Missing final state puzzle inmonopole-fermion scattering
Ryutaro Matsudo
KEK Theory Center
Oct. 20, 2021
In collaboration with Ryuichiro Kitano
arXiv:2103.13639
Seminar at Kyoto University

Ryutaro Matsudo (KEK)

arXiv:2103.13639

Oct. 20, 2021

1 / 29Rubakov-Callan effect[V. A. Rubakov 1982, C. G. Callan 1982]
When a proton collides with a GUT monopole, it decays into a positron andmesons.
The effect has been used to set limits on the monopole flux in the Universe.
Ryutaro Matsudo (KEK)

arXiv:2103.13639

Oct. 20, 2021

2 / 29Helicity flip
When there are a magnetic monopole and a unit charge, theelectromagnetic field has the angular momentum with magnitude 1/2.
Z1~ × B)
~ = 1 r̂0~
d3 x ~x × (EJEM =4π
2
Then the total angular momentum, the sum of J~EM and the spin of thefermion, can be zero.
When a charged fermion collides with a monopole, its helicity has to flip.

Ryutaro Matsudo (KEK)

arXiv:2103.13639

Oct. 20, 2021

3 / 29Puzzle: Two-flavor massless case

The final state has to have the opposite helicity to the initial state and thesame flavor charge.
When there are two or more flavors of massless fermion, the left and righthanded particles have different flavor charge.

There are no candidates of the final state, which areconsistent with the helicity flip and the flavor charge
conservation.
Ryutaro Matsudo (KEK)

arXiv:2103.13639

Oct. 20, 2021

4 / 29Outline

1

Missing final state puzzle

2

Review: S-wave approximation

3

The solution to the puzzle

Ryutaro Matsudo (KEK)

arXiv:2103.13639

Oct. 20, 2021

5 / 29Outline

1

Missing final state puzzle

2Review: S-wave approximation

3The solution to the puzzle

Ryutaro Matsudo (KEK)

arXiv:2103.13639

Oct. 20, 2021

6 / 29Set up
We consider an SU (2) gauge theory with an adjoint Higgs and 4 flavors of
Weyl fermions, where SU (2) is spontaneously broken down to U (1).
The global symmetry is SU (4).
SU (4) quadruplets (fund. rep.)z(   + }| +   + {a+a2a3a4
1
SU (2) doublets, −, −, −b−bbb4
1
2
3
The theory can be regarded as an approximation of SU (5) GUT:   +e
a1= L3 ,b1dL

Ryutaro Matsudo (KEK)   3d¯La2= −,b2eL   1a3uL=,b3ū2LarXiv:2103.13639

   2a4ūL=.
b4ū1L

Oct. 20, 2021

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The low energy effective theory

We approximate the theory as the gauge theory of the unbroken U (1).
a : The U (1) gauge field,
f = da,aj : The left-handed Weyl fermions with charge +1,
bj : The left-handed Weyl fermions with charge −1.
The ’t Hooft-Polyakov monopole is approximated by the (background) Diracmonopole.
X, Y bosons, GUT Higgs bosons are considered to be infinitely heavy.

Ryutaro Matsudo (KEK)

arXiv:2103.13639

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The s-wave scattering
In a monopole background, there are s-wave states of fermions (the total
angular momentum is zero) due to the contribution from the EM field.
According to the analysis of the Dirac equation, the higher partial waves
cannot reach the monopole core. [R. Jackiw & C. Rebbi, 1976]
When the energy of the incoming particle is sufficiently small, we can neglect
the effect of the higher partial waves.
In our setup, for s-wave fermions, only aj and āj can be incoming particles,
and only bj and b̄j can be outgoing particles.

Ryutaro Matsudo (KEK)

arXiv:2103.13639

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The missing final state puzzle
The helicity, the U (1) charge and the representation of SU (4) areaj : (L, +1, ),bj : (L, −1, ),āj : (R, −1, ),b̄j : (R, +1, )

If the initial state is a1 , the quantum number of the final state has to be(R, +1, ). However, there are no particles with this quantum number.
When we consider the multi-particle state, it has to consist of bj and b̄jbecause the s-wave states of aj and āj cannot be out-going particles and thehigher partial waves cannot reach the core of the monopole.
The only possible answer is “the semiton state”.

bj /2: “Semiton”, the state with bj number 1/2.
Ryutaro Matsudo (KEK)

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What is the ”semiton state”

b1 b̄2 b̄3 b̄4| , , , , Mi ?
2 2 2 2
RThe3 state0 has half fermion numbers. (The state is the eigenstate ofd xb̄1 σ b1 with the eigenvalue 1/2. )The state is orthogonal to any multi-particle state of aj , bj , āj , b̄j .
By projecting the fields to the s-wave component (the s-waveapproximation), it was found that the state is actually the final state.
[C. G. Callan 1984, J. Polchinski 1984, J. M. Maldacena & A. W. W. Ludwig 1997]

The problem is what an appropriate interpretation of the state is. A non-particlestate? A multi-particle state?
Our claim is that it is a new single-particle state that can exist only in themonopole background and only when the fermions are massless.
Ryutaro Matsudo (KEK)

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Interpretation of the final state
Probabilistic interpretation [C. G. Callan (1984), V. A. Rubakov (1988)]:
In the SU (5) GUT,3

a1 = e+L,

a2 = dL ,a3 = u1L ,a4 = u2L ,b1 = d3L ,b2 = e−L,

b3 = u2L ,b4 = u1L .

“The semiton state” is1 + 1 1
1
1e + u + u2 + d3
2 R 2 R 2 R 2 L
The state is interpreted as1
11 2 3√ |e+R , Mi + √ |uR uR dL , Mi ?
22This is problematic because e+R = b̄2 is in the antifundamental representationof SU (4), i.e., the flavor charge is not conserved.

Ryutaro Matsudo (KEK)

arXiv:2103.13639

Oct. 20, 2021

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Outline

1

Missing final state puzzle

2

Review: S-wave approximation

3The solution to the puzzle

Ryutaro Matsudo (KEK)

arXiv:2103.13639

Oct. 20, 202113 / 29

S-wave approximation and bosonization
The final state was obtained as the soliton in the bosonized theory in thes-wave approximation, where we only consider the spherically symmetricfields.

Z4X

1
S4d = d4 x − 2 f ?f +iaj σ µ Dµ aj + ibj σ µ Dµ bj  ,4gj=1

S-wave approximation, bosonization, soving Gauss’s law−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−→#Z ∞ Z ∞ "441 X
g4 1 X 222drdt((∂t φi ) − (∂r φi ) ) −(φi ) ,S2d =8π i=132π 3 r2 i=10−∞
The correspondence between the fermion and boson fields is roughly
a1 b2 ↔ eiφ1 ,Ryutaro Matsudo (KEK)

a2 b1 ↔ eiφ2 ,a3 b4 ↔ eiφ3 ,arXiv:2103.13639

a4 b3 ↔ eiφ4 .

Oct. 20, 2021

14 / 29

Fermions are kinks
The fermions correspond to the kink solitons.
Kinks = Fermions with the charge +1
Anti-kinks = Fermions with the charge −1
Incoming (anti-)kinks = aj (āj )Outgoing (anti-)kinks = b̄j (bj )The kink corresponds to the s-wave state of the fermion in 4d.

Ryutaro Matsudo (KEK)

arXiv:2103.13639

Oct. 20, 2021

15 / 29Currents and boundary conditions
The currents of U (1) and the maximal torus of SU (4) is (α, β = t, r)
1 X
4πr2 J α =εαβ ∂β φj ,2π j
1εαβ ∂β (φ1 − φ2 ),2π1α=4πr2 J(0,0,1,−1)εαβ ∂β (φ3 − φ4 ),2π1
α4πr2 J(1,1,−1,−1)=∂α (φ1 + φ2 − φ3 − φ4 ).
2παJ(...)
is the current of U (1) generated by diag(...).
The boundary condition should be imposed so that any charge does not flow
into the infinitesimal region around the monopole for the conservation:α=4πr2 J(1,−1,0,0)

4πr2 J r = 0,r4πr2 J(...)= 0 at r = 0,

which implies∂t (φ1 + φ2 + φ3 + φ4 ) = ∂t (φ1 − φ2 ) = ∂t (φ3 − φ4 ) = 0,∂r (φ1 + φ2 − φ3 − φ4 ) = 0.
Ryutaro Matsudo (KEK)

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Scattering process
Let the initial state be a1 . In g → 0 limit, the solution is1
φ1 = f (t + r) − f (t − r),
2f (∞) = 2π, f (−∞) = 0.

φ2 =1f (t − r),2

1
φ3,4 = − f (t − r)2

The final state corresponds toThe final state seems to be b1 /2 + b̄2 /2 + b̄3 /2 + b̄4 /2, which is hardly
interpreted.

Ryutaro Matsudo (KEK)

arXiv:2103.13639

Oct. 20, 2021

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Massive case: No puzzle
In the massive case, this puzzle disappears.
Let us introduce the mass term,
m1 (a1 b2 + a2 b1 + h.c.) + m2 (a3 b4 + a4 b3 + h.c.),
The global symmetry reduces to U (1) × U (1), which is generated by
H1 = diag(1, −1, 0, 0),

H2 = diag(0, 0, 1, −1),and the charge correspond to
H3 = diag(1, 1, −1, −1)
is not conserved.
There is a candidate of the final state
a1 : (L, +1, Q(1,−1,0,0) = 1, Q(0,0,1,−1) = 0)
→ b̄2 : (R, +1, Q(1,−1,0,0) = 1, Q(0,0,1,−1) = 0).
Ryutaro Matsudo (KEK)

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Numerical result in massive case
The numerical simulation was done by [S. Dawson & A. N. Schellekens (1983)].

Due to the mass termµ(m1 cos φ1 + m1 cos φ2 + m2 cos φ3 + m2 cos φ4 )φ1 = φ2 = −φ3 = −φ4 = π is no longer the vacuum configuration. Thus thepotential energy increases as the semiton state goes out.
At some point, the following pair of semiton states are produced:ās2 :=11111111
a1 + ā2 + a3 + a4 and bs1 := b̄1 + b2 + b3 + b4
2
2
2222
22

The state 12 a1 + 21 ā2 + 12 a3 + 12 a4 collides with the monopole,and then b̄2 appears.
Ryutaro Matsudo (KEK)

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What happens when we take the massless limit?
In the scattering process, when the semitons reach r ∼ 1/m, the values of φinear the core start to change.
This means that, in the region where r  1/m, the theory can be regarded
as the massless theory.
In the massless limit, every point is near the monopole.

Ryutaro Matsudo (KEK)

arXiv:2103.13639

Oct. 20, 2021

20 / 29

Outline

1Missing final state puzzle

2

Review: S-wave approximation

3The solution to the puzzle

Ryutaro Matsudo (KEK)

arXiv:2103.13639

Oct. 20, 2021

21 / 29

The fermion condensate

Fact: The following fermionic operators have nonzero expectation values:h(ai1 bi2 )(ai3 bi4 )i =1r6

c3 εi1 i2 i3 i4 .

To reproduce the helicity flip, operators that mixes aj and bj have to have
non-zero expectation value.
The operator should be SU (4) invariant to maintain the SU (4) symmetry
without NG bosons with the SU (4) charges.

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The “effective” theory of the phases of thecondensates
Let us consider the effective theory of the operators’ phases.
It is convenient to express them using the phases of the fermion fields.
Let αj be the phase of aj and βj be that of bj .
There are four independent variables, e.g.,
XθA =(αj + βj ), θ1j = α1 − β1 − αj + βjfor j = 2, 3, 4.

jTo reproduce the chiral anomaly, the effective Lagrangian contains
θA f ∧ f /(8π 2 ) + θ12 f ∧ F12 /(4π 2 ) + · · · ,F12 : U (1) ∈ SU (4) field strength corresponding to diag(1, −1, 0, 0)11µ⇒ j µ = εµνρσ 2 ∂ν θA fρσ , j12= εµνρσ 2 ∂ν θ12 fρσ ...
4π4π(cf. Axions)Ryutaro Matsudo (KEK)

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Strings of the phase is fermions
Let us consider a string configuration of the phase, around which the phasewinds. (c.f. axion strings)
Due to the topological coupling to the gauge field, there have to be edge
modes on the string. [C. G. Callan & J. A. Harvey 1985]
There is an edge excitation that has the unit charge and the same flavour
charges as the fermion.
The edge mode gives the object spin 1/2.⇒ The object can be regarded
as a fermion. (c.f. Nf = 1 QCD [Z. Komargodski 2018] η0 strings = baryons)
We will see that the object is suitable to the final state of the scattering

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Monopole bags and strings
A monopole bag has the electric and flavor charges:
Z ∞ZZ
1~ · dS~ = 1.
dr∂
θQ = d3 x j 0 =Br A
4π 2 0
Sr2

Because α1 ∼ α1 + 2π, the wall of α1 can have a boundary, which is astring of α1 .
The edge modes contribute to the charge so that the total charge is aninteger.

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The final state of the scattering
The objects can be regarded as fermions.
Some of them have opposite helicity to the original fermions.
⇒ New fermions!

The final state of the monopole-fermion
scattering is identified with the new fermions.

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The boundary condition at the core of the monopole
In the process of the scattering, the boundary condition at the core of themonopole plays an important role.
The boundary condition is determined so that the monopole does not havethe electric and flavor charges:
ZZ~ = 0,~ = 0,~j · dS~j1k · dS
Sε2

Sε2

which implies that∂t θA |r=0 = 0,∂t θ1k |r=0 = 0,∀k,i.e., the phases of the condensations can not change at the core of themonopole.
Ryutaro Matsudo (KEK)

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A soliton picture of the scattering
As the soliton approaches the monopole, the charge moves from the edge tothe bulk due to the Witten effect.
When the wall reaches the monopole, the wall bounces back due to theboundary condition, and the string goes throw the monopole. Then the string
can shrink and disappear because it has no charge.
The wall can break by creating the string.
The final state corresponds to the new particle with (R, +1, ).

(cf. Scattering of an axion string and a monopole [I. Kogan 1993])
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Summary

When a charged fermion collides with a monopole, the helicity of the s-wave
component of the fermion has to flip.
Puzzle: If there are two flavors of massless Dirac fermions, any fermions in
the action cannot be the final state of the monopole-fermion
scattering, which should be consistent with the flavor charge
conservation and the helicity flip.
We solve this puzzle by identifying the final state as a new fermion, which
can be described as a soliton of the fermion condensates.

Ryutaro Matsudo (KEK)

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Backup

Ryutaro Matsudo (KEK)

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The edge state
By substituting the 2π jump of αj into the effective Lagrangian
P2j (αj + βj )f ∧ f /(8π ), we obtain the Chern-Simons theory as the theoryon the wall:Z1a∧f4πWhen the wall has the boundary, the CS theory is not gauge invariant.
⇒ There has to be a chiral edge mode:
ZZ
1
1a∧f +(Dx φ(Dt φ + vDx φ) dxdt − φf ),4π R×D2
4π ∂D2
Dφ := dφ − a,a → a + dλ,

φ → φ + λ.

φ is a 2π-periodic scalar, thus we can define the winding number of φ,Z
1dφ.
2π ∂D2The pancake with the exited edge state with this winding number ±1 can beconsidered as a fermion.
Ryutaro Matsudo (KEK)

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31 / 29The charge of the edge state
The U (1) charge is
1
Q=2πZ1(dφ − a) +2π∂D 2Zf.
D2In the gauge where the Dirac string does not penetrate the pancake, thecharge is given as a winding number of φ around the edge.
RThe quantum eigenstate of ∂D2 dφ/2π with the eigenvalue +1 (−1) in theedge theory has the charge ±1.
Classically, the state corresponds to the solution of the eq. of motion of theedge theory. If we neglect the gauge fields, it is
φ = ±2π(x − vt)/L.
By introducing the background fields of the maximal torus of SU (4), we canconfirm that the edge state also have the flavor charge corresponding to thefundamental representation.

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The spin of the pancake
The spin of the object is given as the generator of the translation along theedge:
Jz =Z LLL
Px =2π8π 2dx (∂x φ)2 ,0

where we neglect the gauge fields.
By substituting the solution φ = ±2π(x − vt)/L, we obtain
Jz =1.
2The direction of the spin depends only on the orientation of the wall, anddoes not depend on the charge.

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