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Take a particle of mass $m$ trapped in an infinite potential well between $0$ and $a$. The energy spectrum and the wave functions are

$$\displaystyle E_n = \frac{\hbar^2\pi^2}{2ma^2} n^2$$

$$\displaystyle \psi_n(x) = \sqrt{\frac{2}{a}} \sin\frac{n\pi x}{a}$$

Now if the same particle is trapped in 3D rectangle of sides $a$, $b$, and $c$ we have

$$\displaystyle E_n = \frac{\hbar^2\pi^2}{2m} \left(\frac{n_x^2}{a^2} + \frac{n_y^2}{b^2} + \frac{n_z^2}{c^2}\right)$$

$$\displaystyle \psi_n(x,y,z) = \sqrt{\frac{8}{abc}} \sin\frac{n_x\pi x}{a} \sin\frac{n_y\pi y}{b} \sin\frac{n_z\pi z}{c}$$

Since the 3D case is more general than the 1D case, in principle one can take the limit to get the 1D case. Taking b and c tends to infinity will reproduce the 1D spectrum but it will kill the wave function.

How to take the limit on $\psi_n(x,y,z)$ to get $\psi_n(x)$ ?

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1 Answer

The right limit is obviously $b,c\to 0$, not infinity. That's the limit in which the box becomes very thin in two directions, i.e. a line interval.

In that limit, one finds out that the $n_y^2$ and $n_z^2$ terms have a huge, infinite effect on the energy. So only if those terms have the minimal possible value, $$ n_y=n_z=1, $$ we get states of acceptable energy. All other states, i.e. those with higher values of $n_y,n_z$, have infinitely higher energies and decouple in the limit.

In that limit, $\psi_n$ factorizes. The dependence on $y$ and $z$ becomes unievrsal, $$ \psi_{n_x,n_y,n_z}^{3D} (x,y,z) = \psi^{1D}_{n\equiv n_x} (x) \cdot\sqrt{\frac{2}{b}}\sin \frac{\pi y}b\cdot \sqrt{\frac{2}{c}}\sin \frac{\pi z}c.$$ The last two sine-factors are just normalization constants and the Schrödinger equation for $\psi^{3D}$ reduces to that for $\psi^{1D}$, with $E^{1D}\equiv E^{3D}-E_0$ where $E_0$ is the constant shift of the energy we have to subtract (which changes nothing physical).

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