zero-state
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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
1-qubit
| Gate | Result | Comment |
|---|---|---|
| $I$ gate (Identity gate) | $I\lvert 0\rangle = \lvert 0\rangle$ | The identity gate leaves $\lvert 0\rangle$ unchanged. |
| $X$ gate (NOT 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. |
| $R_x(\theta)$ | $R_x(\theta)\lvert 0\rangle = \cos\tfrac{\theta}{2}\lvert 0\rangle - i\sin\tfrac{\theta}{2}\lvert 1\rangle$ | Tilts the state from $\lvert 0\rangle$ toward $\lvert 1\rangle$ with an imaginary phase on the $\lvert 1\rangle$ component. |
| $R_y(\theta)$ | $R_y(\theta)\lvert 0\rangle = \cos\tfrac{\theta}{2}\lvert 0\rangle + \sin\tfrac{\theta}{2}\lvert 1\rangle$ | Real amplitudes; at $\theta=\pi/2$ gives $\lvert +\rangle$, the same result as the Hadamard gate up to global phase. |
| $R_z(\theta)$ | $R_z(\theta)\lvert 0\rangle = e^{-i\theta/2}\lvert 0\rangle$ | Global phase only; $\lvert 0\rangle$ is on the $z$-axis so a $z$-rotation has no observable effect. |
| $U(\theta,\phi,\lambda)$ | $U\lvert 0\rangle = \cos\tfrac{\theta}{2}\lvert 0\rangle + e^{i\phi}\sin\tfrac{\theta}{2}\lvert 1\rangle$ | $\lambda$ drops out; every single-qubit state is reachable from $\lvert 0\rangle$ with appropriate $(\theta,\phi)$. |
2-qubits
| Gate | Result | Comment |
|---|---|---|
| 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 (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 | $\text{SWAP}\lvert 0\rangle\lvert\psi\rangle = \lvert\psi\rangle\lvert 0\rangle$ | Exchanges the two qubit states; the $\lvert 0\rangle$ moves to the second register. |
| iSWAP | $\text{iSWAP}\lvert 00\rangle = \lvert 00\rangle$; $\text{iSWAP}\lvert 01\rangle = i\lvert 10\rangle$ | Adds a phase of $i$ when states are exchanged; trivial when both qubits are $\lvert 0\rangle$. |
3-qubits
| Gate | Result | Comment |
|---|---|---|
| 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 (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 $\lvert 1\rangle$. |
Reaching other states
1-qubit
| 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. |
2-qubits
| State | Gates | Comment |
|---|---|---|
| $\lvert\Phi^+\rangle$ | $\text{CNOT}(H\otimes I)\lvert 00\rangle = \lvert\Phi^+\rangle$ | H puts qubit 1 into superposition; CNOT entangles qubit 2 conditionally. The simplest Bell state circuit, starting from two $\lvert 0\rangle$ qubits. |
| $\lvert\Phi^-\rangle$ | $(Z\otimes I)\,\text{CNOT}(H\otimes I)\lvert 00\rangle = \lvert\Phi^-\rangle$ | Same as $\lvert\Phi^+\rangle$ with a final $Z$ on qubit 1 to flip the relative sign between $\lvert 00\rangle$ and $\lvert 11\rangle$. |
| $\lvert\Psi^+\rangle$ | $\text{CNOT}(H\otimes X)\lvert 00\rangle = \lvert\Psi^+\rangle$ | X first flips qubit 2 to $\lvert 1\rangle$; H and CNOT then entangle in the anti-correlated subspace $\tfrac{1}{\sqrt{2}}(\lvert 01\rangle + \lvert 10\rangle)$. |
| $\lvert\Psi^-\rangle$ | $(Z\otimes I)\,\text{CNOT}(H\otimes X)\lvert 00\rangle = \lvert\Psi^-\rangle$ | Same as $\lvert\Psi^+\rangle$ with a final $Z$ to introduce the minus sign. The singlet state, antisymmetric under qubit exchange. |
List of code implementations
zero-state.1781304221.txt.gz · Last modified: by Ivan Janevski