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I'm trying to determine the degeneracy of states given by $g(\epsilon)=g_{0} \epsilon$ for a system that is trapped in a quite specific potential.

In two dimensions, the particle has a potential as in the infinite square well, while on the third dimension, the potential behaves as if a harmonic oscillator.

The energy for a single particle is given by

$$\epsilon = \frac{h^{2}}{8mL^{2}}(n_{x}^{2} + n_{y}^{2}) + hf \biggl(n_{z} - \frac{1}{2}\biggr)$$

I'm having some difficulty with this problem because of the broken symmetry. Is there any analytic way to determine the degeneracy as a function of energy?

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Is that second equals sign supposed to be a plus sign? –  David Z May 13 '11 at 2:08
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That they're bosons is irrelevant. The statistics affects the occupation of the states, not the degeneracy of the states. –  Ramashalanka May 13 '11 at 4:03
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Also, you give the degeneracy, $g(\epsilon)=g_0\epsilon$, as a density of states. Are you looking for an answer in the continuous energy limit (not valid for small $\epsilon$)? For small $\epsilon$ the density of states is a sum of delta functions (the first corresponding to $n_x=n_y=n_z=1$) with degeneracy 1. –  Ramashalanka May 13 '11 at 4:09
    
Are you looking for multi-particle states? Is there an interaction potential of some form? Why can't you just solve for the single particle states as below, and then symmetrize your total wave function? And if you're not doing this, why is it important that the particle is a boson? –  Jerry Schirmer May 13 '11 at 16:17
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2 Answers

up vote 1 down vote accepted

The Hamiltonian is

$$ H=-\frac{\hbar^2}{2m}\vec{\nabla}^2 + V(\vec{r}) = \sum_{j=1}^3 H_{x^j}, $$

with potential

$$ V(\vec{r}) = V_x(x) + V_y(y) + V_z(z), $$

where

$$ V_x(x)= \left\{\begin{array}{ccc} 0 &\mathrm{for}& |x|<\frac{L}{2} \\ \infty &\mathrm{for}& |x|\geq \frac{L}{2}\end{array}\right\}, $$ $$ V_y(y)= \left\{\begin{array}{ccc} 0 &\mathrm{for}& |y|<\frac{L}{2} \\ \infty &\mathrm{for}& |y|\geq \frac{L}{2}\end{array}\right\}, $$ $$ V_z(z)= \frac{m\omega^2}{2}z^2.$$

Note that

$$H_{x^j}=-\frac{\hbar^2}{2m}\partial^2_{x^j} + V_{x^j}(x^j) $$

only depends on one coordinate $x^j$, where $j=1,2,3$. Thus we can use separation of variables, as suggested by dmckee, and first solve the time-independent Schroedinger equation for three well-known one-dimensional problems. It is not hard to see that the 3D energy-levels $E_{\vec{n}}$ becomes a sum of the 1D energy-levels,

$$E_{\vec{n}} = E_{xy}(n^2_x+n^2_y)+E_z(n_z+\frac{1}{2}),$$

where the states are labeled by three integers $\vec{n}=(n_x,n_y,n_z)$ with

$$n_x\geq 1,\qquad n_y\geq 1,\qquad \mathrm{and}\qquad n_z\geq 0, $$

and

$$ E_{xy}:= \frac{1}{8m} \left(\frac{h}{L}\right)^2 \qquad \mathrm{and}\qquad E_z :=\hbar\omega=hf$$

are two constants.

Note that there is one state per unit-volume in $\vec{n}$ space. For large enough energy $E$, where $E\gg E_{xy}$ and $E\gg E_z$, the zero-point energy $\frac{1}{2}E_z$ and some other discrete features become irrelevant, and we can approximate the total number $N(E_{\vec{n}}\leq E)$ of states $\vec{n}$ with energy $E_{\vec{n}}\leq E$ by the volume of (a quarter of) a paraboloid with finite height in the following way.

$$ N(E_{\vec{n}}\leq E) \approx\mathrm{Volume}\left\{(n_x,n_y,n_z) \in \mathbb{R}^3_+\mid E_{xy}(n^2_x+n^2_y)+E_z n_z \leq E \right\} $$ $$= \int_0^{\frac{E}{E_z}} dn_z \ \mathrm{Area}\left\{(n_x,n_y) \in \mathbb{R}^2_+\mid n^2_x+n^2_y\leq \frac{E - E_z n_z}{E_{xy}}\right\}$$ $$= \int_0^{\frac{E}{E_z}} dn_z \ \frac{\pi}{4} \frac{E - E_z n_z}{E_{xy}} = \frac{\pi}{4E_{xy}} \left[E n_z - E_z \frac{n_z^2}{2} \right]^{n_z=\frac{E}{E_z}}_{n_z=0} = \frac{\pi}{8} \frac{E^2}{E_{xy}E_z}, $$ where we used that the area of a quarter disc with radius $r$ is $\frac{1}{4}\pi r^2$. Thus the density of states $g(E)$ at energy $E$ is then approximately

$$g(E) = \frac{dN(E_{\vec{n}}\leq E)}{dE} \approx \frac{\pi}{4} \frac{E}{E_{xy}E_z}=g_0 E, $$

where

$$ g_0 = \frac{\pi}{4E_{xy}E_z} =\frac{2\pi m L^2}{h^3f} $$

is a constant.

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This seems to scream out for a Separation of Variables approach. In addition to those links you can find it in any book on math methods in physics or any reasonably advanced book on differential equations.

The short--short version is you will write you solution in parts that depend only on the independent bits

$$ \Psi(x,y,z) = W(x,y) Z(z) $$

and after applying the partial derivatives you will find that you have two much simpler equations to work with.

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