# How quickly does an atomic electron decay?

I've looked in lots of books on atomic physics, but I've not been able to find an answer to:

Question: Take a hydrogen in empty space whose electron is excited. How long will it take for the electron to emit a photon and move to a lower energy level?

For simplicity assume that the electron begins in the first excited state $$|1\rangle$$, and that the only other state is the ground state $$|0\rangle$$. So the question becomes about the precise form of the wavefunction $$\psi(t)$$ which begins at $$|1\rangle$$. I guess that it shows oscillatory behavior for small $$t$$ and looks something like $$|0\rangle+e^{-t}|1\rangle$$ for large $$t$$; is this correct? I am very interested in seeing a derivation.

Also,

Question: What is the experimental evidence that $$|\psi(t)|$$ takes the form of the answer to the above question?

• Regarding your first question... The mean lifetime of the $2p$ state is only 1.6 nanoseconds. But the mean lifetime of the $2s$ state is much longer: about 1/8 of a second. – G. Smith Jul 20 at 22:18
• This type of problem is typically approached using time-dependent perturbation theory that leads to a formula called Fermi's golden rule. Time-resolved measurement of a property of hydrogen gas (such as reflectivity or transmissivity) after exciting the gas by a femtosecond laser pulse would be one way to probe this timescale. This method is referred to as a pump-probe spectroscopy – kevinkayaks Jul 20 at 22:46
• @kevin You don't need femtosecond pulses to probe a nanosecond lifetime. A run of the mill picosecond laser will work just fine. – Emilio Pisanty Jul 21 at 0:25
• have a look at the hydrogen wavefunctions for hydrogen hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydsch.html . In general all lines have a width., consistent with measurements even to the fine structure constants. – anna v Jul 21 at 5:15

Take a hydrogen in empty space whose electron is excited. How long will it take for the electron to emit a photon and move to a lower energy level?

The decay is exponential (at least, with respect to the population in the excited state). If you want to find the lifetimes, the simplest way is to use the NIST Atomic Spectra Database: go to Lines and enter H I for neutral hydrogen; the transition rates are the $$A_{ki}$$, which for the $$2p→2s$$ transition in hydrogen equals $$A_{ki} = 6.2649\times 10^8\:\rm s^{-1}$$, corresponding to a lifetime of about $$1.59\:\rm ns$$.

I guess that it shows oscillatory behavior for small $$t$$ and looks something like $$|0\rangle+e^{-t}|1\rangle$$ for large $$t$$; is this correct?

Not quite. If you want to do this in full rigour, then you cannot just quantize the atom $$-$$ you also need to quantize the electromagnetic field. In rough lines, if you start off in an excited state $$|e\rangle$$ that will decay to a ground state $$|g\rangle$$, the total quantum state is an entangled state between the atom and the field, where the latter starts off in the vacuum $$|0\rangle$$ and ends up in a one-photon wavepacket $$|{1;\chi(t)}\rangle$$, which may (depending on the configuration) end up depending on time. Thus, the full quantum state reads something like $$|\Psi(t)\rangle = e^{-\gamma t}|e\rangle|0\rangle + |g\rangle|{1;\chi(t)}\rangle.$$