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--- zero-state [June 12, 2026 at 22:15] – Ivan Janevski
+++ zero-state [Unknown date] (current) – removed - external edit (Unknown date) 127.0.0.1
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-# $\lvert 0\rangle$ (Zero state)
-The **zero state** $\lvert 0\rangle$ is one of the two computational basis states of a qubit. It is the quantum analogue of a classical `0` bit, and it is the standard initial state used in most quantum circuits. The other computational basis state is [[one-state|$\lvert 1\rangle$]].
-$$\lvert 0\rangle = \begin{pmatrix}1\\0\end{pmatrix}$$
-On the [[bloch-sphere]], $\lvert 0\rangle$ corresponds to the north pole at coordinates $(0, 0, 1)$. It is an eigenstate of the Pauli-Z gate with eigenvalue $+1$, meaning $Z\lvert 0\rangle = \lvert 0\rangle$. Applying the [[h-gate|Hadamard gate]] to $\lvert 0\rangle$ produces the equal superposition state $\lvert +\rangle = (\lvert 0\rangle + \lvert 1\rangle)/\sqrt{2}$.
-## Applying gates
-^ Gate ^ Result ^ Comment ^
-| [[i-gate]] | $I\lvert 0\rangle = \lvert 0\rangle$ | The identity gate leaves $\lvert 0\rangle$ unchanged. |
-| [[x-gate]] | $X\lvert 0\rangle = \lvert 1\rangle$ | Flips $\lvert 0\rangle$ to $\lvert 1\rangle$; quantum analogue of classical NOT. |
-| [[y-gate]] | $Y\lvert 0\rangle = i\lvert 1\rangle$ | Bit flip with an imaginary phase factor. |
-| [[z-gate]] | $Z\lvert 0\rangle = \lvert 0\rangle$ | $\lvert 0\rangle$ is an eigenstate of $Z$ with eigenvalue $+1$. |
-| [[h-gate]] | $H\lvert 0\rangle = \lvert +\rangle$ | Rotates the north pole to the $+x$ equatorial point of the Bloch sphere. |
-| [[s-gate]] | $S\lvert 0\rangle = \lvert 0\rangle$ | Phase only affects the $\lvert 1\rangle$ component; $\lvert 0\rangle$ is unchanged. |
-| [[t-gate]] | $T\lvert 0\rangle = \lvert 0\rangle$ | Phase only affects the $\lvert 1\rangle$ component; $\lvert 0\rangle$ is unchanged. |
-| [[rx-gate|$R_x(\theta)$]] | $R_x(\theta)\lvert 0\rangle = \cos\tfrac{\theta}{2}\lvert 0\rangle - i\sin\tfrac{\theta}{2}\lvert 1\rangle$ | Tilts the state from $\lvert 0\rangle$ toward $\lvert 1\rangle$ with an imaginary phase on the $\lvert 1\rangle$ component. |
-| [[ry-gate|$R_y(\theta)$]] | $R_y(\theta)\lvert 0\rangle = \cos\tfrac{\theta}{2}\lvert 0\rangle + \sin\tfrac{\theta}{2}\lvert 1\rangle$ | Real amplitudes; at $\theta=\pi/2$ gives $\lvert +\rangle$, the same result as the Hadamard gate up to global phase. |
-| [[rz-gate|$R_z(\theta)$]] | $R_z(\theta)\lvert 0\rangle = e^{-i\theta/2}\lvert 0\rangle$ | Global phase only; $\lvert 0\rangle$ is on the $z$-axis so a $z$-rotation has no observable effect. |
-| [[u-gate|$U(\theta,\phi,\lambda)$]] | $U\lvert 0\rangle = \cos\tfrac{\theta}{2}\lvert 0\rangle + e^{i\phi}\sin\tfrac{\theta}{2}\lvert 1\rangle$ | $\lambda$ drops out; every single-qubit state is reachable from $\lvert 0\rangle$ with appropriate $(\theta,\phi)$. |
-| [[cnot-gate|CNOT]] (as control) | $\text{CNOT}\lvert 0\rangle\lvert t\rangle = \lvert 0\rangle\lvert t\rangle$ | Control is off; target qubit $\lvert t\rangle$ is always left unchanged. |
-| [[cnot-gate|CNOT]] (as target) | $\text{CNOT}\lvert c\rangle\lvert 0\rangle = \lvert c\rangle\lvert c\rangle$ | Flipped to $\lvert 1\rangle$ only when the control qubit is $\lvert 1\rangle$. |
-| [[swap-gate|SWAP]] | $\text{SWAP}\lvert 0\rangle\lvert\psi\rangle = \lvert\psi\rangle\lvert 0\rangle$ | Exchanges the two qubit states; the $\lvert 0\rangle$ moves to the second register. |
-| [[iswap-gate|iSWAP]] | $\text{iSWAP}\lvert 00\rangle = \lvert 00\rangle$; $\text{iSWAP}\lvert 01\rangle = i\lvert 10\rangle$ | Adds a phase of $i$ when states are exchanged; trivial when both qubits are $\lvert 0\rangle$. |
-| [[toffoli-gate|Toffoli]] (as control) | $\text{CCX}\lvert 0\rangle\lvert c_2\rangle\lvert t\rangle = \lvert 0\rangle\lvert c_2\rangle\lvert t\rangle$ | Control is off; target is always unchanged regardless of the second control. |
-| [[toffoli-gate|Toffoli]] (as target) | $\text{CCX}\lvert 1\rangle\lvert 1\rangle\lvert 0\rangle = \lvert 1\rangle\lvert 1\rangle\lvert 1\rangle$ | Flipped to $\lvert 1\rangle$ only when both controls are $\lvert 1\rangle$. |
-## Derivation
-^ Gate ^ Matrix form ^ Derivation ^
-| [[i-gate]] | $I = \begin{pmatrix}1&0\\0&1\end{pmatrix}$ | $I\lvert 0\rangle = \begin{pmatrix}1&0\\0&1\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}1\\0\end{pmatrix} = \lvert 0\rangle$ |
-| [[x-gate]] | $X = \begin{pmatrix}0&1\\1&0\end{pmatrix}$ | $X\lvert 0\rangle = \begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}0\\1\end{pmatrix} = \lvert 1\rangle$ |
-| [[y-gate]] | $Y = \begin{pmatrix}0&-i\\i&0\end{pmatrix}$ | $Y\lvert 0\rangle = \begin{pmatrix}0&-i\\i&0\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}0\\i\end{pmatrix} = i\lvert 1\rangle$ |
-| [[z-gate]] | $Z = \begin{pmatrix}1&0\\0&-1\end{pmatrix}$ | $Z\lvert 0\rangle = \begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}1\\0\end{pmatrix} = \lvert 0\rangle$ |
-| [[h-gate]] | $H = \tfrac{1}{\sqrt{2}}\begin{pmatrix}1&1\\1&-1\end{pmatrix}$ | $H\lvert 0\rangle = \tfrac{1}{\sqrt{2}}\begin{pmatrix}1&1\\1&-1\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \tfrac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix} = \lvert +\rangle$ |
-| [[s-gate]] | $S = \begin{pmatrix}1&0\\0&i\end{pmatrix}$ | $S\lvert 0\rangle = \begin{pmatrix}1&0\\0&i\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}1\\0\end{pmatrix} = \lvert 0\rangle$ |
-| [[t-gate]] | $T = \begin{pmatrix}1&0\\0&e^{i\pi/4}\end{pmatrix}$ | $T\lvert 0\rangle = \begin{pmatrix}1&0\\0&e^{i\pi/4}\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}1\\0\end{pmatrix} = \lvert 0\rangle$ |
-| [[rx-gate|$R_x(\theta)$]] | $R_x(\theta) = \begin{pmatrix}\cos\tfrac{\theta}{2}&-i\sin\tfrac{\theta}{2}\\-i\sin\tfrac{\theta}{2}&\cos\tfrac{\theta}{2}\end{pmatrix}$ | $R_x(\theta)\lvert 0\rangle = \begin{pmatrix}\cos\tfrac{\theta}{2}&-i\sin\tfrac{\theta}{2}\\-i\sin\tfrac{\theta}{2}&\cos\tfrac{\theta}{2}\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}\cos\tfrac{\theta}{2}\\-i\sin\tfrac{\theta}{2}\end{pmatrix}$ |
-| [[ry-gate|$R_y(\theta)$]] | $R_y(\theta) = \begin{pmatrix}\cos\tfrac{\theta}{2}&-\sin\tfrac{\theta}{2}\\\sin\tfrac{\theta}{2}&\cos\tfrac{\theta}{2}\end{pmatrix}$ | $R_y(\theta)\lvert 0\rangle = \begin{pmatrix}\cos\tfrac{\theta}{2}&-\sin\tfrac{\theta}{2}\\\sin\tfrac{\theta}{2}&\cos\tfrac{\theta}{2}\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}\cos\tfrac{\theta}{2}\\\sin\tfrac{\theta}{2}\end{pmatrix}$ |
-| [[rz-gate|$R_z(\theta)$]] | $R_z(\theta) = \begin{pmatrix}e^{-i\theta/2}&0\\0&e^{i\theta/2}\end{pmatrix}$ | $R_z(\theta)\lvert 0\rangle = \begin{pmatrix}e^{-i\theta/2}&0\\0&e^{i\theta/2}\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = e^{-i\theta/2}\begin{pmatrix}1\\0\end{pmatrix}$ |
-| [[u-gate|$U(\theta,\phi,\lambda)$]] | $U = \begin{pmatrix}\cos\tfrac{\theta}{2}&-e^{i\lambda}\sin\tfrac{\theta}{2}\\e^{i\phi}\sin\tfrac{\theta}{2}&e^{i(\phi+\lambda)}\cos\tfrac{\theta}{2}\end{pmatrix}$ | $U\lvert 0\rangle = \begin{pmatrix}\cos\tfrac{\theta}{2}&-e^{i\lambda}\sin\tfrac{\theta}{2}\\e^{i\phi}\sin\tfrac{\theta}{2}&e^{i(\phi+\lambda)}\cos\tfrac{\theta}{2}\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}\cos\tfrac{\theta}{2}\\e^{i\phi}\sin\tfrac{\theta}{2}\end{pmatrix}$ |
-## Relationship with other states
-^ State ^ Relationship ^
-| [[one-state|$\lvert 1\rangle$]] | Orthogonal companion: $\langle 0\vert 1\rangle = 0$; together they span the qubit Hilbert space (north and south poles of the Bloch sphere). The X gate converts between them: $X\lvert 0\rangle = \lvert 1\rangle$ and $X\lvert 1\rangle = \lvert 0\rangle$. |
-| [[plus-state|$\lvert +\rangle$]] | $H\lvert 0\rangle = \lvert +\rangle$. Conversely, $\lvert 0\rangle = \tfrac{1}{\sqrt{2}}(\lvert +\rangle + \lvert -\rangle)$, so $\lvert 0\rangle$ is the symmetric superposition of the two X-basis eigenstates. |
-| [[minus-state|$\lvert -\rangle$]] | $ZH\lvert 0\rangle = \lvert -\rangle$ ($H$ then $Z$). $\lvert 0\rangle$ decomposes in the X basis as $\tfrac{1}{\sqrt{2}}(\lvert +\rangle + \lvert -\rangle)$ — equal weight on both eigenstates. |
-| [[i-state|$\lvert +i\rangle$]] | $SH\lvert 0\rangle = \lvert +i\rangle$: Hadamard maps to the $+x$ equatorial point, S rotates 90° around $z$ to the $+y$ pole. |
-| [[minus-i-state|$\lvert -i\rangle$]] | $S^\dagger H\lvert 0\rangle = \lvert -i\rangle$. Together, $\lvert 0\rangle = \tfrac{1}{\sqrt{2}}(\lvert +i\rangle + \lvert -i\rangle)$, so $\lvert 0\rangle$ has equal overlap with both Y-basis states. |
-| [[phi-plus-state|$\lvert\Phi^+\rangle$]] | $\tfrac{1}{\sqrt{2}}(\lvert 00\rangle + \lvert 11\rangle)$, prepared from $\lvert 00\rangle$ by $H\otimes I$ then CNOT. Measuring either qubit and finding $\lvert 0\rangle$ collapses the other to $\lvert 0\rangle$. |
-| [[phi-minus-state|$\lvert\Phi^-\rangle$]] | $\tfrac{1}{\sqrt{2}}(\lvert 00\rangle - \lvert 11\rangle)$, prepared from $\lvert 00\rangle$ by $H\otimes I$, CNOT, then $Z$. Measuring and finding $\lvert 0\rangle$ collapses the partner to $\lvert 0\rangle$, same as $\lvert\Phi^+\rangle$ but with a relative minus sign. |
-| [[psi-plus-state|$\lvert\Psi^+\rangle$]] | $\tfrac{1}{\sqrt{2}}(\lvert 01\rangle + \lvert 10\rangle)$, prepared from $\lvert 01\rangle$. Measuring qubit 1 and finding $\lvert 0\rangle$ collapses qubit 2 to $\lvert 1\rangle$ — anti-correlated. |
-| [[psi-minus-state|$\lvert\Psi^-\rangle$]] | $\tfrac{1}{\sqrt{2}}(\lvert 01\rangle - \lvert 10\rangle)$, the singlet Bell state. Measuring qubit 1 and finding $\lvert 0\rangle$ collapses qubit 2 to $\lvert 1\rangle$, same anti-correlation as $\lvert\Psi^+\rangle$ but with a minus sign. |
-## List of code implementations
-- [[zero-state-qiskit|Qiskit]]