Spin and the Bloch Sphere

Every pure spin-$\tfrac{1}{2}$ state lives in a two-dimensional complex Hilbert space, and every state in that space has a natural home on a unit sphere called the Bloch sphere. Understanding this map turns abstract state vectors into geometric objects you can visualize and reason about intuitively.

The general spin-½ state

The most general normalized single-qubit (spin-$\tfrac{1}{2}$) state can be written

where $\theta \in [0, \pi]$ and $\phi \in [0, 2\pi)$ are real angles. The phase factors are chosen so that the coefficient of $|0\rangle$ can always be taken as real and non-negative, which removes the unobservable global phase from the parametrization. You can verify normalization directly:

Mapping angles to the sphere

The pair $(\theta, \phi)$ are precisely the polar and azimuthal angles of a point on the unit sphere. We define the Bloch vector $\hat{n} = (\sin\theta\cos\phi,\,\sin\theta\sin\phi,\,\cos\theta)$ and assign:

• North pole $(\theta = 0)$: $|\psi\rangle = |0\rangle$ — spin-up along $z$, eigenstate of $S_z$ with eigenvalue $+\hbar/2$.
• South pole $(\theta = \pi)$: $|\psi\rangle = e^{i\phi}|1\rangle \sim |1\rangle$ — spin-down along $z$, eigenvalue $-\hbar/2$.
• Equator $(\theta = \pi/2)$: equal-weight superpositions, spin pointing along a horizontal axis.

The key points on the equator have direct physical meaning:

| $\phi$ | State | Physical interpretation | |--------|-------|------------------------| | $0$ | $\tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle) = |{+x}\rangle$ | Spin-up along $+x$ | | $\pi$ | $\tfrac{1}{\sqrt{2}}(|0\rangle - |1\rangle) = |{-x}\rangle$ | Spin-down along $+x$ | | $\pi/2$ | $\tfrac{1}{\sqrt{2}}(|0\rangle + i|1\rangle) = |{+y}\rangle$ | Spin-up along $+y$ | | $3\pi/2$ | $\tfrac{1}{\sqrt{2}}(|0\rangle - i|1\rangle) = |{-y}\rangle$ | Spin-down along $+y$ |

Why a sphere and not a circle?

A general vector $\alpha|0\rangle + \beta|1\rangle$ has two complex amplitudes, i.e. four real degrees of freedom. Normalization $|\alpha|^2 + |\beta|^2 = 1$ removes one, leaving three. The remaining reduction comes from the fact that the overall global phase $e^{i\gamma}$ of a state vector is unobservable (it cancels in every Born-rule probability), so two vectors related by a global phase represent the same physical state. Dividing out this equivalence removes one more degree of freedom, leaving two. Those two free real parameters — equivalently the modulus and argument of the global-phase-invariant ratio $\beta/\alpha$ — are exactly the angles $\theta$ and $\phi$. The result is a two-sphere $S^2$ — the Bloch sphere.

Spin eigenstates as Bloch vectors

When you measure $S_{\hat{n}} = \hat{n} \cdot \mathbf{S}$ (spin along an arbitrary unit vector $\hat{n}$), the eigenstates with eigenvalue $+\hbar/2$ and $-\hbar/2$ correspond to the Bloch vectors $+\hat{n}$ and $-\hat{n}$ respectively. In other words, antipodal points on the Bloch sphere are orthogonal states. This is a distinguishing feature of spin-$\tfrac{1}{2}$: in ordinary $3$-d space, orthogonal vectors are $90°$ apart, but orthogonal spin states are $180°$ apart on the Bloch sphere.

Measurement probabilities from the Bloch vector

Given a state $|\psi\rangle$ with Bloch vector $\hat{n} = (\sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta)$, the probability of obtaining spin-up ($+\hbar/2$) when measuring along the $z$-axis is

For $\theta = 0$ (north pole, $|0\rangle$) this gives $P(+z) = 1$, as expected. For $\theta = \pi/2$ (equator) it gives $P(+z) = \cos^2(\pi/4) = 1/2$, the 50/50 result for a spin state pointing along a horizontal axis and measured along $z$. For $\theta = \pi$ (south pole, $|1\rangle$) it gives $P(+z) = \cos^2(\pi/2) = 0$.

Try it

The equator of the Bloch sphere represents spin states pointing along horizontal axes. Prepare the spin eigenstate $|{+x}\rangle$ (spin-up along the $x$-axis), which has $\theta = \pi/2$ and $\phi = 0$:

Starting from $|0\rangle$ (north pole), apply the single gate that lands your Bloch vector on the $+x$ point of the equator.

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