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Table of Contents
$\lvert 0 \rangle$ (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 $\lvert 1\rangle$.
$$\lvert 0\rangle = \begin{pmatrix}1\\0\end{pmatrix}$$
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 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 $+1$. |
| Hadamard 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. |
| Rotation-X 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. |
| Rotation-Y 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. |
| Rotation-Z 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. |
| Unitary 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. |
| $\lvert 1\rangle$ | $X\lvert 0\rangle = \lvert 1\rangle$ | X flips the qubit; the quantum analogue of classical NOT. $X$ is its own inverse: $X^2 = I$. |
| $\lvert +\rangle$ | $H\lvert 0\rangle = \lvert +\rangle$ | Hadamard rotates the north pole to the $+x$ equatorial point, creating the equal superposition $\tfrac{1}{\sqrt{2}}(\lvert 0\rangle + \lvert 1\rangle)$. |
| $\lvert -\rangle$ | $ZH\lvert 0\rangle = \lvert -\rangle$ | H first produces $\lvert +\rangle$, then Z flips its relative phase to give $\tfrac{1}{\sqrt{2}}(\lvert 0\rangle - \lvert 1\rangle)$. |
| $\lvert +i\rangle$ | $SH\lvert 0\rangle = \lvert +i\rangle$ | H rotates to the $+x$ equatorial point, then S rotates 90° around $z$ to land on the $+y$ pole. |
| $\lvert -i\rangle$ | $S^\dagger H\lvert 0\rangle = \lvert -i\rangle$ | Same as $\lvert +i\rangle$ but with the inverse phase rotation, landing on the $-y$ pole instead. |
Qiskit
# Run: python main.py # Apply each single-qubit gate to |0> and print the resulting statevector. import numpy as np from qiskit import QuantumCircuit from qiskit.quantum_info import Statevector def apply(op): qc = QuantumCircuit(1) op(qc) return Statevector(qc).data print(apply(lambda q: q.id(0))) # [1.+0.j, 0.+0.j] I|0> = |0> print(apply(lambda q: q.x(0))) # [0.+0.j, 1.+0.j] X|0> = |1> print(apply(lambda q: q.y(0))) # [0.+0.j, 0.+1.j] Y|0> = i|1> print(apply(lambda q: q.z(0))) # [1.+0.j, 0.+0.j] Z|0> = |0> print(apply(lambda q: q.h(0))) # [0.707+0.j, 0.707+0.j] H|0> = |+> print(apply(lambda q: q.s(0))) # [1.+0.j, 0.+0.j] S|0> = |0> print(apply(lambda q: q.t(0))) # [1.+0.j, 0.+0.j] T|0> = |0> print(apply(lambda q: q.rx(np.pi/2, 0))) # [0.707+0.j, 0.-0.707j] Rx(pi/2)|0> print(apply(lambda q: q.ry(np.pi/2, 0))) # [0.707+0.j, 0.707+0.j] Ry(pi/2)|0> = |+> print(apply(lambda q: q.rz(np.pi/2, 0))) # [0.707-0.707j, 0.+0.j] Rz(pi/2)|0> = e^{-i*pi/4}|0> print(apply(lambda q: q.u(np.pi/2, 0, 0, 0))) # [0.707+0.j, 0.707+0.j] U(pi/2,0,0)|0>
Density matrix
Matrix
Expanding with the column vector $\lvert 0\rangle$ and its conjugate transpose $\langle 0 \rvert$, the density matrix of $\lvert 0\rangle$ is the outer product $\rho = \lvert 0\rangle\langle 0\rvert$.
$$\rho = \lvert 0\rangle\langle 0\rvert = \begin{pmatrix}1\\0\end{pmatrix}\begin{pmatrix}1 & 0\end{pmatrix} = \begin{pmatrix}1 & 0\\0 & 0\end{pmatrix}$$
Purity
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, 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.