Dear jaskaran,

Go through "Quantum optics" by  Marlan O. Scully and M. suhail Zubairy(in short Scully and Zubairy)....page no 206 "weisskopf-wigner theory".

Spontaneous emission is done by vacuum field modes, which are perturbing your system (especially excited state) and consequently uncertainty will be there in energy and in time. According to Heisenberg uncertainty relation, this uncertainty in energy will give rise to life time of the excited state.

Dear Jaskaran, although it goes against the general theoretical frameworks , you should read the keyring theory of particle mechanics on this page

http://www.particlemechanics.com/discussion_theory_solids_liquids_gases_key_ring_atom.htm

also see the youtube videos on this theory for explanations to some of your questions.

regards Gerro

http://www.particlemechanics.com/discussion_theory_solids_liquids_gases_key_ring_atom.htm

Spontaneous emission is the process by which a quantum system such as an atom, molecule, nanocrystal or nucleus in an excited state undergoes a transition to a state with a lower energy (e.g., the ground state) and emits quanta of energy. Light or luminescence from an atom is a fundamental process that plays an essential role in many phenomena in nature and forms the basis of many applications, such as fluorescent tubes, older television screens (cathode ray tubes), plasma display panels, lasers, and light emitting diodes. Lasers start by spontaneous emission, and then normal continuous operation works by stimulated emission.

If a light source ('the atom') is in the excited state with energy E_2, it may spontaneously decay to a lower lying level (e.g., the ground state) with energy E_1, releasing the difference in energy between the two states as a photon. The photon will have 

E_2 - E_1 = Planck constant x Angular frequency

References

1. R. Loudon, The Quantum Theory of Light, 3rd ed. (Oxford University Press Inc., New York, 2001).

Rate of spontaneous emission 

The rate of spontaneous emission (i.e., the radiative rate) can be described by Fermi's golden rule.[2] The rate of emission depends on two factors: an 'atomic part', which describes the internal structure of the light source and a 'field part', which describes the density of electromagnetic modes of the environment. The atomic part describes the strength of a transition between two states in terms of transition moments. In a homogeneous medium, such as free space, the rate of spontaneous emission in the dipole approximation is given in following Ref.:

The rate of spontaneous emission (and thus the lifetime of the excited level) is determined both by the properties of the atom and by the mode structure of the surrounding medium. For an atom in free space (or in a homogeneous medium), transition cross sections can be used for calculating the rate of spontaneous emission events (→ radiative lifetime). Typical upper-state lifetimes of atoms are a few nanoseconds if there are allowed transitions to lower levels; much longer values can occur for forbidden transitions.

2. B. Henderson and G. Imbusch, Optical Spectroscopy of Inorganic Solids (Clarendon Press, Oxford, UK, 1989).

There are a good number of papers related to experimental investigations on spontaneous emissions

Experimental detection of photons emitted during inhibited

spontaneous emission by David Branning,  Alan L. Migdall, and Paul G. Kwiat

2. Decoherence by spontaneous emission in atomic-momentum transfer experiments by Peter Goetsch and Robert Graham, The American Physical Society, 54(6), 1996 

.

That is all well and good, but my question still stands: why does it have to occur??

I don't want to know 'how' it occurs, but rather 'why' it occurs??

It is simple in my point of view. In the nature everything prefer to be in the ground state (the lower energy level) than if  an atom  is not in the ground state it will decay, and then it emits the extra energy as a photon, what we know as spontaneous emission, because we do not need to stimulate it. The system always try to stay in the lower energy state, and it does it by itself.

I hope helped.

First, due to the energy conservation, an atom in the excited state has exactly the same energy as the atom in the ground state plus the emitted photon. In this sense, the system does not at all go to the lower-energy state. Second, the energy conservation is the direct and immediate consequence of the homodeneity of time. The latter is not that obvious however, but so far there has not been any indication of the violation of the energy conservation of an ISOLATED system. Third, if an excited state of a system is its eigenstate, the system will remain in this state forever by definition. Otherwise, it is not an eigenstate but rather a superposition of several true eigenstates. In this case, the system evolves between these true eigenstates and has finite probability to be found in the state different from the initial one.

If you perform an experiment on an "isolated" atom, you actually put it in contact with your device and the atom stops being isolated. The eigenstates of the system "atom plus device" are not the same as just the eigenstates of the atom times the eigenstates of the device. So, the state of your atom will evolve. I'm not experienced in vacuum fluctuations but in my opinion it is just another way of saying that the system is not exactly isolated from the environment. Finally, I would repeat that the system remains in its true eigenstate forever, irrespective of whether the state is excited or ground. There are no any intrinsic reasons to de-excite.

In each particular case you should consider whether your excited state is actually an approximately eigenstate (like radioactive nuclei).

If you take into consideration interaction between the atom and the electromagnetic field then the state

|A> = |an excited state of atom, no photons>

is not an eigenstate of the system. It is a so called quasi-stationary state. Without interaction there is continuum of states

|B>-s = |the ground state of atom, a photon>

with energies equal or very close to the energy of |A>. If you take into consideration interaction, then the non-stationary state   |A> evolves into superposition of |A> and |B>-s. The amplitude of the state |A> decreases, and the amplitudes of the states |B>-s increase. This is a spontaneous decay.

Essentially consider all the modes of the radiation field as quantum oscillators with energy levels (n+1/2)hf. In the visible, all these modes are most often empty so n=0. However, there is the half photon per mode on average. This  means that due to fluctuations, there sometimes is a photon and sometimes not. When there is a photon in this mode, this can stimulate the atom to decay by stimulated emission. Now at some level, this is not formally correct, but if you do the calculation you end up with the correct spontaneous emission rate. So like all physics, it is correct within approximations.

I have to agree with Vladimir. Let me put his statement in a bit different form: a photon with a certain energy is the electromagntic field (actually, a wave), which lasts infinitely in time and space (let's forget relativistic effects for a while). An atom interacting with this electromagnetic field exhibits persistent Rabi oscillations between the ground and excited states absorbing and emitting photons periodically. If it is in the pure excited state at the moment, it will be in the pure ground state after a half of the Rabi period. The Rabi frequency is proportional to the amplitude of the electromagnetic field of the photon. Very roughly, the atom is stimulated to emit a photon by the very emitten photon itself or by the photon it absorbed a Rabi period ago to get to the excited state, which is the same.

Depending of the ordering of operators, it may be interpreted as solely induced by vacuum fluctuations or solely induced by radiation reaction or by a combination of them. 

In my opinion fluctuations have only a role to trigger the process. Once triggered, radiation reaction causes it to complete. 

http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.31.95

http://www.opticsinfobase.org/josab/abstract.cfm?uri=josab-2-3-447

Sorry the first kill was incorrect. 

http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.31.958

Spontaneous emission is a decay process just like radioactive decay. Like the latter since we we don't know the cause, we say it's a random process. Decays from different levels have different half lives. Decays with long half-lives are said to be "inhibited." Whereas with radioactive decay , the internal workings of the nucleus are completely unknown, we do have some knowledge of the dynamics of electrons around a nucleus. This can lead to finding reasons for some "inhibitions."

so spontaneous emission is like a decay. then why dont we define a half life for it?? instead we define the lifetime of a state. Is it just standard notation or does have some physical significance??

The state half life partitions into a number of emission half lives depending on the final state . Unless you detect the emitted photon, you don't know the final state.

In standard quantum electrodynamics, vacuum fluctuations (equivalent to half a photon in each mode) cause half of the spontaneous emission and retarded self action causes the other half.  The latter gives rise to the time asymmetry involved.

Alternatively in quantised Wheeler-Feynman absorber theory, the spontaneous emission is caused by the advanced reaction from the absorber.

B.H. Bransden & C.J. Joachain in "Quantum Mechanics" (second edition) p 515 :"The process of spontaneous emission takes place in the absence of any external field and can be looked upon as the quantum analogue of the effect predicted by classical electromagnetism that any accelerated charged particle emits radiation"

Classical analogy quoted by Joris is misleading in my opinion. An electron in a 1s (groud) state of a hydrogen atom is an accelerated charged particle from the classical point of view. But it does not emit. On the contrary, an electron in an excited state (also an accelerated charged particle) does. What is the analogy then?

Quantum theory of "accelerated charges" accommodates for a stable ground state, whereas from a classical point of view this is not the case. So the classical analogue obviously has its limits and is only a way of looking upon this quantum effect. Hopefully it can help in trying to grasp the underlying physical phenomenon.

Alexander's analogyt is the motion of the wave packet namely the time dependence of the expectation value of the positing that corresponds to classical motion. In this sense since all the stationary states of the

I don't know why the first parts of the message appears distorted and second half of my messages has disappeared completely. 

Continued. (sorry for inconvenience)

Alexander's description of relationship between classical and quantum is misleading because according to Ehrenfest theorem it is the motion of the wave packet namely the time dependence of the expectation value of position that corresponds to classical motion. 

The square of the electronic wave function and therefore expectation value of position of the electron are time independent in all the electronic energy eigenstates of the atom including the ground state. Thus there is no acceleration in ground state. 

We can understand the spontaneous emission in the following way semi classically:

Assume that the atom is in an energy eigenstate at t= t0. Because of the fluctuations in the environment, at time t = t0 + Δt the Hamiltonian has changed slightly and the atom is not in an energy eigenstate and therefore not stationary anymore . Assume that we have a small contribution from a lower state. If we calculate the time dependence of , we discover that it oscillates with a frequency that is exactly the difference of the eigenfrequences of two states. Thus the situation is similar to an oscillating dipole that radiates with this frequency. Since energy dissipates to radiation field, the energy of the atom must decrease. This can happen only if the contribution of the lower energy state increases and the contribution of upper energy state decreases. The process starts slowly but while the contribution from lower energy state increases, at the same time the amplitude of the oscillation increases. It becomes a maximum when contributions of bot states are equal to each other. After this point the process slows down giving an exponential tail. 

This model was first developed by Fermi very early. It was then improved by E.T. Jayne's and coworkers in the sixties known as semiclassical theory which triggered controversies regarding whether quantization of electromagnetic field is necessary to understand spontaneous emission. 

In my opinion the vacuum photons have the effect that the process doesn't start start slowly as in semiclassical model but it starts as exponential decay right at the beginning. 

@Atilla, this looks to me as an interesting point! Enlightening :-) so to speak. Thanks!

Dear jaskaran,

Go through "Quantum optics" by  Marlan O. Scully and M. suhail Zubairy(in short Scully and Zubairy)....page no 206 "weisskopf-wigner theory".

Spontaneous emission is done by vacuum field modes, which are perturbing your system (especially excited state) and consequently uncertainty will be there in energy and in time. According to Heisenberg uncertainty relation, this uncertainty in energy will give rise to life time of the excited state.

Thank you all for your inputs. I read about Jaynes-Cummings model and Scully too. It was a great help indeed, but I am somehow stuck with the idea of a statistical description of spontaneous emission like radioactive decay. Is it possible??

Go to Quantum statistical properties of radiation by W. L. Louisell John Wiley 1973.

You wil find all the tools to understand in the more efficient way all what you need to know to go on to think about your problem.

 Dear Jaskaran,

As suggested by D. T. Pegg above, the emission is not spontaneous, but is actuated by the ticklings from the advanced waves from future that impinge on the atom in the excited state. Any transition is a transaction (see Cramer's transactional interpretation).

The decays of other systems are also not spontaneous, nor is spontaneous symmetry breaking of particle physics. It is against the inquisitive spirit of science to take a process as spontaneous (i. e without a reason) or random.

Secondly, the atom is not an isolated system, even when it is isolated from all else! Because it is fundamentally a result of coulomb or electromagnetic interaction between the positively charged nucleus and the negatively charged electrons. The EM field is already there, built into the "isolated" system called the atom.

It was good to go through the posts on this thread.

Consider neutron decay as perhaps the simplest example of "spontaneous" decay. I'm putting spontaneous in quotes because I intend to let this process interpret what this word means.  We have this process n -> p + beta(-) + anti-neutrino with a half-life of  611s. We know of no external effects that have any measurable effect on this half-life so we are left with internal processes involving the motions of 3 quarks and many gluons. Calculations have indeed been done on this internal motion but without getting into detail, it's obvious that this half-life would entail some huge number of these motions. I suggest then that we say spontaneous to simply cover our ignorance of whatever might be the cause.. I also think this usage extends to the decay of excited electronic states where it seems we can be more inventive of possible influences. But as far as I know, none really  remove the inherent statistical aspect of spontaneous decay.

I fully agree with your observations James. The statistical/probabilistic nature remains.

For the relavance of radiation reaction not only in spontaneous emisson but aldı in measurement problem

https://www.researchgate.net/publication/2190816_Source_Field_Effects_and_Wave_Function_Collapse

Atilla Gürel Well explained.