# What does the subscript of atomic orbital mean?

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.

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.
• I have deleted my post, if I get correct relevance I will repost. But, you can notice that d_xy is symmetric about x+y axis. – Immortal Player Mar 25 '14 at 0:46
• @GODPARTICLE Do you mean x+y axis is x=y axis? – Ben Mar 25 '14 at 0:53
• Yes. The vector x+y will be $45^0$ with respect to both x and y axis. – Immortal Player Mar 25 '14 at 5:11
• But it does not work for all （╯_╰） – Ben Mar 25 '14 at 11:28
• It does for all xy, yz, zx involved subscripts. – Immortal Player Mar 25 '14 at 11:31

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.
• 1)What do you mean "you will find that they contain the cartesian axes as $x−y$, $z$ and $x+y$."? 2)I want know why the subscripts are written as $z$, $z^2$, $x^2-y^2$ and so on respectively. 3)You say "it is carefully crafted in a way that the formulas are so simple.", and I'm not clear about "the formulas". Can you explain it more specificly? – Ben Mar 24 '14 at 9:57
• I know how to combine spherical to obtain $p_x$, $p_y$...But I do not know why choose x, y, z and so on as subscripts. – Ben Mar 24 '14 at 14:36
• The orientation of the orbitals. For $p$, one is oriented in the x direction one is oriented in the y direction and one is one the z direction. – 1110101001 Mar 25 '14 at 4:17
• Bach: We chose the orientation for immense convenience. It is great, if you expand some equation and magically 95% of the terms disappear just because you have chosen good coordinate system. And the labels $p_x$,... just show, which axes and in which power appear in the spherical harmonic expression. – ssavec Mar 25 '14 at 8:03