Peeking into the θ vacuum of
4d SU(2) Yang-Mills theory
Norikazu Yamada (KEK/SOKENDAI)
in collaboration with
Ryuichiro Kitano (KEK/SOKENDAI)
Ryutaro Matsudo (KEK)
Masahito Yamazaki (Kavli IPMU)
Based on 2102.08784 [hep-lat], JHEP02, 073 (2021)@Kyoto Univ. 2021/06/30

Introduction• Many interesting subjects related to the θ term or topology of YM theorye.g.) strong CP, axion, EDM, fate of UA(1), …

Focus on the θ dependence of free energy density in 4d SU(2) YM

2Free energy density:

f(θ)

Z(θ)
−Vf(θ)
e
=
Z(0)

Where

Z(θ) = 𝒟U e∫

−SYM+iθQ1
a a4
Q = d x q(x) and q(x) =
ϵ
F
F
μνρσ
μν
ρσ
2
∫
64π

For SU(N) YM theory,

Q ∈ ℤ ⇒ Z(θ) = Z(θ + 2π) ⇒ f(θ) = f(θ + 2π)
SYM is CP even ⇒ Z(θ) = Z(−θ) ⇒ f(θ) = f(−θ)
3}f(π − θ′) = f(π + θ′)θ dependence and CP violation
Dilute instanton gas approximation (DIGA)f(θ) = χ(1 − cosθ)
E. WITTENf(θ)• a single branch• smooth everywhere

1. The vacuum energy as a function as a function of 0 may have a single branch (first drawing)4
y have several branches, which cross at particular values of 0 (second drawing).

θ dependence and CP violation
Dilute instanton gas approximation (DIGA)f(θ) = χ(1 − cosθ)
E. WITTEN

Large N argument [Witten (1980, 1998)]22f(θ) = χ/2 min(θ + 2πk) + O(1/N )
k∈ℤ

f(θ)

f(θ)

-π

0

2π

π

3π

θ

• a single branch• smooth everywhere• several branches crossing• spontaneous CPV (1st order PT) at θ = π withthe order parameter df(θ)/dθ = − i⟨q(x)⟩1. The vacuum energy as a function as a function of 0 may have a single branch (first drawing)4
y have several branches, which cross at particular values of 0 (second drawing).

θ dependence and CP violation
Dilute instanton gas approximation (DIGA)f(θ) = χ(1 − cosθ)
E. WITTEN

Large N argument [Witten (1980, 1998)]22f(θ) = χ/2 min(θ + 2πk) + O(1/N )
k∈ℤ

f(θ)

f(θ)

-π

0

2π

π

3π

θ

• a single branch• smooth everywhere• several branches crossing• spontaneous CPV (1st order PT) at θ = π withthe order parameter df(θ)/dθ = − i⟨q(x)⟩Interested in f(θ) around θ ≈ π in 4d SU(N) YM theory.

1. The vacuum energy as a function as a function of 0 may have a single branch (first drawing)4
y have several branches, which cross at particular values of 0 (second drawing).

Learning from 2d CP

N−1

model

N
ℒ=
Dμz Dμz − iθq2gz : N-component complex scalar field with z̄z = 1
Dμ = ∂μ + iAμ ,Aμ = iz̄∂μz

1iq(x) =
ϵμν∂μ Aν =ϵμνDμz Dμz
2π2π• Good testing ground for 4d SU(N) because of many similarities[asymptotic freedom, dynamical mass gap, instanton, 1/N expandable, …]• Gapped and CP broken at θ = π for N ≥ 3.
1
• But CP ( i.e. N = 2 ) is exceptional !

gapless and no CPV at θ = π (⟺Haldane conjecture)5f(θ) in 2d CP

N−1

prediction (4) throughout the range 0 < #< 2$. On theother hand, for CP5 (h’s) and CP9 (& ’s), as shown in
Fig. 2, the coefficient extracted from a simple area-law fit
to the Wilson loops continues to rise beyond #¼ $, violating the expected symmetry (7). We will argue that the
behavior of CP5 and CP9 for #> $ is an effect of having

model (lattice results)[Keith-Hynes and Thacker (2008)]

f(θ)/χ vs θ/2π

20

9

○ : CP
5: CP
1× : CPf(θ)/χ
E/χ t

15tion of "ð#Þ.
The results for "ð#Þ for CP1 are plotted for 0 < #< $ in
Fig. 3. With the topological susceptibility fixed to the value
obtained from the fluctuation of the integer-valued global2
topological charge, !t ¼ h% i=V, the solid line is a zeroparameter fit to the dilute instanton gas formula (4). Also
plotted (dotted line) is the leading-order large N prediction,

1
1
2
−S
−iθQ
−Vsub fsub(θ)
CP(N−1)
sub
"ð#
Þ
%
"ð0Þ
¼
!
#
:
(8)
t
e
=
𝒟z𝒟z̄
e
2
∫
Z[0]
5
9
The corresponding results for CP and CP are shown in
Figs. 4 and 5. The−iθQsolid and dotted curves are again the

= ⟨e

⟩

sub

1
−i
lat
Qsub5 =d 2xϵμν∂μ Aν → Qsub
=
ln Px
∑
∫x∈V
2π
2π
sub
x∈Vsub(Px : plaquette)
410CP1 E/χt

5

0
0

0.2

0.4

0.6

0.8

1

CP is indeed consistent with the DIGA,f(θ) = χ(1 − cos θ)

3

1

2
while
others indicate CPV.

1

1

θ/2π

FIG. 2 (color online). The free energy density "ð#Þ for CP 6(!’s), CP5 (h’s), and CP9 (& ’s) as measured from fractionallyPrevious Lattice calculations of 4d SU(N)
1 a aℒθ = FμνFμν − iθq
4
2

g
a a
q(x) =ϵFFμνρσ μν ρσ
264π• Sign problem makes direct latticecalculation difficult/impossible.

7

Previous Lattice calculations of 4d SU(N)
1 a aℒθ = FμνFμν − iθq4• Relies on Taylor expansion around θ = 0

χ 2
2
4
f(θ) = θ (1 + b2θ + b4θ + ⋯)
2

2

g
a a
q(x) =ϵFFμνρσ μν ρσ264π

and determines each coefficient on the lattice by⟨Q 2⟩θ=0χ=V⟨Q 4⟩θ=0 − 3⟨Q 2⟩2θ=0
b2 = −
12⟨Q 2⟩θ=0• Sign problem makes direct latticecalculation difficult/impossible.

⟨Q 6⟩θ=0 − 15⟨Q 2⟩θ=0⟨Q 4⟩θ=0 + 30⟨Q 2⟩3θ=0
b4 =360⟨Q 2⟩θ=0⋮

7

First two coefficients

0.04

c
2
2
4
χ/σ = C∞ + 2 + O(1/N )
N

0.03

C
0.02

0.01

0.00
0.00

N=8 N=6 N=5

0.02

0.04

N=4

0.06

1/N

2

N=3

0.08

0.10

0.12

1: Results for C ≡ χ/σ 2 versus 1/N 2 , for N = 3, 4, 5, 6, 8. We show MC data taken from Refs. [195]rcle, by cooling), [375] (square, by cooling), [379] (diamonds, by cooling), [164] (left triangle, by[191] (cross, by overlap), [30] (triangle, by heating), [222] (open circle, by HYP smoothing). Then the y-axis, indicated by an asterisk, shows the extrapolations to N = ∞ obtained8 in Ref. [195].
ta have been slightly shifted along the x-axis to make them more visible.

[Review by Vicari and Panagopoulos (2018)]

First two coefficients

0.04[Review by Vicari and Panagopoulos (2018)]c224
χ/σ = C∞ + 2 + O(1/N )Nb̄2
4b2 = 2 + O(1/N )
N
0.4

0.03

0.0
2

C

N b2

0.02

-0.4

0.01

N=8 N=6 N=5

N=4

N=3

-0.8

-1.2

0.00
0.00

0.02

0.04

0.06

1/N

2

0.08

0.10

0.12

3

4

5

6

N

13:dataPlot
offrom
N 2 bRefs.
1: Results for C ≡ χ/σ 2 versus 1/N 2 , for N = 3, 4, 5, 6, 8. WeFigure
show MCtaken[195] N , for N = 3, 4, 6. We show MC data taken from Ref. [195]2 versus
rcle, by cooling), [375] (square, by cooling), [379] (diamonds,
by cooling),
triangle,by by heating), from Ref. [277] (diamond, by overlap). Some dat
cooling),from[164]
Ref.(left[199](square,[191] (cross, by overlap), [30] (triangle, by heating), [222] (open
HYP smoothing).
The
havecircle,
been byslightly
shifted along
the x-axis to make them more visible.
n the y-axis, indicated by an asterisk, shows the extrapolations to N = ∞ obtained8 in Ref. [195].
ta have been slightly shifted along the x-axis to make them more visible.

First two coefficients

0.04[Review by Vicari and Panagopoulos (2018)]c224
χ/σ = C∞ + 2 + O(1/N )Nb̄2
4b2 = 2 + O(1/N )
N
0.4

0.03

0.0
2

C

N b2

0.02

-0.4

0.01

N=8 N=6 N=5

N=4

N=3

-0.8

-1.2

0.00
0.00

0.02

0.04

0.06

1/N

2

0.08

0.10

0.12

3

4

5

N

N≥3

6

small
corrections
the
large
N
limit
indicates
CPV
for
.
13:to
of
N 2 bRefs.
versus
N
,
for
N
=
3,
4,
6.
We
show
MC
data
taken from Ref. [195]
1: Results for C ≡ χ/σ 2 versus 1/N 2 ,only
for N =
3, 4, 5, 6,
8. WeFigure
show MC
dataPlot
taken
from
[195]
2
rcle, by cooling), [375] (square, by cooling),by cooling),
triangle,by by heating), from Ref. [277] (diamond, by overlap). Some dat(No [379] (diamonds,
calculation)
cooling),from[164]
Ref.(left[199](square,[191] (cross, by overlap), [30] (triangle, by heating), [222] (open
HYP smoothing).
The
havecircle,
been byslightly
shifted along
the x-axis to make them more visible.
n the y-axis, indicated by an asterisk, shows the extrapolations to N = ∞ obtained8 in Ref. [195].
ta have been slightly shifted along the x-axis to make them more visible.

SU(2)Finite temperature
8π
11N
11N
4− 3
DIGA : f(θ) ∼ T 4 exp −3 (1 − cos θ)
(1
−
cos
θ)
∼
T
Λ
( g 2(T) )∝ χ(T )
= θ 2 /2(1 + b θ 2 + b θ 4 + ⋯)2

2

⇒ b2 = − 1/12,

9

4

⋯

Finite temperature
8π
11N
11N
4− 3
DIGA : f(θ) ∼ T 4 exp −3 (1 − cos θ)
(1
−
cos
θ)
∼
T
Λ
( g 2(T) )∝ χ(T )
= θ 2 /2(1 + b θ 2 + b θ 4 + ⋯)2

2

R(T ) = χ(T )/χ(T = 0)

⇒ b2 = − 1/12,[Del Debbio, Panagopoulos, Vicari (2004)]
1.0
0.8

R

0.6
0.4

N=4, Lt=6
N=4, Lt=8
N=6, Lt=6

0.2
0.0

-0.1

0.0

0.1

t = T/Tc − 1

ratio R(T ) ≡ χ(T )/χ(T = 0) as a function of the reduced temperature t ≡ T /Tc − 1.
gauge theory, these curves are expected to converge to a continuum 9curve when Lt → ∞,ngth, in units of the lattice spacing, of the lattice along the Euclidean time direction. From

4⋯

Finite temperature
8π
11N
11N
4− 3
DIGA : f(θ) ∼ T 4 exp −3 (1 − cos θ)
(1
−
cos
θ)
∼
T
Λ
( g 2(T) )∝ χ(T )
= θ 2 /2(1 + b θ 2 + b θ 4 + ⋯)2

2

R(TVICARI) = χ(T )/χ(T = 0)
BONATI, D’ELIA, ROSSI, andb2 = −D1/12,⋯ (2016)
PHYSICAL ⇒REVIEW
94, 085017[Del Debbio, Panagopoulos, Vicari (2004)]

related to the fact that the full-fledged dependence on θ of1.0
the vacuum energy
for any finite value of N must exhibit a
2π periodicity which disappears in the large-N limit, thus
0.8
implying a noncommutativityof the expansions and avanishing radius of convergence in the variable θ̄ ≡ θ=N.
0.6
R mention
We finallythat the large-N behavior (26) of the
topological susceptibility has been confirmed by numerical
0.4
N−1 L =6
results of lattice CPN=4,
models[23,72–75]. Instead,
t
N=4,
Lt=8
numerical resultsfortheθ-expansion
coefficients
b
have
2n
0.2
N=6, Lt=6
never been obtained yet.
0.0

-0.1

0.0

IV. CONCLUSIONS
t = T/Tc − 1

4

0.1

0

T=0, N=6

N=3, Lt=10
N=6, Lt=6

T=0, N=3-0.025-0.05

b2-0.0751−
12

instanton gas-0.1[Bonati, et al. (2016)]-0.125

-0.05

0

0.05

t

0.1

0.15

We study the large-N scaling behavior of the θ dependFIG.
11.
Behaviorofbacrossthedeconfinementtransitionfor2
ratio R(T ) ≡ χ(T )/χ(T = 0) as a function of N−1
the reduced temperature t ≡ T /Tc − 1.
ence
of
4D
SUðNÞ
gauge
theories
and2DCP
models,
SUð3Þ
and
SUð6Þ
[t ∞,
is the reduced temperature defined by
9gauge theory, these curves are expected to converge to a continuum
curve
when
Lt →
t ¼ timeðT − direction.
T c Þ=T c ]. The
horizontal bands denote the zero temperwherein θunits
is theparameter
associatedwiththe along
Lagrangian
ngth,of thelattice spacing,of thelatticethe Euclidean
FromFinite temperature
8π
11N
11N
4− 3
DIGA : f(θ) ∼ T 4 exp −3 (1 − cos θ)
(1
−
cos
θ)
∼
T
Λ
( g 2(T) )∝ χ(T )
= θ 2 /2(1 + b θ 2 + b θ 4 + ⋯)2

2

R(TVICARI) = χ(T )/χ(T = 0)
BONATI, D’ELIA, ROSSI, andb2 = −D1/12,⋯ (2016)
PHYSICAL ⇒REVIEW
94, 085017[Del Debbio, Panagopoulos, Vicari (2004)]

related to the fact that the full-fledged dependence on θ of1.0
the vacuum energy
for any finite value of N must exhibit a
2π periodicity which disappears in the large-N limit, thus
0.8
implying a noncommutativityof the expansions and avanishing radius of convergence in the variable θ̄ ≡ θ=N.
0.6
R mention
We finallythat the large-N behavior (26) of the
topological susceptibility has been confirmed by numerical
0.4
N−1 L =6
results of lattice CPN=4,
models[23,72–75]. Instead,
t
N=4,
Lt=8
numerical resultsfortheθ-expansion
coefficients
b
have
2n
0.2
N=6, Lt=6
never been obtained yet.
0.0

-0.1

0.0

IV. CONCLUSIONS
t = T/Tc − 1

4

0.1

0

T=0, N=6

N=3, Lt=10
N=6, Lt=6

T=0, N=3-0.025-0.05

b2-0.0751−
12

instanton gas-0.1[Bonati, et al. (2016)]-0.125

-0.05

0

0.05

t

0.1

0.15

We study the large-N scaling behaviorof the forθ dependFIG.
11.
Behaviorofbacrossthedeconfinementtransition
for
DIGA
works
(No
calculation)
T
>
1.15
T
SU(2)
2
ratio R(T ) ≡ χ(T )/χ(T = 0) as a function of N−1
the reduced temperaturet ≡ T /Tc − 1.
c
ence
of
4D
SUðNÞ
gauge
theories
and2DCP
models,
SUð3Þ
and
SUð6Þ
[t ∞,
is the reduced temperature defined by
9gauge theory, these curves are expected to converge to a continuum
curve
when
Lt →
t ¼ timeðT − direction.
T c Þ=T c ]. The
horizontal bands denote the zero temperwherein θunits
is theparameter
associatedwiththe along
Lagrangian
ngth,of thelattice spacing,of thelatticethe Euclidean
FromSummary of previous results on f(θ)• Large N argument seems robust

CPV at θ = π and large N• Formal arguments tell “For general N, CP has to be broken atθ = π if the vacuum is in the confining phase.”

SU(N) for N ≫ 1

T
Tc[Gaiotto, et al.(2017)], [Kitano, Suyama, NY(2017)]

deconfineconfineCP• Numerical evidences of CPV for N ≥ 3• What happens to the possible smallest N, i.e. SU(2) ?

π1Is it like “large N” or “2d CP ” ?

2πθ

Lattice numerical simulations

SU(2)?

T
Tc

deconfineconfine

π

θ

10

2π

χ and b2 in SU(2) at T = 0
0.1

[14]
[15]
[16]
[37], [43], [44], [45]this work[14]
ph fitχ / σ2

0.08

0.1
0.06

χ/σ

2

[15]
0.04
0.08
[16]
[37], [43], [44], [45]
0.02
0.06
this work
ph fit0.04 00
0.1• smoothly connected to large N limit
N ∈ ℤ can be analytically continued to ℝf(θ) would be smooth function of N

0.2

0.3

1
(i.e. not instanton-like)• b2 ≠ −12

conjectured that SU(2) belongs to large N class and

1/N2

0.02

CPV takes places at θ = π.

0
0

0.1

0

b2

b2

0.3

0.2

0.3

DIGA
[15]
[16]
this work
ph fit

0

-0.1

-0.05

0.2
1/N2

-0.05

0

DIGA 0.1[15][16]
this work
ph fit2[Kitano, NY, Yamazaki (2021)]

1/N2
11

χ and b2 in SU(2) at T = 0
0.1

[14]
[15]
[16]
[37], [43], [44], [45]this work[14]
ph fitχ / σ2

0.08

0.1
0.06

• smoothly connected to large N limit
N ∈ ℤ can be analytically continued to ℝf(θ) would be smooth function of N

χ/σ

2

[15]
0.04
0.08
[16]
[37], [43], [44], [45]
0.02
0.06
this work
ph fit0.04 0
0
0.1

0.2

0.3

1
(i.e. not instanton-like)• b2 ≠ −12

conjectured that SU(2) belongs to large N class and

1/N2

0.02

CPV takes places at θ = π.
T

0
0

0.1

0

0.2

0.3

SU(2) ?

deconfine

2

1/N

Tc

b2

confine

0

-0.1

-0.05

0

CP

DIGA
[15]
[16]
this work
ph fit-0.05

b2[Kitano, NY, Yamazaki (2021)]

DIGA 0.1[15][16]
this work
ph fit
2

0.2

0.3

1/N2
11

π
θ

2π

New method without any expansion[Kitano,Matsudo,NY,Yamazaki(2021)]

Introduce sub-volume Vsub = l

Qsub =∑

q(x) ∉ ℤ

4

and“sub-volume method”

x∈VsubZ(θ)
1−Sg+iθQsub−Vsub fsub(θ)
iθQsub
e
=
=
𝒟U e
= ⟨e
⟩θ=0
Z(0) Z(0) ∫1
fsub(θ) = −ln⟨ cos(θQsub) ⟩θ=0
Vsub
s(θ)
2
f(θ) = lim fsub(θ) = lim f(θ)++ O(1/l )}lVsub→∞l→∞ {4with ldyn ≪ Vsub ≪ Vfull(ldyn : dynamical length scale)s(θ) : surface tension
12

θ≠0
Vsub

θ=0
Vfull

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Expected behavior of fsub(θ) as a function of Vsub
4
• Vsub ≫ ldyn would have to be satisfied.

0
0
.
0

00
0
.
0

1

<q(

)>
0
(
x) q

4
• As long as Vsub ≫ ldyn , fsub(θ) is expected to show the scalings(θ)
2behavior, fsub(θ) = f(θ)++ O(1/l ).

l• Buch a behavior will end as Vsub → Vfull , where
Qsub → Qfull ∈ ℤ . Thus, Vsub ≪ Vfull is required.

• On the other hand, the method fails when | θ Qsub | ∼ π because
fsub(θ) ∝ ln⟨ cos(θQsub) ⟩ becomes ill-defined.
• Crucial question:
4
Vsub satisfying ldyn ≪ Vsub ≪ Vfull and | θ Qsub | < π exists ?

13

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Vsub

θ≠0

θ=0

a
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fo

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B. Simulation in quenched QCD

While many calculations of the hadron spectrum havebeen performed in quenched QCD, comparisons between ourfull QCD results and quenched results from other simulations may be subject to systematic uncertainties due to thedifference in the simulation details. We therefore carry out aset of quenched calculations of the hadron spectrum using
the same lattice actions and simulation parameters as thoseIn the static potential calculation,
Wilsonfor full QCD
runs. loop is inserted.
Our simulations are performed at ! "6.0, where the lat0.1074#14$
fm.
We taketicespacingfrom m " equals−Sfixed
∮
i∮ A
i
A
−V(𝒜)
QCD
c SW"1.769 which is the value determined nonperturbatively by the ALPHA Collaboration %35&. Three lattice sizes 123 !48, 163 !48, and 203 !48 are employed ini∮ Aorder to investigate finite-size effects.
Gauge configurations are generated with a combination of𝒜→∞
the heat-bath and over-relaxation algorithms. We call fourheat-bath sweeps with a succeeding over-relaxation step an
In sub-volume method, insteada operator
extendingiteration.
We accumulate
statistics of over
60 000 iterations on

In measurements in full QCD, we use six values of thevalence quark mass corresponding to the hopping parameter
K val,i (i"1, . . . ,6)"0.1340, 0.1343, 0.1346, 0.1350,
0.1355, and 0.1358, which cover the range of m PS,val /m V,val!0.5–0.8. At each sea quark mass, therefore, there is one
value of K val,i which equals K sea and is identified as the lightquark mass. Other five values of K val,i correspond to themass of strange quarks treated in the quenched approximation. In the following, we use the abbreviation ‘‘diagonal
data’’ to represent hadron correlators or masses with a quark
mass combination in which all valence quark masses areequal to the sea quark mass.
We employ meson operators defined by

Similarity to the static potential calculation
Z( □ )
1=
𝒟U Tr[eZ(1)Z(1) ∫

V(𝒜) = − lim ln⟨Tr[e]e= ⟨Tr[e]⟩ → e]⟩ = σ𝒜 + ⋯M #x $"q̄ (xf ) 'q (g)x ,T'"I, ( 5 , ( ) , ( 5 ( ) ,where f and g are flavor indices and x is the coordinates on
the lattice. Meson correlatorsR*M (x)M (0) † + are calculated
for the following 11 combinations of valence quark masses:

subvolume is inserted.
f(θ) is analogous to σ in the static potential.

[JLQCD (2003)

3#6$

FIG. 7. Static quark potential on
14 20 !48. Left and right figures show data at K sea

R"0.1340 and 0.1355, respectively.

About smearing
1
a a
q(x)
=
ϵ
F
F
Need
to
numericallycalculateonthelattice
μνρσ
μν
ρσ
•
2
64π

• Raw configurations are contaminated by local lumps.
• Smearing (= smoothing a configuration) removes such short-distance artifacts.
• However, at the same time, smearing may alter relevant topological excitations, too.
• We studied this point and developed the procedure to restore relevant information.
[Kitano, NY, Yamazaki (2021)]- calculate an observable every 5 steps of the smearing
extrapolate
those
back
to
,
n
→
0
⟨O⟩
=
lim
⟨O(n
)⟩
APE
APE
nAPE→015

Lattice parameters and observables• SU(2) YM theory by Symanzik improved gauge action

4
β
=
=
1.975
1/(aT
)
=
9.50
(relatively
fine:
)
c
•
2
g
3
• Vfull = 24 × {48, 6, 8} (T = 0, 1.2Tc, 1.6Tc)

Vfull

θ=0

• Periodic boundary condition in all directions• # of configs = { 68000 , 5000 , 5000 }• Calculate Qsub =∑θ≠0q(x) and estimatex∈Vsub1ln⟨ cos(θQsub) ⟩✓ f(θ) = − lim
Vsub→∞ Vsubdf(θ)
1 ⟨ Qsub sin(θQsub) ⟩
=
lim
✓
dθ
⟨ cos(θQsub) ⟩Vsub→∞ Vsub

Vsub

which are used to crosscheck each other
16

bbe
of
=
ze
is
he
nt
of
es
ge

4
3
2
1
0

a fsub(θ) × 10

a f(θ) × 10

5

5

4

a fsub(θ) × 10

5

6

θ=π/2
π
3π/2
5

θ=π/2, nAPE=25
θ=π, nAPE=25
θ=3π/2, nAPE=25

6
5.5
5
4.5
4
3.5
3
2.5
2
1.5
1

4

7

0

0

0.02

0.04

0.06

0.08

0.1

1/l

10

20

30

40

50

60

nAPE

FIG. 2: The linear extrapolation of f (θ) to nAPE = 0.

3

Vsub = l × 6 with l ∈ {12, ⋯, 20}•
FIG. 1: Examples of subvolume• dependence of f (✓) andLinear
fit
works
well
in
eitherextrapolations.
5the extrapolation to the infinite volume.
• Not monotonic function,4 f(π) > f(3π/2)

FIG. 4volume

The fit range is fixed in advance by examining the response of the global topological charge to the smearing
as done in Ref. [12]. The resulting fit ranges and thesimple

sub

17

)/χ

to

2

4

he
he
ue
re

Finite temperature (T = 1.2 Tc)

3

1.2 Tc2
θ /2
1-cosθ∫dθ df/dθ
full volume (nAPE=45)θ dependence of f(θ) at T = 1.2Tc
3• Systematic error due to ambiguity of the scaling region

654f(θ) / χis large for θ > π

1.2 Tc2
θ /2
1-cosθ∫dθ df/dθ
full volume (nAPE=45)

3• Within large uncertainty, consistent with the DIGA.
≈ 0 no CPV above Tc• df(θ)/dθ
θ=πdf• Numerical consistency with ∫ dθ dθ

2
1• Similar results at T = 1.6 Tc

0
-1
0

π/2

π

3π/2

2π

θ
FIG. 3: The ✓ dependence of f (✓) at T = 1.2 Tc .
18

π/2

0

π

3π/2

2π

l → ∞ limit at T = 0
θFIG. 3: The ✓ dependence of f (✓) at T = 1.2 Tc .

4

• Vsub = l with l ∈ {10, 12, ⋯, 20}
4• Data in the range of ldyn ≪ Vsub ≪ Vfull are

θ=π/2, nAPE=20
θ=π, nAPE=20θ=3π/2, nAPE=20

20
15

as(θ)
fsub(θ) = f(θ) +l10• Linear extrapolation works well.

4

a fsub(θ) × 10

5

fitted to

5
0
0

0.02

0.04

0.06

0.08

0.1

1/l
FIG. 4: The linear extrapolation of fsub (✓) to the infinite
volume limit.
19

nAPE → 0 limit at T = 0
•

18
16

• Linear fit works well.
20• Monotonic function f(π) < f(3π/2)

12s1(θ) / a χ

a f(θ) × 10

5

14

4

40
Fit range nAPE = [20,determined in
T = 40]0[Kitano, NY, Yamazaki
1.2 T(2021)].
c
30∫dθ ds1/dθ

10
8

θ=π/2
π
3π/2

6
4

10
0
-10

2
0

10

20

30

40

50

60

-20
0

nAPE

FIG. 5: The linear extrapolation of a f (θ) to nAPE = 0.
4

20

π/2

π

3π/2

2

θ
FIG. 7: Comparison of θ dependence of the surface
s1 (θ) at T = 0 and 1.2 Tc .

θ dependence of f(θ) at T = 0
4

14

T=0
2
θ /2
1-cosθ∫dθ df/dθ
full volume (nAPE=40)1210f(θ) / χ• Succeed to calculate up to θ ∼ 3π/2
• Monotonically increasing function• Inconsistent with DIGA
• f(π − θ) ≠ f(π + θ) requires explanation.
• Re-weighting (=full volume) method works

8

only around θ = 0.

6

df
• Numerical consistency with ∫dθ dθ

4
2
0
0

π/2

π

3π/2

2π

θ
5
4

T=0
θ
sin θ

21

at
df(θ)/dθ
T
=
0
0
π/2
π
3π/2
0

2π

θ
5

T=0
θ
sin θ
∆f
full volume (nAPE=40)

df(θ)/dθ / χ

43• Order parameter is non-zero
df(θ)/dθθ=π= − i⟨ q(x) ⟩θ=π ≠ 0

spontaneous CPV at θ = π

2
1
0
-1
0

π/2

π

3π/2

2π

θ
FIG. 6: f (✓) (top) and df (✓)/d✓ (bottom).

22

Discussion• At T > Tc ,• At T = 0 ,f(π − θ) ≠ f(π + θ) and it is not likeconsistent with 1 − cos θ and no CPV as
expected (though non-trivial).

f(θ)

-π

0

2π

π
θ

Why ?

23

3π

Interpretationf(θ)• Sub-volume method seems to trace an original

f0(θ)

f1(θ)

branch even after the crossing point is passed.

1. The vacuum energy as a function as a function of 0 may have a single branch (firstor it may have several branches, which cross at particular values of 0 (second drawing).
FIG.

8 = T CP invariance is more subtle and must be defined as & -+ -&where the nb are arbitrary integers such that C ni = 1.
If Eq. (21) has only one solution for given 8, this solution must be CP con
whenever CP is a symmetry of the equation. However, if there are, for examp
solutions, it can happen that at 8 = v neither solution is CP invariant, but
CP transformation exchanges them. In this case, the CP symmetry is sponta
broken at 19= T. The two solutions must, because of the symmetry, be degenenergy at 19= T, and they are quite likely to cross in energy near 0 = rr.
Let us see how this works in the realistic case

Actually, 24an equivalent problem was analyzed by Dashen some years before Q
discovered [ 161,

tions: the energy of unbroken state [given by the total
mass 2mQ of the two heavy (or, test) quark sources Q
and Q̄ and the energy of the string (3)] is larger compared to the energy of the broken string state (given by
the mass 2mQq̄ of the created light-heavy mesons Qq̄ andQ̄q). The critical string-breaking distance Rbr is thus determined by the energy balance:

Interpretationf(θ)

2mQ + σRbr = 2mQq̄

• Sub-volume method seems to trace an original
branch even after the crossing point is passed.

•(5)(here we neglect a weak Coulomb interaction between
the quarks in the unbroken state and we also disregard
0
exponentially suppressed van der Waals interaction
between the heavy-light mesons in the broken state). If
R > Rbr , then the string breaks and light-heavy mesons
are formed, Fig. 1 (we consider here a simplest case ignoring a multiple pair production via string fragmentation).

f (θ)ωn,s∥ (p∥ ) = ±p2∥ + m2q + (2n + 1 − 2sz )|eq |B , (7)where mq and eq are, respectively, the mass and the electric charge of the fermion, p∥ ≡ pz is the momentumof the fermion along the direction of the magnetic field,
and s∥ = ±1/2 is the projection of the fermion’s spin
onto the axis of the magnetic field. The integer number n = 0, 1, 2, . . . labels the Landau levels. The signs“±” in front of the square root in Eq. (7) refer to, respectively, particle and antiparticle branches of the energy spectrum. The electric charges of the light u and dquarks are, respectively:f1(θ)

qu = +2e,3eqd = − .
3(8)In a strong magnetic field the lowest Landau level(LLL) with n = 0 and sz = 1/2 plays a dominant rolein the quark’s motion because the excited states are too
heavy. Indeed, for the soft (low-momentum) fermions thespin flips, sz → −sz and jumps to the higher states withn ! 1 are energetically suppressed by the typical gap√(9)δEq ∼ 2/lq ,1. The vacuum energy as a function as a function of 0 may have a single branch (firstor it may have several branches, which cross at particular values of 0 (second drawing).
Similar to the calculation of the static potential,
FIG.

where“string breaking” should happen but neverwhere
occurs.
8 = T CP invariance is more subtle and must be defined
as & -+ -&!

where the nb are arbitrary integers such that C ni = 1.
is the magnetic length of the quark q.
If Eq. (21) has only one solution for given 8,
this
must
bebecomes
CP es-conThus,
at thesolution
LLL the motionof quarks
sentially (1+1) dimensional. The quark moves along theof the magneticiffield
and the are,longitudinal
whenever CP is a symmetry of the equation. axis
However,
there
fordynamics
examp
of a free quark is governed by one-dimensional relativistic
solutions, it can happen that Chernodub
at 8 =(2010)
v neitherdispersion
solutionbut
relation: is CP invariant,!
FIG. 1. The conventional string breaking: the QCD string
CP transformationexchanges
Inantiquark
this case, the ωCP(p )symmetry= ± p + m . is sponta
spanned between
the static quark Qthem.
and the staticQ̄ breaks due to light q q̄ pair creation.
broken at 19= T. The two solutions must, becauseof dynamics
the symmetry,degen
The transverse(i.e., the motion of be
the quarkinthe plane orthogonal to the magnetic field) is restricted
stringbreaking
wasareobservedin latticesimulations
energy at 19=of The
T,
and
they
quite
likelytocross
inenergy
near
0
=
rr.
to
a
regionofatypical
size
of
the
order
of
the
magnetic
QCD with two flavors of equal-mass quarks [16]. An
length (10),of thelattice results
tothe real
QCD gives caseLet us see extrapolation
how
this
works
in
the
realistic
the following string breaking distance [16]:lq (B) = 1/

LLL

Rbr ≈ 1.13 fm ,(6)

where statistical and systematical errors are of the orderof 0.1 fm each.
QCD string breaking at nonzero magnetic field.
A suﬃciently strong background magnetic field should∥|eq B| ,q = u, d

2
∥

(10)

2
q

|δr|q " lq (B) .

(11)

In QCD the influence of the magnetic field should become significant when the magnetic length (10) becomescomparable with a typical QCD length scale ΛQCD ∼1 fm−1 ∼ 200 MeV,

Actually, 24an equivalent problem was analyzed by Dashen some years before(12)Qdiscovered [ 161,lq (B) ≃ Λ−1
QCD ≃ Rbr .

potential.

Interpretation

22

p∥Hadron+ mq masses+ (2n + 1 − 2sz )|eq |B ,A.

(7)tions: the energy of unbroken state [given by the total
mass 2mQ of the two B.
heavy(or, test) quark sources Q
in full QCD, we use six values of the
Simulation in quenched QCD
whereInmmeasurements
q and eq are, respectively, the mass and the elecvalence quark mass corresponding to the hopping parameter
and Q̄ and the energy of the string (3)] is larger comcharge
fermion, 0.1343,
p∥ ≡ pz 0.1346,
is the momentum
While many calculations of the hadron spectrum have tricK val,i(i"1,of. . .the,6)"0.1340,
0.1350,pared to theenergyofthebrokenstringstate(givenbybeen performed in quenched QCD, comparisons between our of 0.1355,
the fermion
alongwhich
the coverdirectionof thefield,
and 0.1358,
the rangeof mmagnetic
PS,val /m V,val
the mass 2mofthecreatedlight-heavy
mesonsQq̄
andQq̄
full QCD results and quenched results from other simula- and!0.5–0.8.
At eachmass, therefore,
is one spins∥ = ±1/2
is seathequark
projectionof the there
fermion’sQ̄q). The criticalstring-breaking
distance
Rbr is thus
tions may
be subject to systematic
uncertainties
duedeto theequals
K sea andfield.
is identified
asinteger
the lightnumvalueof axisK val,i which
onto
the
of
the
magneticThe
differencein the balance:
simulation details. We therefore carry out atermined by
the energy
quark mass. Other five values of K val,i correspond to then = 0, 1, 2, . . . labels the Landau levels. The signs
set of quenched calculations of the hadron spectrum using bermassof strange quarks treated in the quenched approxima“±”
inof the squarein Eq. (7)‘‘diagonalrefer to, rethe same2m
latticeactionsandsimulation
parameters
as
those
(5)
tion. Infront
the following,
we useroot
the abbreviation
Q + σRbr = 2mQq̄
for full QCD runs.
spectively,
particlehadronand correlators
antiparticle
branches
the endata’’ to represent
or masseswith aofquark
the lat- ergy
Our simulations
are performed
at ! "6.0, where(here we neglecta weak Coulombinteraction
betweenmass
combinationin electric
which allcharges
valence ofquark
spectrum.
The
themasses
light uareand d
0.1074#14$
fm. We takefixed fromm " equals
equal are,to therespectively:
sea quark mass.
the quarksticein spacing
the unbroken
state
and wealso disregard
quarksdetermined nonc SW"1.769 which is the value 0
We 1
employ meson operators defined byexponentially suppressed van der Waals interaction beperturbatively by the ALPHA Collaboration %35&. Three late2etween the ticeheavy-lightmesonsin thestate).
If in(=f ) +(g) ,
q
=
−
.
q
sizes 123 !48,
163 !48,and broken
203 !48 are
employeddu#6$ (8)M #x $"q̄ x 'q x 3 , '"I, ( 5 , ( ) , (5( ) ,
3
R > Rbr , thenstring breaks
andeffects.
light-heavy mesons
order theto investigate
finite-size
Gauge
configurations
are generated
with casea combination
are formed, Fig.
1 (weconsider herea simplest
ignor- of In a strong magnetic field the lowest Landau level
the heat-bathand over-relaxation
We call four (LLL)
wherewithf and ng areis the coordinateson
ing a multiple
pair production
via stringalgorithms.
fragmentation).
= 0flavor
andindicessz = and
1/2x plays† a dominant roleheat-bath sweeps with a succeeding over-relaxation step anthe lattice. Meson correlators *M (x)M (0) + are calculatedin
the quark’s motion because the excited states are too
iteration. We accumulate statistics of 60 000 iterations on
for the following 11 combinations of valence quark masses:
heavy. Indeed, for the soft (low-momentum) fermions thespin flips, sz → −sz and jumps to the higher states withn ! 1 are energetically suppressed by the typical gap
Static potential√(9)δEq ∼ 2/lq ,f(θ)• Sub-volume method seems to trace an original
branch even after the crossing point is passed.

•

ωn,s∥ (p∥ ) = ±

f (θ)

f (θ)

1. The vacuum energy as a function as a function of 0 may have a single branch (firstor it may have several branches, which cross at particular values of 0 (second drawing).
Similar to the calculation of the static potential,
FIG.

where“string breaking” should happen but neverwhere
occurs.
8 = T CP invariance is more subtle and must be defined
as & -+ -&!

where the nb are arbitrary integers such that C ni = 1.
is the magnetic length of the quark q.
If Eq. (21) has only one solution for given 8,
this
must
bebecomes
CP es-conThus,
at thesolution
LLL the motionof quarks
sentially (1+1) dimensional. The quark moves along theof the magneticiffield
and the are,longitudinal
whenever CP is a symmetry of the equation. axis
However,
there
fordynamics
examp
R by one-dimensional relativisticof a free quark is governed
solutions, it can happen that Chernodub
at 8 =(2010)
v neitherdispersion
solutionbut
relation: is CP invariant,!
FIG. 1. The conventional string breaking: the QCD string
CP transformationexchanges
Inantiquark
this case, the ωCP(p )symmetry= ± p + m . is sponta
spanned between
the static quark Qthem.
and the staticQ̄ breaks due to light q q̄ pair creation.
broken at 19= T. The two solutions must, becauseof dynamics
the symmetry,degen
The transverse(i.e., the motion of be
the quarkinthe plane orthogonal to the magnetic field) is restricted
stringbreaking
wasareobservedin latticesimulations
energy at 19=of The
T,
and
they
quite
likelytocross
inenergy
near
0
=
rr.
to
a
regionofatypical
size
of
the
order
of
the
magnetic
QCD with two flavors of equal-mass quarks [16]. An
length (10),of thelattice results
tothe real
QCD gives caseLet us see extrapolation
how
this
works
in
the
realistic
the following string breaking distance [16]:lq (B) = 1/

|eq B| ,q = u, d(10)FIG. 7. Static quark potential on 203 !48. Left and right figures show data at K sea"0.1340 and 0.1355, respectively.
054502-6

LLL

Rbr ≈ 1.13 fm ,(6)

where statistical and systematical errors are of the orderof 0.1 fm each.
QCD string breaking at nonzero magnetic field.
A suﬃciently strong background magnetic field should

∥

2
∥

2
q

|δr|q " lq (B) .

(11)

In QCD the influence of the magnetic field should become significant when the magnetic length (10) becomescomparable with a typical QCD length scale ΛQCD ∼1 fm−1 ∼ 200 MeV,

Actually, 24an equivalent problem was analyzed by Dashen some years before(12)Qdiscovered [ 161,lq (B) ≃ Λ−1
QCD ≃ Rbr .

potential.

Interpretation2

branch even after the crossing point is passed.

(7)
tions: the energy of unbroken state [given by the total
mass 2mQ of the two B.
heavy(or, test) quark sources Q
in full QCD, we use six values of the
Simulation in quenched QCD
whereInmmeasurements
q and eq are, respectively, the mass and the elecvalence quark mass corresponding to the hopping parameter
and Q̄ and the energy of the string (3)] is larger comcharge
fermion, 0.1343,
p∥ ≡ pz 0.1346,
is the momentum
While many calculations of the hadron spectrum have tricK val,i(i"1,of. . .the,6)"0.1340,
0.1350,pared to theenergyofthebrokenstringstate(givenbybeen performed in quenched QCD, comparisons between our of 0.1355,
the fermion
alongwhich
the coverdirectionof thefield,
and 0.1358,
the rangeof mmagnetic
PS,val /m V,val
the mass 2mofthecreatedlight-heavy
mesonsQq̄
andQq̄
full QCD results and quenched results from other simula- and!0.5–0.8.
At eachmass, therefore,
is one spins∥ = ±1/2
is seathequark
projectionof the there
fermion’sQ̄q). The criticalstring-breaking
distance
Rbr is thus
tions may
be subject to systematic
uncertainties
duedeto theequals
K sea andfield.
is identified
asinteger
the lightnumvalueof axisK val,i which
onto
the
of
the
magneticThe
differencein the balance:
simulation details. We therefore carry out atermined by
the energy
quark mass. Other five values of K val,i correspond to then = 0, 1, 2, . . . labels the Landau levels. The signs
set of quenched calculations of the hadron spectrum using bermassof strange quarks treated in the quenched approxima“±”
inof the squarein Eq. (7)‘‘diagonalrefer to, rethe same2m
latticeactionsandsimulation
parameters
as
those
(5)
tion. Infront
the following,
we useroot
the abbreviation
Q + σRbr = 2mQq̄
for full QCD runs.
spectively,
particlehadronand correlators
antiparticle
branches
the endata’’ to represent
or masseswith aofquark
the lat- ergy
Our simulations
are performed
at ! "6.0, where(here we neglecta weak Coulombinteraction
betweenmass
combinationin electric
which allcharges
valence ofquark
spectrum.
The
themasses
light uareand d
0.1074#14$
fm. We takefixed fromm " equals
equal are,to therespectively:
sea quark mass.
the quarksticein spacing
the unbroken
state
and wealso disregard
quarksdetermined nonc SW"1.769 which is the value 0
We 1
employ meson operators defined byexponentially suppressed van der Waals interaction beperturbatively by the ALPHA Collaboration %35&. Three late2etween the ticeheavy-lightmesonsin thestate).
If in(=f ) +(g) ,
q
=
−
.
q
sizes 123 !48,
163 !48,and broken
203 !48 are
employeddu#6$ (8)M #x $"q̄ x 'q x 3 , '"I, ( 5 , ( ) , (5( ) ,
3
R > Rbr , thenstring breaks
andeffects.
light-heavy mesons
order theto investigate
finite-size
Gauge
configurations
are generated
with casea combination
are formed, Fig.
1 (weconsider herea simplest
ignor- of In a strong magnetic field the lowest Landau level
the heat-bathand over-relaxation
We call four (LLL)
wherewithf and ng areis the coordinateson
ing a multiple
pair production
via stringalgorithms.
fragmentation).
= 0flavor
andindicessz = and
1/2x plays† a dominant roleheat-bath sweeps with a succeeding over-relaxation step anthe lattice. Meson correlators *M (x)M (0) + are calculatedin
the quark’s motion because the excited states are too
iteration. We accumulate statistics of 60 000 iterations on
for the following 11 combinations of valence quark masses:
heavy. Indeed, for the soft (low-momentum) fermions thespin flips, sz → −sz and jumps to the higher states withn ! 1 are energetically suppressed by the typical gap
Static potential√(9)δEq ∼ 2/lq ,

f (θ)

f (θ)

1. The vacuum energy as a function as a function of 0 may have a single branch (firstor it may have several branches, which cross at particular values of 0 (second drawing).
Similar to the calculation of the static potential,
FIG.

where“string breaking” should happen but neverwhere
occurs.
8 = T CP invariance is more subtle and must be defined
as & -+ -&!

where the nb are arbitrary integers such that C ni = 1.
is the magnetic length of the quark q.
If Eq. (21) has only one solution for given 8,
this
must
bebecomes
CP es-conThus,
at thesolution
LLL the motionof quarks
sentially (1+1) dimensional. The quark moves along theof the magneticiffield
and the are,longitudinal
whenever CP is a symmetry of the equation. axis
However,
there
fordynamics
examp
In the present case, the domain to domain-wall
R by one-dimensional relativisticof a free quark is governed
solutions, it can happen that Chernodub
at 8 =(2010)
v neitherdispersion
solutionbut
relation: is CP invariant,transition should occur but does not in this!
FIG. 1. The conventional string breaking: the QCD string
CP transformationexchanges
Inantiquark
this case, the ωCP(p )symmetry= ± p + m . is sponta
spanned between
the static quark Qthem.
and the static
method.
Q̄ breaks due to light q q̄ pair creation.
broken at 19= T. The two solutions must, becauseof dynamics
the symmetry,degen
The transverse(i.e., the motion of be
the quarkinthe plane orthogonal to the magnetic field) is restricted
stringbreaking
wasareobservedin latticesimulations
energy at 19=of The
T,
and
they
quite
likelytocross
inenergy
near
0
=
rr.
to
a
regionofatypical
size
of
the
order
of
the
magnetic
QCD with two flavors of equal-mass quarks [16]. An
length (10),of thelattice results
tothe real
QCD gives caseLet us see extrapolation
how
this
works
in
the
realistic
the following string breaking distance [16]:lq (B) = 1/•

2

p∥Hadron+ mq masses+ (2n + 1 − 2sz )|eq |B ,
A.

f(θ)

• Sub-volume method seems to trace an original•

ωn,s∥ (p∥ ) = ±|eq B| ,q = u, d(10)FIG. 7. Static quark potential on 203 !48. Left and right figures show data at K sea"0.1340 and 0.1355, respectively.
054502-6

LLL

Rbr ≈ 1.13 fm ,(6)

where statistical and systematical errors are of the orderof 0.1 fm each.
QCD string breaking at nonzero magnetic field.
A suﬃciently strong background magnetic field should

∥

2
∥

2
q

|δr|q " lq (B) .

(11)

In QCD the influence of the magnetic field should become significant when the magnetic length (10) becomescomparable with a typical QCD length scale ΛQCD ∼1 fm−1 ∼ 200 MeV,

Actually, 24an equivalent problem was analyzed by Dashen some years before(12)Qdiscovered [ 161,lq (B) ≃ Λ−1
QCD ≃ Rbr .

potential.

Interpretation2

branch even after the crossing point is passed.

(7)
tions: the energy of unbroken state [given by the total
mass 2mQ of the two B.
heavy(or, test) quark sources Q
in full QCD, we use six values of the
Simulation in quenched QCD
whereInmmeasurements
q and eq are, respectively, the mass and the elecvalence quark mass corresponding to the hopping parameter
and Q̄ and the energy of the string (3)] is larger comcharge
fermion, 0.1343,
p∥ ≡ pz 0.1346,
is the momentum
While many calculations of the hadron spectrum have tricK val,i(i"1,of. . .the,6)"0.1340,
0.1350,pared to theenergyofthebrokenstringstate(givenbybeen performed in quenched QCD, comparisons between our of 0.1355,
the fermion
alongwhich
the coverdirectionof thefield,
and 0.1358,
the rangeof mmagnetic
PS,val /m V,val
the mass 2mofthecreatedlight-heavy
mesonsQq̄
andQq̄
full QCD results and quenched results from other simula- and!0.5–0.8.
At eachmass, therefore,
is one spins∥ = ±1/2
is seathequark
projectionof the there
fermion’sQ̄q). The criticalstring-breaking
distance
Rbr is thus
tions may
be subject to systematic
uncertainties
duedeto theequals
K sea andfield.
is identified
asinteger
the lightnumvalueof axisK val,i which
onto
the
of
the
magneticThe
differencein the balance:
simulation details. We therefore carry out atermined by
the energy
quark mass. Other five values of K val,i correspond to then = 0, 1, 2, . . . labels the Landau levels. The signs
set of quenched calculations of the hadron spectrum using bermassof strange quarks treated in the quenched approxima“±”
inof the squarein Eq. (7)‘‘diagonalrefer to, rethe same2m
latticeactionsandsimulation
parameters
as
those
(5)
tion. Infront
the following,
we useroot
the abbreviation
Q + σRbr = 2mQq̄
for full QCD runs.
spectively,
particlehadronand correlators
antiparticle
branches
the endata’’ to represent
or masseswith aofquark
the lat- ergy
Our simulations
are performed
at ! "6.0, where(here we neglecta weak Coulombinteraction
betweenmass
combinationin electric
which allcharges
valence ofquark
spectrum.
The
themasses
light uareand d
0.1074#14$
fm. We takefixed fromm " equals
equal are,to therespectively:
sea quark mass.
the quarksticein spacing
the unbroken
state
and wealso disregard
quarksdetermined nonc SW"1.769 which is the value 0
We 1
employ meson operators defined byexponentially suppressed van der Waals interaction beperturbatively by the ALPHA Collaboration %35&. Three late2etween the ticeheavy-lightmesonsin thestate).
If in(=f ) +(g) ,
q
=
−
.
q
sizes 123 !48,
163 !48,and broken
203 !48 are
employeddu#6$ (8)M #x $"q̄ x 'q x 3 , '"I, ( 5 , ( ) , (5( ) ,
3
R > Rbr , thenstring breaks
andeffects.
light-heavy mesons
order theto investigate
finite-size
Gauge
configurations
are generated
with casea combination
are formed, Fig.
1 (weconsider herea simplest
ignor- of In a strong magnetic field the lowest Landau level
the heat-bathand over-relaxation
We call four (LLL)
wherewithf and ng areis the coordinateson
ing a multiple
pair production
via stringalgorithms.
fragmentation).
= 0flavor
andindicessz = and
1/2x plays† a dominant roleheat-bath sweeps with a succeeding over-relaxation step anthe lattice. Meson correlators *M (x)M (0) + are calculatedin
the quark’s motion because the excited states are too
iteration. We accumulate statistics of 60 000 iterations on
for the following 11 combinations of valence quark masses:
heavy. Indeed, for the soft (low-momentum) fermions thespin flips, sz → −sz and jumps to the higher states withn ! 1 are energetically suppressed by the typical gap
Static potential√(9)δEq ∼ 2/lq ,

f (θ)

f (θ)

1. The vacuum energy as a function as a function of 0 may have a single branch (firstor it may have several branches, which cross at particular values of 0 (second drawing).
Similar to the calculation of the static potential,
FIG.

where“string breaking” should happen but neverwhere
occurs.
8 = T CP invariance is more subtle and must be defined
as & -+ -&!

where the nb are arbitrary integers such that C ni = 1.
is the magnetic length of the quark q.
If Eq. (21) has only one solution for given 8,
this
must
bebecomes
CP es-conThus,
at thesolution
LLL the motionof quarks
sentially (1+1) dimensional. The quark moves along theof the magneticiffield
and the are,longitudinal
whenever CP is a symmetry of the equation. axis
However,
there
fordynamics
examp
In the present case, the domain to domain-wall
R by one-dimensional relativisticof a free quark is governed
solutions, it can happen that Chernodub
at 8 =(2010)
v neitherdispersion
solutionbut
relation: is CP invariant,transition should occur but does not in this!
FIG. 1. The conventional string breaking: the QCD string
CP transformationexchanges
Inantiquark
this case, the ωCP(p )symmetry= ± p + m . is sponta
spanned between
the static quark Qthem.
and the static
method.
Q̄ breaks due to light q q̄ pair creation.
broken at 19= T. The two solutions must, becauseof dynamics
the symmetry,degen
The transverse(i.e., the motion of be
the quarkinthe plane orthogonal to the magnetic field) is restricted
stringbreaking
wasareobservedin latticesimulations
energy at 19=of The
T,
and
they
quite
likelytocross
inenergy
near
0
=
rr.
to
a
regionofatypical
size
of
the
order
of
the
magnetic
QCD with two flavors of equal-mass quarks [16]. An
length (10),of thelattice results
tothe real
QCD gives caseLet us see extrapolation
how
this
works
in
the
realistic
the following string breaking distance [16]:4d SU(N) YM has an topological object
lq (B) = 1/•

2

p∥Hadron+ mq masses+ (2n + 1 − 2sz )|eq |B ,
A.

f(θ)

• Sub-volume method seems to trace an original•

ωn,s∥ (p∥ ) = ±|eq B| ,q = u, d(10)FIG. 7. Static quark potential on 203 !48. Left and right figures show data at K sea"0.1340 and 0.1355, respectively.
054502-6LLL

f0(π + ϵ)f0(π + ϵ)Rbr ≈ 1.13 fm ,→ f1(π + ϵ)∥2∥

|δr|q " lq (B) .

2
q

(11)

called a bag or a domain-wall(6)
In QCD the influence of the magnetic field should be[Luscher(1978)].
comesignificant
when the magnetic length (10) becomes
where statistical and systematical errors are of the ordercomparable with a typical QCD length scale ΛQCD ∼of 0.1 fm each.
1 fm−1 ∼ 200 MeV,QCD string breaking at nonzero magnetic field.
lq (B) ≃ Λ−1(12)
A suﬃciently strong background magnetic field should
QCD ≃ Rbr .

Actually, 24an equivalent problem was analyzed by Dashen some years before Q
discovered [ 161,

potential.

Interpretation2

branch even after the crossing point is passed.

(7)
tions: the energy of unbroken state [given by the total
mass 2mQ of the two B.
heavy(or, test) quark sources Q
in full QCD, we use six values of the
Simulation in quenched QCD
whereInmmeasurements
q and eq are, respectively, the mass and the elecvalence quark mass corresponding to the hopping parameter
and Q̄ and the energy of the string (3)] is larger comcharge
fermion, 0.1343,
p∥ ≡ pz 0.1346,
is the momentum
While many calculations of the hadron spectrum have tricK val,i(i"1,of. . .the,6)"0.1340,
0.1350,pared to theenergyofthebrokenstringstate(givenbybeen performed in quenched QCD, comparisons between our of 0.1355,
the fermion
alongwhich
the coverdirectionof thefield,
and 0.1358,
the rangeof mmagnetic
PS,val /m V,val
the mass 2mofthecreatedlight-heavy
mesonsQq̄
andQq̄
full QCD results and quenched results from other simula- and!0.5–0.8.
At eachmass, therefore,
is one spins∥ = ±1/2
is seathequark
projectionof the there
fermion’sQ̄q). The criticalstring-breaking
distance
Rbr is thus
tions may
be subject to systematic
uncertainties
duedeto theequals
K sea andfield.
is identified
asinteger
the lightnumvalueof axisK val,i which
onto
the
of
the
magneticThe
differencein the balance:
simulation details. We therefore carry out atermined by
the energy
quark mass. Other five values of K val,i correspond to then = 0, 1, 2, . . . labels the Landau levels. The signs
set of quenched calculations of the hadron spectrum using bermassof strange quarks treated in the quenched approxima“±”
inof the squarein Eq. (7)‘‘diagonalrefer to, rethe same2m
latticeactionsandsimulation
parameters
as
those
(5)
tion. Infront
the following,
we useroot
the abbreviation
Q + σRbr = 2mQq̄
for full QCD runs.
spectively,
particlehadronand correlators
antiparticle
branches
the endata’’ to represent
or masseswith aofquark
the lat- ergy
Our simulations
are performed
at ! "6.0, where(here we neglecta weak Coulombinteraction
betweenmass
combinationin electric
which allcharges
valence ofquark
spectrum.
The
themasses
light uareand d
0.1074#14$
fm. We takefixed fromm " equals
equal are,to therespectively:
sea quark mass.
the quarksticein spacing
the unbroken
state
and wealso disregard
quarksdetermined nonc SW"1.769 which is the value 0
We 1
employ meson operators defined byexponentially suppressed van der Waals interaction beperturbatively by the ALPHA Collaboration %35&. Three late2etween the ticeheavy-lightmesonsin thestate).
If in(=f ) +(g) ,
q
=
−
.
q
sizes 123 !48,
163 !48,and broken
203 !48 are
employeddu#6$ (8)M #x $"q̄ x 'q x 3 , '"I, ( 5 , ( ) , (5( ) ,
3
R > Rbr , thenstring breaks
andeffects.
light-heavy mesons
order theto investigate
finite-size
Gauge
configurations
are generated
with casea combination
are formed, Fig.
1 (weconsider herea simplest
ignor- of In a strong magnetic field the lowest Landau level
the heat-bathand over-relaxation
We call four (LLL)
wherewithf and ng areis the coordinateson
ing a multiple
pair production
via stringalgorithms.
fragmentation).
= 0flavor
andindicessz = and
1/2x plays† a dominant roleheat-bath sweeps with a succeeding over-relaxation step anthe lattice. Meson correlators *M (x)M (0) + are calculatedin
the quark’s motion because the excited states are too
iteration. We accumulate statistics of 60 000 iterations on
for the following 11 combinations of valence quark masses:
heavy. Indeed, for the soft (low-momentum) fermions thespin flips, sz → −sz and jumps to the higher states withn ! 1 are energetically suppressed by the typical gap
Static potential√(9)δEq ∼ 2/lq ,

f (θ)

f (θ)

1. The vacuum energy as a function as a function of 0 may have a single branch (firstor it may have several branches, which cross at particular values of 0 (second drawing).
Similar to the calculation of the static potential,
FIG.

where“string breaking” should happen but neverwhere
occurs.
8 = T CP invariance is more subtle and must be defined
as & -+ -&!

where the nb are arbitrary integers such that C ni = 1.
is the magnetic length of the quark q.
If Eq. (21) has only one solution for given 8,
this
must
bebecomes
CP es-conThus,
at thesolution
LLL the motionof quarks
sentially (1+1) dimensional. The quark moves along theof the magneticiffield
and the are,longitudinal
whenever CP is a symmetry of the equation. axis
However,
there
fordynamics
examp
In the present case, the domain to domain-wall
R by one-dimensional relativisticof a free quark is governed
solutions, it can happen that Chernodub
at 8 =(2010)
v neitherdispersion
solutionbut
relation: is CP invariant,transition should occur but does not in this!
FIG. 1. The conventional string breaking: the QCD string
CP transformationexchanges
Inantiquark
this case, the ωCP(p )symmetry= ± p + m . is sponta
spanned between
the static quark Qthem.
and the static
method.
Q̄ breaks due to light q q̄ pair creation.
broken at 19= T. The two solutions must, becauseof dynamics
the symmetry,degen
The transverse(i.e., the motion of be
the quarkinthe plane orthogonal to the magnetic field) is restricted
stringbreaking
wasareobservedin latticesimulations
energy at 19=of The
T,
and
they
quite
likelytocross
inenergy
near
0
=
rr.
to
a
regionofatypical
size
of
the
order
of
the
magnetic
QCD with two flavors of equal-mass quarks [16]. An
length (10),of thelattice results
tothe real
QCD gives caseLet us see extrapolation
how
this
works
in
the
realistic
the following string breaking distance [16]:4d SU(N) YM has an topological object
lq (B) = 1/•

2

p∥Hadron+ mq masses+ (2n + 1 − 2sz )|eq |B ,
A.

f(θ)

• Sub-volume method seems to trace an original•

ωn,s∥ (p∥ ) = ±|eq B| ,q = u, d(10)FIG. 7. Static quark potential on 203 !48. Left and right figures show data at K sea"0.1340 and 0.1355, respectively.
054502-6LLL

f0(π + ϵ)f0(π + ϵ)Rbr ≈ 1.13 fm ,→ f1(π + ϵ)∥2∥

|δr|q " lq (B) .

2
q

(11)

called a bag or a domain-wall(6)
In QCD the influence of the magnetic field should be[Luscher(1978)].
comesignificant
when the magnetic length (10) becomes
where statistical and systematical errors are of the ordercomparable with a typical QCD length scale ΛQCD ∼of 0.1 fm each.
1 fm−1 ∼ 200 MeV,QCD string breaking at nonzero magnetic field.
lq (B) ≃ Λ−1(12)
A suﬃciently strong background magnetic field should
QCD ≃ Rbr .

Actually, 24an equivalent problem was analyzed by Dashen some years before Q
discovered [ 161,

Summary and conclusion• We have developed a sub-volume method, which enables us to calculate f(θ) up toθ ∼ 3π/2 in SU(2) Yang-Mills theory.

• Combining with the theory requirement f(π − θ) = f(π + θ) , our result provides

with the evidence for spontaneous CPV at θ = π and at T = 0 and the existence ofa bag-like object.
1
N=2 belongs to large N class (not like CP model).

• The same method roughly reproduces the DIGA result, f(θ) ∼ χ(1 − cos θ) , above
Tc , which makes the above result more confident.

25

Future studies• exploring the location of Tc(θ)• Suppose that Tc(θ) depends on θ. Then, does it meanthat the β-function depend on θ ?

• Also interesting to apply the sub-volume method to theTcdeconfineconfineCPfinite density system.

T

SU(N) with N = 2, ⋯ ∞

π
θ

26

2π

Intuitive understanding of periodic behavior of f(θ)
11−iθQsubf(θ) = − limln⟨e⟩ = − limln⟨ cos(θQsub) ⟩Vsub→∞ Vsub
Vsub→∞ Vsub
: instanton: anti-instanton

Vsub

Vsub

Qsub = + 1

Vsub

Qsub = 0

Qsub = + 1

4
In this case, Qsub is almost always integer if ρinstanton ≪ Vsub .

f(θ)

θ≈2π

∼0

2π-periodicity can be expected.
27

θ-vacuum• The vacuum can have an integer winding number, labeled by | n⟩.
• But, this label is changed by gauge transformation, e.g. U(1) | n⟩ → | n + 1⟩.
+∞

inθ

−iθ

n−∞
inθ −imθiθQ• Define | θ⟩ = ∑ e | n⟩ ⟺ U(1) | θ⟩ = e | θ⟩|⟨θ
• + θ−⟩J =

∑

e e

m,n

⟨m+ | n−⟩J =

∑
Q

e

∑
m

⟨m+ | m− + Q⟩Jg−Sg+iθQ+ ∫ J⋅A
4
=
𝒟A e
δ Q−
d
xG
G̃
2
∑∫
(
)
∫
32π
Q ∈Q
= 𝒟A e∫−Sg+iθQ+ ∫ J⋅A

28

2