What does the subscript of atomic orbital mean?

Question

As everyone knows, the atomic orbital can be classified as $s, p_z, p_x, p_y, d_{z^2},d_{xz},d_{yz},d_{xy},d_{x^2-y^2}$ and so on. I want to know the meaning of $z^2,x^2-y^2$ and so on. Maybe this is a fundamental question, but I'm not familiar with chemistry.

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I have some ideas about this:

1. For example, we consider the $p$ orbital: $p_z$ is symmetric about $z$ axis; $p_x$ is symmetric about $x$ axis; $p_y$ is symmetric about $y$ axis. But what's the meaning of $z^2$ or $x^2 -y^2$?
2. We consider the symmetry of atomic orbital with group theory. The change of orbital under the symmetry operation is the same as the orbital subscript.

Answer 1

For each azimutal quantum number $l$, the magnetic quantum number $m$ ranges from $-l$ to $+l$. Those give you the number of atomic orbitals "subsripts" you should obtain. Those are then expressed using spherical harmonics. If you look at the table of spherical harmonics at the $l=1$, you will find that they contain the cartesian axes as $x-y$, $z$ and $x+y$. Therefore it is advisable to form the linear combination out of them, so you get $x$, $y$, $z$. If you go to higher angular momentum, things get more complicated, but there is some tradition what linear combinations to use.

To answer your second question, all this formulas are not some inevitable result you obtain by juggling the numbers. Contrary, it is carefully crafted in a way that the formulas are so simple.

Answer 2

Late, but OP seemed unhappy with any of the answers and this is a great question I had for years as a chemistry student, suffering what I felt were vague explanations when I just wanted a precise formula. I wanted to see that subscript as part of the polynomial in the wavefunction. After all that time, here's my answer:

The orbitals have the same symmetry as the function $\psi(x,y,z) = $ "subscript".

p-orbitals $p_x$, $p_y$, and $p_z$ have the same symmetry as $\psi = x$, $\psi = y$, and $\psi = z$. They have nodes where that function is 0, and they flip sign where that function does, just dampened by a decreasing exponential $e^{-r}$, rather than going off to infinity. Basically $\psi = xe^{-r}$ (Not exactly the solution but very close).

d-orbitals have same symmetry as $\psi = xy$, $\psi = yz$, $\psi = xz$, $\psi = x^2-y²$, ... and the tricky one, $z²$ is short for $z² - (x² + y²)$, which gives you conical nodes and a sign flip with the horizontal radius $\sqrt{x^2 + y²} > z$.

The same holds true for the f-orbitals, with again one of them being a short-hand. Slightly annoying, I know, but once someone TELLS you, it's ok.

Hope that helps.