Multiple-qubit systems
Assume we have 2 qubits for now. Generalizing from there is straightforward.
Intuitively, given 2 classical bits, the possible configurations are 4:
$$00$$$$11$$$$10$$$$01$$As in the single-qubit case, we make a notational trick and write the states:
$$\ket{0}\ket{0}$$$$\ket{1}\ket{1}$$$$\ket{1}\ket{0}$$$$\ket{0}\ket{1}$$Now, take a 2-qubits register:
Once measured, the possible configurations they may assume are the 4 we just saw. So, following the same idea of the single-qubit state, we can define the state $\ket{\Phi}$ of a 2-qubit register as
$$ \ket{\Phi}= \alpha_1\ket{0}\ket{0}+\alpha_2\ket{0}\ket{1}+\alpha_3\ket{1}\ket{1}+\alpha_1\ket{1}\ket{1}$$with the usual condition
$$|\alpha_1|^2 + |\alpha_2|^2+ |\alpha_3|^2 +|\alpha_4|^2 = 1$$and $\alpha_i\in \mathbb{C}$ for every $i=1,2,3,4$.
Why didn’t we write $\ket{x}\ket{y}$ instead of $\ket{\Phi}$? This is the first taste of weirdness for this topic: not every 2-qubit system can be expressed as a “product”, but we don’t need to dive further in this here.
Tensor product
We can express $\ket{0}\ket{0}$ as a vector as well. If a qubit can assume 2 possible configurations and thus its states can be represented by 2-dimensional complex vectors, it comes naturally that 2-qubit systems have states that can be represented by 4-dimensional complex vectors, as they have 4 possible configurations at the end of the day (i.e. after measurement). In general, $n$-qubit systems’state can be represented by $2^n$-dimensional complex vectors. We can use the same principle as before: we just perform a correspondence between states and vectors of the canonical basis of $\mathbb{C}^{2^n}$. So in our case:
$$\ket{0}\ket{0}\leftrightarrow \begin{pmatrix}1\\0\\0\\ 0\end{pmatrix}$$$$\ket{0}\ket{1}\leftrightarrow\begin{pmatrix}0\\1\\0\\ 0\end{pmatrix}$$$$\ket{1}\ket{0}\leftrightarrow\begin{pmatrix}0\\0\\1\\ 0\end{pmatrix}$$$$\ket{1}\ket{1}\leftrightarrow\begin{pmatrix}0\\0\\0\\ 1\end{pmatrix}$$No need to memorize which vector corresponds to which state, as we are about to introduce an operation that is behind the curtains of this correspondence: the tensor product.
Given two vectors
$$\begin{pmatrix} a_1\\ \vdots \\ a_n \end{pmatrix}, \begin{pmatrix} b_1\\ \vdots \\ b_m \end{pmatrix},$$we define their tensor product as:
$$\begin{pmatrix} a_1\\ \vdots \\ a_n \end{pmatrix} \otimes \begin{pmatrix} b_1\\ \vdots \\ b_m \end{pmatrix} = \begin{pmatrix}a_1b_1\\ \vdots \\ a_1b_m \\ a_2b_1\\ \vdots \\ a_2b_m \\ \vdots \\ a_nb_1 \\ \vdots \\ a_nb_m\end{pmatrix}.$$The result is an $(n\times m)$-dimensional vector.
Properties of the tensor product
Definition
Let $A \in \mathbb{K}^{m \times n}$ and $B \in \mathbb{K}^{p \times q}$ be two matrices over a field $\mathbb{K}$.
The tensor product (also called the “Kronecker product”) of $A$ and $B$ is the matrix $A \otimes B \in \mathbb{K}^{mp \times nq}$ defined as:
Given $x, y, z \in \mathbb{C}^{2^n}$, $a,b \in \mathbb{C}$ and let $A, B, C, D$ be $2^n\times 2^n$ complex matrices. Then
- $(x + y) \otimes z = x \otimes z + y \otimes z$
- $z \otimes (x + y) = z \otimes x + z \otimes y$
- $a x \otimes b y = ab (x \otimes y)$
- $(A \otimes B)(C \otimes D) = (AC) \otimes (BD)$
- $(A \otimes B)(x \otimes y) = (Ax) \otimes (By)$
- $(A \otimes B)^{\dagger} = A^{\dagger} \otimes B^{\dagger}$
Now we are ready to uncover the truth:
$$\ket{0}\ket{0} = \begin{pmatrix}1\\0\\0\\ 0\end{pmatrix} = \begin{pmatrix}1\\0\end{pmatrix} \otimes \begin{pmatrix}1\\0\end{pmatrix}= \ket{0}\ket{0}= \ket{0} \otimes \ket{0}$$and so on with the others.
With no surprise, the intuition behind the use of this notation is actually supported by a rigorous algebraic system. One can think of the tensor product as a natural way to encode the possible configurations of a system composed of $n$ qubits.