측정, 아다마르 게이트와 CNOT 게이트Measurement, the Hadamard Gate and the CNOT Gate

A matrix $\Omega \in M_n(\mathbb{C})$ is called Hermitian if it equals its own conjugate transpose, i.e.,

(equivalently, $\omega_{ji} = \overline{\omega_{ij}}$ for all $i, j$). The properties of Hermitian matrices used in the notes are: a Hermitian matrix is diagonalizable, every eigenvalue is real, and it admits an orthonormal basis of eigenvectors.

For example, the following matrix is Hermitian.

1. The matrix $\Omega$ related to measurement is a Hermitian matrix on $H = \mathbb{C}^n$. Hence $\Omega$ has real eigenvalues $\lambda_i$ and a corresponding orthonormal eigenbasis $\{v_i\}_{1 \le i \le n}$ (for convenience, assume the $\lambda_i$’s are distinct) satisfying $\Omega v_i = \lambda_i v_i$.

2. Writing $v = \Sigma_{i=1}^{n} a_i v_i$ with $\Sigma_{i=1}^{n} |a_i|^2 = 1$, measuring $v$ using $\Omega$ yields the eigenvalue $\lambda_i$ as the measurement, with probability $|a_i|^2$.

3. After the eigenvalue $\lambda_i$ is measured, the state $v$ collapses to the corresponding eigenvector $v_i$. In summary,

The eigenvectors of $\Omega$ are $|0\rangle = (1, 0)$ (eigenvalue $2$) and $|1\rangle = (0, 1)$ (eigenvalue $1$), and $\{|0\rangle, |1\rangle\}$ is already orthonormal. Since $v = \frac{1}{\sqrt{5}}|0\rangle + \frac{2}{\sqrt{5}}|1\rangle$,

1. the probability of measuring $2$ is $\left(\frac{1}{\sqrt{5}}\right)^2 = \frac{1}{5}$, with new state $|0\rangle$.

2. the probability of measuring $1$ is $\left(\frac{2}{\sqrt{5}}\right)^2 = \frac{4}{5}$, with new state $|1\rangle$.

The two probabilities sum to $\frac{1}{5} + \frac{4}{5} = 1$.

The Hadamard gate is a 1-qubit gate given by the matrix

Let $|+\rangle := \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ and $|-\rangle := \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$. Then

and in general $\alpha|0\rangle + \beta|1\rangle \mapsto \alpha|+\rangle + \beta|-\rangle$. Since $H\overline{H}^{T} = I$, $H$ is a unitary matrix, and since $H^2 = I$, $H$ is its own inverse. The Hadamard gate turns a basis state into an equal superposition; for instance, measuring $H|0\rangle$ gives $0$ with probability $\frac{1}{2}$ and $1$ with probability $\frac{1}{2}$. Because of this, the Hadamard gate is used to create superpositions at the start of quantum algorithms.

Note. The figure file is not included with the notes. Consistently with the preceding note, where [Figure.1] was the symbol for the CNOT gate and [Figure.2] for the SWAP gate, we take [Figure.3] to be the CNOT gate and [Figure.4] the SWAP gate, each acting on the two-qubit input $|A\rangle|B\rangle$ (the first qubit $A$ as control, the second qubit $B$ as target), and solve under this reading.

[Figure.3] (CNOT gate). As in the truth table of Example 4.10, the CNOT gate leaves the control qubit $A$ unchanged and replaces the target qubit by $B \oplus A$, so the output state is $|A\rangle|B \oplus A\rangle$.

Letting $|00\rangle = (1, 0, 0, 0)$, $|01\rangle = (0, 1, 0, 0)$, $|10\rangle = (0, 0, 1, 0)$, $|11\rangle = (0, 0, 0, 1)$, the matrix of the CNOT gate from Example 4.10 is

Hence a general superposition $a|00\rangle + b|01\rangle + c|10\rangle + d|11\rangle$ maps as

[Figure.4] (SWAP gate). The SWAP gate exchanges the two qubits, so the output state is $|B\rangle|A\rangle$.

In the same basis order $|00\rangle, |01\rangle, |10\rangle, |11\rangle$, the matrix of the SWAP gate is

Hence a general superposition maps as

Incidentally, the SWAP gate can be implemented as three CNOT gates with alternating control and target.