While solving a potential well problem we get scattering states and bound states (if exist). Number of the bound states we get depends on the potential profile. What I want to ask is, what is the importance of number of bound states we are getting? For example for Dirac delta potential well we have only one bound state. What more information we can derive by this knowledge?

  • $\begingroup$ Comment to the question (v2): From the potential profile/shape and the Schrödinger eq., we can in principle find the bound state spectrum, in particular the number of bound states. Conversely, knowledge of the bound state spectrum can be used to gain knowledge of the potential profile/shape. See e.g. this Phys.SE post and links therein for further details and semiclassical formulas. $\endgroup$ – Qmechanic Apr 16 '15 at 12:29
  • $\begingroup$ Do you want to say that they serve as signature of the potential shape? Is that all? $\endgroup$ – seeking_infinity Apr 17 '15 at 6:09

Modern electronic devices like quantum well lasers, resonant tunneling diodes, quantum cascade lasers and detectors heavily rely on the spatial and energetic position of such bound states. This defines their transport and optical properties.

On a separate notice: any well, no matter how shallow or narrow, has at least one bound state.

  • $\begingroup$ I want a more basic answer. I know these things but I want to know how "number" of bound states ply a role? $\endgroup$ – seeking_infinity Apr 16 '15 at 9:23
  • $\begingroup$ For certain applications that involve optical transitions between condiction band states, you need two or more bound states. Under that respect their number matters. I admit, this is still a very applied answer. $\endgroup$ – engineer Apr 16 '15 at 9:30
  • $\begingroup$ "any well, no matter how shallow or narrow, has at least one bound state." This is partially wrong. There are potential wells in 3d, which have no bound states. $\endgroup$ – Gonenc Jun 16 '15 at 18:55

Generally speaking, bound states in a system give rise to resonant behaviour. The number of states defines the number of different resonances that can be observed.

For example, consider the absorption of photons by an electron trapped in a quantum well:

If the well only contains one bound state then it could absorb any photon with any energy greater than the binding energy of the state (i.e., a transition from state $|1\rangle$ to anywhere in the continuum), so a broad spectrum would be observed.

If the well contains two bound states with energies $E_1$ and $E_2$, then an incident photon with energy $\hbar\omega_{21} = E_2 - E_1$ would resonate with the $|2\rangle\to|1\rangle$ transition, and this would give rise to an additional strong, and narrow peak in the absorption spectrum.

If there were three bound states, then there could be three resonant peaks in the spectrum corresponding to $\hbar\omega_{32} = E_3 - E_2$, $\hbar\omega_{31}=E_3 - E_1$ and $\hbar\omega_{21} = E_2 - E_1$.

So, the number of bound states determines the number of possible resonant interactions. Note, though, that not all of these transitions are necessarily permitted... symmetry rules can often forbid light from interacting with certain pairs of states.


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.