Matrix Multiplication

In quantum mechanics every operation on a state vector — every quantum gate, every time-evolution operator, every measurement projector — is a matrix. To apply two operations in sequence you multiply their matrices. Matrix multiplication is therefore not optional bookkeeping; it is the language in which quantum dynamics is written.

The rule

Given two $2 \times 2$ matrices

A = \begin{pmatrix} a_{00} & a_{01} \\ a_{10} & a_{11} \end{pmatrix}, \qquad B = \begin{pmatrix} b_{00} & b_{01} \\ b_{10} & b_{11} \end{pmatrix},

their product $C = AB$ is the $2 \times 2$ matrix whose $(i,j)$ entry is

c_{ij} = \sum_{k} a_{ik}\, b_{kj}.

In words: take row $i$ of $A$ and column $j$ of $B$ , multiply them term by term, and sum the results. This is the dot product of a row with a column.

Writing it out for a $2 \times 2$ product:

AB = \begin{pmatrix} a_{00}b_{00} + a_{01}b_{10} & a_{00}b_{01} + a_{01}b_{11} \\ a_{10}b_{00} + a_{11}b_{10} & a_{10}b_{01} + a_{11}b_{11} \end{pmatrix}.

Worked example: the Hadamard gate squared

The Hadamard matrix is

H = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}.

Computing $H^2 = H \cdot H$ :

H^2 = \frac{1}{2}\begin{pmatrix} 1\cdot 1 + 1\cdot 1 & 1\cdot 1 + 1\cdot(-1) \\ 1\cdot 1 + (-1)\cdot 1 & 1\cdot 1 + (-1)\cdot(-1) \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I.

So $H^2 = I$ : applying the Hadamard gate twice brings a qubit back to its original state. This identity, which follows purely from matrix multiplication, tells you that $H$ is its own inverse.

Order matters: AB does not equal BA in general

Consider the Pauli matrices

X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.

Multiplying $XZ$ :

XZ = \begin{pmatrix} 0\cdot 1 + 1\cdot 0 & 0\cdot 0 + 1\cdot(-1) \\ 1\cdot 1 + 0\cdot 0 & 1\cdot 0 + 0\cdot(-1) \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}.

Multiplying $ZX$ :

ZX = \begin{pmatrix} 1\cdot 0 + 0\cdot 1 & 1\cdot 1 + 0\cdot 0 \\ 0\cdot 0 + (-1)\cdot 1 & 0\cdot 1 + (-1)\cdot 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.

$XZ \neq ZX$ : the product depends on the order. In quantum computing this non-commutativity is not an annoyance but a fundamental physical fact — it is the algebraic origin of the uncertainty principle.

How matrix products act on state vectors

A matrix acts on a column vector by the same row-times-column rule. For a $2 \times 2$ matrix $M$ and a two-component ket

M|\psi\rangle = \begin{pmatrix} m_{00} & m_{01} \\ m_{10} & m_{11} \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} m_{00}\alpha + m_{01}\beta \\ m_{10}\alpha + m_{11}\beta \end{pmatrix}.

This is just matrix multiplication with the vector treated as a single-column matrix. Everything you practice here carries over directly to applying gates in quantum circuits.

Try it

This is a numerical exercise — your code should return a number, not a circuit. Compute the $(0,1)$ entry — that is, row 0, column 1 — of the product $XZ$ , where

X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.

Run your code to see the quantum state.