There are $N$ non interacting electrons in a potential well: \begin{align} H&= -{1 \over 2 } \nabla^2 + U(x,y,z) \\ U(x,y,z)&={1\over2}\omega^2z^2 \; \mbox{for} \; (x,y) \in [0,L]\times [0,L]; \; U(x,y,z)=\infty \; \mbox{otherwise} \end{align} I need to calculate the single particle density of states (in the thermodynamic limit) in some particular conditions, but I don't understand how to work the things out. The hamiltonian is separable, so the single particle energy eigenstates are (in atomic units): \begin{align} \epsilon_{\vec{n}} = {1\over2}\left(\pi \over L \right)^2 \left({n_x}^2+{n_y}^2\right) + \omega \left(n_z + {1 \over 2} \right) \end{align} The conditions are the following:

  1. $\epsilon < {1 \over 2}\omega$;
  2. ${1 \over 2}\omega < \epsilon < {3 \over 2}\omega$;
  3. ${3 \over 2}\omega < \epsilon < {5 \over 2}\omega$;

For 1) I tried by saying that, for that condition, there are no energy eigenstates, since the ground state has higher energy than $\omega /2$. Then $g(\epsilon)=0$.

In the second case I don't know what to do, since I know nothing about the value of $\omega$ or $L$, so I don't know how many states there are in that range of frequencies. Is there a way to calculate a density of states "per direction" so that I could, for example, calculate the 2D DOS in $x$ and $y$ and a 1D DOS in $z$ and then put them together in some way?

Thank you for your time.

  • $\begingroup$ You only need a 2D DOS, for both (2) and (3). For (2) notice that $n_z=0$, since $n_z =1$ already means $\epsilon_{\bf n} > 3/2$. For (3) you only have $n_z =0$ and $n_z = 1$, since $n_z =2$ gives $\epsilon_{\bf n} > 5/2$. In all cases the 2D DOS is of the form $n^2_x + n^2_y \le R^2$, which you should be able to handle in the thermodynamic limit. $\endgroup$ – udrv Jun 14 '16 at 20:34
  • $\begingroup$ Thank you for your comment. I don't understand why in x,y we have the continuum limit while in z we don't. Is it because the factor 1/L appears only for x and y? $\endgroup$ – PPeg Jun 14 '16 at 20:43
  • $\begingroup$ Yes, the thermodynamic limit usually means $L$ large relative to some relevant scale. In this case it will be something of the form $\omega L^2>>1$ (restore all constants for proper units). $\endgroup$ – udrv Jun 14 '16 at 20:54

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