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--- two-qubits [June 13, 2026 at 02:57] – external edit 127.0.0.1
+++ two-qubits [June 13, 2026 at 03:23] (current) – Ivan Janevski
@@ Line 6: / Line 6: @@
 A general two-qubit state is a superposition over the four computational basis states, with the normalization condition saying the total probability must be 1:
-$$\lvert\psi\rangle = c_{00}\lvert 00\rangle + c_{01}\lvert 01\rangle + c_{10}\lvert 10\rangle + c_{11}\lvert 11\rangle, \qquad |c_{00}|^2 + |c_{01}|^2 + |c_{10}|^2 + |c_{11}|^2 = 1$$
+$$\lvert\psi\rangle = \begin{pmatrix}c_{00}\\c_{01}\\c_{10}\\c_{11}\end{pmatrix}, \qquad \lvert\psi\rangle = c_{00}\underbrace{\begin{pmatrix}1\\0\\0\\0\end{pmatrix}}_{\lvert 00\rangle} + c_{01}\underbrace{\begin{pmatrix}0\\1\\0\\0\end{pmatrix}}_{\lvert 01\rangle} + c_{10}\underbrace{\begin{pmatrix}0\\0\\1\\0\end{pmatrix}}_{\lvert 10\rangle} + c_{11}\underbrace{\begin{pmatrix}0\\0\\0\\1\end{pmatrix}}_{\lvert 11\rangle}, \qquad |c_{00}|^2 + |c_{01}|^2 + |c_{10}|^2 + |c_{11}|^2 = 1$$
 ## Product states
@@ Line 32: / Line 32: @@
 | [[ket-11|$\lvert 11\rangle$]] | Both qubits are $\lvert 1\rangle$. |
-### X-basis
+### X-basis (Hadamard basis)
 Pick [[ket-plus|$\lvert +\rangle$]] and [[ket-minus|$\lvert -\rangle$]] for each qubit instead. These states look "spread out" in the computational basis, because [[ket-plus|$\lvert +\rangle$]] $= (\lvert 0\rangle + \lvert 1\rangle)/\sqrt{2}$ mixes both. But in the X basis they are perfectly sharp, just like $\lvert 00\rangle$ is sharp in the Z basis. You get from the computational basis to this one by applying a Hadamard to every qubit.
@@ Line 48: / Line 48: @@
 | $\lvert --\rangle$ | $\tfrac{1}{2}(\lvert 00\rangle - \lvert 01\rangle - \lvert 10\rangle + \lvert 11\rangle)$ | Both qubits in [[ket-minus|$\lvert -\rangle$]]. |
-### Y-basis
+### Y-basis (Phase basis)
 Pick [[ket-plus-i|$\lvert +i\rangle$]] and [[ket-minus-i|$\lvert -i\rangle$]] for each qubit. These are the eigenstates of the Y gate. The coefficients are complex, so the Y-basis states are the first ones on this page with imaginary entries. You get from the computational basis to the Y basis by applying $SH$ to every qubit ($H$ first, then $S$).
@@ Line 81: / Line 81: @@
 ^ State ^ Definition ^ Correlations ^
-| [[ket-phi-plus|$\lvert\Phi^+\rangle$]] | $\tfrac{1}{\sqrt{2}}(\lvert 00\rangle + \lvert 11\rangle)$ | Same-value in Z and X; anti-correlated in Y. Prepared from $\lvert 00\rangle$ by $H\otimes I$ then CNOT. |
+| [[ket-phi-plus|$\lvert\Phi^+\rangle$]] | $\tfrac{1}{\sqrt{2}}(\lvert 00\rangle + \lvert 11\rangle)$ | Same-value in Z and X; anti-correlated in Y. Prepared from $\lvert 00\rangle$ by $H\otimes I$ then CX. |
 | [[ket-phi-minus|$\lvert\Phi^-\rangle$]] | $\tfrac{1}{\sqrt{2}}(\lvert 00\rangle - \lvert 11\rangle)$ | Same-value in Z; anti-correlated in X. The minus sign is invisible in Z-basis measurements. |
-| [[ket-psi-plus|$\lvert\Psi^+\rangle$]] | $\tfrac{1}{\sqrt{2}}(\lvert 01\rangle + \lvert 10\rangle)$ | Anti-correlated in Z; same-value in X. Prepared from $\lvert 01\rangle$ by $H\otimes I$ then CNOT. |
+| [[ket-psi-plus|$\lvert\Psi^+\rangle$]] | $\tfrac{1}{\sqrt{2}}(\lvert 01\rangle + \lvert 10\rangle)$ | Anti-correlated in Z; same-value in X. Prepared from $\lvert 01\rangle$ by $H\otimes I$ then CX. |
 | [[ket-psi-minus|$\lvert\Psi^-\rangle$]] | $\tfrac{1}{\sqrt{2}}(\lvert 01\rangle - \lvert 10\rangle)$ | Anti-correlated in every basis. The only antisymmetric Bell state; also called the singlet. |
@@ Line 99: / Line 99: @@
 ## Gates
-Two-qubit gates are $4\times 4$ unitary matrices acting on $\mathbb{C}^4$. The three most common ones are CX (CNOT), CZ, and SWAP. Row and column ordering follows the computational basis: $\lvert 00\rangle, \lvert 01\rangle, \lvert 10\rangle, \lvert 11\rangle$.
+Two-qubit gates are $4\times 4$ unitary matrices acting on $\mathbb{C}^4$. The three most common ones are CX (CX), CZ, and SWAP. Row and column ordering follows the computational basis: $\lvert 00\rangle, \lvert 01\rangle, \lvert 10\rangle, \lvert 11\rangle$.
 $$CX = \begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix}
@@ Line 118: / Line 118: @@
 qc = QuantumCircuit(2, 2)
 qc.h(0)       # Hadamard on qubit 0 → superposition
-qc.cx(0, 1)   # CNOT: control=qubit 0, target=qubit 1 → entanglement
+qc.cx(0, 1)   # CX: control=qubit 0, target=qubit 1 → entanglement
 qc.measure([0, 1], [0, 1])