By using time-dependent perturbation theory and some simplifying assumptions, we know that a transition for a charged particle $q$ from a quantum-mechanical state $m$ to $n$ is allowed if and only if \begin{align} \newcommand{\braket}[1]{\langle #1 \rangle} \braket{\psi_m|\boldsymbol{\mu}|\psi_n} \neq 0 \\ \sqrt{|\braket{\psi_m|\hat{\mu}_x|\psi_n}|^2 + |\braket{\psi_m|\hat{\mu}_y|\psi_n}|^2 + |\braket{\psi_m|\hat{\mu}_z|\psi_n}|^2} \neq 0 \tag{1} \end{align} Eq. (1) is written in Cartesian coordinates where the dipole moment operator has the form $\boldsymbol{\mu} = \mu_x\mathbf{i} + \mu_x\mathbf{j} + \mu_x\mathbf{k}$.
For instance, it is possible to prove that for the one-dimensional PIB, a transition from state $n_1 \to n_2$ is allowed when $n_2 - n_1$ is an odd number or $\Delta n = \pm1,\pm3,\pm5,...$. For example, if $n_1 = 1$ and $n_2 = 3$, then $n_2 - n_1 = 2$ is even and the transition is forbidden.
I am obtaining an absurd result for the three-dimensional PIB. Let us study the general transition of the charge $q$ from the three quantum numbers $(n_{x1},n_{y1},n_{z1}) \to (n_{x2},n_{y2},n_{z2})$. Eq. (1) shows that the three integrals must vanish in order to have a forbidden transition. We write down the first one and take advantage of the fact that this triple integral is separable in Cartesian coordinates. We get \begin{align} \newcommand{\braket}[1]{\langle #1 \rangle} \braket{\psi_{n_{x1},n_{y1},n_{z1}}|\hat{\mu}_x| \psi_{n_{x2},n_{y2},n_{z2}}} = &\sqrt{\frac{2}{l}} \int_0^l \sin\left(\frac{n_{x1}\pi x}{l}\right)(qx) \sin\left(\frac{n_{x2}\pi x}{l}\right)\mathrm{d}x \\ &\color{red}{\sqrt{\frac{2}{l}}\int_0^l\sin\left(\frac{n_{y1}\pi y}{l}\right) \sin\left(\frac{n_{y2}\pi y}{l}\right)\mathrm{d}y} \\ &\color{blue}{\sqrt{\frac{2}{l}} \int_0^l \sin\left(\frac{n_{z1}\pi z}{l}\right) \sin\left(\frac{n_{z2}\pi z}{l}\right)\mathrm{d}z } \tag{2} \end{align} We know that the eigenfunctions for the particle in a box form a complete set of orthogonal wave functions. Thus, if we want a nonzero integral, the red and blue terms force me to set $\color{red}{n_{y1} = n_{y2}}$ and $\color{blue}{n_{z1} = n_{z2}}$. In other words, the quantum numbers related to $\color{red}{y}$ and $\color{blue}{z}$ cannot change.
With some imagination and symmetry we can predict that the second integral in Eq. (1) will force me to set $n_{x1} = n_{x2}$ and $n_{z1} = n_{z2}$ (quantum numbers related to $x$ and $z$). By this point, no quantum number can change. Thus, the 3D-PIB cannot change its initial state (at least regarding electromagnetic radiation), which is ridiculous and I am making a mistake, but I cannot find why.
