Reading the Bloch Display

After learning what the Bloch sphere represents mathematically, the next step is reading it as a live display while you run circuits. The simulator renders the Bloch sphere in real time and reports three numbers: the x, y, z components of the Bloch vector. Knowing how to decode those numbers back into the qubit state is the skill this lesson develops.

The Bloch vector and what it shows

Every single-qubit pure state $|\psi\rangle = \cos\!\tfrac{\theta}{2}|0\rangle + e^{i\phi}\sin\!\tfrac{\theta}{2}|1\rangle$ maps to a point on the unit sphere with Cartesian coordinates

x = \sin\theta\cos\phi, \qquad y = \sin\theta\sin\phi, \qquad z = \cos\theta.

These three numbers are what the Bloch display shows. You can read the state back from them:

z tells you about vertical position — $z = +1$ is $|0\rangle$ , $z = -1$ is $|1\rangle$ , $z = 0$ means the qubit is on the equator (equal probability of each outcome).
x and y describe the azimuthal position on the equator. If $z = 0$ , then $x = \cos\phi$ and $y = \sin\phi$ , so you can recover the relative phase directly from the angle $\phi = \arctan(y/x)$ .

Reading specific states

Checking your Bloch display after a circuit is a quick sanity test:

| State | x | y | z | |---|---|---|---| | $|0\rangle$ | $0$ | $0$ | $+1$ | | $|1\rangle$ | $0$ | $0$ | $-1$ | | $|{+}\rangle = H|0\rangle$ | $+1$ | $0$ | $0$ | | $|{-}\rangle = H|1\rangle$ | $-1$ | $0$ | $0$ | | $|{+i}\rangle = SH|0\rangle$ | $0$ | $+1$ | $0$ | | $TH|0\rangle = \tfrac{1}{\sqrt{2}}\bigl(|0\rangle+e^{i\pi/4}|1\rangle\bigr)$ | $\tfrac{1}{\sqrt{2}}$ | $\tfrac{1}{\sqrt{2}}$ | $0$ |

The last row shows a state halfway between the $+x$ and $+y$ axes. Its Bloch vector is exactly $\hat{n} = (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0)$ , with $x = y \approx 0.707$ and $z = 0$ . You can derive this from $\theta = \pi/2$ , $\phi = \pi/4$ :

x = \sin\!\tfrac{\pi}{2}\cos\!\tfrac{\pi}{4} = 1 \cdot \tfrac{1}{\sqrt{2}} = \tfrac{1}{\sqrt{2}}, \qquad y = \sin\!\tfrac{\pi}{2}\sin\!\tfrac{\pi}{4} = 1 \cdot \tfrac{1}{\sqrt{2}} = \tfrac{1}{\sqrt{2}}.

Try it

Prepare the state $\tfrac{1}{\sqrt{2}}\bigl(|0\rangle + e^{i\pi/4}|1\rangle\bigr)$ — the equator point midway between the $+x$ and $+y$ axes. After running, your Bloch display should show $x \approx 0.707$ , $y \approx 0.707$ , $z = 0$ .

Run your code to see the quantum state.

Inspect all three Bloch components. If the display shows $x = y \approx 0.707$ and $z = 0$ , you have correctly placed the Bloch vector at azimuthal angle $\phi = \pi/4$ on the equator — and you know how to read the display.