---
title: Two-Level Systems and the Bloch Sphere
module: Spin
moduleNumber: 8
lessonNumber: 3
order: 803
summary: >
  Every two-state quantum system is a spin-½ in disguise. Its Hamiltonian is an
  effective magnetic field, its pure states are points on the Bloch sphere, and
  its unitary evolution is a rigid rotation of that sphere. The same structure
  produces avoided level crossings, the ammonia inversion doublet and its maser,
  and the qubit.
topics: [Spin]
sources:
  - book: Cohen-Tannoudji, Diu & Laloë
    ref: "Ch. IV §C — Two-Level Systems; Complements on the Ammonia Molecule"
  - book: Sakurai & Napolitano
    ref: "§3.2 Spinor Geometry; §2.1 Two-State Dynamics"
  - book: Griffiths & Schroeter
    ref: "§4.4 Spin; §11.1 Two-Level Systems"
draft: false
---

The [precession lesson](/quantum-mechanics/spin/spin-in-a-magnetic-field-precession-and-resonance)
solved a spin in a magnetic field. That problem is not special to spin. Any
quantum system with two relevant states — two atomic levels near resonance, the
two positions of a tunneling atom, two coupled molecular configurations — has a
two-dimensional Hilbert space and a $2\times 2$ Hamiltonian, and every such
Hamiltonian is a spin in an effective field. The geometry of spin-½, the Pauli
matrices, and the Bloch sphere therefore describe every two-level system at once.

## The generic two-level Hamiltonian

Any Hermitian $2\times 2$ matrix expands in the identity and the three Pauli
matrices with real coefficients,

$$
H = d_0\,I + \vec d\cdot\vec\sigma
 = d_0\,I + d_x\sigma_x + d_y\sigma_y + d_z\sigma_z,
$$

because $\{I,\sigma_x,\sigma_y,\sigma_z\}$ is a basis for the real vector space of
Hermitian $2\times 2$ matrices.[^ct-twolevel] The identity term $d_0 I$ shifts both
eigenvalues equally and never affects dynamics or measurement probabilities;
discard it. The remaining traceless part $\vec d\cdot\vec\sigma$ is $|\vec d|$
times $\hat d\cdot\vec\sigma$, formally identical to the spin coupling
$-\gamma\vec B\cdot\vec S$ with an effective field along $\hat d$. Using
$(\hat d\cdot\vec\sigma)^2 = I$, the eigenvalues are

$$
E_\pm = d_0 \pm |\vec d|,
\qquad
|\vec d| = \sqrt{d_x^2 + d_y^2 + d_z^2},
$$

and the eigenstates are the spinors pointing along $\pm\hat d$, the
$\chi_\pm(\hat d)$ built in the
[Stern–Gerlach lesson](/quantum-mechanics/spin/spin-half-pauli-matrices-and-stern-gerlach).
Every two-level problem reduces to reading off the vector $\vec d$ and applying
the spin-½ results. The gap between the levels is $2|\vec d|$, and it closes only
if all three components of $\vec d$ vanish at once.

## The Bloch sphere

A normalized state of a two-level system, up to an overall phase, is

$$
\lvert\psi\rangle = \cos\tfrac{\theta}{2}\,\lvert 0\rangle
 + e^{i\varphi}\sin\tfrac{\theta}{2}\,\lvert 1\rangle,
\qquad
0\le\theta\le\pi,\quad 0\le\varphi<2\pi,
$$

with $\lvert 0\rangle,\lvert 1\rangle$ the eigenstates of $\sigma_z$. The two real
angles $(\theta,\varphi)$ are the polar and azimuthal coordinates of a point on a
unit sphere, the **Bloch sphere**. The correspondence is faithful: distinct points
give physically distinct states, and the two removed real parameters (normalization
and global phase) carry no observable content.

The point's Cartesian coordinates are the expectation values of the Pauli
operators, the **Bloch vector**

$$
\vec r = \langle\vec\sigma\rangle
 = \big(\sin\theta\cos\varphi,\ \sin\theta\sin\varphi,\ \cos\theta\big),
\qquad
|\vec r| = 1 .
$$

The north and south poles are $\lvert 0\rangle$ and $\lvert 1\rangle$; the equator
holds the equal superpositions, which differ only by the phase $\varphi$.
Orthogonal states sit at antipodal points: $\lvert 0\rangle$ and $\lvert 1\rangle$
are a full $\pi$ apart on the sphere though only $\pi/2$ apart as spinors, the
half-angle again. The density matrix packages the same information,

$$
\rho = \lvert\psi\rangle\langle\psi\rvert = \tfrac12\big(I + \vec r\cdot\vec\sigma\big),
$$

with $|\vec r| = 1$ for the pure states considered here; mixed states fill the
interior $|\vec r| < 1$ and are developed with the
density matrix.

$$
% caption: A pure two-level state maps to a point on the Bloch sphere at polar
% angle theta and azimuth phi; the Bloch vector's components are the expectation
% values of sigma_x, sigma_y, sigma_z, and the poles are the sigma_z eigenstates.
\begin{tikzpicture}[>=stealth, font=\small, scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) circle (2.1);
\draw[black, dashed] (-2.1,0) arc (180:360:2.1 and 0.58);
\draw[black] (2.1,0) arc (0:180:2.1 and 0.58);
% axes
\draw[->, black] (0,-2.5) -- (0,2.7) node[above, font=\footnotesize] {$z$};
\draw[->, black] (-2.6,-0.75) -- (2.7,0.78) node[right, font=\footnotesize] {$x$};
\draw[->, black] (-1.0,-1.05) -- (1.15,1.2) node[above, font=\footnotesize] {$y$};
% poles
\node[font=\footnotesize, anchor=south] at (-0.25,2.12) {$0$};
\node[font=\footnotesize, anchor=north] at (-0.25,-2.12) {$1$};
% state vector
\draw[->, very thick, acc] (0,0) -- (1.35,1.35);
\node[circle, fill=acc, inner sep=1.6pt] at (1.35,1.35) {};
\node[text=acc, font=\footnotesize, anchor=south west] at (1.32,1.38) {state};
% projection to equatorial plane
\draw[dashed, black] (1.35,1.35) -- (1.35,-0.02);
\draw[dashed, black] (0,0) -- (1.35,-0.02);
% polar angle arc
\draw[thick] (0,1.0) arc (90:45:1.0);
\node[font=\footnotesize, text=black, anchor=west] at (0.35,0.82) {polar};
\end{tikzpicture}
$$

## Evolution as rotation

Drop the identity term and write the Hamiltonian as
$H = \tfrac{\hbar\Omega}{2}\,\hat n\cdot\vec\sigma$, where $\hbar\Omega/2 = |\vec d|$
and $\hat n = \hat d$. The evolution operator exponentiates directly, using
$(\hat n\cdot\vec\sigma)^2 = I$ to sum the series:

$$
U(t) = e^{-iHt/\hbar}
 = e^{-i(\Omega t/2)\,\hat n\cdot\vec\sigma}
 = \cos\tfrac{\Omega t}{2}\,I - i\sin\tfrac{\Omega t}{2}\,\hat n\cdot\vec\sigma .
$$

> **Proof.** Let $A = \hat n\cdot\vec\sigma$, so $A^2 = I$. Then $A^{2k} = I$ and
> $A^{2k+1} = A$. Splitting the exponential series into even and odd powers,
> $$
> e^{-i\beta A} = \sum_k\frac{(-i\beta)^{2k}}{(2k)!}I + \sum_k\frac{(-i\beta)^{2k+1}}{(2k+1)!}A
> = \cos\beta\,I - i\sin\beta\,A,
> $$
> with $\beta = \Omega t/2$, since the even series is $\cos\beta$ and the odd
> series is $-i\sin\beta$.

Acting on the Bloch vector, this unitary rotates $\vec r$ by the angle $\Omega t$
about the axis $\hat n$, at angular rate $\Omega$. The spinor picks up half that
angle, the $SU(2)$-to-$SO(3)$ two-to-one map: a $2\pi$ rotation of the Bloch
vector returns it to itself while multiplying the spinor by $-1$, and only
$\Omega t = 4\pi$ restores the spinor. Every closed-form two-level dynamics from
the previous lessons is a special case. Larmor precession is rotation about $\hat
z$; on-resonance Rabi flopping is rotation about $\hat x$ in the rotating frame;
general driving is rotation about the tilted effective-field axis.

$$
% caption: Unitary evolution rotates the Bloch vector rigidly about the axis n at
% angular rate Omega; the state sweeps a cone around n, which is Larmor precession
% when n is the z-axis and Rabi flopping when n lies in the equatorial plane.
\begin{tikzpicture}[>=stealth, font=\small, scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) circle (2.1);
\draw[black, dashed] (-2.1,0) arc (180:360:2.1 and 0.58);
\draw[black] (2.1,0) arc (0:180:2.1 and 0.58);
% rotation axis n
\draw[->, very thick, black] (0,0) -- (0.7,2.4);
\node[text=black, font=\footnotesize, anchor=south] at (0.7,2.4) {axis $n$};
% state vector and its orbit (cone around n)
\draw[->, very thick, acc] (0,0) -- (1.9,1.05);
\node[circle, fill=acc, inner sep=1.6pt] at (1.9,1.05) {};
\node[text=acc, font=\footnotesize, anchor=west] at (1.9,1.05) {state};
% orbit ellipse tilted about n
\draw[thick] (1.9,1.05) arc (-8:150:1.35 and 0.5);
\draw[->, thick] (0.05,1.62) arc (150:200:1.35 and 0.5);
\node[font=\footnotesize, text=black] at (0.15,0.55) {rotation};
\end{tikzpicture}
$$

## Avoided crossings and level repulsion

Take a two-level system whose diagonal energies depend on a control parameter
$\lambda$ and would cross at some $\lambda_0$, with a constant coupling $v$ off the
diagonal:

$$
H(\lambda) = \begin{pmatrix} \varepsilon_1(\lambda) & v \\ v^\ast & \varepsilon_2(\lambda) \end{pmatrix},
\qquad
E_\pm(\lambda) = \bar\varepsilon \pm \tfrac12\sqrt{(\varepsilon_1-\varepsilon_2)^2 + 4|v|^2},
$$

with $\bar\varepsilon = \tfrac12(\varepsilon_1+\varepsilon_2)$. Where the bare
energies cross, $\varepsilon_1 = \varepsilon_2$, the exact eigenvalues are
separated by $E_+ - E_- = 2|v|$. The coupling prevents the levels from touching:
they approach, repel, and exchange character. The two eigenstates at the crossing
are the symmetric and antisymmetric combinations of the bare states, not the bare
states themselves. A genuine crossing of two levels requires both the diagonal
difference and the coupling to vanish at once — two conditions on the parameters,
so in a single-parameter family crossings are avoided generically.[^sn-repel]

$$
% caption: Two diabatic levels that would cross (dashed) are split by a coupling v
% into adiabatic levels (solid) that repel with a minimum gap 2|v|; the eigenstates
% exchange character as the parameter sweeps through the crossing.
\begin{tikzpicture}[>=stealth, font=\small, scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (-0.2,0) -- (6.4,0) node[right, font=\footnotesize] {parameter};
\draw[->, black] (0,-0.3) -- (0,3.4) node[above, font=\footnotesize] {energy};
% diabatic lines (dashed) crossing at (3, 1.7)
\draw[dashed, black] (0.3,0.3) -- (5.7,3.1);
\draw[dashed, black] (0.3,3.1) -- (5.7,0.3);
% adiabatic (avoided): upper and lower branches
\draw[very thick, acc] plot[domain=0.3:5.7, samples=80]
  ({\x}, {1.7 + 0.5*sqrt((0.5185*(\x-3))^2 + 0.36)});
\draw[very thick, black] plot[domain=0.3:5.7, samples=80]
  ({\x}, {1.7 - 0.5*sqrt((0.5185*(\x-3))^2 + 0.36)});
% gap arrow
\draw[<->, black!70] (3,1.4) -- (3,2.0);
\node[font=\footnotesize, anchor=west] at (3.1,1.7) {min gap};
\end{tikzpicture}
$$

## The ammonia molecule and the maser

The ammonia molecule NH$_3$ is a nitrogen atom joined to a triangle of three
hydrogens. The nitrogen has two equilibrium positions, one on each side of the
hydrogen plane, separated by a potential barrier. Call the localized states
$\lvert L\rangle$ and $\lvert R\rangle$; by symmetry they have the same energy
$E_0$, and quantum tunneling through the barrier couples them with amplitude $A$:

$$
H = \begin{pmatrix} E_0 & -A \\ -A & E_0 \end{pmatrix},
\qquad
\lvert S\rangle = \tfrac{1}{\sqrt2}\big(\lvert L\rangle + \lvert R\rangle\big),
\qquad
\lvert A\rangle = \tfrac{1}{\sqrt2}\big(\lvert L\rangle - \lvert R\rangle\big).
$$

The symmetric and antisymmetric combinations are the stationary states, with
energies $E_0 - A$ and $E_0 + A$. Tunneling lifts the degeneracy and splits the
level by $2A$. For ammonia the splitting corresponds to a transition frequency of
about $23.87\ \mathrm{GHz}$ (a microwave wavelength near $1.25\ \mathrm{cm}$), the
inversion line that drove the first maser.[^ct-ammonia] Placing molecules in the
upper state $\lvert A\rangle$ and stimulating the $23.87\ \mathrm{GHz}$ transition
gives coherent microwave amplification — the ammonia maser.

$$
% caption: The nitrogen atom sees a symmetric double-well potential; tunneling
% through the barrier splits the degenerate localized states into a lower
% symmetric level and an upper antisymmetric level separated by 2A.
\begin{tikzpicture}[>=stealth, font=\small, scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% double well potential
\draw[thick, black] plot[domain=-2.6:2.6, samples=120]
  ({\x}, {0.55*(\x*\x - 2.2)*(\x*\x - 2.2)/2.4 + 0.2});
% levels inside wells (both below the barrier top ~1.31)
\draw[very thick, dashed, black] (-2.0,1.16) -- (2.0,1.16);
\node[text=black, font=\footnotesize, anchor=west] at (2.2,1.16) {antisymmetric};
\draw[very thick, acc] (-1.75,0.82) -- (1.75,0.82);
\node[text=acc, font=\footnotesize, anchor=west] at (2.2,0.82) {symmetric};
% splitting arrow
\draw[<->, black!70] (1.15,0.82) -- (1.15,1.16);
\node[font=\footnotesize, anchor=west] at (1.25,0.99) {$2A$};
% well labels
\node[font=\footnotesize] at (-1.45,-0.15) {left};
\node[font=\footnotesize] at (1.45,-0.15) {right};
\end{tikzpicture}
$$

A molecule prepared in $\lvert L\rangle$ is not stationary. Writing it as
$\lvert L\rangle = \tfrac{1}{\sqrt2}(\lvert S\rangle + \lvert A\rangle)$ and letting
the two stationary states accumulate their phases, the probability of finding the
nitrogen still on the left is

$$
P_L(t) = \cos^2\!\Big(\frac{At}{\hbar}\Big),
$$

a complete oscillation between the two wells at angular frequency $2A/\hbar$. This
is the two-level flopping of the previous lesson with the tunneling amplitude
playing the role of the drive: the nitrogen tunnels back and forth through the
barrier, and the same $\cos^2$ law that governed Rabi inversion governs the
inversion of the molecule.

$$
% caption: A nitrogen atom started on the left tunnels coherently to the right and
% back; the probability of finding it on the left oscillates as cos^2(At/hbar) with
% period pi*hbar/A, the two-level flopping driven by the tunneling amplitude.
\begin{tikzpicture}[>=stealth, font=\small, scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (7.0,0) node[right, font=\footnotesize] {time};
\draw[->, black] (0,0) -- (0,3.0) node[above, font=\footnotesize] {left probability};
\node[font=\footnotesize, anchor=east] at (0,2.6) {$1$};
\node[font=\footnotesize, anchor=east] at (0,1.3) {$\tfrac{1}{2}$};
\draw[dashed, black] (0,2.6) -- (7.0,2.6);
\draw[very thick, acc] plot[domain=0:6.8, samples=140]
  ({\x}, {2.6*(cos(deg(3.1416*\x/3.4)))^2});
\node[text=acc, font=\footnotesize, anchor=south west] at (3.6,0.15) {tunnels to right};
\end{tikzpicture}
$$

## The qubit

A two-level system with a controllable Hamiltonian is a **qubit**. Its state is a
point on the Bloch sphere; the computational basis is the pair of poles
$\lvert 0\rangle,\lvert 1\rangle$; and a measurement in that basis returns $0$ or
$1$ with probabilities $\cos^2(\theta/2)$ and $\sin^2(\theta/2)$. Any single-qubit
operation is a unitary, hence a Bloch-sphere rotation, realized physically by
turning on a Hamiltonian $\tfrac{\hbar\Omega}{2}\hat n\cdot\vec\sigma$ for a chosen
duration — exactly the driven-spin control of magnetic resonance. The spin-½
algebra assembled across this module is therefore the operational core of a single
qubit; entangling two of them turns the single Bloch vector into a joint state whose
correlations no pair of individual vectors can reproduce.

[^ct-twolevel]: Cohen-Tannoudji, Diu & Laloë, _Quantum Mechanics_, Vol. I (Wiley,
1977), Ch. IV §C. The decomposition $H = d_0 I + \vec d\cdot\vec\sigma$ is exact
for any Hermitian $2\times 2$ matrix; the physics lives entirely in the real
three-vector $\vec d$.

[^sn-repel]: Sakurai & Napolitano, _Modern Quantum Mechanics_, 3rd ed. (Cambridge,
2021), §3.2 and the discussion of the von Neumann–Wigner non-crossing rule: a
crossing of two nondegenerate levels requires tuning two independent parameters,
so it does not occur generically in a one-parameter family.

[^ct-ammonia]: Cohen-Tannoudji, Diu & Laloë, _Quantum Mechanics_, Vol. I (Wiley,
1977), Ch. IV complements on the ammonia molecule. The inversion transition near
$23.87\ \mathrm{GHz}$ was the basis of the first maser (Gordon, Zeiger & Townes,
1954); the value is a measured molecular constant.