Is energy conserved in  decay of hydrogen atom in superposed state? This looks like a paradox.
Let's say we have an hydrogen atom. Superposition of states could be possible for electrons. But if an electron is in a superposition, I guess it could decay into a lower state by emitting a photon.
Different initial and final energy expectance would make possible photons of arbitrary energy levels. But that is not the case.
So that, what is wrong? Is energy conservation violated?
 A: Quantum system can emit only photons with energy equal (within the uncertainty) to the difference between two energy states.
Even if the atom is in a superposition of energy states
$$
\left|\Psi\right> = C_0 \left|0\right> + C_1 \left|1\right> + C_2 \left|2\right> + \ldots \qquad (1)
$$
with average energy somewhere between the levels, it can emit only certain set of photons: $E_1 - E_0$, $E_2 - E_0$, $E_2 - E_1$ etc.
Emission of a photon is an act of measurement since the energy of the emitted particle contains information about the atom. If the energy of the photon is $E_2 - E_1$ then the energy of the electron in the atom is $E_1$ - the energy of the final state of the transition. The next photon emitted by this atom will have energy equal to $E_1 - E_0$ for sure.
If one observe photons emitted by an ensemble of atoms in state (1) he will see $E_1 - E_0$ photons with probability $\left|C_1\right|^2$, any of $E_2 - E_0$ and $E_2 - E_1$ with probability $\left|C_2\right|^2$ and so on.
The total energy emitted by the system while it is coming to ground state is equal to average energy of state (1) multiplied by the number of atoms in the ensemble.
Energy conservation is not violated.
The same is true for mixed states for which the probability of certain photon is determined by the density matrix of the system.
A: To start with a hydrogen atom has one electron. This electron will be in a specific energy level, it will not be mixed as you seem to assume. It is possible that it is not at the lowest energy level because it might have interacted before our observation and the electron was kicked to a higher energy level, it will still be a unique level;  it will decay to the lowest energy level emitting a photon of appropriate energy.
Energy conservation is not in question .
A: I am always amazed at the number of people who will say that a hydrogen atom cannot exist in a mixed state. This claim has no basis in the equations of quantum mechanics. When we solve a differential equation, e.g. the Schroedinger equation for a hydrogen atom, we may choose the technique of separation of variables to give us a set of basis states, which we can then use to express arbitrary states by combining those basis states in linear superpositions. There is no theoretical or experimental grounds for maintaining that an actual hydrogen atom can only exist in one of those basis states.
As for the other part of the question, there is no experimental means to show that light can only be emitted from atoms in discrete quantities. Copenhagen wants us to talk about atoms in discrete states making quantum leaps from one state to another, and a whole machinery of mathematics has been worked out to analyze physical systems using this picture. But that doesn't mean it's the only viable picture; in fact, there is no experimental way to distinguish this picture from an alternage paradigm whereby system of atoms exists in any mixture of states, radiating and absorbing energy smoothly according to Maxwell's equations.
EDIT: There is what amounts to a mathematical proof of what I am saying here in this earlier stackexchange discussion: Distinguishability in Quantum Ensembles
