Quantum State Tomography

A quantum state is described by a density matrix $\rho$, but a single measurement returns only one classical bit per qubit. State tomography is the procedure for reconstructing the entire matrix $\rho$ from the statistics of many repeated measurements in carefully chosen bases.

Expanding $\rho$ in the Pauli basis

Any single-qubit density matrix can be written as a real combination of the identity and the three Pauli matrices:

The three numbers $(r_x, r_y, r_z)$ are exactly the Bloch vector of the state. Reconstructing $\rho$ therefore reduces to measuring the three expectation values $\langle X\rangle$, $\langle Y\rangle$, and $\langle Z\rangle$. Each is bounded in $[-1, 1]$, and a valid pure state has $r_x^2 + r_y^2 + r_z^2 = 1$.

Measuring in three bases

The simulator — like real hardware — only measures in the computational ($Z$) basis. To estimate the other two expectation values we rotate the desired axis onto the $Z$ axis before measuring:

• $\langle Z\rangle$: measure directly. With outcome probabilities $p_0, p_1$ we have $\langle Z\rangle = p_0 - p_1$.
• $\langle X\rangle$: apply a Hadamard $H$ first (it maps the $X$ eigenstates $|\pm\rangle$ to $|0\rangle, |1\rangle$), then measure.
• $\langle Y\rangle$: apply $S^\dagger$ then $H$ (this maps the $Y$ eigenbasis onto the computational basis), then measure.

Running each rotated circuit for many shots gives an estimate of the corresponding expectation value, and from the three values you assemble $\rho$. With finite shots the estimate has statistical error $\sim 1/\sqrt{N}$, and the raw reconstruction can even fall slightly outside the Bloch ball — which is why practitioners follow up with a maximum-likelihood fit that forces $\rho$ to be a valid (positive, unit-trace) density matrix.

The measurement-rotation primitive

The single most important mechanical step in tomography is the basis rotation. Consider the state $|+\rangle = \tfrac{1}{\sqrt 2}(|0\rangle + |1\rangle)$. Measured directly in the $Z$ basis it is maximally random ($p_0 = p_1 = \tfrac12$), revealing only that $\langle Z\rangle = 0$. But $|+\rangle$ is the $+1$ eigenstate of $X$, so if we first apply $H$ we get

$H|+\rangle = |0\rangle,$

and the measurement is now deterministic: outcome $0$ with probability $1$, i.e. $\langle X\rangle = +1$. That single rotated circuit nails the $X$ component of the Bloch vector exactly.

Try it

Prepare $|+\rangle$ on one qubit, then perform an $X$-basis measurement by rotating the $X$ axis onto $Z$ before measuring. The grader checks the resulting computational-basis distribution: a correct $X$-basis measurement of $|+\rangle$ yields outcome $0$ with probability $1$.

After running, open the Probabilities tab: the bar for 0 should reach $1.0$, confirming $\langle X\rangle = +1$.

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