion theory. We illustrate the procedure here by estimating the light and
strange sigma terms for the baryon octet.
1

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1 Introduction
Hadron sigma terms, σ(H)
(for example the nucleon) coming from the vacuum connected expectation value
of the up (u) down (d) and strange (s) quark mass terms in the QCD Hamiltonian,
l
, σ(H)
s
are defined1as that part of the mass of the hadron
σ(H)
l
= mR
l?H|(uu + dd)R|H?,σ(H)
s
= mR
s?H|(ss)R|H?, (1)
where we have taken the u and d quarks to be mass degenerate, mu = md ≡
ml. (The superscript
to the hadron mass come from the chromo-electric and chromo-magnetic gluon
pieces and the kinetic energies of the quarks, [2]. Sigma terms are interesting
because they are sensitive to chiral symmetry breaking effects. Experimentally
the value for σ(N)
l
has been deduced from low energy π-N scattering. A delicate
extrapolation to the chiral limit [1] gives a result for the isospin even amplitude of
σπN/f2
l
), from which the sigma term may be found. The precise
value obtained this way has been under discussion for many years. However
within the limits of our lattice calculation, this will not concern us here and for
orientation we shall just quote a range of results from earlier analyses of [3, 4]
of 45(8)MeV while a later dispersion analysis [5] suggested a much higher value
64(7)MeV. An estimation using heavy baryon chiral perturbation theory gave
45MeV, [6]. Even less is known about the nucleon strange sigma term. Eq. (1)
is usually written (in particular for the nucleon) as
Rdenotes a renormalised quantity.) Other contributions
π(with σπN≡ σ(N)
σ(N)
l
=mR
l?N|(uu + dd − 2ss)R|N?
1 − y(N)R
,y(N)R=
2?N|(ss)R|N?
?N|(uu + dd)R|N?, (2)
(i.e. we consider σ(N)
culation, e.g. [1] (which we will discuss in more detail later) uses first order in
SU(3) flavour symmetry (octet) breaking to give
l
and y(N)Rrather than σ(N)
l
and σ(N)
s
.) The simplest cal-
σ(N)
l
=
mR
l
mR
s− mR
l
MΞ+ MΣ− 2MN
1 − y(N)R
∼
26
1 − y(N)RMeV,(3)
and
σ(N)
s
=mR
mR
s
l
1
2y(N)Rσ(N)
l
∼ 325
y(N)R
1 − y(N)RMeV, (4)
where mR
leading order PCAC formula for this ratio gives
s/mR
lis the ratio of the strange to light quark masses, which using the
mR
s/mR
l= (2M2
K− M2
π)/M2
π∼ 25. (5)
1Or more accurately as the matrix element of the double commutator of the Hamiltonian
with two axial charges. However this is equivalent to the definition given in eq. (1), see for
example [1].
2

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The Zweig rule, ?N|(ss)R|N? ∼ 0 would then give
σ(N)
l
∼ 26MeV,σ(N)
s
∼ 0MeV, (6)
while any non-zero strangeness content, y(N)R> 0 would increase this value of
σ(N)
l
, σ(N)
s
(and indeed due to the large coefficient, σ(N)
Determination of the strange sigma term (and in particular y(N)R) is impor-
tant in constraining the cross section for the detection of dark matter. WIMPs
would be scattered off nuclei by the exchange of scalar particles, such as the Stan-
dard Model Higgs particle, which will interact more strongly with heavier quark
flavours. This coupling can be parameterised in terms of the fractional contribu-
tion of a quark flavour q to the nucleon’s mass MN, fTq= mR
While the contributions of the charm and heavier flavours approach a constant
that is proportional to the gluonic contribution fTg, there is a strong dependence
of the cross section on the value of fTs, see e.g. [7, 8] and references therein.
Computing the sigma terms from lattice QCD has a long history from quenched
to 2 flavour and more recently 2 + 1 flavour simulations, e.g. [9, 10, 11, 12, 13,
14, 15, 16, 17, 18, 19], with a status report being given in [20]. In general more
recent results tend to give a lower σ(N)
s
term than earlier determinations.
In this article, we shall investigate this simple picture as described in eqs. (3),
(5) and in particular test the linearity assumption of SU(3) flavour symmetry
breaking.
s
quite rapidly).
q?N|(qq)R|N?/MN.
2 Flavour symmetry expansions
Lattice simulations start at some point in the (mR
the physical point (mR∗
physical point with a∗.) As we shall be considering flavour symmetry breaking
then we shall start here at a point on the flavour symmetric line mR
then consider the path keeping the average quark mass constant, m = const..
The SU(3) flavour group (and quark permutation symmetry) then restricts the
quark mass polynomials that are allowed, [21], giving for the baryon octet
s,mR
l) plane and then approach
s,mR∗
l) along some path. (In future we shall denote the
l= mR
sand
MH= M0(m) + cHδml+ O(δm2
l), (7)
with
cH=







3A1
3A2
−3A2
H = N
H = Λ
H = Σ
H = Ξ−3(A1− A2)
(8)
where
δml= ml− m,m =1
3(2ml+ ms), (9)
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and A1and A2are unknown coefficients. So to linear order in the quark mass,
we only have two unknowns (rather than four). A similar situation also holds for
the pseudoscalar and vector octets (one unknown) and baryon decuplet (also one
unknown). These functions highly constrain the numerical fits. (At O(δm2
the baryon decuplet has a further constraint.)
Permutation invariant functions of the masses XS, (or ‘centre of mass’ of the
multiplet) can be defined which have no linear dependence on the quark mass.
For example for the baryon octet we have
l) only
XN=1
3(MN+ MΣ+ MΞ) = M0(m) + O(δm2
l).(10)
(The corresponding result for the pseudoscalar octet is given later in eq. (29).)
Furthermore expanding about a specific fixed point, ml= ms= m0on the
flavour symmetric line and allowing m to vary, we then have
M0(m) = M0(m0) + M′
0(m0)(m − m0) + O((m − m0)2). (11)
We will see that A1, A2give all the non-singlet hyperon terms and M′(m0) the
singlet sigma terms.
As an example of the quark mass expansion from a point on the flavour
symmetric line in Fig. 1 we plot the baryon octet MNO/XN for NO= N, Λ, Σ,
0.00 0.250.500.75 1.001.25
Mπ
2/Xπ
2
0.8
0.9
1.0
1.1
1.2
MNO/XN [Octet]
experiment
N(lll)
Λ(lls)
Σ(lls)
Ξ(lss)
sym. pt.
0.000.250.50 0.751.001.25
Mπ
2/Xπ
2
0.8
0.9
1.0
1.1
1.2
MNO/XN [Octet]
experiment
N(lll)
Λ(lls)
Σ(lls)
Ξ(lss)
sym. pt.
Figure 1: MNO/XN(NO= N, Λ, Σ, Ξ) against M2
symmetric line given by κ0= 0.12090, left panel and κ0= 0.12092, right panel. The
323× 64 lattices are filled circles, while the 243× 48 lattices are open triangles. Also
shown is the combined fit of eq. (33) (the dashed lines) to the 323× 64 lattice data.
The fit results are the open circles, while the experimental points are the (red) stars.
π/X2
πfor initial point on the flavour
Ξ against M2
2 + 1 O(a) improved clover fermions at β = 5.50, [22] using two starting values
for the quark mass on the flavour symmetric line, namely κ0= 0.12090, 0.12092.
π/X2
πtogether with a linear fit, eq. (8) and implicitly eq. (29) using
4

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All the points have been arranged in the simulation to have constant m. We
see that a linear fit provides a good description of the numerical data from the
symmetric point (where Mπ∼ X∗
In a little more detail, the bare quark masses are defined as
π= 410.9MeV) down to the physical pion mass.
amq=1
2
?1
κq
−
1
κ0;c
?
, withq = l,s,0, (12)
(with the index q = 0 denoting the common quark along the flavour symmetric
line) and where vanishing of the quark mass along the SU(3) flavour symmetric
line determines κ0;c. (The explicit value is not needed in the following.) Keeping
m = constant ≡ m0gives
κs=
1
3
κ0−
2
κl
. (13)
So once we decide on a κlthis then determines κs. These initial κ0values chosen
here, namely κ0= 0.12090 and 0.12092 are close to the path that leads to the
physical point (κ0= 0.12092 being slightly closer). (This is discussed in more
detail in [21], which also contains numerical tables and phenomenological values
for the hadron masses. Results not included there are given in the Appendix B.)
This path is also illustrated later in section 4.3, Fig. 4. Although finite size effects
tend to cancel in ratios of quantities from the same multiplet, we nevertheless
fit just to the results from the 323× 64 lattices (filled circles) using the linear
fit of eq. (7). Finally note that we also have a similar flavour expansion for the
pseudoscalar octet as for the baryon octet, as will be discussed in section 4.3.
3 (Hyperon) scalar matrix elements
Scalar matrix elements can be determined from the gradient of the hadron mass
(with respect to the quark mass) by using the Feynman–Hellman theorem which
is true for both bare and renormalised quantities. So if we take the derivative
with respect to the bare quark mass we get the bare qq matrix element,
∂MH
∂ml
= ?H|(uu + dd)|H?,
∂MH
∂ms
= ?H|ss|H?,(14)
while if we take the derivative with respect to the renormalised quark mass we
get the renormalised matrix element. In the left panel of Fig. 2, we show the
nucleon (green) and the flavour symmetric nucleon (magenta) against 1/κl, 1/κ0
respectively (from eq. (12) these are proportional to the bare quark mass). From
the Feynman–Hellmann theorem, the slope of the magenta curve gives the total
5

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8.25 8.268.278.28
1/κl, 1/κ0
0.35
0.40
0.45
0.50
0.55
aMN
sym. pts. 32
sym. pts. 24
32
24
3x64
3x48
3x64
3x48
0.00 0.010.020.03
2
0.040.05
(aMπ)
0.35
0.40
0.45
0.50
0.55
aMN
sym. pts. 32
sym. pts. 24
32
24
3x64
3x48
3x64
3x48
Figure 2:
m = const. points, green circles with κ0= 0.12090) and versus 1/κ0(for the flavour
symmetric points, magenta). The common points are denoted by red circles. The
243×48 volume results are open circles together with a dashed line for the (linear) fit,
while the 323× 64 volume results are filled circles and lines. Similarly the right panel
shows the nucleon mass aMN, versus (aMπ)2(same notation as for the left panel).
The left panel shows the nucleon mass, aMN, versus 1/κl, (for the
?
bution2. The difference between the two contributions gives the disconnected
contribution. Because here all three quark masses are equal, the disconnected
contribution for all three quarks will be the same. The two slopes thus give the
estimates
?
?
?N|ss|N?
?
for bare lattice quantities.
To look at renormalised matrix elements, we need a plot against the renor-
malised mass, (aMπ)2(as in leading order PCAC, M2
renormalised quark mass, eq. (31)). This is shown in the right panel of Fig. 2.
The slopes are now much closer to each other. We now find the estimates
?
?
?N|(ss)R|N?
?
2Eq. (7) can be extended to the ‘partially quenched’ case, [21], where the sea quark masses
remain constrained by m = const. but the valence quark masses µl, µs are unconstrained.
Defining δµq= µq−m then for the nucleon, the leading change is particularly simple, cNδml→
cNδµl. For the other members of the octet, Λ, Σ, Ξ, both δµl, δµsoccur, [21].
q=u,d,s?N|qq|N?, while the slope of the green curve gives the valence contri-
q?N|qq|N?con
q?N|qq|N?
∼
4.0
9.7∼ 0.41
q?N|qq|N?
∼
1
3
?9.7 − 4.0
9.7
?
∼ 0.19,(15)
πis proportional to the
q?N|(qq)R|N?con
q?N|(qq)R|N?
∼
3.2
4.3∼ 0.74
q?N|(qq)R|N?
∼
1
3
?4.3 − 3.2
4.3
?
∼ 0.085, (16)
6

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for renormalised lattice quantities, giving y(N)R∼ 2 × 0.085/(1 − 0.085) ∼ 0.19.
So although for bare matrix elements, there is a significant strange quark content
this is reduced in the renormalised matrix element.
We shall now try to make these considerations a little more quantitive.
4 (Hyperon) σ equations
4.1Renormalisation
For Wilson (clover) fermions under renormalisation the singlet and non-singlet
pieces of the quark mass renormalise differently [23, 24]. We have
mR
q= ZNS?mq+ αZ1
3(2ml+ ms)?,
qmqqq =?
αZ=ZNS− ZS
ZNS
. (17)
In the action the term?
invariant or RGI quantity. Upon writing this in a matrix form and inverting gives
qmR
q(qq)Ri.e. is a renormalisation group
(qq)R=
1
ZNS
?
qq −
αZ
1 + αZ
1
3(uu + dd + ss)
?
, (18)
so for αZ?= 0 then there is always mixing between bare operators.
As an example of where this manifests itself, the relation between the bare,
y(H), and renormalised y(H)R, cf. eq. (2), is then given by
y(H)R=y(H)−2
3αZ(1 − y(H))
3αZ(1 − y(H))
1 +1
, (19)
so we see that y(H)R?= y(H)for clover fermions. Additionally, since αZ> 0 and
y(H)∼
> 0 we find that y(H)R< y(H), i.e. is reduced.
Useful quark combinations are the octet and singlet combinations, namely
(uu + dd)R− 2(ss)R
=
1
ZNS
?(uu + dd) − 2(ss)?
1
ZNS(1 + αZ)
(uu + dd)R+ (ss)R
=
?(uu + dd) + (ss)?. (20)
Furthermore, using the Feynman-Hellman theorem, eq. (14) and with the hadron
flavour expansion, eq. (7) together with eq. (11) gives
?H|(uu + dd)R− 2(ss)R|H? =
1
ZNScH
1
ZNS
(21)
?H|(uu + dd)R+ (ss)R|H? =
M′
0
1 + αZ
. (22)
Eq. (21), the equation for the matrix element of an octet operator, only involves
cH(the hadron mass expansion keeping the singlet quark mass constant), while
7

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eq. (22), the matrix element of a singlet operator, only involves M′
changing the singlet quark mass). Eq. (21) also leads to eq. (3) as discussed in
the introduction3.
Finally from eq. (21), (22) note that the quantities
0(occuring when
(ms− ml)?H|(uu + dd) − 2ss|H?, (2ml+ ms)?H|(uu + dd) + ss|H?, (23)
are RGI, all Z factors cancel when they are renormalised. Linear combinations of
these two quantities are also RGI in particular the combination used previously
of σ(H)
l
+ σ(H)
s
=?
are not RGI, see eq. (18). The renormalised quantities are mixtures of the two
lattice quantities, and αZis needed to relate lattice values to continuum values.
Refering back to Fig 2 we see that the bare lattice strange sigma term is much
larger that the renormalised strange sigma term, due to a cancellation between
the two terms in eq. (18).
qmq?H|qq|H?. However, σ(H)
l
and σ(H)
s
considered separately
4.2σ equations
Multiplying the renormalised quark mass, eq. (17), together with eqs. (21), (22)
(or more generally with eq. (18)) we can find RGI combinations (i.e. a form where
the renormalisation constant ZNScancels). In particular we find
σ(H)
l
− 2rσ(H)
s
=
3r
1 + 2r(1 + αZ)m0cH
3r
1 + 2rm0M′
(24)
σ(H)
l
+ rσ(H)
s
=
0(m0),(25)
where r is the ratio of quark masses
r ≡mR
l
mR
s
. (26)
Thus we have to find the (fixed) coefficients (1+αZ)m0cH, m0M′
determine the physical values of the sigma terms by extrapolating to the point
where the quark mass ratio takes its physical value, i.e. r = r∗.
We observe that we have two simultaneous equations, which can be easily
solved to give
0(m0). We then
σ(H)
l
=
r
1 + 2r[(1 + αZ)m0cH+ 2m0M′
1
1 + 2r[−(1 + αZ)m0cH+ m0M′
0(m0)]
σ(H)
s
=
0(m0)] . (27)
3The RHS of eq. (21) can be re-written as cN/ZNS= 3A1/ZNS. Together with MΞ+MΣ−
2MN= −9A1δml= 3A1(mR
that also picks out the A1coefficient is MΞ− MΛ= −3A1δml.
s− mR
l)/ZNSthis gives eq. (3). An alternative mass combination
8

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We see that the smallness of σ(H)
the presence of an additional r in its numerator. As σ(H)
M′
y(H)R, as can be seen either directly from eq. (27) or from eq. (22),
l
in comparison to σ(H)
s
is certainly guaranteed by
> 0 we must also have
s
0(m0) > (1 + αZ)maxcH. These coefficients are also sufficient to determine
y(H)R= 2−(1 + αZ)m0cH+ m0M′
(1 + αZ)m0cH+ 2m0M′
0(m0)
0(m0). (28)
Again, as seen in section 3, y(H)Ronly depends on gradients and not on the
physical point.
It is now convenient to normalise the coefficients by XN so we now need to
find the coefficients (1 + αZ)m0cH/XN(m0) and m0M′
0(m0)/XN(m0).
4.3Determination of the coefficients
The hint for determining the coefficients from our lattice data is given in section 3,
where we consider gradients with respect to a renormalised or physical quantity
– here taken as the pion mass. As in eq. (7) we also have a similar expansion for
the pseudoscalar octet,
M2
π= M2
0π+ 2αδml+ O(δm2
l), (29)
(together with M2
Analogously to eq. (10) we can define a flavour singlet quantity
K= M2
0π− αδml+ O(δm2
l), M2
ηs= M2
0π− 4αδml+ O(δm2
l)).
X2
π=1
3(2M2
K+ M2
π) = M2
0π+ O(δm2
l). (30)
However, as well as eq. (7), we have the the additional constraint from PCAC
M2
π= 2BR
0mR
l,(31)
(together with M2
K= BR
0(mR
l+ mR
s), M2
ηs= 2BR
0mR
s) which implies that
M2
0π= 2α(1 + αZ)m,α = BR
0ZNS.(32)
If we now consider an expansion in the (physical) pion mass then eliminating δml
between eq. (7) and eq. (29) gives
MH
XN
=
?
1 − [(1 + αZ)m0
cH
XN]
?
+ [(1 + αZ)m0
cH
XN]M2
π
X2
π
, (33)
from the point on the symmetric line m0= m. Thus if we plot MH/XN versus
M2
diately yields (1 + αZ)m0cH/XN. The only assumption is that the ‘fan’ plot
π/X2
π(holding the singlet quark mass, m constant) then the gradient imme-
9

Page 10

splittings remain linear in δmldown to the physical point. In Fig. 1 we show this
plot giving the results
(1 + αZ)m03A1
XN
=0.1899(55),0.2066(68),
(1 + αZ)m03A2
XN
= 0.03942(314), 0.04164(431), (34)
for κ0= 0.12090, 0.12092 respectively.
Alternatively on the flavour symmetric line, ml= m (i.e. δml= 0), so varying
m from a point m0gives
M2
π(m) = M2
0π(m) = M2
0π(m0) + M2′
= 2α(1 + αZ)(m0+ (m − m0)),
0π(m0)(m − m0)
(35)
which gives M2′
eqs. (11), (35) gives
0π(m0) = 2α(1 + αZ). So now eliminating (m − m0) between
XN(m)
XN(m0)=
?
1 − [m0M′
0(m0)
XN(m0)
]
?
+ [m0M′
XN(m0)
0(m0)
]X2
X2
π(m)
π(m0).(36)
Again in a plot of XN(m)/XN(m0) versus X2
ately gives the required ratio m0M′
XN and M2
particular m). In Fig. 3 we plot XN(m)/XN(m0) versus X2
π(m)/X2
π(m0) the gradient immedi-
0(m0)/XN(m0). We have also replaced MNby
π(which allows us to use all the 323× 64 data available for a
πby X2
π(m)/X2
π(m0). From
0.40 0.500.60 0.70
Xπ
0.80
2(m0)
0.90 1.00 1.10
2(m)/Xπ
0.80
0.85
0.90
0.95
1.00
1.05
1.10
XN(m)/XN(m0)
Figure 3: XN(m)/XN(m0) versus X2
together with the linear fit from eq. (36).
π(m)/X2
π(m0) along the flavour symmetric line,
eq. (36) this gives
m0M′
XN(m0)
0(m0)
= 0.273(32). (37)
10

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Finally the quark mass ratio, r, must be estimated. In Fig. 4 we plot (2M2
K−
0.00 0.050.10
2
0.15
Mπ
2/XN
0.0
0.1
0.2
0.3
0.4
(2MK
2−Mπ
2)/XN
2
κ0=0.12090, 24
κ0=0.12090, 32
3x48
3x64
0.000.05 0.10
2
0.15
Mπ
2/XN
κ0=0.120920, 24
κ0=0.120920, 32
3x48
3x64
Figure 4: (2M2
0.12092 (right panel). The 323× 64 volume results are given by the filled symbols,
while the 243× 48 volume results are shown using empty triangles. The fit is given in
eq. (38). Experimental points are denoted by (red) stars.
K− M2
π)/X2
Nversus M2
π/X2
Nfor κ0= 0.12090 (left panel) and κ0=
M2
π)/X2
Nversus M2
π/X2
N. From eq. (29) we have
2M2
K− M2
X2
N
π
= 3M2
0π
X2
N
− 2M2
π
X2
N
.(38)
As in section 2, we see that for constant m the data points lie on a straight line
(i.e. there is an absence of significant non-linearity). Furthermore the gradient is
fixed at −2. (Indeed leaving the gradient as a fit parameter for the κ0= 0.12090
confirms that this gradient is very close to −2.) Together with PCAC, eq. (31)
this gives the x-axis is proportional to mR
and thus the ratio gives r. Taking our physical scale to be defined from M2
(i.e. from the x-axes of Fig. 4) gives
lwhile the y-axis is proportional to mR
s
π/X2
N|∗
1
r∗=mR
s
mR
l
????
∗
=
?27.28(16) κ0= 0.12090
26.23(24) κ0= 0.12092
. (39)
What can we say about corrections to the linear terms? The simple linear
fit describes the data well, from the symmetric point to our lightest pion mass,
both along the m = const. line and the flavour symmetric line.To see the
11

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possible influence of curvature we compare linear fits with quadratic fits as briefly
discussed in Appendix A. These will be used in section 5 to estimate possible
systematic effects.
4.4 Some approximate relations between the σ terms
We shall now discuss some relations between the sigma terms within a multi-
plet (here taken to be the baryon octet) which are exact within the linear case
discussed here, but which we might expect to always be approximately true.
The singlet relation eq. (22) or eq. (25) is the same for every hadron. So in
terms of sigma terms this becomes
σ(H)
l
+ rσ(H)
s
≈ σ(H′)
l
+ rσ(H′)
s
.(40)
At the flavour symmetric point it follows from group theory that a singlet operator
has the same value for every member of a multiplet, so eq. (40) must hold. But
this can change if we move away from the symmetric point. (We shall briefly
discuss this at the end of this section.)
We can find another collection of near identities by summing over a singlet
combination of hadrons — this can be either a singlet of S3or a singlet of SU(3).
If we do this, the expectation values of uu, dd and ss will be exactly equal at the
flavour symmetry point, and stay again nearly equal away from the symmetry
point. By this argument we expect
σ(Λ)
l
+ σ(Σ)
+ σ(Ξ)
l
≈ 2r?σ(Λ)
≈ 2r?σ(N)
s
+ σ(Σ)
s
?
σ(N)
l
+ σ(Σ)
l
l
s
+ σ(Σ)
s
+ σ(Ξ)
s
?. (41)
(Again this relation, as with the other relations discussed here, is exactly true for
the linear case.)
Other relations come from the Gell-Mann–Okubo relation, [25, 26] that the
27-plet mass combination is very small,
2MN− 3MΛ− MΣ+ 2MΞ≈ 0, (42)
for all values of ml,ms. In our approach, its derivatives are also near zero. We
therefore expect
2σ(N)
l
2σ(N)
s
− 3σ(Λ)
− 3σ(Λ)
l
− σ(Σ)
− σ(Σ)
l
+ 2σ(Ξ)
+ 2σ(Ξ)
l
≈ 0
≈ 0.
sss
(43)
We obtain an even stronger version of these relations by taking the singlet com-
bination, proportional to (uu + dd)R+ (ss)R,
2σ(N)
l
− 3σ(Λ)
l
− σ(Σ)
l
+ 2σ(Ξ)
l
+ r?2σ(N)
s
− 3σ(Λ)
s
− σ(Σ)
s
+ 2σ(Ξ)
s
?≈ 0.(44)
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There is also a relation between the sigma terms and the hadron masses,
[2] as the constants A1and A2which occur in the mass splittings also occur in
the leading order expressions for the sigma terms. So there will be connections
between masses and sigma terms. One particularly simple relation is
MH− σ(H)
l
− σ(H)
s
≈ MH′ − σ(H′)
l
− σ(H′)
s
. (45)
(i.e. the baryon mass difference is closely accounted for by the sigma terms.) For
the linear case this is again exact, with this equation being equal to M0(m0) −
m0M′
see that this is just the common hadron mass in the chiral limit along the flavour
symmetric line, when ml= 0 = msor m = 0. σ(H)
as that part of the hadron mass which is due to mland msrespectively. The
remnant, M0(m0) − m0M′
and gluon kinetic energy, interaction energy, etc., [2], i.e. the part of the hadron
mass which is not due to the coupling with the Higgs vacuum expectation value.
We can use the higher order mass equations in [21] to estimate how well the
relations in this section hold. Most of the relations have violations proportional
to the first power of the SU(3) breaking parameter, δml. The corrections to
eqs. (40) and (41) and the first relation in eq. (43) are O(mlδml). The σsrelation
in eq. (43) has corrections O(msδml). When we combine these two relations
to form eq. (44), the leading violation terms cancel, and we have a relation with
corrections O(mlδm2
and O(δm2
0(m0) for all the octet baryons (upon using eqs. (7), (27)). From eq. (11) we
l
and σ(H)
s
can be thought of
0(m0), is the part of the hadron mass due to the quark
l). The corrections to the mass relation eq. (45) are O(mδml)
l).
5 Results
We can now numerically determine y(H)Rand σ(H)
We start with y(H)R. From eq. (28), together with eqs. (34), (37) and eq. (8)
gives the results in Table 1. The first error is the fit error (in this case dominated
by the error in eq. (37)), while the second error indicates possible effects from
higher order terms in the flavour expansion, as indicated at the end of section 4.3.
We see that there is an order of magnitude increase in the fraction of ?H|(ss)R|H?
compared to ?H|(uu + dd)R|H? as we increase the strangeness content of the
baryon from the nucleon (no valence strange quarks) to the Ξ (two valence strange
quarks).
Turning to the sigma terms themselves, from eq. (24) we can find an indication
of the magnitude of σ(N)
l
as approximately (with XN= 1.1501GeV),
l
, σ(H)
s
.
σ(N)∗
l
∼ [22 ∼ 25] +σ(N)∗
s
13
MeV > [22 ∼ 25]MeV, (46)
(for κ0= 0.12090, 0.12092 respectively). The last inequality follows as obviously
13

Page 14

NΛΣΞ
κ0= 0.12090
0.80(14)(28)
23(3)(4)
250(34)(68)
κ0= 0.12092
0.79(14)(28)
24(3)(4)
247(34)(69)
y(H)R∗
σ(H)∗
l
σ(H)∗
s
0.22(9)(15)
29(3)(4)
89(34)(59)
1.23(20)(41)
20(3)(4)
334(34)(68)
2.14(38)(64)
16(3)(5)
453(34)(58)
[MeV]
[MeV]
y(H)R∗
σ(H)∗
l
σ(H)∗
s
0.18(9)(15)
31(3)(4)
71(34)(59)
1.25(20)(42)
21(3)(4)
336(34)(69)
2.30(42)(68)
16(3)(4)
468(35)(59)
[MeV]
[MeV]
Table 1: Results for the baryon octet for y(H)R∗, σ(H)∗
for κ0= 0.12090, 0.12092.
l
, σ(H)∗
s
with H = N, Λ, Σ, Ξ
σ(N)∗
s
to this result.
The results for σ(H)∗
the statistical error, while the second error is due to possible systematic effects
as discused in Appendix A.1 and A.2 (in quadrature). While the data for κ0=
0.12090 is more complete than for κ0 = 0.12092 (cf. the plots in Fig. 1) and
demonstrates linear behaviour, as the path starting at κ0= 0.12092 is closer to
the physical point (cf. Fig. 4) we shall use these values as our final values. These
results are illustrated in Fig. 5 for y(H)R∗for H = N, Σ, Ξ, Λ. By varying r in
> 0. Indeed this shows that a non-zero σ(N)∗
s
> 0 can only add a few MeV
l
and σ(H)∗
s
are also given in Table 1. The first error is
N
ΛΣΞ
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
y
(H)R*
Figure 5: y(H)R∗for H = N, Λ, Σ, Ξ using the results from Table 1 for κ0= 0.12092.
14

Page 15

0.00 0.25 0.500.75 1.001.25
Mπ
2/Xπ
2
0
50
100
150
200
250
300
350
400
450
500
550
σl and σs [MeV] [Octet]
N(lll)
Λ(lls)
Σ(lls)
Ξ(lss)
Figure 6: σ(H)
H = N, Λ, Σ, Ξ for κ0= 0.12092. The physical and symmetric lines are denoted by
vertical dashed lines.
l
(decreasing from the symmetric point x = 1) and σ(H)
s
(increasing) for
eq. (27)4, we plot in Fig. 6 σ(H)
Ξ from the symmetric point (vertical dashed line at x = 1) to the physical point
(left vertical dashed line). σlis rapidly decreasing while σsis increasing as we
decrease the quark mass. Also, as expected σ(N)
σ(N)
s
is smallest. Finally in Fig. 7 we plot σ(H)∗
Ξ, again using Table 1.
l
and σ(H)
s
for the baryon octet, H = N, Λ, Σ and
l
is largest for the nucleon, while
, σ(H)∗
s
against H = N, Λ, Σ and
l
6 Conclusions
Keeping the average quark mass constant gives very linear ‘fan’ plots from the
flavour symmetric point down to the physical point. This implies that an expan-
sion in the quark mass from the flavour symmetric point will give information
about the physical point. In this article we have applied this to estimating the
sigma terms (both light and strange) of the nucleon octet. There has been no
use of a chiral perturbation expansion (indeed this is an opposite expansion to
4Using, for example, the results from the left panel of Fig. 4, r may be re-written as
r =
M2
π/X2
π/X2
π
3 − 2(M2
π).
15