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Suppose a single hydrogen atom is in a superposition of energy eigenstates:

$$ \psi = \frac{1}{\sqrt{2}}\psi_{100} + \frac{1}{\sqrt{2}}\psi_{200} \,.$$

Then energy will be $E = \frac{1}{2}(13.6\,\mathrm{eV}) + \frac{1}{2}(3.4\,\mathrm{eV}) = 8.5\,\mathrm{eV}$.

But there is no spectral line at $8.5\,\mathrm{eV}$. Why not?

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  • $\begingroup$ Why would there be a spectral line at 8.5eV? When you perform the overlap integral to determine transition probabilities, it will, by necessity of the wavefunctions, resolve to transitions between specific states, not mixed states. $\endgroup$
    – Jon Custer
    Jan 6, 2016 at 15:37
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    $\begingroup$ Note that you should carefully distinguish between superpositions and mixed states, which are more complicated. $\endgroup$
    – Emilio Pisanty
    Jan 6, 2016 at 15:37
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    $\begingroup$ We have MathJax running on the site, so you can write math in a LaTeX-math-mode-alike language. You can find a brief explanation of this and our other markup in the help center. In any case, I've edited this post to improve the markup. $\endgroup$
    – dmckee --- ex-moderator kitten
    Jan 6, 2016 at 16:48
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    $\begingroup$ If there are two people in a room and their ages are 10 and 30, their average age is 20, but no one in the room is age 20. The same thing is happening here. What you calculated is the expectation value of the energy, i.e. the average you'll get if you could measure the energy over and over a large number of times. $\endgroup$
    – elifino
    Jan 6, 2016 at 17:15

1 Answer 1

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8.5 eV is not the observed energy. That is only the expected value of the observed energy. What you have instead is a 50% chance of observing 13.6eV and a 50% chance of observing 3.4eV. But you will never observe 8.5eV; you will only observe one or the other.

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  • $\begingroup$ The atomic system before emitting a photon has a total Energy of -8.5 eV. After emitting a photon it is in the ground state with energy -13.6 eV and the emitted photon has an energy of the respective Lyman alpha line of 10.2 eV. This means that the total energy is not conserved. How can the atom emit a photon without violating energy conservation. $\endgroup$
    – Jannick
    Jan 6, 2016 at 18:01
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    $\begingroup$ @Jannick, no, only the expected energy is -8.5eV. You don't know what the "actual" energy is. $\endgroup$
    – Paul
    Jan 6, 2016 at 18:07
  • $\begingroup$ Change energy to expected energy in my post. Shouldn't the expected energy be conserved then? $\endgroup$
    – Jannick
    Jan 6, 2016 at 18:13
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    $\begingroup$ That's not quite accurate - the energy expectation value of a closed system is constant. In the case of photon emission from this superposition, the system goes from |ground state atom⟩ + |excited atom⟩ to |ground state atom⟩$\otimes$(|no photon present⟩ + |photon present⟩), with the same expectation value for the energy. $\endgroup$
    – Emilio Pisanty
    Jan 6, 2016 at 18:46
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    $\begingroup$ @EmilioPisanty, fair enough. An observation of the energy will see only one or the other, but then it is no longer a closed system. $\endgroup$
    – Paul
    Jan 6, 2016 at 18:57

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