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For a particle trapped inside a rectangular box of side lengths $l_x$ $l_y$ and $l_z$, the energies are $E_{n_x,n_y,n_x}=\frac{\pi^2\hbar^2}{2m}(\frac{n_x^2}{l_x^2}+\frac{n_y^2}{l_y^2}+\frac{n_z^2}{l_z^2})$. It would appear as though if we take the limit as one of the lengths goes to zero, then the energy would blow up to infinity. But in this limit, the 3-D box becomes a 2-D rectangle, and the energies of a particle in a 2-D rectangle are well-defined and finite. This appears to be a contradiction. Do we simply neglect the excitations in the third dimension when we look at the 2-D case?

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  • $\begingroup$ In the flattened 3D box $l_z$ is getting smaller, but the associated standing-wave frequency is increasing. So the energy is increasing. In the 2D box there is no $l_z$ and no frequency associated with it. $\endgroup$ – John Duffield Apr 20 '15 at 18:50
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The three directions $x$, $y$ and $z$ are separable for the particle-in-a-box problem - the behaviour in each is independent of that in the others. Thus, each direction when separately considered only gives the contribution to the energy due to the limits of the box, or equivalently, the 'part' of the wavefunction, in that direction. In the case of $l_z \rightarrow 0$, only the contribution to the energy due to the '$z$ part' of the solution is infinite, and the rest contribute finite amounts (as you would expect by considering them as the solution to the corresponding 2-D rectangle problem). There is therefore no contradiction.

If you really want to extend a 2-D rectangle (say in $x, y$) to a 3-D box, the natural method of extension in this case is arguably to have no dependence of any physically relevant quantity (such as the potential) on $z$ i.e. $l_z \rightarrow \infty$.

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