Quantum oscillons may be long-lived

: Hertzberg has constructed a quantum oscillon that decays into pairs of relativistic mesons with a power much greater than the radiation from classical oscillon decay. This result is often construed as a proof that quantum oscillons decay quickly, and so are inconsequential. We apply a construction similar to Hertzberg’s to the quantum kink. Again it leads to a rapid decay via the emission of relativistic mesons. However, we find that this is the decay of a squeezed kink state to a stable kink state, and so it does not imply that the quantum kink is unstable. We then consider a time-dependent solution, which may be an oscillon, and we see that the argument proceeds identically.


Introduction
Violent events in field theory, ranging from first order phase transitions to kink-antikink collisions, excite the available excitations.After some time, only the longest lived excitations remain, and these dominate the dynamics.A fairly generic result [1,2,3] in classical field theory is that these longest lived excitations are oscillons [4].However, it is widely believed that in quantum field theory the oscillon lifetime is much shorter, and so they never dominate the dynamics, radically affecting the phenomenology of such events.This belief arises largely from two studies.These considered a lift to quantum field theory of an oscillon [5] and a breather [6] solution in classical field theory.The authors found that the quantum state emits radiation, in particular pairs of relativistic mesons, at a much faster rate than the radiation emission in classical field theory.Extrapolating the radiation power from the initial state to later times, one may conclude that the oscillon lifetime is much shorter in quantum field theory than in classical field theory.
These results are quite general and are derived robustly.Ref. [5] uses standard [7] quantum field theory arguments in the Heisenberg picture, while Ref. [6] uses the classicalquantum correspondence of Refs.[8,9], which is reliable at order O(g 0 ) in the coupling g.Nonetheless, they both rely on a set of approximations.Given the striking phenomenological consequences of this result, it is worth understanding the role played by these approximations, and their applicability to problems of interest.
Most of the approximations are well-motivated.For example, processes which affect the decay amplitude at O(g) are ignored, and indeed these are subleading to the decays which are found.However, both studies chose a specific lift of the classical field theory solution of interest.In the case of Ref. [5] this choice is encoded in Eqs. ( 30) and (31), which, as the author writes is "its unperturbed vacuum state."Similarly in Ref. [6], the initial condition is given in Eqs.(17) and (18) which corresponds to the case in which, "The quantum harmonic oscillators ... are taken to be in their ground state initially ... this corresponds to the quantum field ψ being in its noninteracting vacuum."Furthermore, it is difficult to connect the quantum evaporation of the sine-Gordon breathers [6] with the results obtained via the integrability [10,12,13].Indeed, it is rigorously proven that, depending on the value of the coupling constant g, there are n stable breathers at the quantum level, where this number is the integer part of 1 − g 2 /g.All these states are bound states of the first breather which exists until g = 1/ √ 2. In the semiclassical limit, where g → 0, the number of stable breathers tends to infinity.However, one should be aware that these states, being Hamiltonian eigenstates, do not approach the Hertzberg states in the semiclassical limit.Nonetheless the existence of stable quantum breather states gives a hope that there is quantum lift which may lead to stable semiclassical breather and a long living quantum oscillon.
The goal of the present note is to investigate how this choice of lift affects the decay of a quantum state into pairs of relativistic mesons.If indeed the choice of lift affects the decay rate, it is important to understand how this choice may affect the oscillon decay rate reported in Ref. [5].
We begin in Sec. 2 by constructing a quantum state corresponding to a classical kink.It is not the usual lift of Ref. [11], but rather is constructed by shifting the noninteracting vacuum.We see that its one-loop energy is higher than that of the kink ground state and it decays into pairs of relativistic mesons, just like the quantum oscillon of Ref. [5].Needless to say, the end point of this decay is a lower energy, stable quantum kink, and so the strong initial radiation did not imply that the kink is unstable.Next, in Sec. 3, we generalize this discussion to time-dependent solutions.Now we no longer know whether there is a stable quantum ground state corresponding to the classical solution.However, again we observe the same rapid decays into pairs of relativistic mesons.

The Classical Theory
Let us consider a (1+1)-dimensional theory with a scalar field ϕ(x) and its conjugate momentum π(x).The classical Hamiltonian is characterized by a coupling constant g and a potential V with two degenerate minima.Consider a stationary kink solution ϕ(x, t) = f (x) of the classical equations of motion that interpolates between these two minima.It has a classical mass which can be written, for example, as We define the meson mass m by We only are interested in potentials such that these two limits agree, as otherwise one-loop corrections will break the vacuum degeneracy and the kink will be accelerated from the false vacuum towards the true vacuum [14].

Plane Wave Expansion
We pass to the quantum theory by imposing the canonical commutation relations on the Schrodinger picture operators ϕ(x) and π(x).We will work exclusively in the Schrodinger picture.
We expand the field and its conjugate momentum in the plane wave basis (2.5) This expansion is easily inverted which, using (2.4), implies the operators A ‡ and A are creation and annihilation operators Note that ϕ(x) and π(x) are Hermitian, and so Now that our operators do not commute, we need an ordering prescription to define our Hamiltonian H. Thus we define the quantum Hamiltonian to be (2.9) where :: is the usual normal ordering that places all A ‡ p to the left of A p .

The Hertzberg State
Let us define the perturbative vacuum |H⟩ to be that which satisfies for all p.Note that and so ⟨H|ϕ(x)|H⟩ = 0. (2.12) We will define the Hertzberg kink state to be where the displacement operator D f is defined to be and f (x) is a stationary kink solution of the classical equations of motion for ϕ(x, t).This is called a displacement operator because The expectation value of the ϕ(x) field on this state is Thus the Hertzberg kink state indeed describes a kink.

Time Evolution
Let the Hertzberg state evolve to a time t and call it |K where the first equality defines |H(t)⟩.Then Act on both sides with where we have defined the kink Hamiltonian [15] This situation may be interpreted using the language of passive transformations, as the quantum version of the passive transformation ϕ(x, t) → ϕ(x, t) − f (x) in classical field theory 1 .The state itself is |K H (t)⟩, and like all Schrodinger picture states it evolves via the action of the Hamiltonian H.However, as a result of the definition in the first equality of Eq. (2.17), it is completely characterized by another ket, |H(t)⟩, which evolves via the action of the kink Hamiltonian H ′ .Thus the same kink state can be equivalently described using two different kets, |K H (t)⟩ and ⟩, but if one uses the later than one must conjugate all operators by D f .

More abstractly, |ψ⟩ → D †
f |ψ⟩ is a passive transformation, changing the ket as a vector in the Hilbert space but still representing the same state.Like all passive transformations, it must be accompanied by a transformation of all functions acting on the state, such as the Hamiltonian.We refer to the |K H (t)⟩ representation of the state as the defining frame, and the |H(t)⟩ representation as the kink frame.This passive transformation viewpoint is intuitive, but is not essential for any calculations below.
Using Eq. (2.9), one can show [15] that where H ′ n contains n powers of fields when normal ordered, one finds that is the classical energy of the kink.H ′ 1 vanishes as f (x) solves the classical equations of motion.The terms are all suppressed by powers of the coupling g.Thus the only operator which is not a c-number and not suppressed by powers of g is H ′ 2 .

Energy of the Hertzberg State
Even though we have not yet evaluated H ′ 2 , we already have enough information to evaluate the energy of the Hertzberg state.It is The key step, the second equality, follows because each H ′ n is normal ordered, and so has all A on the right, which annihilate |H⟩ and A † on the left, which annihilate ⟨H|.The only exception is H ′ 0 which is a c-number and contains neither A nor A † .Thus the quantum energy of the Hertzberg state exactly equals its classical energy Q 0 .In contrast, it has long been known [11,16] that the kink ground state has a negative quantum correction to its mass, of order O(m).Thus the Hertzberg state D f |H⟩ has more energy than the kink ground state.If it were an eigenstate of H it would nonetheless be stable.We will now show that it is not an eigenstate, and so is not stable.

The Kink Hamiltonian in the Plane Wave Expansion
We have argued that the evolution of a kink state is dominated by the action of H ′ 2 , which consists of all terms in Eq. (2.22) that are bilinear in the fields.Substituting Eq. (2.1) into Eq.(2.22) this is We begin with the first term The second term is (2.29) The third term (2.30) So, if we act with this Hamiltonian on the perturbative vacuum |H⟩, defined by A p |H⟩ = 0 we get Using (2.20) this implies that, up to corrections of order O(g) is not proportional to |H⟩.Instead pairs of mesons are created, corresponding to A † A † |H⟩ in Eq. (2.31).Of these, some, before clearing the kink, will be reabsorbed via the A2 term in H ′ 2 .To avoid this complication, and so learn whether radiation escapes to infinity, we should work in the normal mode basis.
In conclusion, acting with the Hamiltonian on the Hertzberg state we produce pairs of mesons i.e., radiation.So, we could (wrongly) conclude that the quantum kink is unstable.But this is only because we chose a wrong ground state.The correct ground state is |0⟩ which is annihilated by B operators, which are annihilation operators of the discrete and continuum normal modes [16].

The Normal Mode Expansion
With this motivation, following Refs.[16,15], we consider the expansion of the field and its conjugate momentum in the normal mode basis 2 (2.33)Here the normal modes g(x) correspond to solutions of the linearized classical equations of motion of the form f (x)+g(x)e −iωt .There are three kinds of normal modes, classified by the value of the frequency ω.First there is the zero-mode g B (x) with frequency ω B = 0. Next, there are continuum modes g k (x) with ω k = √ m 2 + k 2 .Finally, some kinks have discrete, real shape modes g S (x) with 0 < ω S < m.The symbol dk/(2π) represents an integral dk/(2π) over continuum modes plus a sum S over shape modes.They are normalized so that discrete modes integrate to unity and continuum modes integrate to Dirac delta functions.
As the normal modes are a basis, the transformation (2.33) is invertible.Therefore the canonical commutation relations (2.4) lead to the commutation relations of the normal mode operators Thus, B ‡ and B are creation and annihilation operators for continuum and shape modes.
On the other hand, ϕ 0 / √ Q 0 and √ Q 0 π 0 are the position and momentum operators of the kink center of mass.
Inserting (2.33) into (2.6)we arrive at the Bogoliubov transformation where g is the Fourier transform of g.In the last line we have introduced an abstract matrix notation, defined by the previous line, so that In particular, the contraction of two p indices represents integration over p with a factor of 1/2π while the contraction of two k indices represents an integral over continuum modes with a factor of 1/2π plus a sum over discrete nonzero modes.The B index appears on the right side of the M matrix, and its contraction is just ordinary multiplication.
The formal inverse of the matrix M then satisfies Note that if k is a discrete index, representing a shape mode, then 2πδ(p − p ′ ) should be replaced with δ pp ′ .This follows from the fact that M −1 M is the identity matrix.

Radiation from the Hertzberg State
In this basis, the O(g 0 ) part of the kink Hamiltonian is just [16,15] Here Q 1 is negative and is the one-loop correction to the kink mass.The ground state of this leading order Hamiltonian is the leading order kink ground state |0⟩, defined by Physically, π 0 |0⟩ = 0 implies that the kink has no momentum, while B k |0⟩ = 0 implies that the normal modes are in their ground states.
The state |0⟩ is the perturbative ground state of the one-kink sector.Therefore a basis of the one-kink sector, in the kink frame, consists of the state |0⟩ acted upon with any number m of zero modes and n meson creation operators We may expand |H⟩ in this basis In other words, |H⟩ is not the kink ground state, it is a superposition of the ground state with various excited states.
We can be a bit more precise.Recall that |H⟩ is annihilated by A p , while |0⟩ is annihilated by π 0 and B k .Recall also that π 0 and B k are related to A p by a Bogoliubov transformation.We conclude that |H⟩ is the squeezed state constructed by acting on |0⟩ with the squeeze operator that performs this Bogoliubov transformation.Thus the decay of |H⟩ is the usual decay of a squeezed state, and the fact that it decays into pairs of mesons results from the fact that the squeeze operator is the exponential of a bilinear in the creation operators.
We will be interested in the ground state component of |H⟩ and the 2-meson component

.44)
With this motivation, we will calculate the following quantity, which is independent of the normalizations of |0⟩ and |H⟩ This quantifies the fraction of the |H⟩ state that contains two unbound mesons, which proceed immediately to escape.
What have we gained by using normal modes instead of plane waves?Now our evolution operator (2.39) contains only B ‡ k B k , no terms which change the number of normal modes or even their frequencies.Therefore any normal modes present in the initial state |H⟩ will persist.If they are continuum modes, this means that they will escape to infinity.Thus the quantity in Eq. (2.45) characterizes how much two-meson radiation will escape to infinity.
The matrix element is easily evaluated using Eq.(2.38) Again using (2.35) one easily evaluates the commutators (2.47) These two matrices can be contracted with the (M −1 ) k 1 p 1 in Eq. (2.46) using (2.37), to yield This quantity is not quite the amplitude for the emission of two mesons, because of the peculiar ⟨0|H⟩ in the denominator.However, it is proportional to this amplitude, with a k 1 and k 2 -independent constant of proportionality, and also it is independent of the normalization of |0⟩ and |H⟩.More precisely, this matrix element is the ratio of the 2-meson part The fact that it is nonzero illustrates that |H⟩ is a superposition of states, at least one of which B ‡ k 1 B ‡ k 2 |0⟩ 0 contains pairs of mesons.These mesons travel away from the kink at relativistic speeds, and so escape in a time 1/m.One exception is the case where k 1 and k 2 are both shape modes.The doubly-excited shape mode is unstable in models such as the ϕ 4 double well, and decays with a lifetime [17] of order O(1/g 2 m 2 ), which at weak coupling is much longer than the other modes.Also, it may be that either k 1 or k 2 is a shape mode and the other is a continuum mode, in which case, after relaxing this summand in the state |H⟩, one arrives at an excited kink.We conclude that |H⟩ decays to a superposition of the ground state kink with its stable excitations.

The Inverse of M
From Eq. (2.13) of Ref. [18] gB (p)g Ignoring the zero-mode, this means that the inverse of gk (p) is g−k (−p).
But how do we invert M ?We first decompose while the zero-mode column is Eq. (2.49) implies (2.52) The matrix gk (p) is supported near k ∼ p, with an exponential falloff in (k − p)/m.In this sense, A is much larger than B for ultrarelativistic k ≫ m.This justifies the expansion Here This yields the usual geometric series sum of the propagator, where one recognizes BA −1 as the Green's function defined in Ref. [19].
3 Time-Dependent Solutions

The Defining Frame
Again, we consider a (1+1)-dimensional field theory with operators ϕ(x) and π(x) satisfying canonical commutation relations (2.4) Define a theory using the Hamiltonian We no longer demand that V have degenerate minima.The classical Hamilton equations are Fix a classical solution ϕ(x, t) = f (x, t) to this equation Now consider a nearby solution ϕ(x, t) = f (x, t) + g(x, t).The classical equations of motion are Subtracting Eq. (3.4) yields At linear order in g this simplifies to

The Comoving Frame
Define the Schrodinger picture displacement operator Recall that the Schrodinger picture operators ϕ(x) and π(x) are independent of t.The displacement operator performs the passive transformation which does depend on t, although we leave the t-dependence of ϕ ′ (x) and π ′ (x) implicit so that these fields are not confused with Heisenberg picture fields.
Similarly we define which is unitarily equivalent to H and so has the same spectrum.It is easily evaluated We may again represent each state with two distinct kets, corresponding to two frames.If a state at time t is represented in the defining frame by the ket |ψ(t)⟩, then we define a comoving frame in which the same state is denoted by the ket D (t) † f |ψ(t)⟩.This ket is comoving in the sense that the expectation value of the field ϕ(x) is shifted by −f (x) and so it is comoving with the solution in field space.
How do comoving frame kets evolve?As defining frame kets |ψ(t)⟩ evolve via the action of H, one finds that comoving frame kets In other words, in the comoving frame, time translations are generated by The additional commutator term is the standard partial derivative term in the Hamiltonian in classical mechanics that arises when performing a canonical transformation with explicit time dependence.
It is interesting to remark that transition (3.15) from the defining Hamiltonian H to the comoving Hamiltonian H′ (t) which generates time translation resembles the covariant derivative.
The commutator is easily calculated Thus we conclude that the time translation operator in the comoving frame is the comoving Hamiltonian Let us decompose this into the terms H′(t) n in H′(t) with n powers of ϕ(x) and π(x).Now H′(t) 0 is arbitrary, as it may be modified by adding a phase to the displacement operator.The first interesting term is the tadpole where the last equality follows from Eq. (3.4).At all higher n, H′ and H ′ agree, and so the comoving Hamiltonian formally has the same form as the kink Hamiltonian.For example H′(t) To robustly show that these mesons escape to infinity requires an understanding of the normal modes, which is an open problem for unstable and time-dependent solutions such as the oscillon.Nonetheless, it illustrates a counter example to the tempting logic that a large initial decay rate into pairs of relativistic mesons implies a short lifetime.If a Bogoliubov transformation exists here that can transform the A operators into at least the classically stable normal modes, then only the classically unstable ones will lead to an instability of the transformed state.These unstable modes have wavelengths of order the oscillon size, and so correspond to highly nonrelativistic mesons, which are quite different from those of the fast decay process described by Hertzberg.This lends hope to a conjecture that an oscillon state exists which decays much more slowly than the quantum oscillon of Ref. [5] or the breather of [6].

Summary and Speculations
We have considered lifts of solutions of classical field theory to quantum states.We have found that the decay rate of the quantum state to pairs of relativistic mesons depends not only on the classical solution, but also depends strongly on the choice of lift.This implies that the determination of the quantum decay rate of a given classical solution should be phrased as a minimization problem, where one selects the quantum lift with the lowest decay rate.This lift is the state that one expects to find after the oscillon has had time to relax.
In light of this result, it would be interesting to examine the dependence of the decay rates found in Refs.[5] and [6] on the choice of lift.It is possible that, as in the case of the quantum kink, a lift of the oscillon may be found whose decay rate is qualitatively smaller than that of the lift considered in Ref. [5].
Wilder speculations are possible.The oscillon is not the first system where a Bogoliubov transformation leads to an instability consisting of emission of pairs of particles.In the case of an accelerating frame, such an argument leads to the Unruh effect [20].When this frame corresponds to a timelike Killing vector beyond an event horizon, the corresponding radiation is called Hawking radiation [21].
It is commonly believed that Hawking radiation continues until the black hole mass decreases either to zero, or at least until it becomes extremal.This leads to numerous potential problems such as information loss [22], naked singularities [23] and remnants [24].Curing these problems has led to conjectures such as the weak gravity conjecture [25], the no global symmetry conjecture [26] and the completeness hypothesis [27], which have become the pillars of modern thinking regarding quantum gravity.However what if instead, like the Hertzberg kink, the black hole state described by Hawking only radiates a very small amount, of order the Planck scale perhaps, before settling into its proper one-loop ground state?
We recognize this as the same time translation operator as in the case of the kink in Eq. (2.27).Now we are done.The calculation has been reduced to that of the kink.One may define a Hertzberg state for any time-dependent solution, including an oscillon, as D 2 = 1 2 dx : π 2 (x) : +V ′′ (gf ) : ϕ 2 (x) : .(3.20)