Production of massive bosons from the decay of a massless particle beam
PProduction of matter from the decay of a massless particle clump
Ariel Arza Institute for Theoretical and Mathematical Physics (ITMP),Lomonosov Moscow State University, 119991 Moscow, Russia
Abstract
We study the production of scalar particles χ coming from the decay of a highly occupied scalar field φ in the framework ofa toy model with an interaction gφχ . We explore the parametric resonant instabilities coming from Bose enhancement in twoopposite cases; (1) massive φ and massless χ , which has been vastly discussed in the literature, and (2) massless φ and massive χ . For a momentum p of the decaying field and a mass m χ of the produced particles, we find that in case (2) the decay of themassless clump is allowed by Bose enhancement if the energy density of φ exceeds the instability threshold p m χ / (2 g ) . Wealso compute the spontaneous decay rate for both cases. The decay of a particle into other species is one of thesimplest and most relevant effects in relativistic field the-ories. From the theoretical point of view, the decay ratefor a process φ i → φ j + φ k can be defined in perturbationtheory as [1] Γ = V (cid:90) d p j (2 π ) V (cid:90) d p k (2 π ) d | S fi | dt (1)where V is the volume where the theory is defined, t thetime and | S fi | the transition probability of the process.When the initial and final states are asymptotic, i.e. freeof interactions, the matrix element S fi ≡ (cid:104) f | S | i (cid:105) is S fi = i (2 π ) δ ( p i − p j − p k ) (cid:112) ω p i V (cid:112) ω p j V (cid:112) ω p k V M fi (2)where M fi is the Feynman amplitude. In this formula,the delta function imposes that the momentum and en-ergy conservation must be exactly fulfilled in the transi-tion.One interesting case, that inspired us to write this let-ter, is the axion decay into two photons. For this processEq. (1) gives us Γ a → γ = g aγγ m a π (3)in the axion rest frame. Here g aγγ is the axion to twophotons coupling and m a the axion mass. On the otherhand, when the axion field is highly occupied in a particu-lar state, one can find that the photon field suffers a para-metric resonance which leads to an exponential growthof the photon occupancy number [2–8]. Parametric res-onance is a well known effect with many applications incosmology, for instance in the physics of inflation [9–15](see Ref. [16] for a review). Let’s discuss briefly thisparametric resonance in more detail for the axion case.In a homogeneous and free axion field background withenergy density ρ a the emission of pairs of photons withmomenta (cid:126)k and − (cid:126)k from axion decay obeys the followingequations of motion for the photon creation and annihi-lation operators ∂ t a γ,(cid:126)k = σ a e i(cid:15) (cid:126)k t a † γ, − (cid:126)k . (4) Here σ a = g aγγ (cid:114) ρ a (5)and (cid:15) (cid:126)k = 2 k − m a , (6)where k = | (cid:126)k | . Moreover, we have used a circular po-larization basis where both polarizations are decoupled,then Eq. (4) holds for any of them. Eq. (4) was foundunder the assumption of a highly occupied axion field,so this analysis does not consider back-reactions over theaxions. The general solution of (4) is a γ,(cid:126)k ( t ) = e i(cid:15) (cid:126)k t/ (cid:20) a γ,(cid:126)k (0) (cid:18) cosh( s k t ) − i (cid:15) (cid:126)k s k sinh( s k t ) (cid:19) + σ a s k a γ, − (cid:126)k (0) † sinh( s k t ) (cid:21) (7)where s k = (cid:113) σ a − (cid:15) (cid:126)k / . For every mode (cid:126)k the occu-pancy number is f γ,k ( t ) = 1 V (cid:104) | a γ,(cid:126)k ( t ) † a γ,(cid:126)k ( t ) | (cid:105) = σ a s k sinh( s k t ) . (8)To get (8) we assumed no initial photon occupancy num-ber, i.e. f γ,k (0) = 0 . For each mode (cid:126)k , we have exponen-tial growth solutions as long as (cid:15) (cid:126)k < σ a or m a − σ a < k < m a σ a . (9)We can compute the photon density using a saddle pointapproximation, for σ a t (cid:29) we find n γ ( t ) = 2 (cid:90) d k (2 π ) f γ,k ( t ) (cid:39) m a σ a (cid:112) (2 π ) e σ a t . (10)From (10) and (9) we see that the photon density growsat the rate γ a → γ = 2 σ a ∼ g aγγ (cid:114) ρ a (11)1 a r X i v : . [ h e p - t h ] O c t ffecting a bandwidth of size δk = 2 σ a (12)centered at k = m a / . Two interesting concerns arisefrom the above results. First, it seems to be a contra-diction between Eqs. (3) and (11). The most importantdiscrepancy is that the former scales as g aγγ while thelatter as g aγγ . Fortunately, references [17, 18] give a verynice explanation about what is happening. Looking atthe Boltzmann equation one finds ˙ n a = − Γ a → γ (cid:0) f γ,m a / (cid:1) n a . (13)We can see that the decay rate is corrected by a factor f γ,m a / . When f γ,m a / > / the decay rate is en-hanced dramatically. This effect is known as stimulateddecay or Bose enhancement. To probe that f γ,m a / > / can be reached easily, let’s analyze Eq. (13) at the be-ginning, when f γ,m a / = 0 . Since ˙ n γ = − n a , after asmall period of time δt we get n γ = 2Γ a → γ n a δt . Fromthe uncertainty principle the photon field is spread in theenergy bandwidth δk ∼ /δt , then f γ,m a / ∼ n γ π ( m a / δk/ (2 π ) = 8 π Γ a → γ n a m a δk . (14)Now, using (3) and taking (12) for δk , we find f γ,m a / ∼ π > (15)It means that the effect coming from Bose enhancementbecomes important instantly. Moreover, using (14), Eq.(13) can be written as ˙ n γ ∼ σ a n γ (16)which is consistent with (10), at least in the order ofmagnitude.The second concern is the energy conservation. All theaxions are at rest and have the same energy ω a = m a ,then by energy momentum conservation one expects thatevery produced photon should have an energy m a / .The modulus of the momentum of the produced photonsspreads over a bandwidth given by (9), so most of themare slightly shifted from m a / . If we would consider theproduced photons as particles in a final free of interac-tions state, clearly the modulus of its momentum shouldbe its energy and we would think about a problem withenergy conservation. However it is not the case becausethe photon states produced from every single decay con-tinue to interact with the background, at least when theyare still inside the axion cloud region. Looking at Eq. (7)we see that a consequence of this interaction is that thefrequency of every photon mode is shifted respect to k byadding an amount − (cid:15) (cid:126)k / . It leads to a photon frequency E k = k − (cid:15) (cid:126)k / m a / as expected. So far we have discussed the particular case of the para-metric resonance in axion physics, nevertheless solutionsof the form (7) are recurrent in decays of highly occupiedfields. While the analog of σ a has in general a momen-tum dependence ( σ a → σ (cid:126)k ) and a different form as theaxion case (it also depends on the form of the coupling),it always scales as the squared root of the energy density ρ φ i of a decaying field φ i . On the other hand, and alsovery important, (cid:15) k has always the same form; i.e. for aprocess φ i → φ j + φ k , if the decaying field φ i is occupiedin a state with momentum (cid:126)p , (cid:15) (cid:126)k = ω ( j ) (cid:126)k + ω ( k ) (cid:126)p − (cid:126)k − ω ( i ) (cid:126)p (17)where ω ( l ) (cid:126)k = (cid:112) k + m l ( l = { i, j, k } ) and m l the corre-sponding mass of each field φ l . The parametric resonanceand depletion of the field φ i takes place when (cid:15) (cid:126)k < σ (cid:126)k . (18)It leads to ask ourself, what if φ i is massless? . What arethe conditions for Eq. (18) to be satisfied in the masslesscase?. Let’s first discuss the one particle case. When φ i iscomposed by one particle or a few of them, all the abovediscussion is not appropriate because it does not considerback-reactions on φ i , however the analysis is quite simple.Since the initial and final states are free of interactions,they have energies ω ( i ) (cid:126)p and ω ( j ) (cid:126)k + ω ( k ) (cid:126)p − (cid:126)k , respectively andenergy conservation implies that the kinematic conditionfor the particle to decay is (cid:15) (cid:126)k = 0 . For m i = 0 , (cid:15) (cid:126)k can bezero if m j = m k = 0 . This case has been studied in detailin [20–23]. When the produced particles are massive (orone of them) (cid:15) (cid:126)k can never be zero and the single parti-cle can not decay. When φ i is highly populated in somemomentum mode, the condition for the decay is actuallyEq. (18). The cloud starts to decay into particles withmomentum modes satisfying Eq. (18) even if (cid:15) (cid:126)k = 0 hasno solutions, then the decay into massive particles couldbe allowed for a massless φ i . As σ (cid:126)k scales as √ ρ φ i , it isclear that we approach to the kinematic condition (cid:15) (cid:126)k = 0 in the limit ρ φ i → . With respect to the energy of theproduced particles, as in the axion case, the time evo-lution of the creation and annihilation operators gives acorrection − (cid:15) (cid:126)k / in the frequency of the fields as a con-sequence that the produced particles continue to interactwith the decaying field. For each produced particle φ j the energy is E ( j ) (cid:126)k = ω ( j ) (cid:126)k − (cid:15) (cid:126)k / (cid:16) ω ( j ) (cid:126)k − ω ( k ) (cid:126)p − (cid:126)k + ω ( i ) (cid:126)p (cid:17) (19) In models of massless preheating [19] the inflaton field suffersself-interactions due to a λφ potential. It makes the zero momen-tum modes of the inflaton to oscillate harmonically. It finally drivesthe decay of the field into SM particles by parametric resonance.Our question is rather directed at free initial massless particles andtheir possibility to decay into massive species. φ k we have E ( k ) (cid:126)p − (cid:126)k = ω ( k ) (cid:126)p − (cid:126)k − (cid:15) (cid:126)k / (cid:16) ω ( k ) (cid:126)p − (cid:126)k − ω ( j ) (cid:126)k + ω ( i ) (cid:126)p (cid:17) . (20)The sum of both frequencies gives us E ( j ) (cid:126)k + E ( k ) (cid:126)p − (cid:126)k = ω ( i ) (cid:126)p , (21)so for every single decay the energy is conserved.At this point we have taken the properties of the Boseenhancement to explain how a cloud of free massless par-ticles can be depleted by the decay into massive particles.As far as we know this possibility has not been discussedin the literature. In the following we will discuss thismechanism in detail using a simple toy model involvingtwo scalar fields.We consider the following interaction hamiltonian den-sity H I = gφχ (22)where φ represents the decaying particles, χ the producedparticles and g the coupling constant. For now we con-sider both fields to be massive. The equations of motionof the system are ( (cid:3) + m φ ) φ = − gχ (23) ( (cid:3) + m χ ) χ = − gφχ (24)where m φ and m χ are the masses of φ and χ , respectively.We are interested in the case where φ is highly occupiedin a single momentum mode (cid:126)p . We also do not want toconsider back-reactions over φ , so the RHS of Eq. (23)becomes . These considerations lead us to write φ as aclassical field in the form of a monochromatic plane waveas φ ( (cid:126)x, t ) = (cid:112) ρ φ ω p cos( (cid:126)p · (cid:126)x − ω (cid:126)p t ) (25)where ρ φ is the time averaged energy density of the field φ and ω (cid:126)p = (cid:113) p + m φ . We write the quantum field χ interms of creation and annihilation operators as χ ( (cid:126)x, t ) = (cid:90) d k (2 π ) (cid:112) (cid:126)k (cid:16) χ (cid:126)k ( t ) e i(cid:126)k · (cid:126)x + χ (cid:126)k ( t ) † e − i(cid:126)k · (cid:126)x (cid:17) (26)where Ω (cid:126)k = (cid:113) k + m χ and the operators χ (cid:126)k and χ † (cid:126)k satisfy the commutation relations [ χ (cid:126)k , χ (cid:126)k (cid:48) ] = 0 and [ χ (cid:126)k , χ † (cid:126)k (cid:48) ] = (2 π ) δ ( (cid:126)k − (cid:126)k (cid:48) ) . Inserting (25) and (26) into(24) we get the following set of equations for the newoperator A (cid:126)k = χ (cid:126)k + χ †− (cid:126)k (cid:16) ∂ t + Ω (cid:126)k (cid:17) A (cid:126)k = − ω (cid:126)p α (cid:32)(cid:115) Ω (cid:126)k Ω (cid:126)k − (cid:126)p A (cid:126)k − (cid:126)p e − iω (cid:126)p t + (cid:115) Ω (cid:126)k Ω (cid:126)k + (cid:126)p A (cid:126)k + (cid:126)p e iω (cid:126)p t (cid:33) (27) where we have defined α ≡ g (cid:112) ρ φ ω (cid:126)p . (28)Let’s study the resonant solutions of Eq. (27) analyticallyusing the rotating wave approximation (RWA). To doso we first write χ (cid:126)k ( t ) = a (cid:126)k ( t ) e − i Ω (cid:126)k t where a (cid:126)k variatesslowly respect to χ (cid:126)k . Neglecting second derivatives of a (cid:126)k Eq. (27) can be written as ∂ t a (cid:126)k = ∂ t a †− (cid:126)k e i Ω (cid:126)k t − iσ (cid:126)k (cid:16) a (cid:126)k − (cid:126)p e i(cid:15) (1) (cid:126)k t + a † (cid:126)p − (cid:126)k e i(cid:15) (2) (cid:126)k t (cid:17) − iσ (cid:126)k + (cid:126)p (cid:16) a (cid:126)k + (cid:126)p e i(cid:15) (3) (cid:126)k t + a †− (cid:126)p − (cid:126)k e i(cid:15) (4) (cid:126)k t (cid:17) (29)where σ (cid:126)k = g (cid:115) ρ φ / ω (cid:126)p Ω (cid:126)p − (cid:126)k Ω (cid:126)k (30)and (cid:15) (1) (cid:126)k = Ω (cid:126)k − Ω (cid:126)k − (cid:126)p − ω (cid:126)p (31) (cid:15) (2) (cid:126)k = Ω (cid:126)k + Ω (cid:126)p − (cid:126)k − ω (cid:126)p (32) (cid:15) (3) (cid:126)k = Ω (cid:126)k − Ω (cid:126)k + (cid:126)p + ω (cid:126)p (33) (cid:15) (4) (cid:126)k = Ω (cid:126)k + Ω − (cid:126)k − (cid:126)p + ω (cid:126)p . (34)The RWA allows us to only keep terms that oscillateslowly respect to Ω (cid:126)k . We can immediately neglect thefirst term of the RHS of Eq. (29). The recognition of theother terms that can be neglected will depend on (cid:126)k . The (cid:15) ( i ) (cid:126)k defined in (31)-(34) account for the efficiency of someparticular processes. This association is listed as follows (cid:15) (1) (cid:126)k : φ (cid:126)p + χ (cid:126)k − (cid:126)p ↔ χ (cid:126)k (35) (cid:15) (2) (cid:126)k : φ (cid:126)p ↔ χ (cid:126)k + χ (cid:126)p − (cid:126)k (36) (cid:15) (3) (cid:126)k : φ (cid:126)p + χ (cid:126)k ↔ χ (cid:126)k + (cid:126)p (37) (cid:15) (4) (cid:126)k : vacuum ↔ χ (cid:126)k + φ (cid:126)p + χ − (cid:126)k − (cid:126)p (38)When one of them becomes smaller than some threshold,the associated process is excited. Before discussing thosethresholds let us comment an interesting fact of the RWA.The free hamiltonian of the produced field can be writtenas H χ ( t ) = (cid:90) d k (2 π ) (cid:0) Ω (cid:126)k a (cid:126)k ( t ) † a (cid:126)k ( t ) + R (cid:126)k ( t ) (cid:1) (39)where R (cid:126)k ( t ) is a complicated operator that involves firstand second derivatives of a (cid:126)k ( t ) . In the RWA R (cid:126)k is negli-gible respect to Ω (cid:126)k a † (cid:126)k a (cid:126)k , so the form of the hamiltonianallows us to define a number operator n (cid:126)k = a † (cid:126)k a (cid:126)k andtherefore a (cid:126)k and a † (cid:126)k can be interpreted as time depen-dent annihilation and creation operators for χ particles.3 IG. 1: η (cid:126)κ as a function of µ for κ = 0 . , . and . . We used θ = 0 and α = 10 − . We are interested in the case where at t = 0 there isno χ particles. Therefore, at the very beginning the onlyrelevant process to account is φ (cid:126)p → χ (cid:126)k + χ (cid:126)p − (cid:126)k . Defining (cid:15) (cid:126)k ≡ (cid:15) (2) (cid:126)k and neglecting non-relevant terms, Eq. (29)becomes ˙ a (cid:126)k = − iσ (cid:126)k a † (cid:126)p − (cid:126)k e i(cid:15) (cid:126)k t . (40)The general solution of (40) is a (cid:126)k ( t ) = e i(cid:15) (cid:126)k t/ (cid:20) a (cid:126)k (0) (cid:18) cosh( s (cid:126)k t ) − i (cid:15) (cid:126)k s (cid:126)k sinh( s (cid:126)k t ) (cid:19) − i σ (cid:126)k s (cid:126)k a (cid:126)p − (cid:126)k (0) † sinh( s (cid:126)k t ) (cid:21) . (41)For every momentum mode (cid:126)k its occupancy number is f χ,(cid:126)k ( t ) = 1 V (cid:104) | a (cid:126)k ( t ) † a (cid:126)k ( t ) | (cid:105) = σ (cid:126)k s (cid:126)k sinh( s (cid:126)k t ) (42)and the total density of produced particles is given by n χ ( t ) = (cid:90) d k (2 π ) σ (cid:126)k s (cid:126)k sinh( s (cid:126)k t ) . (43)The modes that exhibit parametric resonance are theones that satisfy s (cid:126)k > . Before continuing, we wouldlike to discuss the validity of our calculation. Our resultswere found in a RWA where the amplitudes a (cid:126)k variatesslowly respect to χ (cid:126)k , so Eq. (41) is valid under the con-dition Ω (cid:126)k (cid:29) σ (cid:126)k . Assuming Ω (cid:126)k ∼ Ω (cid:126)p − (cid:126)k ∼ ω (cid:126)p we wouldbe safe for α (cid:28) . (44)Even though condition (44) will be useful in most of theremaining discussion, we have to pay special attentionwhen either Ω (cid:126)k or Ω (cid:126)p − (cid:126)k approaches to zero because σ (cid:126)k could blow up. For instance, when m χ = 0 , Ω (cid:126)p − (cid:126)k = 0 if FIG. 2: η (cid:126)κ as a function of θ for κ = 0 . , . and . . We used α = 10 − and µ = √ α/ . (cid:126)k = (cid:126)p . Now we are going to study the instability condi-tions in two opposite cases; (1) massive φ and massless χ , and (2) massless φ and massive χ . Case (1) has beenstrongly discussed in the literature, so we are using itas a matter of comparison with case (2) which is whatconcerns us. Let’s define the dimensionless quantities (cid:126)κ = (cid:126)k/ω p , µ = m χ /ω p , (cid:126)v = (cid:126)p/ω p and β (cid:126)κ = (cid:112) κ + µ .Let us also define η (cid:126)κ = s (cid:126)κ /ω p , which in terms of theprevious parameters is given by η (cid:126)κ = (cid:115) α β (cid:126)v − (cid:126)κ β (cid:126)κ −
14 ( β (cid:126)κ + β (cid:126)v − (cid:126)κ − . (45)For case (1) we can always choose a reference frame wherethe particles φ are at rest, then we can make v = 0 .Eq. (45) simplifies and the instabilities take place for / − α < κ < / α . For case (2) Eq. (45) mustbe evaluated with v = 1 . Fig. 1 shows η (cid:126)κ as a functionof µ for α = 10 − , θ = 0 and different values of κ . Ofcourse we avoid the extremes κ = 1 and κ = 0 where ourcalculation could break down. We can see instabilitiesonly for µ (cid:28) . Fig. 2 shows η (cid:126)κ as a function of θ for α =10 − , µ = √ α/ and different values of κ . We observethat the instabilities occur for θ smaller than . To havea better look of the instability region, see Fig. 3 where η (cid:126)κ is plotted as a function of κ for a fixed µ and differentvalues of θ . We see that in terms of κ the bandwidth isof the order for θ = 0 , much bigger than case (1) whereit is of the order of α . However we also see that theinstability region becomes smaller as θ increases, whichis the main advantage of case (1) where the instabilitytakes place for any angle. We also see in Fig. 3 thatclose to the extremes, κ ≈ and κ ≈ , η (cid:126)κ is enhanced. Itmay contribute to the efficiency of the process, althoughwe have to be always alert that s (cid:126)k /k (cid:28) is satisfied,otherwise we would enter in regions where our calculationbreaks down. Although these plots give us some generalsfeatures of the instability conditions, we can use the factthat m χ (cid:28) ω (cid:126)p and θ (cid:28) to find analytical expressionsrelated to the instability window. The frequencies can be4 IG. 3: Shape of η (cid:126)κ as a function of κ for θ = 0 , . and . .Here we have used α = 10 − and µ = √ α/ . approximated as Ω (cid:126)k ≈ k + m χ / (2 k ) and Ω (cid:126)p − (cid:126)k ≈ p − k +( pkθ + m χ ) / (2( p − k )) . When computing (cid:15) (cid:126)k the maincontributions of Ω (cid:126)k and Ω (cid:126)p − (cid:126)k cancel with ω (cid:126)p getting (cid:15) (cid:126)k ≈ p ( m χ + k θ )2 k ( p − k ) . (46)In the above result we have assumed that we are farenough from k = 0 and k = p .To find the instability condition we solve the equation η (cid:126)κ = 0 to get the instability limits for κ . As θ = 0 givesthe biggest instability window for κ , we evaluate at thisangle obtaining the limits κ ± = 1 ± (cid:112) − µ /α . (47)The size of the window is δκ = (cid:112) − µ /α , (48)therefore the instability is triggered when µ < √ α orwhen ρ φ > p m χ g (49)in terms of the required energy density.It is not too difficult to check that the instability con-dition (49) is Lorentz invariant. It is, of course, what wewould expect. Now two questions arise; why the instabil-ities were found for p (cid:29) m χ ?. What happen for p ∼ m χ or p (cid:28) m χ ?. We answer the first question as follows. Onthe one hand notice that p (cid:29) m χ is not the instabilitycondition, the instability occurs rather when µ < √ α .On the other hand we are working in a RWA where thevalidity of our analysis is limited by α (cid:28) . These twofacts imply that the instability must satisfy p (cid:29) m χ as aconsequence. In other words, our rotating wave approx-imation holds only for reference frames where p (cid:29) m χ .To answer the second question, we take Eq. (27) in the FIG. 4: Plots of the amplitude squared | A (cid:126)k | as a function of timefor the modes (cid:126)k , (cid:126)k + (cid:126)p , (cid:126)k + 2 (cid:126)p and (cid:126)k + 3 (cid:126)p . See the text for details. limit p (cid:28) m χ . Taking into account only the modes (cid:126)k and (cid:126)k − (cid:126)p we get the system (cid:0) ∂ t + m χ (cid:1) A (cid:126)k = − ω (cid:126)p α A (cid:126)k − (cid:126)p e − iω (cid:126)p t (50) (cid:0) ∂ t + m χ (cid:1) A (cid:126)k − (cid:126)p = − ω (cid:126)p α A (cid:126)k e iω (cid:126)p t (51)For solutions of the form A (cid:126)k = c (cid:126)k e γt e − iω (cid:126)p t/ and A (cid:126)k − (cid:126)p = d (cid:126)k e γt e iω (cid:126)p t/ we find that γ is given by γ = ω (cid:126)p α − m χ . (52)It is clear that the instability is also activated when (49)is fulfilled.As our initial condition is a highly occupied φ and no χ particles are present, we were interested in the behaviorof Eq. (29) at the very beginning. Now we are going todiscuss what happens after the modes (cid:126)k and (cid:126)p − (cid:126)k aresignificantly occupied. For case (1) none of the other (cid:15) ( i ) (cid:126)k become small enough to excite the corresponding process,then φ continue decaying into momentum modes (cid:126)k and (cid:126)p − (cid:126)k of χ particles and Eqs. (41), (42) and (43) are stillvalid until back-reactions are activated, i.e. when theinverse process χ → φ starts to be important. Of course,it happens when φ is significantly depleted. For case (2) (cid:15) (1) (cid:126)k and (cid:15) (4) (cid:126)k are big, of the order of Ω (cid:126)k , but (cid:15) (3) (cid:126)k is as smallas (cid:15) (2) (cid:126)k . It means that if the process φ (cid:126)p → χ (cid:126)k + χ (cid:126)p − (cid:126)k isefficient, once χ (cid:126)k and χ (cid:126)p − (cid:126)k are abundant, the processes χ (cid:126)k + φ (cid:126)p → χ (cid:126)k + (cid:126)p and χ (cid:126)p − (cid:126)k + φ (cid:126)p → χ (cid:126)p − (cid:126)k will becomeefficient too. Following the same reasoning, a cascade isproduced and every process of the form χ (cid:126)k + n(cid:126)p + φ (cid:126)p → χ (cid:126)k +( n +1) (cid:126)p and χ n(cid:126)p − (cid:126)k + φ (cid:126)p → χ ( n +1) (cid:126)p − (cid:126)k ( n = 0 , , , ... )will be eventually important after some period of time.To probe this last claim we compute the classical versionsof Eqs. (27) numerically assuming that only the modes (cid:126)k and (cid:126)p − (cid:126)k are initially occupied. Fig. 4 shows | A (cid:126)k | as a function of time for the modes (cid:126)k , (cid:126)k + (cid:126)p , (cid:126)k + 2 (cid:126)p and (cid:126)k +3 (cid:126)p . We do not plot the modes (cid:126)p − (cid:126)k , (cid:126)p − (cid:126)k , (cid:126)p − (cid:126)k and5 (cid:126)p − (cid:126)k because, according to the initial conditions, thetime evolution of their square amplitudes are identical asthe ones for (cid:126)k , (cid:126)k + (cid:126)p , (cid:126)k + 2 (cid:126)p and (cid:126)k + 3 (cid:126)p , respectively.We evaluate at (cid:126)k = ω (cid:126)p / and we assume that at thebeginning the modes (cid:126)k and (cid:126)p − (cid:126)k have initial conditions | A (cid:126)k (0) | = | A (cid:126)p − (cid:126)k (0) | = 1 . All the other modes havezero initial amplitude squared. We see that the squareamplitudes of modes different from (cid:126)k and (cid:126)p − (cid:126)k begin togrow gradually. In Fig. 4 we also compute numerically inthe rotating wave approximation (see dotted lines). Webasically take Eqs. (29) and solve the system neglectingfast oscillating terms. It is clear that this approximationis in perfectly agreement with the full solutions.Now we are going to find some analytical formulas forthe spontaneous decay rate for case (1) as well as for case(2). At first order in perturbation theory, the scatteringmatrix S fi of the transition between two states | i (cid:105) and | f (cid:105) is given by S fi = (cid:90) T dt (cid:90) d x (cid:104) f | H I | i (cid:105) . (53)Here T is the time during which the system transits fromthe state | i (cid:105) to the state | f (cid:105) . As the transitions that aretypically studied involve single processes where the finalstate is free of interactions, one can just make T → ∞ .However, in some cases this approximation breaks downdue to finite time effects [24–26] or simply if the transitiontime is not big enough to be considered as infinite [27].As in our approach φ is considered as a classical field (see(25)) our Fock space only contains information about χ particles. We are interested in the production of the firstpairs of χ particles, so if (cid:126)k and (cid:126)q are the momenta ofthese produced particles, the initial and final states are | i (cid:105) = | (cid:105) and | f (cid:105) = (cid:12)(cid:12)(cid:12) (cid:126)k, (cid:126)q (cid:69) , respectively. Using (22), (25)and (26) we have at leading order S fi = 1 V σ (cid:126)k (2 π ) δ ( (cid:126)k + (cid:126)q − (cid:126)p ) e i(cid:15) (cid:126)k T/ sin( (cid:15) (cid:126)k T / (cid:15) (cid:126)k / . (54)We compute the decay rate using Eq. (1). Replacing (54)into (1) and applying a extra factor / coming from thefact the final states are identical particles we get Γ φ → χ = V (cid:90) d k (2 π ) σ (cid:126)k sin( (cid:15) (cid:126)k T ) (cid:15) (cid:126)k . (55)Let us first consider the scenario where φ is occupiedby one particle state. In this case the system transitsto a free of interactions state after the decay, so we cantake T → ∞ . In this limit sin( (cid:15) (cid:126)k T ) /(cid:15) (cid:126)k → πδ ( (cid:15) (cid:126)k ) , thenthe usual energy-momentum conservation relation, Ω (cid:126)k +Ω (cid:126)p − (cid:126)k − ω (cid:126)p = 0 , involving the asymptotic initial and finalstates holds. For case (1) we, of course, obtain the usualformula Γ φ → χ = g πm φ . (56) For case (2) we get Γ φ → χ = 0 because (cid:15) (cid:126)k can never bezero. It is the usual result that prohibits the decay ofmassless particles into massive particles due to the lackof final asymptotic states that satisfy energy-momentumconservation. Now we consider the highly occupied φ field scenario. It is clear that there is no asymptoticstates after the first decay, the evolution of the field χ israther described by (41). In parametric resonance, theoccupancy number of produced particles grows exponen-tially as Eq. (42), at least at the beginning, thereforethe transition time of the process is T = (2 s (cid:126)k ) − . Sincethe main contributions for the growth are provided bymodes that fulfill s (cid:126)k (cid:29) (cid:15) (cid:126)k , we can use the approximation sin( (cid:15) (cid:126)k T ) = sin( (cid:15) (cid:126)k / s (cid:126)k ) ≈ (cid:15) (cid:126)k / s (cid:126)k with great accuracy.Eq. (55) becomes Γ φ → χ = V (cid:90) d k (2 π ) σ (cid:126)k (cid:113) σ (cid:126)k − (cid:15) (cid:126)k . (57)For case (1) we can make σ (cid:126)k ≈ m φ α to preserve onlyterms of order g . We find Γ (1) φ → χ = V α m φ π (cid:90) m φ α − m φ α d(cid:15) (cid:113) m φ α − (cid:15) (58) = g N φ πm φ (59)where N φ = V ρ φ /m φ is the number of φ particles. No-tably, for N φ = 1 we recover Eq. (56), although theconnection to the one particle case is not clear becauseEq. (59) was found for N φ (cid:29) . For case (2) the inte-gration is not straightforward as in case (1). In terms ofdimensionless parameters and assuming θ (cid:28) , which istrue on behalf of previous discussions, we get Γ (2) φ → χ = g N φ πp F ( µ/ √ α ) (60)where F ( x ) = 1 π (cid:90) κ + ( x ) κ − ( x ) dκ (cid:32) π − arcsin (cid:32) x (cid:112) κ (1 − κ ) (cid:33)(cid:33) . (61)We recall that κ ± ( µ/ √ α ) are defined in (47). In termsof the quantity ω (cid:126)p which is m φ for case (1) and p for case(2), the ratio between both decay rates is simply Γ (2) φ → χ Γ (1) φ → χ = F ( µ/ √ α ) . (62)The plot of F ( µ/ √ α ) is shown in Fig. 5. As we alreadydiscussed, for case (2) the decay is clearly plausible if µ < √ α but it becomes impossible for µ ≥ √ α . For µ (cid:28) √ α the decay rate approaches its maximum whichis a half of the decay rate for case (1).To finish we will briefly discuss the conditions to havea significantly depletion of the φ field. We already said6 IG. 5: Plot of F as a function of µ/ √ α . See the text for details. that condition (49) triggers the instability, however it isa necessary but not sufficient condition for a substantialdecay of the cloud. The missing condition is, of course,that the time t res during which the instability takes placeis enough to reach the exponential regime. It requires α ω (cid:126)p t res > . To estimate t res let’s assume a beam com-posed of φ particles whose length in the propagation di-rection is d . We need the produced particles to stay insidethe beam region for the parametric resonance to develop,otherwise there is no Bose enhancement. Therefore t res is the time that χ particles take to leave the beam region.As the χ particles have momenta pointing mainly to thesame direction as the beam propagates, t res should begiven by d/ (1 − v χ ) , where v χ is the velocity of χ parti-cles. This velocity is approximately − µ / , which leads to the following depletion condition d > µ αp . (63)Notice that our old condition for the resonance µ < α plus the additional condition d > / (2 p ) are enough tosatisfy requirement (63). As a summary, in this articlewe have taken the simple scalar toy model gφχ to studythe process φ → χ . We were interested in the particularcase of massless φ and massive χ . Although the one par-ticle analysis prohibits the decay by the requirement ofenergy-momentum conservation of asymptotic states, wefound that when φ is highly occupied, Bose enhancementeffects allows the decay as long as the energy density ofthe decaying particles exceeds the threshold defined inEq. (49). The small shift in the frequency of the pro-duced particles allows the energy conservation of eachsingle decay. We also computed the spontaneous decayrate finding that it is at most a half of the rate corre-sponding to the opposite case of massive φ and massless χ . 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