Project: State Tomography (Intro)

Every qubit state corresponds to a point on the Bloch sphere: a unit sphere where the north pole is $|0\rangle$ , the south pole is $|1\rangle$ , and every other point is a superposition. Knowing where a state lives on that sphere means knowing its Bloch vector $\vec{r} = (\langle X \rangle, \langle Y \rangle, \langle Z \rangle)$ , the triple of expectation values of the three Pauli operators.

A single measurement cannot reveal all three components — each measurement collapses the state and gives only a single bit. State tomography gets around this by repeating the preparation many times and measuring in three different bases, extracting one component per basis.

The three basis measurements

Z basis (the default)

Measure the qubit as-is. The Born rule gives

P(0) = \tfrac{1}{2}(1 + \langle Z \rangle), \qquad P(1) = \tfrac{1}{2}(1 - \langle Z \rangle),

\langle Z \rangle = P(0) - P(1).

For $|0\rangle$ this yields $\langle Z \rangle = 1$ (north pole); for $|1\rangle$ it yields $-1$ (south pole); for $|+\rangle$ it gives $0$ (equator).

X basis

Applying a Hadamard before measuring rotates the $X$ axis onto the $Z$ axis:

H\,|+\rangle = |0\rangle, \qquad H\,|-\rangle = |1\rangle.

The measurement outcome then encodes the $X$ component:

\langle X \rangle = P(0\,|\,\text{after }H) - P(1\,|\,\text{after }H).

For $|+\rangle$ , applying $H$ gives $|0\rangle$ , so the outcome is always $0$ and $\langle X \rangle = 1$ . The Bloch vector tip is on the $+x$ axis — exactly as expected.

Y basis

To extract $\langle Y \rangle$ , first apply $S^\dagger$ (the inverse phase gate) and then $H$ . Together the two gates satisfy $(HS^\dagger)\,Y\,(HS^\dagger)^\dagger = Z$ , so the composite rotation maps $Y \to Z$ for measurement purposes. The protocol is identical in structure:

\langle Y \rangle = P(0\,|\,\text{after }S^\dagger H) - P(1\,|\,\text{after }S^\dagger H).

Together the three measurements reconstruct $\vec{r}$ completely. For a pure state $|\vec{r}| = 1$ ; for a mixed state $|\vec{r}| < 1$ .

Try it

Measure $|+\rangle$ in the X basis. Prepare the state with a Hadamard, rotate to the Z basis with a second Hadamard, then measure. The grader checks the output distribution: because $H|+\rangle = |0\rangle$ , the result should be $0$ with probability $1$ , confirming $\langle X \rangle = 1$ .

Run your code to see the quantum state.

After running, open the Probabilities panel. The bar for outcome 0 should be at $1.0$ and the bar for 1 at $0.0$ . That single number — $\langle X \rangle = 1 - 0 = 1$ — tells you the Bloch vector is exactly on the $+x$ axis, which is the definition of $|+\rangle$ .