# Why do the wavefunction wavenumbers for a box and for a periodic potential differ?

Consider a particle in a one-dimensional infinite square well potential of width $$L$$. Solving the Time Independent Schrödinger equation, and considering the boundary conditions, gives wavefunction solutions with wavenumbers: $$k = \frac{n\pi}{L}$$ For $$n=1,2,3,...$$

Solving a system of a periodic potential with infinite potential walls gives wavefunction solutions with wavenumbers: $$k=\frac{2\pi n}{L}$$ for $$n= \pm1, \pm2, \pm3, ...$$

This means that the energy eigenvalues for a periodic potential are four times larger than the ones for a box potential.

1. Why are only positive wavenumbers allowed for the wavefunction of a particle in a box?
2. Intuitively, why do particles in periodic potentials have higher wavenumbers and energies?

1. It is not that negative wavenumbers are not allowed. All the negative wavenumbers correspond to the same solution to the Schroedinger equation for the particle in a box, as the positive wavenumbers, with the only difference being an overall sign. This can be thought of as a phase and hence does not have a physical meaning. So, both positive and negative wavenumber solutions describe the same system. Tossing them corresponds to avoiding double counting... The only time this matters is when we express a function as a linear combination of eigenfunctions of the "particle in a box" problem. Recall that linear combination means expressing an arbitrary function $$\phi(x)$$ as $$\phi(x)=\sum_{n=1}^{\infty}c_n\sin(\frac{n\pi x}{L})$$ where $$L$$ is the "size" of the box. The set of functions $$\{\sin(\frac{n\pi x}{L}),n=1,2,3,...\}$$ are a complete set! So, taking the corresponding negative values of the wavenumbers does not make any sense...
• I think it can be applied to a periodic potential in the very special case of the potential being infinite at $x=L, 2L, 3L,...$. You see, the BCs of the periodic wave function impose the condition $e^{ikL}=1$, which in general does not guarantee that the wavefunction behaves the same for $-n$ as it would for $n$. But, if you further impose $\psi(x=0)=\psi(x=L)=0$, then you are left with sine functions comprising your set of all possible solutions. Then, making the substitution $2n\rightarrow-2n$ does not yield a linearly independent solution! Jul 14, 2022 at 13:43