Equivalent Circuits

Two quantum circuits are equivalent when they perform exactly the same transformation on every input state — that is, when they implement the same unitary matrix $U$ . Recognising equivalent circuits is a practical skill: compilers use it to swap an expensive gate for a cheaper sequence of gates, and hardware providers publish tables of such identities so you can target their native gate set.

Phase gates and how they compose

The $T$ gate (sometimes called the $\pi/8$ gate) advances the phase of $|1\rangle$ by $\pi/4$ :

T = \begin{pmatrix} 1 & 0 \\ 0 & e^{i\pi/4} \end{pmatrix}.

Applying $T$ twice means multiplying the matrix by itself:

T^2 = \begin{pmatrix} 1 & 0 \\ 0 & e^{i\pi/4} \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & e^{i\pi/4} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & e^{i\pi/2} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & i \end{pmatrix}.

That last matrix is the $S$ gate (also written $P(\pi/2)$ ). In other words, two consecutive $T$ gates are circuit-equivalent to a single $S$ gate. Both circuits act identically on every qubit state, so no experiment can distinguish them.

By the same logic, $S^2 = Z$ and $T^4 = Z$ , giving a small family of related identities:

T^2 = S, \qquad S^2 = Z, \qquad T^4 = Z, \qquad T^8 = I.

Each equality is a statement about unitary matrices, and each can be verified by multiplying the $2 \times 2$ matrices together.

What the grader checks

The unitary grader computes the full $2 \times 2$ complex matrix of your circuit and compares it to the expected matrix — in this case $S$ . An empty circuit (identity $I$ ) does not match $S$ , so you must actually apply the gates.

Try it

Implement the $S$ gate using only two $T$ gate calls. The grader will compare the unitary of your circuit against $S$ directly.

Run your code to see the quantum state.

After running, inspect the Matrix tab to see the $2 \times 2$ unitary. The bottom-right entry should read $i \approx 0 + 1i$ , confirming you have realized $S = T^2$ .