Mathematical Introduction: Matrices and how they can be used in quantum computing

By the end of the lesson, students will be able to

• understand what matrices are and how to perform basic operations like addition and multiplication on them;
• discover how matrices are used in quantum mechanics and quantum computing;
• learn about the Kronecker (tensor) product, a special matrix operation used, for example, to describe multi-qubit systems;
• become familiar with the Dirac (bra-ket) notation, a symbolic way of representing quantum states.

A matrix is an ${\displaystyle m\times <!--\; \times \; -->n}$ array of numbers, ${\displaystyle m}$ being the number of rows and ${\displaystyle n}$ the number of columns. The order of a matrix is the number of rows times the number of columns, i.e. ${\displaystyle m\times <!--\; \times \; -->n}$. Let's say if a matrix has 5 rows and 4 columns then the order of the matrix will be 20. 

${\displaystyle M=\left[\begin{array}{cccc}{a}_{11}& a_{12}& a_{13}& a_{14}\\ a_{21}& a_{22}& a_{23}& a_{24}\\ a_{31}& a_{32}& a_{33}& a_{34}\\ a_{41}& a_{42}& a_{43}& a_{44}\\ a_{51}& a_{52}& a_{53}& a_{54}\end{array}\right]}$

A square matrix is a matrix that has the same number of rows and columns.

${\displaystyle M=\left[\begin{array}{ccc}1& 2& -<!--\; -\; -->4\\ 0& 1& 5\\ -<!--\; -\; -->2& 0& 3\end{array}\right]}$

An identity matrix is a square matrix with 1s in the main diagonal and 0s elsewhere.

${\displaystyle M=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]}$

In order to perform an addition or subtraction, matrices must have the same number of rows and the same number of columns. Matrix addition and subtraction are performed element by element. Each element of one matrix is added to or subtracted from the corresponding element of the other matrix.

Example

For two given matrices ${\displaystyle A}$ and ${\displaystyle B}$, calculate ${\displaystyle A+B}$ and ${\displaystyle A-<!--\; -\; -->B}$

${\displaystyle A=\left[\begin{array}{cc}2& 3\\ 5& -<!--\; -\; -->1\end{array}\right]}$ and ${\displaystyle B=\left[\begin{array}{cc}1& 2\\ 3& 0\end{array}\right]}$.

Solution

${\displaystyle A+B=\left[\begin{array}{cc}2& 3\\ 5& -<!--\; -\; -->1\end{array}\right]+\left[\begin{array}{cc}1& 2\\ 3& 0\end{array}\right]=\left[\begin{array}{cc}2+1& 3+2\\ 5+3& -<!--\; -\; -->1+0\end{array}\right]=\left[\begin{array}{cc}3& 5\\ 8& -<!--\; -\; -->1\end{array}\right]}$

${\displaystyle A-<!--\; -\; -->B=\left[\begin{array}{cc}2& 3\\ 5& -<!--\; -\; -->1\end{array}\right]-<!--\; -\; -->\left[\begin{array}{cc}1& 2\\ 3& 0\end{array}\right]=\left[\begin{array}{cc}2-<!--\; -\; -->1& 3-<!--\; -\; -->2\\ 5-<!--\; -\; -->3& -<!--\; -\; -->1-<!--\; -\; -->0\end{array}\right]=\left[\begin{array}{cc}1& 1\\ 2& -<!--\; -\; -->1\end{array}\right]}$

A matrix can be multiplied with another matrix if their dimensions align (i.e., for two matrices A and B, the number of columns in A must match the number of rows in B). The result yields a matrix that has as many rows as the first matrix and as many columns as the second matrix.

${\displaystyle A=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]}$ and ${\displaystyle B=\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]}$

${\displaystyle AB=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]=\left[\begin{array}{cc}{a}_{11}{b}_{11}+a_{12}{b}_{21}& a_{11}{b}_{12}+a_{12}{b}_{22}\\ a_{21}{b}_{11}+a_{22}{b}_{21}& a_{21}{b}_{12}+a_{22}{b}_{22}\end{array}\right]}$

${\displaystyle BA=\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]=\left[\begin{array}{cc}{b}_{11}{a}_{11}+b_{12}{a}_{21}& b_{11}{a}_{12}+b_{12}{a}_{22}\\ b_{21}{a}_{11}+b_{22}{a}_{21}& b_{21}{a}_{12}+b_{22}{a}_{22}\end{array}\right]}$

Example

Take the two matrices from the example in “Addition and subtraction”. The results of ${\displaystyle AB}$ and ${\displaystyle BA}$ are:

$AB=\left[\begin{array}{cc}2& 3\\ 5& -1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 3& 0\end{array}\right]=\left[\begin{array}{cc}2·1+3·3& 2·2+3·0\\ 5·1+-1·3& 5·2+-1·0\end{array}\right]=\left[\begin{array}{cc}11& 4\\ 2& 10\end{array}\right]$

$BA=\left[\begin{array}{cc}1& 2\\ 3& 0\end{array}\right]\left[\begin{array}{cc}2& 3\\ 5& -1\end{array}\right]=\left[\begin{array}{cc}1·2+2·5& 1·3+2·-1\\ 3·2+0·5& 3·3+0·-1\end{array}\right]=\left[\begin{array}{cc}12& 1\\ 6& 9\end{array}\right]$

Example 3

Consider again the two matrices ${\displaystyle A=\left[\begin{array}{cc}2& 3\\ 5& -1\end{array}\right]}$ and ${\displaystyle B=\left[\begin{array}{cc}1& 2\\ 3& 0\end{array}\right]}$. Their Kronecker product is:
${\displaystyle A\otimes <!--\; \otimes \; -->B=\left[\begin{array}{cc}2& 3\\ 5& -<!--\; -\; -->1\end{array}\right]\otimes <!--\; \otimes \; -->\left[\begin{array}{cc}1& 2\\ 3& 0\end{array}\right]=\left[\begin{array}{cc}2\cdot <!--\; \cdot \; -->\left[\begin{array}{cc}1& 2\\ 3& 0\end{array}\right]& 3\cdot <!--\; \cdot \; -->\left[\begin{array}{cc}1& 2\\ 3& 0\end{array}\right]\\ 5\cdot <!--\; \cdot \; -->\left[\begin{array}{cc}1& 2\\ 3& 0\end{array}\right]& -<!--\; -\; -->1\cdot <!--\; \cdot \; -->\left[\begin{array}{cc}1& 2\\ 3& 0\end{array}\right]\end{array}\right]}$ 
${\displaystyle =\left[\begin{array}{cccc}2& 4& 3& 6\\ 6& 0& 9& 0\\ 5& 10& -<!--\; -\; -->1& -<!--\; -\; -->2\\ 15& 0& -<!--\; -\; -->3& 0\end{array}\right]}$.

Example 4

The Kronecker product is distributive:${\displaystyle (\stackrel{\to <!--\; \to \; -->}{u}+\stackrel{\to <!--\; \to \; -->}{v})\otimes <!--\; \otimes \; -->\stackrel{\to <!--\; \to \; -->}{w}=\stackrel{\to <!--\; \to \; -->}{u}\otimes <!--\; \otimes \; -->\stackrel{\to <!--\; \to \; -->}{w}+\stackrel{\to <!--\; \to \; -->}{v}\otimes <!--\; \otimes \; -->\stackrel{\to <!--\; \to \; -->}{w}}$

a) Verify the distributivity relation for the following vectors:
${\displaystyle \stackrel{\to <!--\; \to \; -->}{u}=\left[\begin{array}{c}1\\ 0\end{array}\right],\stackrel{\to <!--\; \to \; -->}{u}=\left[\begin{array}{c}1\\ 0\end{array}\right]}$ and ${\displaystyle \stackrel{\to <!--\; \to \; -->}{w}=\left[\begin{array}{c}1\\ 0\end{array}\right]}$.

b) In ket-notation, these three vectors may be written as:
${\displaystyle \stackrel{\to <!--\; \to \; -->}{u}=|0⟩}$, ${\displaystyle \stackrel{\to <!--\; \to \; -->}{v}=|1⟩}$ and ${\displaystyle \stackrel{\to <!--\; \to \; -->}{w}=|0⟩}$.

${\displaystyle \stackrel{\to <!--\; \to \; -->}{u}\otimes <!--\; \otimes \; -->\stackrel{\to <!--\; \to \; -->}{w}+\stackrel{\to <!--\; \to \; -->}{v}\otimes <!--\; \otimes \; -->\stackrel{\to <!--\; \to \; -->}{w}}$ can therefore also be written as:

${\displaystyle |0⟩\otimes <!--\; \otimes \; -->|0⟩+|1⟩\otimes <!--\; \otimes \; -->|0⟩}$, or in an even shorter notation: ${\displaystyle |00⟩+|10⟩}$.

The Kronecker product is useful because it allows the manipulation of composite systems by treating them as combinations of single components.

Example 5

Consider the two qubit states:
${\displaystyle |0⟩=\left[\begin{array}{c}1\\ 0\end{array}\right]}$ and ${\displaystyle |1⟩=\left[\begin{array}{c}0\\ 1\end{array}\right]}$.

A quantum state of a two-qubit system (composed of these two qubits) is given by the Kronecker product, for example,
${\displaystyle |0⟩\otimes <!--\; \otimes \; -->|0⟩=\left[\begin{array}{c}1\\ 0\end{array}\right]\otimes <!--\; \otimes \; -->\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right]=|00⟩}$

Note: the Kronecker product is not commutative: ${\displaystyle |0⟩\otimes <!--\; \otimes \; -->|1⟩\ne <!--\; \ne \; -->|1⟩\otimes <!--\; \otimes \; -->|0⟩}$.

Example 6

Quantum states cannot always be written as simple Kronecker products. Let us consider the two-qubit state:

${\displaystyle |{\mathrm{\Phi <!--\; \Phi \; -->}}^{+}⟩=\frac{1}{\sqrt{2}}(|00⟩+|11⟩)}$,

which corresponds to the following sum of Kronecker products:

${\displaystyle |{\mathrm{\Phi <!--\; \Phi \; -->}}^{+}⟩=\frac{1}{\sqrt{2}}\left(\left[\begin{array}{c}1\\ 0\end{array}\right]\otimes <!--\; \otimes \; -->\left[\begin{array}{c}1\\ 0\end{array}\right]\right)+\frac{1}{\sqrt{2}}\left(\left[\begin{array}{c}0\\ 1\end{array}\right]\otimes <!--\; \otimes \; -->\left[\begin{array}{c}0\\ 1\end{array}\right]\right)}$

This is equal to:

${\displaystyle |{\mathrm{\Phi <!--\; \Phi \; -->}}^{+}⟩=\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right]+\frac{1}{\sqrt{2}}\left[\begin{array}{c}0\\ 0\\ 0\\ 1\end{array}\right]=\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ 0\\ 0\\ 1\end{array}\right]}$

The resulting state cannot be represented by a Kronecker product of two basic qubit states.

1. B
2. A
3. B
4. A