# How are names for atomic orbitals with high $l$ generated?

I'm looking at Table of Real Spherical Harmonics and trying to understand how the names for the real orbitals are generated. Note that I'm a physicist. In chemistry parlance, I'm trying to come up with an algorithm to generate the names like $$p_x, p_y, p_z, d_{yz}, d_{x^2-y^2}, \ldots, f_{y(3x^2-y^2)}, \ldots$$

I know that the letters go like $$s, p, d, f, g, h, \ldots$$ but I don't know how to generate the subscripts. It is clear that the subscript is somehow extracted by looking at the polynomial expansion of the real spherical harmonic and dropping some terms for concision, but I can't figure out the exact rules.

For example, I'm not sure why $$Y_{3, -3} = f_{y(3x^2-y^2)} \propto (3x^2-y^2)y$$ but $$Y_{4, -3} = g_{zy^3} \propto (3x^2-y^2)yz$$

Direct questions:

• Since the Cartesian expansions are so similar why aren't the subscripts more similar?
• How is the order of terms in the subscript determined? Why do we have $$Y_{3, -1} = f_{yz^2}$$ but $$Y_{4, -3}= g_{zy^3}$$?
• What is a reference that tabulates these things beyond $$l=4$$?

Side notes: Because of this trickiness of notation I tend to prefer the $$Y_{l, m}$$ notation because it is unambiguous and easily generalizable to high orbitals. Unfortunately it has the downside that it only makes sense if you're familiar with the complex $$Y_l^m$$ orbitals and how the real orbitals are the real and imaginary parts of those complex orbitals. You do also lose some of the nice visual intuition from $$p_x, p_y, p_z$$, but I argue this intuition is sort of a fool's dream because it basically breaks down by the $$d$$ orbitals when you see something like $$d_{z^2}$$ or $$d_{x^2-y^2}$$.

Some more context: I'm interested in this because I am trying to make large tables with visualizations for the complex and real orbitals going out to high $$n$$, for example, at least $$n=6$$. Currently I've made tables and labelled them using the $$Y_{lm}$$ convention described here, however, I'm getting pushback and being asked to label instead using the atomic orbital form. Unfortunately this is difficult because (1) I can't just put the formulation into code (2) I don't want to hardcode each name in and (3) even if I did want to hardcode in the atomic orbital names I wouldn't know how beyond what's listed on that Wikipedia page.

• My suspicion is that it's entirely arbitrary and down to the whims of whoever is tabulating these... There's a possible argument for $d_{z^2}$ in that the "correct" term, $2z^2 - x^2 - y^2$, is equal to $3z^2 - r^2$ and we just drop the spherically symmetric part of this ($r^2$) to end up with something that's $\sim z^2$. (We're already dropping the powers of $r$ in the denominator, and that's probably because they are spherically symmetric.) I don't know how much this applies to the higher orbitals, maybe you could find a pattern. Commented Feb 18, 2022 at 18:36
• @orthocresol That's my suspicion too. The tip about ignoring the spherically symmetric part does look like it might be helpful. I'll do some more looking with that in mind. I'll add a edit with some more context about why I'm curious about this. Commented Feb 18, 2022 at 20:21

Here's a stab at what is going on. A note upfront: The approach I lay out here only gives the sames answers up to the $$f$$ orbitals as shown at Wikipedia and The Orbitron. It is just the best I could come up with. I would still greatly appreciate any other answers that can further resolve the issue.

Following the quantum mechanics convention on wikipedia, the complex spherical harmonics are given by

\begin{align} Y_l^m(\theta, \phi) =& (-1)^m \sqrt{\frac{(2l + 1)}{4\pi}\frac{(l-m)!}{(l+m)!}}P_l^m(\cos\theta) e^{im\phi}\\ =& N_{lm}P_l^m(\cos(\theta))e^{im\phi} \end{align}

Where $$N_{lm}$$ is a scalar normalization factor and $$P_l^m$$ is an Associated Legendgre Polynomial. These polynomials have a closed form

\begin{align} P_l^m(x) =& (-1)^m 2^l (1-x^2)^{m/2} \sum_{k=m}^l \frac{k!}{(k-m)!} \binom{l}{k}\binom{\frac{l+k-1}{2}}{l} x^{k-m}\\ =& M_{lm} (1-x^2)^{m/2} G_{l-m}(x)\\ =& M_{lm}K_{lm} (1-x^2)^{m/2} x^{l-m} H_{l-m-1} \end{align} Where $$M_{lm}$$ is a normalization constant, $$G_{l-m}$$ is a polynomial of order $$l-m$$ and $$G_{l-m} = K_{lm}(x^{l-m} + H_{l-m-1}(x))$$ Where $$K_{lm}$$ is a constant and $$H_{l-m-1}$$ is a polynomial of order $$l-m-1$$.

If $$x = \cos(\theta)$$ then $$(1-x^2)^{m/2} = (\sin(\theta))^m$$. Let $$C_{lm} = N_{lm}M_{lm}K_{lm}$$ We then have

$$Y_l^m(\theta, \phi) = C_{lm} e^{im\phi}(\sin(\theta))^m\left((\cos(\theta))^{l-m} + H_{l-m-1}(\cos\theta)\right)$$

We now re-express in cartesian components. For that we have \begin{align} e^{i\phi} =& \frac{x+i y}{\sqrt{x^2 + y^2}}\\ \sin(\theta) =& \frac{\sqrt{x^2 + y^2}}{\sqrt{x^2+y^2+z^2}} = \frac{\sqrt{x^2 + y^2}}{r}\\ \cos(\theta) =& \frac{z}{\sqrt{x^2 + y^2 + z^2}} = \frac{z}{r} \end{align}

Importantly, we can see that $$e^{i\phi}\sin(\theta) = (x+iy)/r$$. With this we get \begin{align} Y_l^m(x, y, z) =& C_{lm} \frac{(x+iy)^m}{r^m}\left(\frac{z}{r^{l-m}} + H_{l-m-1}\left(\frac{z}{r}\right)\right)\\ =& \frac{C_{lm}}{r^l}(x+iy)^m\left(z^{l-m} + r^{l-m}H_{l-m-1}\left(z/r\right)\right) \end{align}

From this complex spherical harmonic we can extract two real spherical harmonics. These will be related to (the signs of the actual harmonics may differ from the expressions below because of various conventions)

\begin{align} &\frac{C_{lm}}{r^l}\text{Re}\left\{(x+iy)^m\right\}\left(z^{l-m} + r^{l-m}H_{l-m-1}\left(z/r\right)\right)\\ &\frac{C_{lm}}{r^l}\text{Im}\left\{(x+iy)^m\right\}\left(z^{l-m} + r^{l-m}H_{l-m-1}\left(z/r\right)\right)\\ \end{align}

These are the expressions from which extract the subscript for the atomic orbital name.

Focusing on $$(x+iy)^m$$. For every term in this polynomial the exponents for $$x$$ and $$y$$ add up to $$m$$. $$\text{Re/Im}\left\{(x+iy)^m\right\}$$ will in general include terms with both $$x$$ and $$y$$ but there is one term (or zero terms) that has the biggest exponent for $$x$$ and one term (or zero terms) that has the biggest exponent for $$y$$.

The procedure for coming up with the subscript is as follows.

• $$z^{l-m}$$ will appear in the subscript. This means we ignore all of $$H_{l-m-1}(z/r)$$.
• Take $$\text{Re/Im}\left\{(x+iy)^m\right\}$$ depending on which orbital you are considering. Find the term with the biggest $$x$$ exponent and the term with the biggest $$y$$ exponent and add these two terms together. You will have something like $$Ax^ay^b + B x^cy^d$$. If one of these terms is missing then one of $$A$$ or $$B$$ is zero.
• Now take $$Ax^ay^b + B x^c y^d$$ and factor out any possible scalars and factors of $$x$$ and $$y$$. You will now have something like $$A x^a y^b(x^c + B y^c)$$ or $$A x^a y^b(Bx^c + y^c)$$ (these are not the same $$A, B, a, b, c$$ as just before).
• Strip off $$A$$ from your final expression so you now have $$x^a y^b (B x^c + y^c)$$. This expression will appear in the subscript. There are two parts here, the factored out part $$x^a y^b$$ and the remaining prat $$(Bx^c + y^c)$$ or $$(x^c + By^c)$$
• We now put together the three parts.
• First put in the factored out part like $$x^a y^b$$ with $$x$$ preceding $$y$$.
• Then put the $$z^{l-m}$$ part.
• Finally put the remaining part $$(B x^c + y^c)$$ or $$(x^c + B y^c)$$.
• This should result in a subscript $$x^a y^b z^{l-m} (B x^c + y^c)$$ or $$x^a y^b z^{l-m} (x^c + B y^c)$$

A few notes:

First, and importantly, this scheme does not give the same answer as Wikipedia for two orbitals listed on that page.

• Wikipedia says $$Y_{4, -3} = g_{zy^3} \propto (3x^2-y^2) yz$$. I would say this should be $$g_{yz(3x^2-y^2)}$$.
• Similarly, Wikipedia says $$Y_{4, 3} = g_{zx^3} \propto (x^2-3y^2)xz$$. I would say this should be $$g_{xz(x^2-3y^2)}$$.
• My ordering also differs from that on Wikipeda. But honestly, i don't think they were following a pattern for the order on Wikipedia.

Wikipedia uses the same notation as The Orbitron, and I wonder if Wikipedia got their notation from this webpage. The Orbitron goes out to $$i$$ orbitals and I see similar discrepencies between the scheme used on that page and my approach for $$g$$ and $$h$$ orbitals. The author stops quoting abbreviated polynomials for $$i$$ orbitals for the relevant $$m$$ quantum numbers so I can't compare past that. I've reached out to the author of the Orbitron for further comment and I'm also trying to reach out to Wikipedia folks.

One advantage of my approach over whatever is going on on Wikipedia and the Orbitron is that the $$x, y$$ part of the coefficient is the same for orbitals with the same $$m$$ quantum number. They only differ in the factor of $$z$$.

Note that $$a + b + c + l-m = l$$, so all the exponents always add up to $$l$$ which is a nice spherical harmonic like result.

Finally, the issue remains of determining the $$a$$, $$b$$, $$c$$ and $$B$$ factors arising from $$\text{Re/Im}\left\{(x+iy)^m\right\}$$ alone. A computer could work these out without too much trouble but it would be nice to have closed form expressions for them. It should be possible to work it out using the binomial theorem.

Finally an alternative approach would be to simply take the terms with the biggest exponents in $$\text{Re\Im}\{(x+iy)^m\}$$ and add them if there max $$x$$ and $$y$$ exponents are the same, but this also doesn't agree with Wikipedia as it would replace something like $$f_{y(3x^2-y^2)}$$ with $$f_{y^3}$$, but it does explain wikipedia terms like $$g_{zy^3}\propto (3x^2-y^2)yz$$.

More notes: I've done a little more research and I really can't find any authoritative reference that gives names for orbitals higher than $$f$$. If anyone is familiar with one please let me know. I'm going to personally consider orbitals $$g$$ and higher to NOT have official cartesian polynomial abbreviation names and pursue and advertise the $$l, m$$ notation as being the proper way to refer to these higher orbitals. More motivation for this below.

For orbitals $$f$$ and below the $$xy$$ part of "abbreviated polynomial" subscript DOES allow you to identify the azimuthal nodal planes, for example, $$p_x$$ is zero when $$x=0$$ and $$d_{xy}$$ is zero when $$x=0$$ or $$y=0$$. This even holds for the intimidating $$f_{x(x^2-3y^2)}$$ which is zero if $$x=0, \pm \sqrt{3}y$$. However, when you move to orbitals with $$m>4$$ (for example $$g$$ orbitals with $$(l, m) = (4, \pm)$$), there are now 3 nodal planes, and any subscript that identifies all of them is going to require multiple terms and it is really not convenient to have expressions like that. It is likely for this reason authors don't typically give subscripts for orbitals at this order.

Frankly, for anyone reading this, my opinion is that the $$l, m$$ notation is far superior above basically the $$p$$ orbitals. The number $$|m|$$ tells you how many azimuthal nodal planes there are and the number $$l - |m|$$ tells you how many polar nodla planes there are. If $$m$$ is positive then the orbital has a set of lobes aligned along the $$x$$ axis. if $$m$$ is negative then the orbital is just like the one with $$+m$$ but rotated about the $$z$$ axis by $$\pm 360/(4|m|)$$ degrees ($$\pm$$ depending on if $$m$$ is positive or negative by the Condon-Shortley phase). No need to abbreviate a polynomial to convey this geometric information when it is already communicated by the $$l, m$$ numbers when properly understood.

• "No need to abbreviate a polynomial to convey this geometric information when it is already communicated by the l,m numbers when properly understood." By the same logic, there's no need to name elements when the atomic number is properly understood, but names are still useful. Commented Feb 19, 2022 at 19:46
• @Andrew I see your point. I guess the point is that, just like we only have "proper" names for atomic species up to $Z\approx 120$, we only have "cartesian polynomial names" for atomic orbitals up to $\ell=3$. It's a little annoying because there seems to be some usage for larger $\ell$, but it is not well documented or reference in the literature. So it gives the impression of "mystical" high order orbitals that have wacky names where noone knows where they come from. This is the impression which I am trying to eliminate. Commented Feb 20, 2022 at 0:20