---
title: Expectation Values, the Virial Theorem, and Scaling
module: The Quantum Hydrogen Atom
moduleNumber: 2
lessonNumber: 5
order: 205
summary: >
  The radial matrix elements ⟨r^k⟩ of hydrogenic states are the raw material of
  every later correction. This lesson derives ⟨1/r⟩ from the virial theorem,
  builds the full family ⟨r⟩, ⟨r²⟩, ⟨1/r²⟩, ⟨1/r³⟩ from Kramers' recursion and
  the Feynman-Hellmann theorem, and reads off their scaling with n, ℓ, and Z.
  The virial balance ⟨T⟩ = −½⟨V⟩ = −E fixes the energy budget of every bound
  state.
topics: [The Quantum Hydrogen Atom]
sources:
  - book: Griffiths & Schroeter
    ref: "Ch. 4 — Quantum Mechanics in Three Dimensions; §4.2; Kramers' relation (problems)"
  - book: Bransden & Joachain
    ref: "Ch. 3 — The Hydrogen Atom; expectation values of powers of r"
  - book: Foot
    ref: "Ch. 2 — The Hydrogen Atom; §2.2"
draft: false
---

The [hydrogen wave functions](/atomic-physics/quantum-hydrogen-atom/radial-equation-in-full)
are exact, so every quantity that a later correction needs is a definite number,
not an estimate. Fine structure, hyperfine structure, the Stark and Zeeman
shifts, and quantum defects all reduce, at first order, to expectation values of
powers of $r$ taken in the unperturbed hydrogenic states. Relativistic kinetic
energy weights $\langle 1/r^2\rangle$; spin-orbit coupling weights
$\langle 1/r^3\rangle$; the Darwin and contact terms weight $|\psi(0)|^2$. This
lesson computes the family $\langle r^k\rangle$ once and states the scaling
laws that follow, and it establishes the virial balance that fixes the split of
each energy into kinetic and potential parts.

Throughout, write the hydrogenic length scale as

$$
a = \frac{a_0}{Z},
\qquad a_0 = \frac{\hbar^2}{\mu ke^2},
$$

so that every radius carries $Z$ through $a$ and every energy through
$E_n = -Z^2\mathrm{Ry}/n^2$.

## The virial theorem

For a stationary state the expectation value of any operator that lacks explicit
time dependence is constant, so its time derivative vanishes. Applying this to
the virial operator $\hat{\vec r}\cdot\hat{\vec p}$ and using
$\d\langle \hat G\rangle/\d t = (i/\hbar)\langle[\hat H,\hat G]\rangle$ gives

$$
0 = \frac{\d}{\d t}\langle \hat{\vec r}\cdot\hat{\vec p}\rangle
= \frac{i}{\hbar}\big\langle[\hat H,\,\hat{\vec r}\cdot\hat{\vec p}]\big\rangle.
$$

Evaluating the commutator with $\hat H = \hat p^2/2\mu + V(\hat r)$ produces the
kinetic energy on one side and the radial derivative of the potential on the
other:

$$
2\langle T\rangle = \big\langle \hat{\vec r}\cdot\nabla V\big\rangle.
$$

> **Theorem (Quantum virial theorem).** In a bound stationary state of a
> potential homogeneous of degree $s$ (so that $\vec r\cdot\nabla V = sV$),
> $$
> 2\langle T\rangle = s\langle V\rangle.
> $$
> For the Coulomb potential $V=-kZe^2/r$, $s=-1$ and
> $$
> 2\langle T\rangle = -\langle V\rangle,
> \qquad
> \langle T\rangle = -E_n,
> \qquad
> \langle V\rangle = 2E_n.
> $$

The last line follows from $\langle T\rangle + \langle V\rangle = E_n$ combined
with $2\langle T\rangle = -\langle V\rangle$. Every hydrogenic level therefore
carries a kinetic energy equal to its binding energy and a potential energy
twice as negative:

$$
\langle T\rangle_n = \frac{Z^2\,\mathrm{Ry}}{n^2},
\qquad
\langle V\rangle_n = -\frac{2Z^2\,\mathrm{Ry}}{n^2}.
$$

$$
% caption: The virial balance in hydrogen: the mean potential energy is 2E_n, the
% mean kinetic energy is −E_n, and their sum is the binding energy E_n < 0.
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,-2.6) -- (0,2.2) node[above, black!70] {energy};
\draw[black, dashed] (-0.3,0) -- (5.4,0);
\node[black, anchor=east] at (-0.3,0) {0};
% kinetic bar (positive)
\draw[black, very thick] (0.7,0) -- (0.7,1.4);
\fill[black] (0.5,0) rectangle (0.9,1.4);
\node[black, anchor=south] at (0.7,1.45) {T};
% total energy level
\draw[acc, very thick] (2.3,-0.7) -- (3.3,-0.7);
\node[acc, anchor=west] at (3.35,-0.7) {$E_n$};
% potential bar (negative, twice as deep)
\draw[black, very thick] (4.6,0) -- (4.6,-2.1);
\fill[black] (4.4,0) rectangle (4.8,-2.1);
\node[black, anchor=north] at (4.6,-2.15) {V};
\node[align=center, black] at (2.6,1.6) {kinetic, size $E_n$};
\node[align=center, black] at (4.6,-1.0) {$V=2E_n$};
\end{tikzpicture}
$$

## The mean inverse radius

The virial theorem gives $\langle 1/r\rangle$ with no further integration. Since
$\langle V\rangle = -kZe^2\langle 1/r\rangle = 2E_n$,

$$
\left\langle \frac{1}{r}\right\rangle_{n\ell}
= -\frac{2E_n}{kZe^2}
= \frac{Z}{a_0 n^2}
= \frac{1}{a\,n^2}.
$$

The result depends on $n$ alone, not on $\ell$. It reproduces the Bohr relation:
the mean inverse radius equals the inverse of the $n$-th Bohr radius $n^2 a$.
This single expectation value already fixes the leading Coulomb energy through
$\langle V\rangle$, and it recurs in every hyperfine and volume-shift estimate.

## Kramers' recursion

The remaining moments follow from a recursion among consecutive powers,
obtained by taking expectation values of the radial equation against $r^{s}$ and
$r^{s-1}$. **Kramers' relation** connects three neighbouring moments:[^gs-kramers]

$$
\frac{s+1}{n^2}\langle r^{s}\rangle
- (2s+1)\,a\,\langle r^{s-1}\rangle
+ \frac{s}{4}\Big[(2\ell+1)^2 - s^2\Big]a^2\,\langle r^{s-2}\rangle = 0.
$$

The relation is a finite-difference equation in the power $s$; seeded by
$\langle r^0\rangle = 1$ and $\langle 1/r\rangle = 1/an^2$, it climbs to every
positive moment.

> **Worked example.** Compute $\langle r\rangle$ from Kramers' relation. Setting
> $s = 1$ and using $\langle r^0\rangle = 1$, $\langle 1/r\rangle = 1/an^2$,
> $$
> \frac{2}{n^2}\langle r\rangle - 3a
> + \frac{1}{4}\big[(2\ell+1)^2 - 1\big]a^2\cdot\frac{1}{an^2} = 0.
> $$
> The bracket is $(2\ell+1)^2 - 1 = 4\ell(\ell+1)$, so the last term is
> $\ell(\ell+1)a/n^2$, and
> $$
> \frac{2}{n^2}\langle r\rangle = 3a - \frac{\ell(\ell+1)a}{n^2}
> \quad\Longrightarrow\quad
> \langle r\rangle = \frac{a}{2}\big[\,3n^2 - \ell(\ell+1)\,\big].
> $$

Setting $s=2$ in the same way delivers $\langle r^2\rangle$. The first positive
moments are

$$
\langle r\rangle = \frac{a}{2}\big[3n^2 - \ell(\ell+1)\big],
\qquad
\langle r^2\rangle = \frac{a^2 n^2}{2}\big[5n^2 + 1 - 3\ell(\ell+1)\big].
$$

Both grow as $n^2$ and $n^4$ respectively — the electron of a highly excited
state is far out and its position is broadly spread. The mild $\ell$-dependence
subtracts a little for the low-$\ell$ penetrating orbits, which reach closer to
the nucleus on their eccentric excursions.

## Feynman-Hellmann and the negative moments

Kramers' relation cannot start the negative tower on its own: at $s=0$ the
$\langle r^{-2}\rangle$ coefficient vanishes, so $\langle 1/r^2\rangle$ needs an
independent input. The **Feynman-Hellmann theorem** supplies it. Treating the
orbital quantum number $\ell$ as a continuous parameter of the Hamiltonian, whose
only $\ell$-dependence is the centrifugal term
$\hbar^2\ell(\ell+1)/2\mu r^2$,

$$
\frac{\partial E_n}{\partial \ell}
= \left\langle \frac{\partial \hat H}{\partial \ell}\right\rangle
= \frac{\hbar^2(2\ell+1)}{2\mu}\left\langle \frac{1}{r^2}\right\rangle.
$$

Because $n = n_r + \ell + 1$, a shift in $\ell$ at fixed radial number $n_r$ is a
shift in $n$, so $\partial E_n/\partial\ell = \partial E_n/\partial n = 2Z^2\mathrm{Ry}/n^3$.
Solving,

$$
\left\langle \frac{1}{r^2}\right\rangle_{n\ell} = \frac{Z^2}{a_0^2\,n^3\,(\ell+\tfrac12)}
= \frac{1}{a^2\,n^3(\ell+\tfrac12)}.
$$

With $\langle 1/r^2\rangle$ in hand, Kramers' relation at $s=-1$ closes the last
one used by the fine structure. There the surviving terms give
$a\langle 1/r^2\rangle = \ell(\ell+1)a^2\langle 1/r^3\rangle$, so

$$
\left\langle \frac{1}{r^3}\right\rangle_{n\ell}
= \frac{Z^3}{a_0^3\,n^3\,\ell(\ell+\tfrac12)(\ell+1)}
= \frac{1}{a^3\,n^3\,\ell(\ell+\tfrac12)(\ell+1)},
\qquad \ell > 0.
$$

The $\langle 1/r^3\rangle$ moment diverges for $\ell = 0$, which is why the
spin-orbit interaction — proportional to it — vanishes for s-states and the
Darwin contact term takes over there instead.

## The moment table and its scaling

The five moments and their scaling with $n$, $\ell$, and $Z$ collect into one
table. Each is written in units of the hydrogenic length $a = a_0/Z$.

| moment | value | $n$ | $Z$ |
| --- | --- | --- | --- |
| $\langle r\rangle$ | $\tfrac{a}{2}[3n^2-\ell(\ell+1)]$ | $\sim n^2$ | $\sim Z^{-1}$ |
| $\langle r^2\rangle$ | $\tfrac{a^2 n^2}{2}[5n^2+1-3\ell(\ell+1)]$ | $\sim n^4$ | $\sim Z^{-2}$ |
| $\langle 1/r\rangle$ | $\dfrac{1}{a n^2}$ | $\sim n^{-2}$ | $\sim Z$ |
| $\langle 1/r^2\rangle$ | $\dfrac{1}{a^2 n^3(\ell+\tfrac12)}$ | $\sim n^{-3}$ | $\sim Z^2$ |
| $\langle 1/r^3\rangle$ | $\dfrac{1}{a^3 n^3\,\ell(\ell+\tfrac12)(\ell+1)}$ | $\sim n^{-3}$ | $\sim Z^3$ |

$$
% caption: The mean radius ⟨r⟩ grows almost quadratically with n; for fixed n it
% shrinks slightly as ℓ increases, since low-ℓ orbits are the eccentric ones.
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (6.2,0) node[right, black!70] {$n$};
\draw[->, black] (0,0) -- (0,3.4) node[above, black!70] {mean r};
\foreach \n in {1,2,3,4} \node[black, anchor=north] at ({1.4*\n},0) {\n};
% <r> for l=0 (max): (a/2)(3n^2) -> scale so n=4 near top
% points scaled by 0.09
\fill[acc] (1.4,{0.135}) circle (2pt);
\fill[acc] (2.8,{0.54}) circle (2pt);
\fill[acc] (4.2,{1.215}) circle (2pt);
\fill[acc] (5.6,{2.16}) circle (2pt);
\draw[acc, thick] (1.4,0.135) .. controls (2.8,0.4) and (4.2,1.0) .. (5.6,2.16);
\node[acc, anchor=east] at (5.5,2.2) {s state};
% l = n-1 (circular, min slightly lower): (a/2)(3n^2 - (n-1)n)= (a/2)(2n^2+n)
\fill[black] (2.8,{0.45}) circle (2pt);
\fill[black] (4.2,{1.035}) circle (2pt);
\fill[black] (5.6,{1.89}) circle (2pt);
\draw[black, thick, dashed] (2.8,0.45) .. controls (4.2,0.85) and (5.0,1.4) .. (5.6,1.89);
\node[black, anchor=west] at (5.65,1.85) {circular};
\end{tikzpicture}
$$

The pattern is uniform. Positive powers scale as $Z^{-k}$ and grow with $n$;
negative powers scale as $Z^{k}$ and fall with $n$. A correction built from
$\langle 1/r^3\rangle$ therefore scales as $Z^3/n^3$ and concentrates in the low
shells of heavy ions, exactly where fine structure is largest.

$$
% caption: Each perturbation of the hydrogen atom reduces at first order to one
% radial moment: ⟨1/r²⟩ for the relativistic term, ⟨1/r³⟩ for spin-orbit, and
% the density at the origin for the Darwin and hyperfine contact terms.
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth,
  box/.style={draw, minimum width=30mm, minimum height=9mm, align=center}]
\definecolor{acc}{HTML}{4A6FA5}
\node[box] (m2) at (0,1.6) {second inverse moment};
\node[box] (m3) at (0,0) {third inverse moment};
\node[box] (m0) at (0,-1.6) {density at origin};
\node[box] (rel) at (5.2,1.6) {relativistic term};
\node[box] (so) at (5.2,0) {spin-orbit term};
\node[box] (dar) at (5.2,-1.6) {Darwin, contact};
\draw[black, ->] (m2) -- (rel);
\draw[black, ->] (m3) -- (so);
\draw[black, ->] (m0) -- (dar);
\end{tikzpicture}
$$

## Length contraction with nuclear charge

The single factor $a = a_0/Z$ carries the entire $Z$-dependence of the geometry.
Increasing the nuclear charge from hydrogen to a one-electron ion contracts every
orbital radius by $1/Z$ and deepens every binding energy by $Z^2$, without
changing the shape of any wave function.

$$
% caption: The hydrogenic length scale a₀/Z contracts every orbital as Z grows;
% the wave-function shape is unchanged, only rescaled inward.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
% Z=1
\begin{scope}
\fill[black!70] (0,0) circle (1.4pt);
\draw[black, thick] (0,0) circle (1.5);
\node[black, anchor=north] at (0,-1.7) {$Z=1$};
\end{scope}
% Z=2
\begin{scope}[xshift=4.0cm]
\fill[black!70] (0,0) circle (1.4pt);
\draw[black, thick] (0,0) circle (0.75);
\node[black, anchor=north] at (0,-1.7) {$Z=2$};
\end{scope}
% Z=3
\begin{scope}[xshift=7.4cm]
\fill[black!70] (0,0) circle (1.4pt);
\draw[black, thick] (0,0) circle (0.5);
\node[black, anchor=north] at (0,-1.7) {$Z=3$};
\end{scope}
\end{tikzpicture}
$$

Every moment in the table inherits the same scaling, and every correction built
from those moments does too. The relativistic kinetic term, carrying
$\langle 1/r^2\rangle \sim Z^2$ together with the explicit $\alpha^2 Z^2$ of its
prefactor, grows as $Z^4$; this is why fine-structure splittings, invisible in
hydrogen at the eV scale, dominate the spectra of heavy elements. The moments
computed here are the numbers those
[fine-structure](/atomic-physics/fine-structure-and-the-dirac-atom/spin-orbit-thomas-precession)
corrections evaluate.

[^gs-kramers]: **Griffiths & Schroeter**, _Introduction to Quantum Mechanics_, 3rd ed., §4.2 and the Kramers-relation problem — the three-term recursion for $\langle r^s\rangle$, the virial theorem for the Coulomb potential, and the Feynman-Hellmann route to $\langle 1/r^2\rangle$ and $\langle 1/r^3\rangle$.