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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?

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  • $\begingroup$ A word of caution: Mixed state and superposition should not be confused, cf. en.wikipedia.org/wiki/… $\endgroup$
    – Qmechanic ♦
    Dec 18, 2011 at 18:13

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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.

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  • $\begingroup$ Then, it looks that energy conservation could be violated in an individual atom experiment. Am I right? $\endgroup$
    – bitozoid
    Dec 20, 2011 at 20:03
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    $\begingroup$ @bitozoid, the energy of single atom in superposed state is not determined. There is nothing to be conserved and nothing to be violated. We can measure only the average energy, and the averaging has sense only for ensemble. $\endgroup$
    – Maksim Zholudev
    Dec 20, 2011 at 20:11
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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 .

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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

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    $\begingroup$ This is really really wrong. Firstly the question is about one hydrogen atom, not an ensemble of hydrogen atoms. Secondly there are garden variety laboratory experiments in physics courses chemed.chem.purdue.edu/genchem/topicreview/bp/ch6/bohr.html which demonstrate the existence of the discrete spectra. $\endgroup$
    – anna v
    Dec 18, 2011 at 18:04
  • $\begingroup$ The spectrum is exactly the same in either paradigm. The discretenesss of the spectrum has nothing to do with the supposed discreteness in the emission/absorption processes. $\endgroup$
    – Marty Green
    Dec 18, 2011 at 23:12
  • $\begingroup$ Marty, could you explain more what you mean by "the discretenesss of the spectrum has nothing to do with the supposed discreteness in the emission/absorption processes"? $\endgroup$
    – bitozoid
    Dec 20, 2011 at 20:05
  • $\begingroup$ The spectrum is discrete because the energy level transitions are only driven by exact frequencies. But that has nothing to do with the theory that once a transition is started it has to go all the way. I elaborate on these topics in my blog: this article marty-green.blogspot.com/2011/10/… shows how atoms behave like tiny classical antennas, and this article marty-green.blogspot.com/2011/12/… refrences a very good discussion on this site regarding alternative paradigms. $\endgroup$
    – Marty Green
    Dec 20, 2011 at 20:42
  • $\begingroup$ For a mathematical demonstration that you cannot distinguish a collection of pure eigenstates from a system in mixed superpositions, see this earlier stackexchange discussion: physics.stackexchange.com/questions/8123/… $\endgroup$
    – Marty Green
    Dec 21, 2011 at 11:33

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