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--- 0-state [June 12, 2026 at 23:14] – Ivan Janevski
+++ 0-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 [[1-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]] | $R_x(\theta)\lvert 0\rangle = \cos\tfrac{\theta}{2}\lvert 0\rangle - i\sin\tfrac{\theta}{2}\lvert 1\rangle$ | Tilts the state from $\lvert 0\rangle$ toward $\lvert 1\rangle$ with an imaginary phase on the $\lvert 1\rangle$ component. |
-| [[ry-gate]] | $R_y(\theta)\lvert 0\rangle = \cos\tfrac{\theta}{2}\lvert 0\rangle + \sin\tfrac{\theta}{2}\lvert 1\rangle$ | Real amplitudes; at $\theta=\pi/2$ gives $\lvert +\rangle$, the same result as the Hadamard gate up to global phase. |
-| [[rz-gate]] | $R_z(\theta)\lvert 0\rangle = e^{-i\theta/2}\lvert 0\rangle$ | Global phase only; $\lvert 0\rangle$ is on the $z$-axis so a $z$-rotation has no observable effect. |
-| [[u-gate]] | $U\lvert 0\rangle = \cos\tfrac{\theta}{2}\lvert 0\rangle + e^{i\phi}\sin\tfrac{\theta}{2}\lvert 1\rangle$ | $\lambda$ drops out; every single-qubit state is reachable from $\lvert 0\rangle$ with appropriate $(\theta,\phi)$. |
-## Reaching other states
-^ State ^ Gates ^ Comment ^
-| $\lvert 0\rangle$ | $I\lvert 0\rangle = \lvert 0\rangle$ | The identity gate leaves $\lvert 0\rangle$ unchanged. |
-| [[1-state|$\lvert 1\rangle$]] | $X\lvert 0\rangle = \lvert 1\rangle$ | X flips the qubit; the quantum analogue of classical NOT. $X$ is its own inverse: $X^2 = I$. |
-| [[plus-state|$\lvert +\rangle$]] | $H\lvert 0\rangle = \lvert +\rangle$ | Hadamard rotates the north pole to the $+x$ equatorial point, creating the equal superposition $\tfrac{1}{\sqrt{2}}(\lvert 0\rangle + \lvert 1\rangle)$. |
-| [[minus-state|$\lvert -\rangle$]] | $ZH\lvert 0\rangle = \lvert -\rangle$ | H first produces $\lvert +\rangle$, then Z flips its relative phase to give $\tfrac{1}{\sqrt{2}}(\lvert 0\rangle - \lvert 1\rangle)$. |
-| [[i-state|$\lvert +i\rangle$]] | $SH\lvert 0\rangle = \lvert +i\rangle$ | H rotates to the $+x$ equatorial point, then S rotates 90° around $z$ to land on the $+y$ pole. |
-| [[minus-i-state|$\lvert -i\rangle$]] | $S^\dagger H\lvert 0\rangle = \lvert -i\rangle$ | Same as $\lvert +i\rangle$ but with the inverse phase rotation, landing on the $-y$ pole instead. |
-## Density matrix
-The [[density-matrix|density matrix]] of $\lvert 0\rangle$ is the outer product $\rho = \lvert 0\rangle\langle 0\rvert$. Expanding with the column vector $\lvert 0\rangle$ and its conjugate transpose $\langle 0 \rvert$:
-$$\lvert 0\rangle = \begin{pmatrix}1\\0\end{pmatrix} \qquad \langle 0\rvert = \begin{pmatrix}1 & 0\end{pmatrix} \qquad \rho = \lvert 0\rangle\langle 0\rvert = \begin{pmatrix}1\\0\end{pmatrix}\begin{pmatrix}1 & 0\end{pmatrix} = \begin{pmatrix}1 & 0\\0 & 0\end{pmatrix}$$
-This is a pure state: $\rho^2 = \rho$ and $\text{tr}(\rho^2) = 1$. The diagonal entries are the measurement probabilities — probability $1$ of outcome $0$ and probability $0$ of outcome $1$. The off-diagonal entries are the coherences; both are zero here because $\lvert 0\rangle$ is a basis state with no superposition between $\lvert 0\rangle$ and $\lvert 1\rangle$.
-### Bloch sphere
-On the [[bloch-sphere|Bloch sphere]], every single-qubit density matrix can be written as $\rho = \tfrac{1}{2}(I + \vec{r}\cdot\vec{\sigma})$, where $\vec{r}$ is the Bloch vector and $\vec{\sigma} = (X, Y, Z)$. For $\lvert 0\rangle$ the Bloch vector points to the north pole $\vec{r} = (0, 0, 1)$, giving:
-$$\rho = \tfrac{1}{2}(I + Z) = \tfrac{1}{2}\begin{pmatrix}1&0\\0&1\end{pmatrix} + \tfrac{1}{2}\begin{pmatrix}1&0\\0&-1\end{pmatrix} = \begin{pmatrix}1&0\\0&0\end{pmatrix}$$
-### Expectation values
-Expectation values of the Pauli operators follow directly from the density matrix via $\langle P\rangle = \text{tr}(\rho P)$:
-$$\langle X\rangle = \text{tr}(\rho X) = 0 \qquad \langle Y\rangle = \text{tr}(\rho Y) = 0 \qquad \langle Z\rangle = \text{tr}(\rho Z) = 1$$
-The result $\langle Z\rangle = 1$ confirms that $\lvert 0\rangle$ is the $+1$ eigenstate of $Z$; the vanishing $X$ and $Y$ expectations reflect that the state has no transverse coherence.