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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?

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  • $\begingroup$ 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

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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.

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  • $\begingroup$ 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

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