Noah Graham- Breathers, Q-balls, and Oscillons in Quantum Field Theory

8/3/2019 Noah Graham- Breathers, Q-balls, and Oscillons in Quantum Field Theory

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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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Noah Graham- Breathers, Q-balls, and Oscillons in Quantum Field Theory