# Hydrogen atom in superposition of energy eigenstates

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?

• 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. Jan 6, 2016 at 15:37
• Note that you should carefully distinguish between superpositions and mixed states, which are more complicated. Jan 6, 2016 at 15:37
• 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. Jan 6, 2016 at 16:48
• 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. Jan 6, 2016 at 17:15

• 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. Jan 6, 2016 at 18:46