Checkpoint: Navigate the Bloch Sphere

Throughout this module you learned that every single-qubit gate corresponds to a rotation of the Bloch sphere. Preparing a specific target state is therefore a navigation problem: you must choose a sequence of rotations that carries the starting point $|0\rangle$ (the north pole) to the intended destination.

The target state

The target for this checkpoint is

|\psi_{\text{target}}\rangle = \frac{1}{\sqrt{2}}\bigl(|0\rangle + e^{i\pi/4}|1\rangle\bigr).

In Bloch sphere coordinates this state sits at polar angle $\theta = \pi/2$ (on the equator) and azimuthal angle $\phi = \pi/4$ . That places it exactly 45° east of the $+x$ axis state $|+\rangle$ and 45° west of the $+y$ axis state $|{+i}\rangle$ .

Planning the route

The north pole $|0\rangle$ has $\theta = 0$ , far from the equator. To reach $\theta = \pi/2$ you need a gate that moves the Bloch vector off the $z$ -axis. The Hadamard $H$ does exactly that: it is a 180° rotation about the $(x{+}z)/\sqrt{2}$ axis, sending the north pole straight to the $+x$ equatorial point $|+\rangle$ ( $\phi = 0$ ).

After landing on the equator you still need to advance the azimuthal angle from $0$ to $\pi/4$ . The phase gates $Z$ , $S$ , and $T$ are all z-rotations and act purely on $\phi$ . The $T$ gate rotates by $\pi/4$ (one eighth-turn), which is precisely the remaining distance:

T = \begin{pmatrix} 1 & 0 \\ 0 & e^{i\pi/4} \end{pmatrix}, \qquad T|{+}\rangle = \frac{1}{\sqrt{2}}\bigl(|0\rangle + e^{i\pi/4}|1\rangle\bigr) = |\psi_{\text{target}}\rangle.

The two-gate sequence $T \cdot H$ applied to $|0\rangle$ therefore hits the target exactly.

Confirming the route

You can verify the claim step by step. Starting from $|0\rangle = \binom{1}{0}$ :

H|0\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix},

then

T \cdot \frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix}1 \cdot 1 \\ e^{i\pi/4} \cdot 1\end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ e^{i\pi/4}\end{pmatrix}.

The resulting amplitude vector has $|\alpha|^2 = |\beta|^2 = \tfrac{1}{2}$ , confirming that measurement probabilities remain equal. The relative phase $e^{i\pi/4}$ encodes the azimuthal position on the equator.

Try it

Navigate from $|0\rangle$ to the target state $\frac{1}{\sqrt{2}}\bigl(|0\rangle + e^{i\pi/4}|1\rangle\bigr)$ . The grader checks the full statevector (up to global phase), so leaving the circuit empty will not pass — you must reach the correct point on the sphere.

Run your code to see the quantum state.

After running your circuit, check that the $|0\rangle$ amplitude is $\approx 0.707$ (real) and the $|1\rangle$ amplitude is $\approx 0.707\,e^{i\pi/4} = 0.5 + 0.5i$ (complex). That is the exact signature of the target Bloch vector.