What happens to an electron in a molecule once it has absorbed a photon and transitioned? Say we have a molecule capable of absorbing a photon somewhere in the UV/Vis region of the electromagnetic spectrum. Once this electron has transitioned to a higher energy state, does it just stay there? Or does it transition back down to its ground state in spontaneous emission? If the electron doesn't undergo spontaneous emission, what exactly does give rise to spontaneous emission?
 A: In the general case an excited level will spontaneously fall back to a lower level, or even cascade back if there exists an appropriate level in between. Each quantum mechanical level has a width  which is given by its probability for absorbing an electron if it is empty or emitting if it is full .
Studying how lasers are made, which need an overpopulated level which needs a trigger to go to a lower energy,  is enlightening:

The achievement of a significant population inversion in atomic or molecular energy states is a precondition for laser action. Electrons will normally reside in the lowest available energy state. They can be elevated to excited states by absorption, but no significant collection of electrons can be accumulated by absorption alone since both spontaneous emission and stimulated emission will bring them back down.
A population inversion cannot be achieved with just two levels because the probability for absorption and for spontaneous emission is exactly the same, as shown by Einstein and expressed in the Einstein A and B coefficients. The lifetime of a typical excited state is about 10-8 seconds, so in practical terms, the electrons drop back down by photon emission about as fast as you can pump them up to the upper level.

Laser beams need careful construction as you can see in this link.
A: The electronic states of a molecule are eigenfunctions of the time independent Schrodinger equation (with a few approximations like the Born-Oppenheimer approximation). This means those states are time independent so a ground state will never rise to an excited state and an excited state will never decay back to a ground state.
However in the presence of a photon we have also the oscillating electric field of the photon, and this adds a new term to the Schrodinger equation so the ground and first excited states are no longer eigenfunctions and they can mix. So the combined system of the ground state plus a photon has a proportion of the excited state mixed in, and this means there is a probability that when we observe the molecule we'll find it is in the excited state.
We can calculate this probability using perturbation theory, and the equation is called Fermi's golden rule. Doing this calculation tells us how likely the photon is to promote the atom to the excited state, and we'll find that this depends on the photon energy and the probability is high only when the photon energy matches the energy difference between the states.
This calculation is just the same when done in reverse. If we start with the excited state then we can calculate the probability for it to evolve back into the ground state plus a photon, and this gives us the lifetime of the excited state. This decay process is the spontaneous emission you mentioned in your question. It happens because the excited state contains a proportion of the mixed ground+photon state.
A: In the excited states, electrons are not stayed constant. They fluctuate constantly; however, this fluctuation does not enough to come back to the ground state. There are outside effects which interacts our excited electron. Like cosmic background rays. When cosmic background waves touch the excited states, electrons come back to its ground state. This situation is also answer the meaning of energy time uncertainty principle $\Delta E \Delta t \ge \hbar/2$. Since $\Delta E$ cannot be zero since there is a fluctuation, so $\Delta t$ is not infinite. But in the ground state where fluctuation not exists, electron can stay with infinite $\Delta t$
A: There is no experimental means whereby you can cause an atom to "absorb a photon...transitioning to a a higher energy state." What you can do is shine a light on atoms, and measure the intensity of scattered light. That is the ONLY experiment that has ever been done which addresses the subject matter of this question. The notion that the scattered light (which you can measure) is the result of some of those atoms "absorbing photons", "transitioning to higher energy states", and "transitioning (back) to lower energy states (via) spontaneous emission"...those are all theoretical constructs flowing from certain interpretations of quantum theory. They have never been directly observed, and cannot be directly observed.
There is another way to analyze this experiment without resorting to the notion of photons. You can consider the electron in the atom to be a harmonic oscillator, whose k (spring) constant is readily calculated from Schroedinger's equation. You can convert the incident light intensity to an oscillating electric field. You can apply classical mechanics to the oscillating field and the charged harmonic oscillator to calculate the magnitude of oscillation. And you can then use classical antenna theory to calculate the resultant scattering.
If you do all that (and do it correctly) you will get the correct answer for the scattered radiation. You will get the same result that you observe by experiment. You will get the right answer. You do not need to talk about absorbing photons and spontaneous emission.
By the way, if I am wrong about the notion that you get the right answer, I am certainly NOT wrong about the fact that you can do the calculation which I outlined above. Before anyone downvotes my answer, they ought to actually do the calculation for oh, lets say the hydrogen atom, and show me that they get a different answer than the experiment.
You can see some of my calculations on this through links included in my blogpost, "There Are No Pea-Shooters For Photons". 
