Elementary Decay Process (Kouznetsov)

with corrected misprints and deduction recovered.

Introduction
The simplest quantum-mechanical model of the interaction of an atom with the electromagnetic field field, which still allows the spontaneous decay, was suggested a century ago and then widely applied; in particular, to idealized 3-level systems. The original deduction by Weisscopf and Wigner of the rate of decay (and the resulting spectrum of the emitted photons) is rather euristic than rigorous. The Schroedinger equation for the quantum amplitude with excited atom and that with photons allows the solution in terms of generalized functions. Although the resulting amplitude appear to be smooth funcitons; the product of two generalized functions appear in the deduction. At the time of Wigner, there was no mathematical tool to treat well the product of generalized function. Therefore, the problem was considered at the discrete momentum space; sums appear instead of integrals; this makes the deduction complicate and difficult to follow. In this paper, the simplified formalism is suggested.

Here I use the algebraic properties of generalized functions signum and delta; this makes the deduction simple and compact. Such compact notations are important for the efficient consideration of more complicated cases of few levels of the atom and/or deformed density of bosonic states.

In the simple approximation, the atom and field are described by the Hamiltonian (1) $$ H_0= \int a^{\dagger}_{k}a_{k}\!~k\!~{\rm d} k +E u^{\dagger} u {H0} $$ and their interaction is described by the Hamiltonian (2) $$ H_1= \eta \int \left( a^{\dagger}_{k} u+a_{k} u^{\dagger} \right){\rm d} k ,~{H1} $$ where $$~u^{\dagger}~$$ is operator of creation of excitation of atom and $$~a_{k}^{\dagger}~$$ is operator of creation of photon of the mode with wave-number $$~k~$$, coupled to atom; dagger means the Hermitian conjugation; so, $$~u~$$ is operator of relaxation of excitation. In the free space, the mode of a photon may be just a field emitted by a classical dipole. However, in the special case with non-unifom space, for example, an atom trapped in vicinity of a single-mode nanofiber , the mode may have more complicated form. For the process of the elementary decay, the only commutation properties are important: (3) $$ uu^{\dagger}\!+\!u^{\dagger} u=1~, a_k a^{\dagger}_p-a^{\dagger}_p a_k=\delta(k\!-\!p)~, {commu} $$ and all other commutators $$~a~ $$, $$~a^{\dagger}~$$, $$~u~$$, $$~u^{\dagger}~$$ are equal to zero; in addition, $$~u^{2}=0 $$, which means that we have only one atom and it cannot be double-excited.

Hamiltonian $$~H = H_0 + H_1~$$ describes a two-level atom interacting with continuum of field modes: each of these modes can be treated as a harmonic oscillator. %The specific value of energy of the transition is not important at this stage of the deduction. Therefore I use the system of units, where $$\hbar\!=\!c\!=\!1 $$; in these notations, I make no difference between the frequency, wave-number and energy of a photon.

Parameter $$~\eta~$$ appears as coupling constant. It can be calculated in terms of the dipole (or quadrupole) momentum of an atom, or in terms of the $$~A~$$ Einstein coefficient. In general, the coupling constant depends on wavenumber (it is proportional to cube of wave-number), but in the approximation of rotating wave, its dependence on wave-number is not essential, the only value at resonant frequency is important. Therefore, I treat it as a constant and write it outside of the integral. It happens, that $$~\Gamma=2\pi\eta^2~$$ is the decay rate; so, the coupling constant can be recovered also from the empirical data.

Schroedinger equation and single-photon states
The single-quantum solution of the Schroedinger equation (4) $$ {\rm i} ~{\rm d} \Psi/{\rm d} t = H \Psi ~ {Sch} $$ can be written in the following form: (5) $$ \Psi~=~h(t)~{\rm e}^{-{\rm i} Et}\!~ u^{\dagger} |0\rangle +~\int f_{q}(t)\!~{\rm e}^{\!-{\rm i}(q)t}\!~ a_{q}^{\dagger} |0\rangle\!~ {\rm d} q ,{L} $$

where $$~|0\rangle~$$ is vacuum, i.e. $$~a_{k}|0\rangle= u|0\rangle=0~$$; $$~u^{\dagger}|0\rangle ~$$ is state with excited atom, and $$~a^{\dagger}_{q}|0\rangle ~$$ is state with single photon or waveumber (or frequency, or energy) $$~q~$$; $$~h~$$, and $$~f~$$ are C-number functions of real argument(s).

Substituting (5) into (4), we obtain the system (6) $$ {\rm i} \frac{{\rm d} h(t)}{\rm dt} = \int \eta\!~ f_k(t)\!~ {\rm e}^{-{\rm i}(k-E)t}{\rm d} k ~{dhdt}$$

(7) $$ {\rm i} \frac{{\rm d} f_{k}(t)}{{\rm d} t} = \eta\!~ h(t)\!~ {\rm e}^{{\rm i}(k-E)t} .{dfdt} $$

I assume, $$~h(0)=1~$$, $$~f_q(0)=0~$$, which corresponds to the decay from the excited state into the space without photons. In the next chapter, the solution of equation (\ref{dfdt})(7) is constructed in a straight-forward manner. (The same algorithm of solution can be applied to more complicated systems; but here we show it at the simplest example.)

In order to simplify the equations, introduce the function $$~\tilde f ~$$ with shifted subscript: (8) $$ \tilde f_{k-E}(t)=f_k(t).~{tilde} $$ This allows to eliminate parameter $$~E~$$ from the equation, writing (9) $$ {\rm i} \frac{{\rm d} h(t)}{\rm d t} = \int \eta\!~ \tilde f_q(t)\!~ {\rm e}^{-{\rm i}qt}\!~{\rm d} q ,~{dhdttilde} $$

(10) $$ {\rm i} \frac{{\rm d} \tilde f_{q}(t)}{ \rm d t} = \eta\!~ h(t)\!~ {\rm e}^{{\rm i}qt} ,~{dfdttilde} $$ In the approximation of narrow spectral line, we can extend the integration to the infinite range. Then, the equation allows a beautiful solution considered in the next section.

solution
In this section, I present the analytical solution of system ref(dhdttilde), ref(dfdttilde), assuming the infinite range of the integration. As I use the properties of algebra of generalized functions , no other assumption are necessary, and no limiting transitions (as in ) are required.

Equation ref(dfdttilde) can be integrated as follows: (11) $$ \tilde f_{q}(t)=-{\rm i}\eta \int_{0}^{t}h(\tau) {\rm e}^{iq\tau} {\rm d} \tau ~,~{int} $$ I re-write it as follows: (12) $$ \tilde f_{q}(t)=-{\rm i}\eta \int_{0}^{\infty} h(\tau) {\rm e}^{iq\tau} \theta(t-\tau){\rm d} \tau ~,{inta} $$ where $$~\theta ~$$ is unit-step function.

I treat this the unit step function as (13) $$ \theta(x)=\frac{1}{2}+\frac{1}{2}\varepsilon(x) ~,{theta} $$

where $$~\varepsilon ~$$ is generalized signum function, anticommuting with function delta:

(14) $$ \varepsilon(x)\delta(x)+\delta(x)\varepsilon(x)=0 . {ed}{de} $$

However, in ref(inta), there is no difference, use we generalized signum function $$~\varepsilon~ $$ by or conventional signum function, but the difference appears as we change the order of integration and deal with generalized functions.

Substitution of ref(inta)(12) into ref(dhdttilde)(9) gives: (15) $$ {\rm d} h(t)/{\rm d} t = -\eta^{2}\int_{-\infty}^{\infty}\! {\rm d} k~ \int_{0}^{\infty}\! {\rm d} \tau ~ \left( \frac{1}{2} + \frac{1}{2}\varepsilon(t\!-\!\tau)\right) h(\tau) {\rm e}^{ik(t-\tau)} .~{iint} $$ Taking $$~x=\tau\!-\! t ~$$ as a new variable of integration, we rewrite this equation as follows: (16) $$ {\rm d} h(t)/{\rm d} t = -\eta^{2}\int_{-\infty}^{\infty}\! {\rm d} k~ \int_{-t}^{\infty}\! {\rm d} \tau ~ \left( \frac{1}{2} + \frac{1}{2}\varepsilon(x)\right) h(t\!+\!x) {\rm e}^{-ikx} .{iint1} $$ While $$~h(x)~$$ is supposed to be smooth function, it commutes with $$~\varepsilon(x) ~$$, so, (17) $$ \frac{{\rm d} h(t)}{{\rm d} t} = -\frac{\eta^{2}}{4}\int_{-\infty}^{\infty}\! {\rm d} k~ \int_{-t}^{\infty}\! {\rm d} x ~ \Big(1+\varepsilon(x)\Big) h(t\!+\!x) {\rm e}^{-ikx} ~-~ \frac{\eta^{2}}{4}\int_{-\infty}^{\infty}\! {\rm d} k~ \int_{-t}^{\infty}\! {\rm d} x ~ h(t\!+\!x) {\rm e}^{-ikx} \Big(1+\varepsilon(x)\Big) .{iint2} $$

Changing of order of integrations allows to integrate with respect to $$~k~$$, which leads to the product of generalized functions. These functions can be multiplied and form the associative algebra , but they have no need to commute; their order becomes important: (18) $$ \frac{{\rm d} h(t)}{{\rm d} t} = -\frac{\eta^{2}}{4} \int_{-t}^{\infty}\! {\rm d} x ~ \Big(1+\varepsilon(x)\Big)\!~h(t\!+\!x)\!~2\pi\delta(x) ~-~ \frac{\eta^{2}}{4} \int_{-t}^{\infty}\! {\rm d} x ~ \!~h(t\!+\!x)\!~ 2\pi\delta(x) \Big(1+\varepsilon(x)\Big) .{iint3} $$ Using the associativity of generalized funcitons, I find (19) $$ \frac{{\rm d} h(t)}{{\rm d} t} = -\pi \eta^{2}\!\int_{-t}^{\infty}\!\! {\rm d} x \!~h(t\!+\!x)\!~ \delta(x) -\pi\eta^2 \!\int_{-t}^{\infty}\!\! {\rm d} x ~ h(t\!+\!x)\!~\Big(\!\delta(x)\varepsilon(x)\!+\!\varepsilon(x)\delta(x)\!\Big) .{iint4} $$ Due to ref(de)(14), the second term in ref(iint4)(19) is equal to zero. The first term can be integrated without problem due the the $$~\delta~ $$-function, giving (20) $$ \frac{{\rm d} h(t)}{{\rm d} t} =-\pi \eta^{2} h(t). $$ Taking into account the initial condition $$~h(0)=1~ $$, the integration gives (21) $$ h(t)={\rm e}^{-\pi\eta^{2}t}. {h} $$ Then, using (\ref{int}), we recover function $$~\tilde f ~$$: (22) $$ \tilde f_{q}(t)=-{\rm i} \!~\eta\int_0^{t} {\rm e}^{-\pi\eta^2\tau+{\rm i} q\tau} {\rm d} \tau = -{\rm i}\!~\eta\frac{{\rm e}^{-\pi\eta^2t+{\rm i} qt}-1}{-\pi\eta^{2}+{\rm i} q} .(ftqt) $$ In such a way, for the infinite integration in ref(dhdttilde), the exponential solution for $$~h(t)~$$ appears as an analytic and exact result, not as a special and limiting case of consideration of some discrete momentum space with an additional physical assumptions.

Conclusions
From the solution ref(h), ref(ftqt), we get the distribution of the amplitude of photons (23) $$ f_{k}(\infty)=\frac{{\rm i} ~ \eta}{-\pi\eta^{2}+{\rm i}(k-E)} {fk} $$ and the corresponding Lorentian spectrum (24) $$ |f_{k}(\infty)|^{2}=\frac{\eta^{2}}{(\pi\eta^{2})^{2}+(k-E)^{2}} ,{ffk} $$ which is also well-known result.

The new in this paper is the formalism which can be applied to various quantum systems with deformed density of bosonic states. In particular, this formalism allows to re-write the deduction for the decay of 3-level systems in a rigorous and readable manner. Such an upgrade will be matter for the separate publication.