applying-gates-to-zero-state
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Table of Contents
Applying gates to $\lvert 0\rangle$ (Zero state)
| 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 | $Z = \begin{pmatrix}1&0\\0&-1\end{pmatrix}$ | $Z\lvert 0\rangle = \begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}1\\0\end{pmatrix} = \lvert 0\rangle$ |
| Hadamard 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 | $T = \begin{pmatrix}1&0\\0&e^{i\pi/4}\end{pmatrix}$ | $T\lvert 0\rangle = \begin{pmatrix}1&0\\0&e^{i\pi/4}\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}1\\0\end{pmatrix} = \lvert 0\rangle$ |
| $R_x(\theta)$ | $R_x(\theta) = \begin{pmatrix}\cos\tfrac{\theta}{2}&-i\sin\tfrac{\theta}{2}\\-i\sin\tfrac{\theta}{2}&\cos\tfrac{\theta}{2}\end{pmatrix}$ | $R_x(\theta)\lvert 0\rangle = \begin{pmatrix}\cos\tfrac{\theta}{2}&-i\sin\tfrac{\theta}{2}\\-i\sin\tfrac{\theta}{2}&\cos\tfrac{\theta}{2}\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}\cos\tfrac{\theta}{2}\\-i\sin\tfrac{\theta}{2}\end{pmatrix}$ |
| $R_y(\theta)$ | $R_y(\theta) = \begin{pmatrix}\cos\tfrac{\theta}{2}&-\sin\tfrac{\theta}{2}\\\sin\tfrac{\theta}{2}&\cos\tfrac{\theta}{2}\end{pmatrix}$ | $R_y(\theta)\lvert 0\rangle = \begin{pmatrix}\cos\tfrac{\theta}{2}&-\sin\tfrac{\theta}{2}\\\sin\tfrac{\theta}{2}&\cos\tfrac{\theta}{2}\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}\cos\tfrac{\theta}{2}\\\sin\tfrac{\theta}{2}\end{pmatrix}$ |
| $R_z(\theta)$ | $R_z(\theta) = \begin{pmatrix}e^{-i\theta/2}&0\\0&e^{i\theta/2}\end{pmatrix}$ | $R_z(\theta)\lvert 0\rangle = \begin{pmatrix}e^{-i\theta/2}&0\\0&e^{i\theta/2}\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = e^{-i\theta/2}\begin{pmatrix}1\\0\end{pmatrix}$ |
| $U(\theta,\phi,\lambda)$ | $U = \begin{pmatrix}\cos\tfrac{\theta}{2}&-e^{i\lambda}\sin\tfrac{\theta}{2}\\e^{i\phi}\sin\tfrac{\theta}{2}&e^{i(\phi+\lambda)}\cos\tfrac{\theta}{2}\end{pmatrix}$ | $U\lvert 0\rangle = \begin{pmatrix}\cos\tfrac{\theta}{2}&-e^{i\lambda}\sin\tfrac{\theta}{2}\\e^{i\phi}\sin\tfrac{\theta}{2}&e^{i(\phi+\lambda)}\cos\tfrac{\theta}{2}\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}\cos\tfrac{\theta}{2}\\e^{i\phi}\sin\tfrac{\theta}{2}\end{pmatrix}$ |
Gates
1-qubit
| 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. |
| $R_x(\theta)$ | $R_x(\theta)\lvert 0\rangle = \cos\tfrac{\theta}{2}\lvert 0\rangle - i\sin\tfrac{\theta}{2}\lvert 1\rangle$ | 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 $\lvert 1\rangle$. |
applying-gates-to-zero-state.1781304739.txt.gz ยท Last modified: by Ivan Janevski