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Breathers, Q-balls, and Oscillons
in Quantum Field Theory
Noah Graham
Middlebury College
Work done in collaboration with:E. Farhi (MIT)R. L. Jaffe (MIT)
V. Khemani (MIT, UConn)R. Markov (MIT)M. Quandt (Tubingen)O. Schroder (MIT)H. Weigel (Siegen, Tubingen)
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Sine-Gordon breathers: Stability in an integrable system
S[] =
d2x
1
2()() m
4
1 cos
m
is integrable and has an infinite number of conserved charges.
We have static (anti)soliton solutions: (x) = 4m
arctan emx
Solitons pass right through each other with only a phase shift:
(x, t) =4m
arctan
sinh mut
u cosh mx
where u is the incident speed and = 1/1 u2.
Letting u = i/ we obtain an exact breather:
(x, t) =4m
arctan
sin mt
cosh mx where now = 1/1 + 2
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Q-balls: Stability via conserved charge
Q-balls are time-dependent solutions requiring only a single
global charge, in a nonintegrable theory with no static solitons.
S[] = d4x12
()() 12
M2||2 + A||3 ||4The charge is
Q =1
2i d3x t t
.We fix the charge Q by a Lagrange multiplier and obtain the
Q-ball as a local minimum of the energy
E[] = d3x12|t i|2 +1
2||2
+ U(||) + Qwhere U(||) = 1
2(M2 2)||2 A||3 + ||4
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Q-balls: Stability via conserved charge II
The Q-ball solution has simple time dependence:
(x, t) = eit(x)
We obtain the energy function
E[] =
d3x
1
22 + 1
2U()
+ Q
which is to be minimized over variations of and . Theequation for is just an ordinary second-order equation like we
see in Newtonian mechanics
d2
dr2(
r) +
2
r
d
dr(
r) =
U(
)
where r is playing the role of t, is playing the role of x, the
potential is minus U(), and we also have a time-dependent
friction term.
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Q-balls: Stability via conserved charge III
Then solve for (x) asthe bounce between
degenerate minima of
the upside-down
potential:
0 0.5 1 1.5 2 2.5 3 3.5 40.5
0.4
0.3
0.2
0.1
0
0.1
0.2
0.3
U
()
For a given , we can find the action of the associated solution
by this technique. So we minimize this action over to find
the Q-ball solution. [Coleman]
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4 oscillons in one dimension
Suppose we consider the simpler model
S[] = d2x
1
2
()()
4 2
v2
2
which has static (anti)soliton solutions
(x) = v tanh mx2
but does not have a useful conserved charge (we will consideronly breathers with = v at infinity), and is not integrable. So
there are no simple expressions for exact breathers.
But (approximate) breathers still exist...
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Kink oscillons: forever = a very long time
For the right ranges of initial velocities, numerical simulations
show breathers that live for an indefinitely long time.
[Campbell et. al.]
Breathers are stable to all orders in the multiple scale
expansion. [DHN]
After much debate, the current consensus is that in thecontinuum, such configurations do decay, however, due to
exponentially-suppressed non-perturbative effects.
[Segur & Kruskal]
Numerical techniques are available to distinguish infinite and
finite lifetimes. [Hormuzdiar & Hsu]
For many physical applications this distinction is irrelevant.
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4 oscillons in three dimensions
What about a 3-d 4 theory? Now the theory is not integrable,
has no conserved charges, and no static solitons.
There are still (approximate) breathers!
They live a long time, then suddenly decay. [Gleiser]
0 500 1000 1500 2000 2500 300040
50
60
70energy(t)
0 500 1000 1500 2000 2500 30002
0
2
4
(0,t)
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Breather heuristics
Q: Integrability, conserved charges, and the existence of static
solitons all help us find breathers, but none is necessary for
breathers to exist. What is needed?
A: Nonlinearity and a mass gap. The frequency of oscillation of
the breather is always below the lowest mass, < m.
The picture: nonlinearity allows breathers oscillate with a
characteristic frequency that is too small to couple to the free
dispersive waves in the system.
There are no outgoing modes available to dump their energy
into. [Campbell et. al.]
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Oscillon decay
Q: How does such a configuration decay?
A: By coupling to higher-frequency harmonics: 2, 3, etc.
If we cut off the high frequencies with a lattice such that
2 >
m2 + 4/(x)2, then no such harmonics would exist, and
the breather would be absolutely stable.
Even without this limitation, however, breathers can live for an
unnaturally long time.
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(Almost) The Standard Model in three dimensions II
We have:
3 real massive vector bosons (W and Z, degenerate for us)
with mW = gv2 .
1 real massive scalar (Higgs boson) with mH = v
2.
This is a gauge theory, so any remaining degrees of freedom
are gauge artifacts. There are no physical massless degrees of
freedom (because we have ignored electromagnetism).
To make the problem tractable, we will write an ansatz for this
theory that is invariant as close to spherically symmetric as
possible: it is invariant under simultaneous rotations of real
space and isospin space. [Witten, Ratra and Yaffe]
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Higgs fields in the spherical ansatz
Write as a 2 2 matrix:
= 2 1
1 2 so that
0
1= .
An ansatz for :
(x
, t) =
1
g ((r, t) + i(r, t)x
)
must vanish at the origin since x is not defined there.
Ansatz is preserved under U(1) gauge transformations:
exp [i(r, t) x] with (r = 0, t) = 0.
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Gauge fields in the spherical ansatz
An ansatz for A:
A0(x, t) =1
2ga0(r, t) x
A(x, t) =1
2g
a1(r, t)x( x) +
(r, t)
r( x( x))
+(r, t)
r(x )
Since it is not defined at the origin, a1 r and r must vanishthere. Also must vanish to avoid a singular field
configuration.
Ansatz is preserved under U(1) gauge transformations:
A A ig [(r, t)] ( x)with (r = 0, t) = 0.
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Effective 1-d theory
Form reduced fields in 1 + 1 dimensions:
(r, t) = (r, t) + i(r, t) D = ( i2
a)
(r, t) = (r, t) + i((r, t) 1) D = ( ia)a = ( a0(r, t) a1(r, t) ) f = a awhere now , = 0, 1. A U(1) gauge theory!
has charge 1/2 and mass mH.
has charge 1 and mass mW.
Gauge transformation:
a a i(r, t) ei(r,t)/2 ei(r,t)with (r = 0, t) = 0.
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Effective 1-d theory II
The Standard Model action becomes
S[,,a] =4
g2
dt
0dr
(D)D + r2(D)D
14
r2ff 12r2
(||2 1)2 12
(||2 + 1)||2
Re(i2) g2
r2||2 g
2v2
2
2
Work in a0 = 0 gauge. Still have freedom to maketime-independent gauge transformations. Use this freedom toset a1(r, t = 0) = 0. ta1(r, t = 0) is determined from Gauss
Law once the other fields are specified.
So a configuration is specified by initial values and first timederivatives of the two complex quantities and . (Fourdegrees of freedom, as expected.)
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Results (so far)
Since the Higgs is heavier than the W, Higgs-like solitons are
not stable in our model: Lower-energy decay modes are
available through the coupling to the , since the W is lighter
than the Higgs. But we do find long-lived oscillations
dominated by the the field:
0 200 400 600 800 10000
100
200
300energy(t)
0 200 400 600 800 100020
0
20chern(t)
0 200 400 600 800 10005
0
5
(0,t)
0 200 400 600 800 10002
1
0
1
(5,t)
0 200 400 600 800 10001
0
1a(0,t)
102
101
100
101
102
0
0.1
0.2fft: fields
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Quantum Effects
So far everything has been classical. How to compute quantum
energies? Start with the case of a static soliton.
The quantum correction is the sum of the zero-point energies
of the small oscillations of the quantum field in the background
of our soliton:
E = +j
12j [+ bosons, fermions]
also known as the Casimir Energy or the sum over the Dirac
sea.
It is divergent because of the infinite number of modes.
Just because its infinite doesnt mean its zero!
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Renormalization versus Infinity
A huge constant energy would create a huge gravitational
effect in empty space, which we dont see. So there must be
other terms in the total energy that (almost) exactly cancel it
out. Figuring out why this should happen is the cosmologicalconstant problem.
Without gravity, there is no way to detect a constant change in
the energy, no matter how big. But we can compare thezero-point energy of a configuration with some background
potential to the zero-point energy of a free theory and obtain
an observable prediction:
E =j
12
j freej
contributes to the energy needed to turn on the background
potential.
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Renormalization versus Infinity II
The subtraction of the free modes is an example of
renormalization: we use an experimental physical input (the
energy of empty space is nearly zero according to cosmological
observations) to add an allowed term to the theory (a constant
in the energy).
Like the full calculation, this new term is divergent. Limit both
by a cutoff, such as the dimension of spacetime. Then we take
the limit where we remove the cutoff with the physical input
held fixed.
In general, to render the energy finite, we need several
additional subtractions corresponding to renormalization of
things like mass and charge.
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The Fear of All Sums
Once the small set of counterterms have been fixed, they
render finite calculations of all physical quantities.
Although this procedure is well-defined in principle, it representsa recipe for numerical disaster: We are taking the difference of
the huge Casimir energy and the huge counterterms and trying
to extract the small difference between the two.
Said another way, the final answer is of the order of the
smallest energies in the sum; missing just one mode at high
energies would totally drown out the result.
We need to describe the various sums and subtractions very
precisely to avoid making big mistakes. Luckily, quantum
mechanics has provided us with some powerful tools.
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An Example Calculation
Take the s-wave modes of a particle of mass m obeying the
Klein-Gordon equation in a spherically symmetric potential,
d2
dr2 + V(r) + m
2(r) = (k2 + m2)(r) ,where is the reduced wavefunction obeying (0) = 0 and the
energy is
k2 + m2. V(r) is the potential from the soliton.
-2
-1.5
-1
-0.5
0
0.5
1
0 2 4 6 8 10 12 14 16
V(r)(r)
0
(r)
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Casimir Technology
In the free case V(r) = 0, the solution is 0(r) = sin kr.
Even with the soliton, far away from the origin V(r) still goes
to zero. So there we have the free wavefunction up to a phase
shift: (r) = sin(kr + (k)).
Temporarily impose a wall at r = L, where the wavefunctionmust vanish. Then we have a discrete set of modes given by
solving
...
kn+1L + (kn+1) = (n + 1)
knL + (kn) = n...
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Casimir Technology II
Subtracting,kn+1 kn
L +
(kn+1) (kn)
= .
Divide through by kn+1 kn:
1
L +
(kn+1) (kn)kn+1 kn
=
1
kn+1 kn1
L +d(k)
dk =1
kn+1 kn ,where in the second line we are assuming that L is large so thatkn+1 and kn are getting very close together.
We have arrived at a formula for the inverse distance betweennearby levels, also known as the density of states:
(k) =1
L +
d(k)
dk
.
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Casimir Technology III
Using the density of states (k) = 1
L + d(k)dk
,
1
2 j j 1
2 j j bound states
+
1
2
0
k2 + m2(k)dk continuum states
,
Its easy to subtract the analogous result in the free case, since
then (k) = 0 and there are no bound states. So
1
2
j
(j freej ) 1
2
j
j +1
2
0
k2 + m2
d(k)
dkdk .
Scattering theory made the first subtraction painless!
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Born Subtractions
This subtraction represented the zeroth order in the Born
expansion (the free solution) of the full Casimir energy in
powers of the potential V(r). All the subtractions needed for
renormalization can be expressed in terms of a few terms in the
Born series. Then we add back these contributions, together
with counterterms, as ordinary Feynman diagrams.
By explicitly calculating both the Born approximation and the
Feynman diagrams in dimensional regularization, we have
verified that this procedure has added and subtracted the same
thing, and thus not introduced finite errors.
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Born Subtractions II
The calculation is still exact the Born approximation is just
a tool that lets us implement the subtractions precisely and
robustly. We arrive at expressions of the form
E =
=0
12
j
j +
0
k2 + m2
d
dk((k) Born(k)) dk
+ EFeynman
where now we have summed over all partial waves and
EFeynman is the finite contribution to the energy from the
Feynman diagrams combined with the counterterms.
This procedure works for any renormalizable theory, as long asthe soliton has sufficient symmetry so that we can enumerate
the spectrum of small oscillations around it (e.g. as a sum over
channels and an integral over energies).
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Making Infinity Precise
The phase shifts capture precise, nonperturbative informationabout the entire spectrum. For example, compare the numberof states in the free and interacting case,
j
1 freej
1 j
1
bound states
+ 0
1
ddk
dk
continuum
which is zero by Levinsons theorem:
(0) () = nwhere n is the number of bound states.
Phase shifts count the states that left the continuum andbecame bound states!
In a CP-violating theory, we can perform this sum weighted bythe sign of the energy and obtain the charge of the soliton.
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Extend to Q-balls
Since their time dependence is so simple, this technique
extends easily to Q-balls. Let j be the energies of the small
oscillation modes of the static field (r).
The full Casimir energy is
1
2
j
0
|j + | + |j |
=
j
0
max||, |j|
.
The true spectrum contains 3 zero modes in = 1. These
correspond to modes of the static field with j = . There
must then be exactly one unstable mode in = 0 with
j < ; its contribution is truncated to .
The full Casimir energy of the reduced potential is generally
small, so the dominant contribution comes from zero mode
truncation,
0.
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Future Directions
How quickly do photons, gluons, or light fermions destroyStandard Model oscillons?
Can oscillons stay out of equilibrium and make baryons inthe early universe? As our configuration evolves, sphaleronprocesses change baryon number:
N =
1
20 dr a1 +
i
2(Dr Dr)which oscillates with a large amplitude.
Are quantum effects calculable for general oscillons?
Can we apply oscillon ideas to other situations wherenaturalness is an obstacle?
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