큐비트와 게이트Qubits and Gates

As stated in the introduction to this unit, quantum computers use superposition and collapse to exploit both the advantage of waves (the analog computer) and the advantage of particles (the digital computer). Hence

$\boxed{\ (a)\ }$ = the analog computer, $\boxed{\ (b)\ }$ = the digital computer.

Completed sentence: “the basic idea of quantum computing is based on combining advantage of the analog computer (waves) and the digital computer (particles).”

1. The advantage of analog computers is that operations are efficient. For example, if two wires carrying current $A$ and one wire carrying current $B$ meet at a point, Kirchhoff’s current law makes the current after the meeting point equal to $A+B$ instantly and physically. By contrast, performing the same addition on a digital computer requires expressing each number in bits (a 9-digit number becomes 30 bits, for instance) and paying a cost for every single bit-addition, so in this case the analog computer is more efficient.

2. The advantage of digital computers is that their data are digital, consisting of 0 and 1. This makes it convenient to input and output data.

3. The disadvantage of analog computers is that data are difficult to input and output, and computation requires an ideal environment.

A qubit is the unit of information in a quantum computer, given by a unit vector $(x,y) \in \mathbb{C}^2$, i.e. a vector of the form $a|0\rangle + b|1\rangle$ with $|a|^2+|b|^2=1$ (Definition 4.1). It can be viewed as a superposition of the classical bits 0 and 1: if we measure a qubit $(x,y) = x|0\rangle + y|1\rangle$, we measure 0 with probability $|x|^2$ and 1 with probability $|y|^2$ (the state then collapses to that outcome). The notes print 0 and 1 the other way round in §4.1; see the translator’s note there.

1. $(2, 4) = 2e_1+4e_2 = 2|0\rangle+4|1\rangle$.

2. $(7, 2) = 7|0\rangle+2|1\rangle$.

3. $\left(\frac{1}{\sqrt2},-\frac{1}{\sqrt2}\right) = \frac{1}{\sqrt2}|0\rangle-\frac{1}{\sqrt2}|1\rangle$.

Remark. By Definition 4.1 a qubit must be a unit vector in $\mathbb{C}^2$. Since $\left|\frac{1}{\sqrt2}\right|^2+\left|-\frac{1}{\sqrt2}\right|^2=\frac12+\frac12=1$, only (3) is a unit vector, i.e. an actual qubit. The norm of (1) is $\sqrt{2^2+4^2}=\sqrt{20}$ and of (2) is $\sqrt{7^2+2^2}=\sqrt{53}$, so to turn them into qubits they would need to be normalized as $\frac{1}{\sqrt{20}}(2,4) = \frac{2}{\sqrt{20}}|0\rangle+\frac{4}{\sqrt{20}}|1\rangle$ and $\frac{1}{\sqrt{53}}(7,2)=\frac{7}{\sqrt{53}}|0\rangle+\frac{2}{\sqrt{53}}|1\rangle$, respectively.

$A \in M_{n\times n}(\mathbb{C})$ is unitary if $A\overline{A}^{T} = I_n$, i.e. $A^{-1} = \overline{A}^{T}$ (Definition 4.8 (1)).

Unitary matrices preserve the norm of a vector: for any $v \in \mathbb{C}^n$, $\|Av\|^2 = (\overline{Av})^{T}(Av) = \overline{v}^{T}\overline{A}^{T}Av$, and if $A$ is unitary then $\overline{A}^{T}A = A^{-1}A = I_n$, so $\|Av\|^2 = \overline{v}^Tv = \|v\|^2$. Hence a unitary matrix sends a unit vector to a unit vector, i.e. it sends a qubit to a qubit — this is why quantum gates are defined to be unitary matrices.

1. $\boxed{\ (a)\ }$ = a unitary matrix.

2. $\boxed{\ (b)\ }$ = 1-qubit gates, $\boxed{\ (c)\ }$ = the CNOT gate.

By Definition 4.8, a quantum gate for 1-qubit is a unitary $2\times2$ matrix, and as noted at the end of §4.4, every quantum gate is implemented by using 1-qubit gates and the CNOT gate.

Expanding the tensor product (Definition 4.6 (1)),

Writing $|00\rangle=e_1,|01\rangle=e_2,|10\rangle=e_3,|11\rangle=e_4$ (Example 4.10), the coordinate vector of this state is $\frac{1}{\sqrt2}(x,x,y,y)^T$. Applying the matrix of the CNOT gate found in Example 4.10,

That is, the CNOT gate swaps the coefficients of $|10\rangle$ and $|11\rangle$, but since both coefficients equal $y$, the resulting vector is unchanged. Hence

i.e. this 2-qubit is left unchanged by the CNOT gate (the state is preserved).

The columns of the $2\times2$ matrix corresponding to a quantum gate $U$ are the coordinates of $U|0\rangle$ and $U|1\rangle$ (Definition 4.8 (2), Example 4.9). Since $|0\rangle=(1,0)$ and $|1\rangle=(0,1)$,

Indeed $U\overline{U}^T = \frac12\begin{pmatrix}1&1\\1&-1\end{pmatrix}\begin{pmatrix}1&1\\1&-1\end{pmatrix} = \frac12\begin{pmatrix}2&0\\0&2\end{pmatrix}=I_2$, so $U$ is unitary, confirming that $U$ is a legitimate quantum gate (this matrix is exactly the Hadamard gate of §4.6).

Let $|0\rangle|0\rangle=e_1,|0\rangle|1\rangle=e_2,|1\rangle|0\rangle=e_3,|1\rangle|1\rangle=e_4$. The given correspondence $(0,0)\mapsto(0,1)$, $(0,1)\mapsto(1,0)$, $(1,0)\mapsto(1,1)$, $(1,1)\mapsto(0,0)$ means that

The $i$-th column of the matrix corresponding to $U$ is the coordinate vector of $Ue_i$ (Definition 4.8 (2)), so

This matrix is a permutation matrix cycling the standard basis vectors, so each row and column contains exactly one $1$, and hence $U\overline{U}^T=I_4$, i.e. $U$ is unitary.

Note. The statement is printed incompletely (“Prove that matrices corresponding the $X$ gate and $Z$ gate”); we find the matrices and prove they represent the gates.

By the circuit diagrams in Example 4.10 (2), the $X$ gate is a 1-qubit gate sending $\alpha|0\rangle+\beta|1\rangle$ to $\beta|0\rangle+\alpha|1\rangle$, and the $Z$ gate sends $\alpha|0\rangle+\beta|1\rangle$ to $\alpha|0\rangle-\beta|1\rangle$.

The $X$ gate. Let $X := \begin{pmatrix} 0&1\\1&0\end{pmatrix}$. Then

i.e. $X(\alpha|0\rangle+\beta|1\rangle) = \beta|0\rangle+\alpha|1\rangle$, so $X$ realizes exactly the correspondence required of the $X$ gate.

The $Z$ gate. Let $Z := \begin{pmatrix} 1&0\\0&-1\end{pmatrix}$. Then

i.e. $Z(\alpha|0\rangle+\beta|1\rangle) = \alpha|0\rangle-\beta|1\rangle$, so $Z$ realizes exactly the correspondence required of the $Z$ gate.

We show these matrices are unique. By Definition 4.8, a gate is a linear map (a unitary matrix), so it is determined once its values $U|0\rangle, U|1\rangle$ on the basis $|0\rangle, |1\rangle$ are fixed, and the columns of the matrix corresponding to $U$ are the coordinates of $U|0\rangle$ and $U|1\rangle$ (as in Examples 4.9 and 4.10). The $X$ gate must satisfy $|0\rangle\mapsto|1\rangle$ and $|1\rangle\mapsto|0\rangle$, so the columns of its matrix are $(0,1)^T,(1,0)^T$ and nothing else, i.e. the matrix is $X$. Likewise the $Z$ gate must satisfy $|0\rangle\mapsto|0\rangle$ and $|1\rangle\mapsto-|1\rangle$, so its matrix is $Z$ and nothing else.

Finally, we check that $X, Z$ are unitary (Definition 4.8 (1)):

so $X$ and $Z$ are unitary matrices, and hence each is a legitimate quantum gate.