Differences

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--- ket-psi-plus [June 13, 2026 at 01:59] – Ivan Janevski
+++ ket-psi-plus [June 13, 2026 at 03:13] (current) – external edit 127.0.0.1
@@ Line 4: / Line 4: @@
 $$\lvert\Psi^+\rangle = \frac{1}{\sqrt{2}}(\lvert 01\rangle + \lvert 10\rangle) = \frac{1}{\sqrt{2}}\begin{pmatrix}0\\1\\1\\0\end{pmatrix}$$
-It is prepared from $\lvert 01\rangle$ by applying $H\otimes I$ then CNOT, or equivalently from $\lvert 00\rangle$ by first flipping qubit 1 with $X$, then applying $H\otimes I$ and CNOT. Applying $X$ to either qubit of [[ket-phi-plus|$\lvert\Phi^+\rangle$]] also reaches $\lvert\Psi^+\rangle$. In the X basis, $\lvert\Psi^+\rangle = \tfrac{1}{\sqrt{2}}(\lvert ++\rangle - \lvert --\rangle)$, giving same-value X-basis correlations despite the anti-correlations in Z.
+It is prepared from $\lvert 01\rangle$ by applying $H\otimes I$ then CX, or equivalently from $\lvert 00\rangle$ by first flipping qubit 1 with $X$, then applying $H\otimes I$ and CX. Applying $X$ to either qubit of [[ket-phi-plus|$\lvert\Phi^+\rangle$]] also reaches $\lvert\Psi^+\rangle$. In the X basis, $\lvert\Psi^+\rangle = \tfrac{1}{\sqrt{2}}(\lvert ++\rangle - \lvert --\rangle)$, giving same-value X-basis correlations despite the anti-correlations in Z.
 ## Qiskit
 ```python
-# Prepare |Ψ+⟩ = (|01⟩ + |10⟩)/√2 — X on qubit 1, then H on qubit 0, then CNOT(0→1).
+# Prepare |Ψ+⟩ = (|01⟩ + |10⟩)/√2 — X on qubit 1, then H on qubit 0, then CX(0→1).
 from qiskit import QuantumCircuit
 from qiskit.quantum_info import Statevector
@@ Line 16: / Line 16: @@
 qc.x(1)      # |00⟩ → |01⟩
 qc.h(0)      # |01⟩ → (|01⟩ + |11⟩)/√2
-qc.cx(0, 1)  # CNOT: → (|01⟩ + |10⟩)/√2 = |Ψ+⟩
+qc.cx(0, 1)  # CX: → (|01⟩ + |10⟩)/√2 = |Ψ+⟩
 print(Statevector(qc).data)
@@ Line 30: / Line 30: @@
 | [[y-gate]] | $Y_1\lvert\Psi^+\rangle = i\lvert\Phi^-\rangle$ | Bit-flip with phase; the $-i$ on the $\lvert 00\rangle$ term introduces a relative minus sign, giving $\lvert\Phi^-\rangle$ up to global phase. |
 | [[z-gate]] | $Z_1\lvert\Psi^+\rangle = \lvert\Psi^-\rangle$ | Negates the $\lvert 1\rangle$ component of qubit 1, which appears only in $\lvert 10\rangle$, flipping the relative sign to antisymmetric. |
-| [[h-gate]] | $H_1\lvert\Psi^+\rangle$ (not a Bell state) | Bell measurement circuit: CNOT then $H_1$ maps $\lvert\Psi^+\rangle\to\lvert 01\rangle$. |
+| [[h-gate]] | $H_1\lvert\Psi^+\rangle$ (not a Bell state) | Bell measurement circuit: CX then $H_1$ maps $\lvert\Psi^+\rangle\to\lvert 01\rangle$. |
 | [[s-gate]] | $S_1\lvert\Psi^+\rangle = \tfrac{1}{\sqrt{2}}(\lvert 01\rangle + i\lvert 10\rangle)$ | Adds a phase of $i$ to the $\lvert 10\rangle$ term; not a standard Bell state. |
 | [[t-gate]] | $T_1\lvert\Psi^+\rangle = \tfrac{1}{\sqrt{2}}(\lvert 01\rangle + e^{i\pi/4}\lvert 10\rangle)$ | Adds a phase of $e^{i\pi/4}$ to the $\lvert 10\rangle$ term; not a standard Bell state. |
@@ Line 36: / Line 36: @@
 | [[ry-gate]] | Mixes $\lvert\Psi^+\rangle$ with $\lvert\Phi^-\rangle$ | Y-type rotations couple the $\lvert\Psi^+\rangle$ and $\lvert\Phi^-\rangle$ Bell states. |
 | [[rz-gate]] | $R_z(\theta)_1\lvert\Psi^+\rangle = \tfrac{e^{-i\theta/2}}{\sqrt{2}}(\lvert 01\rangle + e^{i\theta}\lvert 10\rangle)$ | Modifies relative phase only; at $\theta=\pi$ gives $\lvert\Psi^-\rangle$ up to global phase. |
-| [[cnot-gate]] | $\text{CNOT}\lvert\Psi^+\rangle = \lvert +\rangle\lvert 1\rangle$ | Disentangles $\lvert\Psi^+\rangle$ back to the product state used to prepare it. |
+| [[cx-gate]] | $\text{CX}\lvert\Psi^+\rangle = \lvert +\rangle\lvert 1\rangle$ | Disentangles $\lvert\Psi^+\rangle$ back to the product state used to prepare it. |
-| [[cwap-gate]] | $\text{SWAP}\lvert\Psi^+\rangle = \lvert\Psi^+\rangle$ | Symmetric under qubit exchange; eigenstate of SWAP with eigenvalue $+1$. |
+| [[swap-gate]] | $\text{SWAP}\lvert\Psi^+\rangle = \lvert\Psi^+\rangle$ | Symmetric under qubit exchange; eigenstate of SWAP with eigenvalue $+1$. |
 | [[iswap-gate]] | $\text{iSWAP}\lvert\Psi^+\rangle = i\lvert\Psi^+\rangle$ | Both terms pick up a factor of $i$; eigenstate of iSWAP with eigenvalue $i$. |