Section 1.10 Systems of Multiple Qubits
Any system of \(n\) qubits will have \(2^n\) basis states. As we have seen already, in a one-qubit system, the two basis states are \(\ket{0}\) and \(\ket{1}\text{.}\) In a two-qubit system, the four basis states are \(\ket{00}\text{,}\) \(\ket{01}\text{,}\) \(\ket{10}\text{,}\) and \(\ket{11}\text{.}\) These four basis states each have vector representations, which are found by using the tensor product
Subsection 1.10.1 Tensor Product
The tensor product (represented by a \(\otimes\) symbol) is an operation between two matrices (or vectors) that multiplies each entry in the matrix on the left by the matrix on the right. Thus the tensor product between a \(m\times n\) matrix and a \(p \times q\) matrix will be a \(mp \times nq \text{,}\) as shown below.
\begin{equation*}
A \otimes B = \begin{pmatrix} A_{11}B & A_{12}B & A{13}B & \\ A_{21}B & A_{22}B & A_{23}B \\ A_{31}B & A_{32}B & A_{33}B \end{pmatrix}
\end{equation*}
where each \(A_{ij}\) is the entry in the \(i\)-th row and \(j\)-th column of \(A\text{.}\)
For scalars \(\alpha,\beta \in \mathbb{C}\) and vectors \(\ket{\psi},\ket{\phi},\ket{\rho}\in\mathcal{H}\) the tensor product can be distributed as follows
\begin{equation*}
\ket{\psi} \otimes (\alpha \ket{\phi} + \beta \ket{\rho}) = \alpha(\ket{\psi} \otimes \ket{\phi}) + \beta(\ket{\psi} \otimes \ket{\rho}) = \alpha \ket{\psi\phi} + \beta \ket{\psi\rho}
\end{equation*}
Computer the following tensor products
\begin{equation*}
1. \ \begin{pmatrix}1 \\ 2 \end{pmatrix} \otimes \begin{pmatrix}4 \\ 8 \end{pmatrix}
\end{equation*}
\begin{equation*}
2. \ \begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix} \otimes \begin{pmatrix} 4 & 1 \\ 0 & 2 \end{pmatrix}
\end{equation*}
Solution.
\begin{equation*}
1. \ \begin{pmatrix}1 \\ 2 \end{pmatrix} \otimes \begin{pmatrix}4 \\ 8 \end{pmatrix} = \begin{pmatrix} 1 \times 4 \\ 1 \times 8 \\ 2 \times 4 \\ 2 \times 8 \end{pmatrix} = \begin{pmatrix} 4 \\ 8 \\ 8 \\ 16 \end{pmatrix}
\end{equation*}
\begin{equation*}
2. \ \begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix} \otimes \begin{pmatrix} 4 & 1 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 2\begin{pmatrix} 4 & 1 \\ 0 & 2 \end{pmatrix} & 1\begin{pmatrix} 4 & 1 \\ 0 & 2 \end{pmatrix} \\ 1\begin{pmatrix} 4 & 1 \\ 0 & 2 \end{pmatrix} & 3\begin{pmatrix} 4 & 1 \\ 0 & 2 \end{pmatrix} \end{pmatrix} = \begin{pmatrix} 8 & 2 & 4 & 1 \\ 0 & 4 & 0 & 2 \\ 4 & 1 & 12 & 3 \\ 0 & 2 & 0 & 6 \end{pmatrix}
\end{equation*}
Subsection 1.10.2 Systems of Multiple Qubits
The vector representation of a two qubit system \(\ket{\psi \phi}\) would be found by the tensor product \(\ket{\psi} \otimes \ket{\phi}\text{.}\) Thus the four basis states of a two qubit system are defined as:
\begin{equation*}
\ket{00} = \ket{0} \otimes \ket{0} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}
\end{equation*}
\begin{equation*}
\ket{01} = \ket{0} \otimes \ket{1} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}
\end{equation*}
\begin{equation*}
\ket{10} = \ket{1} \otimes \ket{0} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}
\end{equation*}
\begin{equation*}
\ket{11} = \ket{1} \otimes \ket{1} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}
\end{equation*}
This concept can be generalized to systems of any number of qubits. A system of two qubits is referred to as bipartite or composite.
The tensor product can also be applied to vector spaces. The tensor product of a vector that exists within a space \(V\) and a vector that exists within a space \(W\) would exist within the space \(V \otimes W\text{.}\) Thus, a system of two qubits would exist within the space \(\mathcal{H} \otimes \mathcal{H}\) and a system of \(n\) qubits exists within the space \(\mathcal{H}^{\otimes n}\text{,}\) where the \(\otimes n\) superscript means taking the tensor product of \(\mathcal{H}\) with itself \(n\) times. Similarly, \(\ket{\psi}^{\otimes n}\) would represent the state \(\ket{\psi}\) tensored with itself \(n\) times. Suppose the space \(V\) is \(n\) dimensional the space \(W\) is \(m\) dimensional, then the vector space \(V \otimes W\) would be \(n\cdot m\) dimensional.
Example 1.10.2.
\begin{equation*}
\ket{\psi}^{\otimes3} = \ket{\psi} \otimes \ket{\psi} \otimes \ket{\psi}
\end{equation*}
Suppose, we have two vector spaces \(V\) and \(W,\) and we want to know what kind of operator act on the space \(V\otimes W\text{.}\) If \(A\) is an operator on the space \(V\) and \(B\) is an operator on the space \(W\text{,}\) then \(A \otimes B\) is an operator on the space \(V \otimes W\text{.}\) An operator \(A\otimes B\) is linear if
\begin{equation*}
\big(A\otimes B\big) \big(\ket{v}\otimes \ket{w}\big) = A\ket{v}\otimes B \ket{w}
\end{equation*}
For \(\ket{v} \in V\) and \(\ket{w} \in W\text{.}\)
Checkpoint 1.10.3.
For the two qubit system \(\ket{\psi}\) compute the probabilities that it will collapse into each the basis states \(\ket{00}\text{,}\) \(\ket{01}\text{,}\) \(\ket{10}\text{,}\) and \(\ket{11}\)
\begin{equation*}
\ket{\psi} = \frac{1}{2\sqrt{2}} \ket{00} + \frac{\sqrt{3}}{2\sqrt{2}} \ket{01} + \frac{1}{2}\ket{10} + \frac{1}{2} \ket{11}
\end{equation*}
Solution.
\begin{equation*}
P(00) = \frac{1}{8}
\end{equation*}
\begin{equation*}
P(01) = \frac{3}{8}
\end{equation*}
\begin{equation*}
P(10) = \frac{1}{4}
\end{equation*}
\begin{equation*}
P(11) = \frac{1}{4}
\end{equation*}
Checkpoint 1.10.4.
Compute \(\ket{\psi}^{\otimes 4}\)
\begin{equation*}
\ket{\psi} = \frac{1}{\sqrt{3}}\ket{0} + \frac{\sqrt{2}}{\sqrt{3}}\ket{1}
\end{equation*}
Hint.
Solution.
\begin{equation*}
\ket{\psi} = \begin{pmatrix} \frac{1}{\sqrt{3}} \\ \frac{\sqrt{2}}{\sqrt{3}} \end{pmatrix} = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ \sqrt{2} \end{pmatrix}
\end{equation*}
\begin{equation*}
\ket{\psi}^{\otimes 4} = \ket{\psi} \otimes \ket{\psi} \otimes \ket{\psi} \otimes \ket{\psi} = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ \sqrt{2} \end{pmatrix} \otimes \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ \sqrt{2} \end{pmatrix} \otimes \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ \sqrt{2} \end{pmatrix} \otimes \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ \sqrt{2} \end{pmatrix}
\end{equation*}
\begin{equation*}
= \frac{1}{3} \begin{pmatrix} 1 \\ \sqrt{2} \\ \sqrt{2} \\ 2 \end{pmatrix} \otimes \frac{1}{3} \begin{pmatrix} 1 \\ \sqrt{2} \\ \sqrt{2} \\ 2 \end{pmatrix} = \frac{1}{9} \begin{pmatrix} 1 \\ \sqrt{2} \\ \sqrt{2} \\ 2 \\ \sqrt{2} \\ 2 \\ 2 \\ 2\sqrt{2} \\ \sqrt{2} \\ 2 \\ 2 \\ 2\sqrt{2} \\ 2 \\ 2\sqrt{2} \\ 2\sqrt{2} \\ 4 \end{pmatrix}
\end{equation*}
