Solving the Schroedinger equation for the H-atom (or any other system, say a particle in a box, or harmonic oscillator or anything), we obtain the energy eigenvalues are sharp with no spread. However, in reality, there must be a spread for the excited states which follows from uncertainty principle, and the spread of an energy level $\Delta E\sim\frac{\hbar}{\tau}$ where $\tau$ is the lifetime of the particle in that state. Why is this effect not captured in the calculation of energy levels by solving the Schroedinger equation?

Is there any other way to compute the spread other than using the uncertainty principle and match them?

  • $\begingroup$ When the atom is coupled to a radiation field, one must treat the the field and the atom as a single coupled system. The radiation field has a large degeneracy: many wave vectors for a single frequency, and thus so does the coupled system. Like any coupled system, the frequencies split, and with the large degeneracy, the splitting manifests as a broadening of the frequency. $\endgroup$ – garyp May 12 '16 at 11:59

In the Schrödinger treatment of the Hydrogen atom, the excited states are true eigenstates, that is, their lifetime is infinite. This means that, if the Hamiltonian is $$ H=\frac{P^2}{2m}+\frac{\alpha}{r} $$ then $\tau=\infty$ for all the energy levels. What this description misses is the interaction of electrons with the electromagnetic field, that is, there is a $$ j\cdot A $$ term missing in $H$. Without this term, the electron doesn't couple to photons, and therefore the atom cannot emit them and decay. If you include the coupling to the electromagnetic field, for example with perturbation theory, you'll see that $\tau^{-1}\neq 0$ but it is very small (which nicely agrees with experiments by the way).

Finally, the most correct treatment of the Hydrogen atom is with QFT, that is, the Dirac+Maxwell equations (for fields instead of wavefunctions). In this case, you can reproduce the Schrödinger results, with relativistic corrections and fine structure and check that only the ground state is a true eigenstate (and the excited states are quasi-bound states, i.e., they have a small imaginary energy, that is, a finite life-time).

For more details, see this or this.

By the way, $\Delta E\sim\hbar/\tau$ has nothing to do with the uncertainty principle; see e.g. What is $\Delta t$ in the time-energy uncertainty principle?.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.