Complex Vector Spaces $\mathbb{C}^n$

A quantum state is a vector, and a quantum gate is a linear map on vectors. Before we can talk about either we need the precise notion of a vector space over $\mathbb{C}$: a set in which we can add elements and multiply them by complex scalars, subject to rules that make these operations behave like the arrows of elementary geometry. The superposition principle of quantum mechanics, which says that if $\lvert\psi\rangle$ and $\lvert\phi\rangle$ are possible states then so is (after normalisation) $\alpha\lvert\psi\rangle+\beta\lvert\phi\rangle$, is literally the statement that states live in a vector space.

The prototype, and the only example we need for finite quantum systems, is $\mathbb{C}^n$: the set of columns $v=(v_1,\dots,v_n)^T$ with complex entries, added and scaled entry by entry,

$u+v=\begin{pmatrix}u_1+v_1\\\vdots\\u_n+v_n\end{pmatrix},\qquad c\,v=\begin{pmatrix}c\,v_1\\\vdots\\c\,v_n\end{pmatrix}.$

A single qubit lives in $\mathbb{C}^2$ and a register of $n$ qubits in $\mathbb{C}^{2^n}$.

Worked example 1 (dependent over $\mathbb{C}$). Take $b_1=(1,i)^T$ and $b_2=(i,-1)^T$ in $\mathbb{C}^2$. Multiply $b_1$ by the scalar $i$:

$i\,b_1=\begin{pmatrix}i\cdot1\\ i\cdot i\end{pmatrix}=\begin{pmatrix}i\\-1\end{pmatrix}=b_2 .$

Hence $i\,b_1-b_2=0$ is a non-trivial vanishing combination (coefficients $i$ and $-1$), and the list is dependent. Viewed as real vectors in $\mathbb{R}^4$, namely $(1,0,0,1)$ and $(0,1,-1,0)$ (real and imaginary parts interleaved), the same two vectors are independent over $\mathbb{R}$, because no real multiple of one equals the other. Which scalars are allowed matters.

Worked example 2 (independent in $\mathbb{C}^3$). For $b_1=(1,0,i)^T$, $b_2=(0,1,1)^T$, $b_3=(1,1,0)^T$, place the vectors as columns of a matrix and compute its determinant by expansion along the first row:

$\det\begin{pmatrix}1&0&1\\0&1&1\\i&1&0\end{pmatrix}=1\,(1\cdot0-1\cdot1)-0\,(0\cdot0-1\cdot i)+1\,(0\cdot1-1\cdot i)=-1-i\ne0 .$

A square matrix has nonzero determinant exactly when its columns are linearly independent, so $b_1,b_2,b_3$ is a basis of $\mathbb{C}^3$.

A vector is a geometric object; its coordinates depend on the basis we use to describe it. Quantum computing constantly switches bases: a measurement "in the $X$ basis" is a measurement in the basis $\{\lvert+\rangle,\lvert-\rangle\}$, and the Hadamard gate is precisely the matrix that converts coordinates between the computational and the $\pm$ basis. The next derivation gives the general recipe.

Define $\lvert+\rangle=\tfrac1{\sqrt2}(1,1)^T$ and $\lvert-\rangle=\tfrac1{\sqrt2}(1,-1)^T$, i.e. $\lvert\pm\rangle=\tfrac1{\sqrt2}(\lvert0\rangle\pm\lvert1\rangle)$. Their change-of-basis matrix is the Hadamard matrix $H$ shown below.

A second basis that appears constantly is the circular or $\pm i$ basis $\lvert{+i}\rangle=\tfrac1{\sqrt2}(1,i)^T$, $\lvert{-i}\rangle=\tfrac1{\sqrt2}(1,-i)^T$, the eigenbasis of the Pauli $Y$ operator. Its change-of-basis matrix is $Q=\tfrac1{\sqrt2}\begin{pmatrix}1&1\\i&-i\end{pmatrix}$ with $\det Q=\tfrac12(-i-i)=-i\ne0$, so it is a basis too. Solving $Q\,(c_{+i},c_{-i})^T=(v_1,v_2)^T$ gives $c_{\pm i}=\tfrac{1}{\sqrt2}(v_1\mp i v_2)$; for $v=\lvert0\rangle=(1,0)^T$ this is $c_{+i}=c_{-i}=\tfrac1{\sqrt2}$, and indeed $\tfrac1{\sqrt2}(\lvert{+i}\rangle+\lvert{-i}\rangle)=\tfrac12(2,0)^T=\lvert0\rangle$. In the widget below, edit a vector and watch its coordinates in all three bases update simultaneously.