What is the magnetic quantum number for $p_x$, $p_y$, $p_z$? After tons of research, I've come to the conclusion that the $p_x$ and $p_y$ orbitals can have either $m_\ell=+1$ or $-1$, but $p_z$ is $0$.
But my professor said that we can attribute any $m_\ell$ value from $- 1,0,+1$ to $p_x.p_y$ and $p_z$.
So who's wrong?
 A: The $\text{p}_x$, $\text{p}_y$, $\text{p}_z$ orbitals are "real" $\ell=1$ wavefunctions.
The actual angular wavefunctions (i.e., simultaneous eigenfunctions of $\hat{{L}}^2$ and $\hat{L}_z$) are the spherical harmonics:
\begin{align}
\mathrm{Y}_{1+} &= -\sqrt{\frac{3}{8\pi}} \sin\theta \hspace{0.1em} e^{i\phi}
= -\frac{1}{\sqrt{2}} \big( \text{p}_x(\theta,\phi) + i\text{p}_y(\theta,\phi) \big)
\,,\\
\mathrm{Y}_{1-} &= \phantom{-}\sqrt{\frac{3}{8\pi}} \sin\theta \hspace{0.1em} e^{i\phi}
= \phantom{0}\frac{1}{\sqrt{2}} \big( \text{p}_x(\theta,\phi) - i\text{p}_y(\theta,\phi) \big)
\,,\\
\mathrm{Y}_{0} &= \phantom{-}\sqrt{\frac{3}{4\pi}} \cos\theta \quad\,
= \text{p}_z(\theta,\phi)
\,.
\end{align}
These are complex-valued functions, so some people (especially chemists) prefer to visualize them by splitting their real and imaginary parts, and that is the "real" orbitals $\text{p}_x$, $\text{p}_y$, $\text{p}_z$.
This messes up the diagonalization:
\begin{align}
\text{p}_x(\theta,\phi) &= \frac{1}{\sqrt{2}} \big( {-\mathrm{Y}_{1+}(\theta,\phi)} + \mathrm{Y}_{1-}(\theta,\phi) \big)
\,,\\
\text{p}_y(\theta,\phi) &= \frac{i}{\sqrt{2}} \big( {\mathrm{Y}_{1+}(\theta,\phi)} + \mathrm{Y}_{1-}(\theta,\phi) \big)
\,,\\
\text{p}_z(\theta,\phi) &= \mathrm{Y}_{10}(\theta,\phi)
\,.
\end{align}
From this expression, it is clear that $\text{p}_x$ and $\text{p}_y$ orbitals are $\frac{1}{2}:\frac{1}{2}$ superpositions of $m=+1$ and $m=-1$, and $\text{p}_z$ orbital is the $m=0$ eigenstate.
$\text{p}_x$ and $\text{p}_y$ are orthogonal, as they mixes $m=+1$ and $m=-1$ with different relative phase factors.
(Just to be clear—of course, we are speaking about $\ell=1$ states.)
To reiterate, $\text{p}_x$ and $\text{p}_y$ are not eigenstates of $\hat{L}_z$. It is kind of "grammatically wrong" to say that "$\text{p}_x$ has $m=+1$" for instance. $\text{p}_x$ is not an eigenstate of $\hat{L}_z$ and the eigenvalue $m$ is not defined at all. The best we can say is that $\text{p}_x$ is a superposition of $m=+1$ and $m=-1$ eigenstates.
