Slater Orbitals

Quantum Chemistry

A Slater orbital is a good approximation to a hydrogen 1s orbital — it's pretty accurate, and has a simple functional form. That form is a decaying exponential with a distance in the argument, $∣ r - R ∣$ . The absolute value is the interesting part. Think back to algebra — $y = ∣ x ∣$ is the V-shape, symmetric about the y-axis. The 3D version of that trick gives you spherical symmetry around $R$ and a cusp at the nucleus, the same kink as the V's sharp point but spun in 3D.

Slater 1s — e^(-|x|)Gaussian — e^(-x²/2)

The Slater 1s wavefunction along an axis through the nucleus is e^(-|x|). The absolute value makes it symmetric about x=0 — and that's exactly what creates the cusp: a sharp point where the slope flips from -1 on the right to +1 on the left, all at one place.

In general a Slater orbital is two pieces multiplied together: a radial piece and an angular piece. The radial piece tells you how the wavefunction fades as you move out from the nucleus. The angular piece tells you which direction it points — that's where angular momentum lives, packaged inside a spherical harmonic $Y_{ℓ}^{m} (θ, ϕ)$ .

χ_{n ℓ m} (r) R_{n} (r) = R_{n} (r) Y_{ℓ}^{m} (θ, ϕ) = N r^{n - 1} e^{- ζ r}

1s Slater Orbitals

The 1s case is the easy one. Both angular momentum quantum numbers are zero, $ℓ = m = 0$ , and $Y_{0}^{0}$ is just a constant — so the angular piece collapses into the normalization and we're left with nothing but radial decay. Centered on $R_{A}$ , the 1s Slater orbital is:

1 s_{A} (r) = e^{- ζ ∣ r - R_{A} ∣}

Two flavors of radial function

Subtle point that trips up almost every new student: "radial wavefunction" can mean one of two different things, and textbooks usually don't say which. Both are real. They're just different objects.

The first is $R (r)$ — the radial part as it appears in $χ (r, θ, ϕ) = R (r) Y_{ℓ}^{m} (θ, ϕ)$ . For 1s it's $e^{- ζ r}$ up to normalization. The second is $u (r) = r R (r)$ , the "reduced" radial function. Some books call $u$ the radial wavefunction because the radial Schrödinger equation written in $u$ has a much cleaner shape — no centrifugal $1/ r^{2}$ mess.

Here's why it matters. The cusp at the nucleus belongs to $R (r)$ , not $u (r)$ . $R$ is finite at $r = 0$ with a kinked slope; $u$ is zero at $r = 0$ and smooth there, because the extra $r$ factor kills the kink. Same wavefunction, two ways to write it, totally different shapes near the origin.

When someone says "the radial wavefunction has a cusp at the nucleus" they mean $R$ . When they say "the radial wavefunction is zero at the origin" they mean $u$ . The Slater 1s we wrote above is the $R$ form — that's why the cusp is there. Pull an $r$ out front and the cusp goes away. This distinction quietly carries over everywhere: numerical integration of the radial equation is almost always done in $u$ , but plots of "the orbital" are almost always $R$ . If a sentence about radial functions doesn't quite line up with what you expect, check which one the author means.

Slater orbitals aren't just 1s

It's tempting to read "Slater orbital" as "1s exponential" and stop there, but the family is bigger. Plug any $(n, ℓ, m)$ combination into the general form and you get a different orbital. The radial part is always $N r^{n - 1} e^{- ζ r}$ and the angular part is the spherical harmonic $Y_{ℓ}^{m} (θ, ϕ)$ . Three concrete examples make the pattern clearer.

A 2s Slater orbital has $n = 2$ and $ℓ = m = 0$ . The radial part picks up an $r$ factor; the angular part is the same constant $Y_{0}^{0}$ as 1s, so the orbital is still spherically symmetric. The whole thing reduces to:

χ_{2 s} (r) \propto r e^{- ζ r}

A 2p orbital has $n = 2$ and $ℓ = 1$ . The radial factor is the same $r e^{- ζ r}$ , but now $m$ ranges over $- 1, 0, + 1$ — three differently-oriented orbitals. In the Cartesian forms chemists actually use these are $2 p_{x}$ , $2 p_{y}$ , $2 p_{z}$ : same radial profile, three different angular patterns. The z-version looks like:

χ_{2 p_{z}} (r) \propto r e^{- ζ r} cos θ

A 3d orbital has $n = 3$ and $ℓ = 2$ . The radial part becomes $r^{2} e^{- ζ r}$ , and $m$ now ranges over five values — five differently-shaped orbitals with the same radial profile. The $3 d_{z^{2}}$ form is:

χ_{3 d_{z^{2}}} (r) \propto r^{2} e^{- ζ r} (3 cos^{2} θ - 1)

Two patterns to lift out. First, only 1s lacks the $r^{n - 1}$ factor — every other Slater orbital has at least one $r$ out front, so its $R (r)$ vanishes at the origin and the cusp goes with it. The 1s cusp is unique to 1s. Second, all Slater orbitals with $ℓ > 0$ have nontrivial angular dependence; the spherically-symmetric ones — 1s, 2s, 3s, and so on — are exactly the $ℓ = 0$ cases.

n = 1 · ℓ = 0 · m ∈ {0}

R(r) = N · r^0 · e^(-ζr)spherically symmetric (one orbital)

1s is the only Slater orbital that peaks at the nucleus. There's no r factor out front, so R(r) is finite at r=0 — and the absolute value in the exponent gives it the cusp.

Click through the four tabs and the patterns from above show up immediately. 1s peaks at the nucleus with a cusp; 2s and 2p have the same radial profile (because under the Slater convention the radial part only depends on n, not ℓ) and both vanish at the origin; 3d has an even larger $r^{2}$ factor that pushes the radial peak further out. The angular momentum quantum number ℓ controls the multiplicity — one orbital for ℓ=0, three for ℓ=1, five for ℓ=2 — but it doesn't change the radial shape under this convention.

Why this form?

There's something nice about the 1s Slater orbital that the textbook usually buries. For a one-electron atom — hydrogen, $He^{+}$ , $Li^{2 +}$ , and so on — it is the exact 1s wavefunction. Not an approximation to it. In atomic units the answer is $e^{- Z r}$ up to normalization, and setting $ζ = Z$ recovers it on the nose.

ψ_{1 s}^{exact} (r) = \frac{Z ^{3/2}}{π} e^{- Z r} = ζ = Z N e^{- ζ r}

The "approximation" word only enters once you add a second electron. Now each electron sees less of the nuclear pull because the other one is partway in the way, and $ζ$ has to drift below Z to capture that screening. One electron is exact; two electrons is already a model. Quantum chemistry breaks fast.

The orbital exponent ζ

$ζ$ is the dial that sets the orbital's size. Crank it up: the electron gets pulled in tight, the orbital becomes small and dense. Crank it down: the electron spreads out, the orbital balloons. For a one-electron atom you don't have to think — $ζ = Z$ exactly. For everything else the other electrons screen part of the nuclear pull, each electron feels an effective charge less than Z, and $ζ$ has to drop below Z to match.

How do you actually pick a number for $ζ$ ? Two options. The first is variational: write the energy as a function of $ζ$ , take a derivative, pick the $ζ$ that minimizes it. Best $ζ$ this functional form can give you, but you have to do the calculus. The second is Slater's rules — basically a lookup table that tells you how much screening to expect from every other electron in the atom. Quick, do-it-in-your-head fast, less accurate than variational, and entirely sufficient for most back-of-envelope work.

ζ1.00

Exact 1s ground state of hydrogen — ζ = Z = 1. The peak of the radial probability sits at r* = 1/ζ = 1 Bohr radius, the textbook Bohr radius result.

Play with the slider for a minute and two things should jump out. First, the peak slides around — specifically $r^{*} = 1/ ζ$ , so a larger $ζ$ moves it inward and a smaller one moves it out. Second, the area under the curve doesn't change. It's a probability, so it has to integrate to one no matter what $ζ$ you pick — which means as the orbital contracts, the curve has to get taller to compensate. The dashed line is the exact-hydrogen reference at $ζ = 1$ ; everything else is a deformation of it.

Check yourself

Five questions to test whether the page actually landed. Click a question to reveal its answer.

Why does

ζ = Z

give the exact wavefunction for a one-electron atom but break for a multi-electron atom?

When there's just one electron, there's nobody else in the way. The electron sees the full nuclear charge Z, and the exact 1s solution is $e^{- Z r}$ — exactly a Slater orbital with $ζ = Z$ . Add a second electron and now the two of them partially shield each other from the nucleus. Each one feels less than the full Z, and the exact wavefunction stops being a Slater orbital at all. $ζ$ has to drop below Z to roughly account for the screening, but you've lost "exact" as a feature.

What does the absolute value

∣ r - R ∣

in the exponent buy you?

The cusp — the sharp, non-smooth point right at the nucleus. A Gaussian like $e^{- α r^{2}}$ doesn't have one; it's perfectly smooth at $r = 0$ . Real hydrogenic ground states do have a cusp because the Coulomb potential $- Z / r$ blows up at the nucleus and the wavefunction has to compensate by kinking. The absolute value bakes that kink in for free. This is one of the few places where a Slater orbital is genuinely better physics than a Gaussian.

Where does the radial probability density of a 1s Slater orbital peak?

At $r^{*} = 1/ ζ$ . Take $P (r) = 4 ζ^{3} r^{2} e^{- 2 ζ r}$ , set the derivative to zero, and the peak is at $r = 1/ ζ$ . For hydrogen with $ζ = 1$ in atomic units, that's the Bohr radius — the textbook "where the electron actually lives" answer.

Where does angular momentum hide in a Slater orbital, and why does the 1s case look so much simpler?

Angular momentum lives entirely in the spherical harmonic $Y_{ℓ}^{m} (θ, ϕ)$ factor — it's the only piece that knows about direction. The 1s case has $ℓ = m = 0$ , so $Y_{0}^{0}$ is just a number, and the angular piece collapses into the normalization. What's left is a purely radial function, which is why the 1s formula is so much shorter than the general one. Climb up to 2p, 3d, and beyond and the spherical harmonic shows back up — the orbital takes on its familiar lobed shapes.

What are the two standard ways to choose a value for

ζ

Variational optimization. Write the energy as a function of $ζ$ , take a derivative, pick the minimum. Best $ζ$ this form can give you — at the cost of doing the calculus. Slater's rules. An empirical recipe — basically a lookup table — that estimates screening from the quantum numbers of every other electron. You can do it in your head in seconds. Less accurate than variational, but if you just need a number to plug in and move on, Slater's rules win on cost.

For an application of 1s Slater orbitals to solving the hydrogen ground state using the variational principle, see the Variational Principle page.