---
title: The Relativistic Kinetic-Energy Correction
module: Fine Structure and the Dirac Atom
moduleNumber: 3
lessonNumber: 2
order: 302
summary: >
  The Bohr energies treat the electron as slowly moving, but its speed is of order
  αc, so the kinetic energy needs a relativistic correction. Expanding
  √(p²c²+m²c⁴) to order (v/c)² produces the perturbation −p⁴/8m³c², whose
  first-order shift on a hydrogenic state is evaluated with the trick p²=2m(E−V).
  The result depends on n and ℓ, is smaller than the gross structure by α²≈5×10⁻⁵,
  and is one of the three pieces that combine into the fine-structure formula.
topics: [Fine Structure and the Dirac Atom]
sources:
  - book: Griffiths & Schroeter
    ref: "Ch. 7 — Perturbation Theory; §7.3.1 The Relativistic Correction"
  - book: Bransden & Joachain
    ref: "Ch. 5 — Interaction of One-Electron Atoms with Fields; §5.1 Fine Structure"
  - book: Foot
    ref: "Ch. 2 — The Hydrogen Atom; Appendix on relativistic effects"
draft: false
---

The gross structure of hydrogen, the $E_n = -Z^2\,\mathrm{Ry}/n^2$ ladder produced
by [solving the radial equation](/atomic-physics/quantum-hydrogen-atom/radial-equation-in-full),
rests on the non-relativistic kinetic energy $T = p^2/2m$. That expression is the
first term of an expansion. The electron in the ground state moves with a
characteristic speed $v \sim Z\alpha c$, where $\alpha = e^2/4\pi\epsilon_0\hbar c
\approx 1/137$ is the fine-structure constant, so $(v/c)^2 \sim (Z\alpha)^2$ is a
small but nonzero fraction. Keeping the next term in the relativistic kinetic
energy shifts every level by a fractional amount of order $(Z\alpha)^2$, splitting
lines that the gross theory leaves coincident. This lesson computes that shift for
hydrogenic states; the [spin-orbit](/atomic-physics/fine-structure-and-the-dirac-atom/spin-orbit-thomas-precession)
and [Darwin](/atomic-physics/fine-structure-and-the-dirac-atom/darwin-term-fine-structure-formula)
corrections, of the same order, follow in the next two lessons.

## The size of the effect

The Bohr model fixes the electron's speed. In the $n$-th circular orbit the
quantized angular momentum $m v_n r_n = n\hbar$ and the force balance
$m v_n^2/r_n = Z e^2/4\pi\epsilon_0 r_n^2$ combine to

$$
\frac{v_n}{c} = \frac{Z\alpha}{n}.
$$

For hydrogen ($Z=1$) in the ground state, $v_1/c = \alpha \approx 7.3\times10^{-3}$,
so $(v/c)^2 \approx 5.3\times10^{-5}$. The relativistic correction to the kinetic
energy is smaller than the kinetic energy itself by this factor, and the kinetic
energy is comparable to the binding energy, so the level shifts land near
$5\times10^{-5}\times 13.6~\text{eV} \approx 10^{-4}~\text{eV}$. That number sets
the scale of fine structure and explains why the effect is invisible in the coarse
Balmer spectrum and prominent only under high resolution.

$$
% caption: The electron's orbital speed as a fraction of c falls as Zα/n; the
% relativistic correction scales as its square, so it is largest for the tightly
% bound low-n, high-Z states.
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (6.2,0) node[right, black!70] {$n$};
\draw[->, black] (0,0) -- (0,3.2) node[above, black!70] {$\frac{v}{c}$};
\foreach \x/\lab in {1/1,2/2,3/3,4/4,5/5} \node[black, anchor=north] at (\x,0) {\lab};
% v/c = alpha/n scaled: take alpha*100 ~ 0.73, scale by 3.6 for visibility
% points at n: 2.6/n
\draw[acc, very thick] (1,2.6) .. controls (1.4,1.75) and (1.7,1.45) .. (2,1.3)
  .. controls (2.6,1.0) and (3.4,0.75) .. (4,0.65)
  .. controls (4.6,0.57) and (5.4,0.5) .. (5.8,0.45);
\foreach \x/\y in {1/2.6, 2/1.3, 3/0.867, 4/0.65, 5/0.52} \fill[acc] (\x,\y) circle (1.6pt);
\node[acc, anchor=west] at (1.15,2.7) {$Z=1$};
% Z=2 curve higher
\draw[black, very thick, dashed] (1,3.05) .. controls (1.5,2.4) and (1.8,2.0) .. (2,1.85)
  .. controls (2.6,1.5) and (3.4,1.25) .. (4,1.15);
\node[black, anchor=west] at (2.3,1.95) {$Z=2$};
\end{tikzpicture}
$$

## Expanding the relativistic kinetic energy

The exact relativistic energy of a free particle of rest mass $m$ and momentum
magnitude $p$ is

$$
E = \sqrt{p^2c^2 + m^2c^4}.
$$

The kinetic energy is this minus the rest energy $mc^2$. Factor out $mc^2$ and
expand the square root in the small parameter $(p/mc)^2$:[^gs-731]

$$
T = mc^2\sqrt{1 + \left(\frac{p}{mc}\right)^2} - mc^2
= mc^2\left[\frac{1}{2}\left(\frac{p}{mc}\right)^2
- \frac{1}{8}\left(\frac{p}{mc}\right)^4 + \cdots\right].
$$

Written out in powers of $p$,

$$
T = \frac{p^2}{2m} - \frac{p^4}{8m^3c^2} + \mathcal{O}\!\left(\frac{p^6}{m^5c^4}\right).
$$

The leading term is the familiar non-relativistic kinetic energy already in the
hydrogen Hamiltonian. The next term is the lowest relativistic correction,

$$
H'_r = -\frac{p^4}{8m^3c^2},
$$

a perturbation to be added to the unperturbed hydrogen Hamiltonian
$H_0 = p^2/2m + V(r)$. The relative size of the two, $(p^2/2m)$ versus the
correction, is $(p/mc)^2/4 \sim (Z\alpha/n)^2$, confirming the estimate above.

$$
% caption: The exact kinetic energy √(p²c²+m²c⁴)−mc² (solid) bends below the
% parabola p²/2m (dashed); the gap is the −p⁴/8m³c² correction, quadratic in the
% small quantity p²/m²c².
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (6.2,0) node[right, black!70] {$\frac{p}{mc}$};
\draw[->, black] (0,0) -- (0,3.4) node[above, black!70] {$\frac{T}{mc^2}$};
% parabola p^2/2 (nonrel): scaled. at x=5 -> 12.5, too big; scale y by 0.22
\draw[black, very thick, dashed] (0,0) .. controls (1.5,0.25) and (3.0,1.0) .. (4.2,1.94)
  .. controls (4.8,2.53) and (5.2,3.0) .. (5.5,3.3);
\node[black, anchor=south east] at (5.4,3.25) {$\frac{p^2}{2m}$};
% exact sqrt(1+x^2)-1: bends below. at x=5 -> sqrt26-1=4.1 vs 12.5; scale y by 0.22
\draw[acc, very thick] (0,0) .. controls (1.6,0.15) and (3.2,0.7) .. (4.2,1.15)
  .. controls (4.8,1.45) and (5.3,1.72) .. (5.7,1.95);
\node[acc, anchor=north west] at (4.9,1.15) {exact};
% shade the gap with a hint
\draw[black, ->] (4.4,1.25) -- (4.4,1.95);
\node[black, anchor=west] at (4.5,1.75) {$\frac{p^4}{8m^3c^2}$ gap};
\end{tikzpicture}
$$

## First-order shift and the p⁴ trick

Because hydrogen's gross-structure levels are degenerate in $\ell$ and $m_\ell$,
one might worry about degenerate perturbation theory. The saving fact is that $H'_r$
commutes with $L^2$ and $L_z$: it is built from $p^4$, a scalar under rotations, and
carries no spin or angular dependence beyond what $p^2 = 2m(E_0 - V)$ already
respects. The unperturbed eigenstates $\lvert n\ell m_\ell\rangle$ are therefore
already the "good" states that diagonalize $H'_r$ within each degenerate shell, and
ordinary first-order theory applies:[^gs-731]

$$
E_r^{(1)} = \langle n\ell m_\ell \lvert H'_r \rvert n\ell m_\ell\rangle
= -\frac{1}{8m^3c^2}\langle p^4\rangle
= -\frac{1}{8m^3c^2}\langle p^2\,p^2\rangle.
$$

Evaluating $\langle p^4\rangle$ directly requires four derivatives of the wave
function and is laborious. The efficient route uses the unperturbed Schrödinger
equation to trade momentum for energy. Since $H_0\lvert\psi\rangle = E_n\lvert\psi\rangle$
and $H_0 = p^2/2m + V$,

$$
p^2\lvert\psi\rangle = 2m\,(E_n - V)\lvert\psi\rangle.
$$

The operator $p^2$ is Hermitian, so it may act to the left on the bra and to the
right on the ket:

$$
\langle p^4\rangle = \langle\psi\lvert p^2\,p^2\rvert\psi\rangle
= \langle p^2\psi \lvert p^2\psi\rangle
= 4m^2\big\langle (E_n - V)^2\big\rangle.
$$

This replaces the fourth-order differential operator with a simple function of $r$,
at the cost of needing two radial expectation values. Expanding the square,

$$
E_r^{(1)} = -\frac{1}{2mc^2}\Big[E_n^2 - 2E_n\langle V\rangle + \langle V^2\rangle\Big].
$$

> **Definition (The $p^2 \to 2m(E-V)$ substitution).** Inside a matrix element
> taken between exact eigenstates of $H_0 = p^2/2m + V$, the operator $p^2$ may be
> replaced by the multiplication operator $2m(E_n - V)$. This turns any power of
> $p^2$ acting on an eigenstate into a power of a function of position, reducing
> $\langle p^4\rangle$ to $\langle V^2\rangle$, $\langle V\rangle$, and $E_n^2$.

$$
% caption: The p⁴ matrix element is reduced by splitting p²·p², replacing each
% factor with 2m(E−V) using the eigenvalue equation, leaving expectation values
% of powers of the Coulomb potential.
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth,
  bx/.style={draw, minimum width=27mm, minimum height=10mm, align=center}]
\definecolor{acc}{HTML}{4A6FA5}
\node[bx, draw=acc, text=acc, thick] (a) at (0,0) {$p^4$ term};
\node[bx] (b) at (4.6,0) {$V$ and $E_n$ powers};
\node[bx] (c) at (0,-2.2) {$\frac1r$, $\frac{1}{r^2}$ averages};
\node[bx] (d) at (4.6,-2.2) {$E_r^{(1)}$};
\draw[acc, ->] (a) -- (b) node[midway, above] {eigenvalue $H_0$};
\draw[acc, ->] (b) -- (c);
\node[align=left] at (3.7,-0.75) {expand,\\Coulomb $V$};
\draw[acc, ->] (c) -- (d) node[midway, below] {collect};
\end{tikzpicture}
$$

## Evaluating on Coulomb states

For a hydrogenic atom the potential is $V(r) = -Ze^2/4\pi\epsilon_0 r$. Write
$\beta \equiv Ze^2/4\pi\epsilon_0$, so $V = -\beta/r$ and $\langle V\rangle =
-\beta\langle 1/r\rangle$, $\langle V^2\rangle = \beta^2\langle 1/r^2\rangle$. The
two radial expectation values are the standard hydrogenic
[results](/atomic-physics/quantum-hydrogen-atom/expectation-values-virial)

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

with $a_0 = 4\pi\epsilon_0\hbar^2/m e^2$ the Bohr radius. Two identities streamline
the algebra. First, $\beta/a_0 = Ze^2/4\pi\epsilon_0 a_0$, and comparing with
$E_n = -\beta/2a_0 n^2 \cdot Z = -Z^2 e^2/8\pi\epsilon_0 a_0 n^2$ gives
$\beta/a_0 = -2n^2 E_n / Z \cdot Z = -2n^2E_n$ once $Z$ is carried inside $\beta$.
Concretely,

$$
\langle V\rangle = -\beta\frac{Z}{n^2a_0} = 2E_n,
\qquad
\langle V^2\rangle = \beta^2\frac{Z^2}{n^3(\ell+\tfrac12)a_0^2}
= \frac{4n\,E_n^2}{\ell+\tfrac12}.
$$

The first is the [virial theorem](/atomic-physics/quantum-hydrogen-atom/expectation-values-virial)
in disguise: $\langle V\rangle = 2E_n$ and $\langle T\rangle = -E_n$. Substituting
both into the bracket,

$$
E_r^{(1)} = -\frac{1}{2mc^2}\left[E_n^2 - 2E_n(2E_n) + \frac{4n\,E_n^2}{\ell+\tfrac12}\right]
= -\frac{E_n^2}{2mc^2}\left[\frac{4n}{\ell+\tfrac12} - 3\right].
$$

> **Theorem (Relativistic kinetic correction for hydrogen).** The first-order shift
> of the level $E_n$ from the $-p^4/8m^3c^2$ term is
> $$
> E_r^{(1)} = -\frac{E_n^2}{2mc^2}\left[\frac{4n}{\ell+\tfrac12} - 3\right]
> = -\frac{(Z\alpha)^4 mc^2}{2n^4}\left[\frac{n}{\ell+\tfrac12} - \frac34\right].
> $$
> It is negative for every hydrogenic state (the bracket is positive because
> $\ell \le n-1$ forces $4n/(\ell+\tfrac12) \ge 4n/(n-\tfrac12) > 3$), so relativity
> lowers every level, and it depends on $\ell$ as $1/(\ell+\tfrac12)$ at fixed $n$.

The second form uses $E_n = -(Z\alpha)^2 mc^2/2n^2$, which recasts $E_n^2/2mc^2$ as
$(Z\alpha)^4 mc^2/8n^4$. The correction is manifestly of order $(Z\alpha)^2$ times
the gross energy $E_n$, the promised fine-structure scale.

## Dependence on n and ℓ

The shift lowers every level, but not uniformly. At fixed $n$ the magnitude falls
as $\ell$ grows, because $1/(\ell+\tfrac12)$ shrinks and high-$\ell$ states are held
away from the nucleus by the [centrifugal barrier](/atomic-physics/quantum-hydrogen-atom/radial-equation-in-full),
where the electron moves slower and the relativistic correction is weaker. The
penetrating $\ell=0$ states, which sample the region of large $p$ near the origin,
are pulled down the most.

$$
% caption: The relativistic shift within the n=3 shell, in units of
% (Zα)⁴mc²/n⁴; s (ℓ=0) drops furthest, d (ℓ=2) least, tracking 1/(ℓ+½).
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth]
\definecolor{acc}{HTML}{4A6FA5}
% unperturbed line
\draw[black, dashed] (0,0) -- (6.2,0);
\node[black, anchor=south west] at (0,0.05) {$E_3$};
% shift ~ -(n/(l+1/2) - 3/4)/2 ; for n=3: l=0: -(6-.75)/2=-2.625; l=1: -(2-.75)/2=-0.625; l=2:-(1.2-.75)/2=-0.225
% scale by 0.9
\foreach \x/\d/\lab in {1/-2.36/s, 3/-0.56/p, 5/-0.20/d}{
  \draw[acc, very thick] (\x-0.55,\d) -- (\x+0.55,\d);
  \draw[black, dotted] (\x,0) -- (\x,\d);
  \node[acc, anchor=north] at (\x,\d-0.05) {\lab};
}
\node[black, anchor=east] at (6.2,-1.6) {shift grows downward};
\draw[->, black] (0,0.4) -- (0,-2.9) node[left, black!70] {$E_r^{(1)}$};
\end{tikzpicture}
$$

A short table makes the pattern quantitative. Writing the shift as
$E_r^{(1)} = -\tfrac12 (Z\alpha)^4 mc^2 n^{-4}\,f(n,\ell)$ with
$f = n/(\ell+\tfrac12) - \tfrac34$:

| $n$ | $\ell$ | $\ell+\tfrac12$ | $f(n,\ell)$ |
| --- | --- | --- | --- |
| $1$ | $0$ | $\tfrac12$ | $2 - \tfrac34 = \tfrac54$ |
| $2$ | $0$ | $\tfrac12$ | $4 - \tfrac34 = \tfrac{13}{4}$ |
| $2$ | $1$ | $\tfrac32$ | $\tfrac43 - \tfrac34 = \tfrac{7}{12}$ |
| $3$ | $0$ | $\tfrac12$ | $6 - \tfrac34 = \tfrac{21}{4}$ |
| $3$ | $2$ | $\tfrac52$ | $\tfrac65 - \tfrac34 = \tfrac{9}{20}$ |

The bare $\ell$-dependence here is not the full story: the
[spin-orbit correction](/atomic-physics/fine-structure-and-the-dirac-atom/spin-orbit-thomas-precession)
carries its own $\ell$-dependence of the same size, and when the two are added the
combined shift reorganizes to depend only on $n$ and the total angular momentum
$j$. The relativistic term taken alone does not respect that final simplicity, so
its $\ell$-pattern is physical only in combination with the others.

## Where it sits among the corrections

The relativistic kinetic term is one of three corrections of order $(Z\alpha)^2
E_n$ that together make up the fine structure. It is worth fixing the hierarchy of
scales before the pieces are assembled, because the same power counting recurs for
hyperfine structure and the Lamb shift.

$$
% caption: The ladder of energy scales in hydrogen: each rung is smaller than the
% one above by roughly α², so gross structure (~10 eV) sits far above fine
% structure (~10⁻⁴ eV) and hyperfine structure (~10⁻⁶ eV).
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth,
  rung/.style={draw, minimum width=40mm, minimum height=8mm, align=center}]
\definecolor{acc}{HTML}{4A6FA5}
\node[rung, draw=acc, text=acc, thick] (g) at (0,3.0) {gross structure};
\node[rung] (f) at (0,1.4) {f\/ine structure};
\node[rung] (h) at (0,-0.2) {hyperf\/ine};
\draw[black, ->] (2.3,3.0) .. controls (3.5,2.5) and (3.5,1.9) .. (2.3,1.4);
\node[black, anchor=west] at (3.4,2.2) {smaller};
\draw[black, ->] (2.3,1.4) .. controls (3.5,0.9) and (3.5,0.3) .. (2.3,-0.2);
\node[black, anchor=west] at (3.4,0.6) {smaller};
\end{tikzpicture}
$$

Because all three fine-structure terms scale as $(Z\alpha)^4 mc^2$, none can be
neglected relative to the others; the [Darwin term](/atomic-physics/fine-structure-and-the-dirac-atom/darwin-term-fine-structure-formula)
acts only on $\ell=0$ states, the spin-orbit term only on $\ell\ge1$, and the
relativistic term on all of them. The remarkable outcome, derived once all three
are in hand, is that their sum collapses to the single
[fine-structure formula](/atomic-physics/fine-structure-and-the-dirac-atom/darwin-term-fine-structure-formula)
depending on $n$ and $j$ alone, the same expression that the exact
[Dirac equation](/atomic-physics/fine-structure-and-the-dirac-atom/dirac-equation-hydrogen)
reproduces to this order. The relativistic kinetic correction is the piece that
carries the "velocity" content of that agreement: it is what the $-p^4/8m^3c^2$
term of the classical energy becomes once the electron is quantized.

[^gs-731]: **Griffiths & Schroeter**, _Introduction to Quantum Mechanics_, 3rd ed., §7.3.1 — expansion of the relativistic kinetic energy, the perturbation $-p^4/8m^3c^2$, the $p^2 = 2m(E-V)$ reduction of $\langle p^4\rangle$, and the closed result $E_r^{(1)} = -(E_n^2/2mc^2)[4n/(\ell+\tfrac12) - 3]$.
[^bj-51]: **Bransden & Joachain**, _Physics of Atoms and Molecules_, 2nd ed., §5.1 — the mass-velocity correction as the leading relativistic term in the Foldy–Wouthuysen reduction of the Dirac Hamiltonian, and its hydrogenic matrix element. <https://www.pearson.com/en-gb/subject-catalog/p/physics-of-atoms-and-molecules/P200000005386>