In my lecture notes, there is the following graphic:

enter image description here

With the 3D "bulk" configuration, there is clearly a $1/2$ power law, which I am able to explain by myself just by deriving the density of states for a free electron. I assume in this configuration, the "confines" for the electron are so large that the electron effectively feels no potential, and therefore can be treated as being unbound.

However, with the $2D, 1D$ and $0D$ cases, why is there a periodic behavior shown here? Clearly, there is a repeat pattern that increases linearly with amplitude (with the exception of what appears to be a sum of delta functions for the final case) where this doesn't appear for the $3D$ case.

What exactly is causing this behavior?

  • $\begingroup$ Confinement is quantising your energy levels. I know that in 2D since DOS is independent of E you have constant line. And the steps are due to quantisation in the third dimension. Maybe the extension to others is straightforwad $\endgroup$ Feb 1, 2020 at 16:56
  • $\begingroup$ "...why is there a periodic behavior shown here?" The behavior shown is not periodic. Periodic means that it repeats with a fixed period. $\endgroup$
    – hft
    Jul 8, 2022 at 21:26

1 Answer 1


Each of the jumps in the well and quantum wire densities comes from another transverse mode becomeing possible. The quantum wire wavefunctions are, for example, $\psi_{n,m,k}(x,y,z)=\phi_{n,m}(x,y)e^{ikz}$ where $\phi_{n,m}(x,y)$ is a standing wave in the transverse directions obeying $$ -(\hbar^2/2m^*)(\partial_x^2+\partial_y^2)\phi_n(x,y) = E_{n,m}\phi(x,y) $$ with boundary conditions that $\phi(x,y)$ is zero on side boundaries of the wire. Then
$$ -(\hbar^2/2m^*)\nabla^2 \psi_{n,m,k}=\{E_{n,m}+ (\hbar^2k^2/2m^*)\}\psi_{n,m,k}. $$ The $E_{n.m}$ are the energies at which the spike/jumps occur in your figure.

  • $\begingroup$ Is $m^*$ an effective/band mass? Also, can you elaborate how you derived the second equation from the first? $\endgroup$
    – sangstar
    Feb 1, 2020 at 17:21
  • $\begingroup$ The second equation is the full Schrodinger equation for free fermions in the wire with $m^*$ the band mass. The first eq. is simply the transverse momentum contribution to the energy. The $m,n$ integer mode numbers label the mode. If $\phi_{n.m}$ satisfies the first eq, then obviously $\psi_{n,m,k}$ satisfies the second with the continuous wavenumber $k$ giving the momentum along the wire. $\endgroup$
    – mike stone
    Feb 1, 2020 at 22:02
  • $\begingroup$ Ah, as the only energy is kinetic for the free fermions, so the total energy is the sum of transverse and perpendicular directions. It seems $E_{n,m}$ dominates when it is nontrivial (if that makes sense or I'm understanding it correctly). Why is this? $\endgroup$
    – sangstar
    Feb 1, 2020 at 22:08
  • $\begingroup$ Each time the energy gets bigger than a new $E_{n,m}$ the corresponding new conduction channel opens in the quantum wire waveguide. The density of states is the sum of the $D(E)$'s of all open channels, and since $D(E) =m^*/k$ in one dimension and diverges at small $k$, you get the infinite spikes in $D$ as each new channel opens up. $\endgroup$
    – mike stone
    Feb 1, 2020 at 22:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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