Matrix Elements

A linear operator becomes concrete the moment you pick a basis: it turns into a grid of numbers, a matrix. The individual numbers in that grid are called matrix elements, and computing them is the everyday arithmetic of quantum mechanics.

Definition

Let $\{|e_1\rangle, \dots, |e_n\rangle\}$ be an orthonormal basis, so $\langle e_i | e_j\rangle = \delta_{ij}$ . For a linear operator $A$ , the matrix element between basis vectors $|e_i\rangle$ and $|e_j\rangle$ is the complex number

A_{ij} = \langle e_i | A | e_j \rangle.

Read it inside out: first $A$ acts on $|e_j\rangle$ to produce a vector $A|e_j\rangle$ , then you take the inner product of that vector with $|e_i\rangle$ . The result is a single scalar — the entry in row $i$ , column $j$ of the matrix of $A$ .

Why this recovers the whole matrix

Recall from the previous lesson that $A|e_j\rangle = \sum_i A_{ij}|e_i\rangle$ . Take the inner product of both sides with $\langle e_i|$ and use orthonormality:

\langle e_i | A | e_j \rangle = \sum_k A_{kj}\,\langle e_i | e_k\rangle = \sum_k A_{kj}\,\delta_{ik} = A_{ij}.

So the bracket $\langle e_i | A | e_j \rangle$ is exactly the coefficient $A_{ij}$ . Sweeping $i$ and $j$ over the basis fills in the entire matrix. The matrix and the operator carry the same information once a basis is fixed.

Reading off the matrix from gate actions

For a qubit with the computational basis $\{|0\rangle, |1\rangle\}$ (with $|0\rangle$ labeling row/ column $0$ and $|1\rangle$ labeling row/column $1$ ), the three Pauli operators are

X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.

\langle 0 | Y | 1 \rangle = \langle 0 | (-i|0\rangle) = -i\,\langle 0 | 0\rangle = -i.

That $-i$ sits in row $0$ , column $1$ — exactly the top-right entry of the matrix above. Likewise

\langle 1 | Y | 0 \rangle = \langle 1 | (i|1\rangle) = i,

the bottom-left entry. Notice $\langle 0|Y|1\rangle = \overline{\langle 1|Y|0\rangle}$ : the off-diagonal elements are complex conjugates of each other. That is the Hermitian symmetry $A_{ij} = \overline{A_{ji}}$ , a property we will study in detail two lessons from now.

A useful sanity check: the trace

The diagonal matrix elements $A_{ii} = \langle e_i | A | e_i\rangle$ add up to the trace,

\operatorname{Tr}(A) = \sum_i \langle e_i | A | e_i \rangle.

Remarkably, the trace does not depend on the chosen orthonormal basis, even though the individual $A_{ii}$ do. For each Pauli matrix above, the two diagonal entries cancel, so $\operatorname{Tr}(X) = \operatorname{Tr}(Y) = \operatorname{Tr}(Z) = 0$ .

Try it

Compute the matrix element $\langle 1 | Y | 0 \rangle$ of the Pauli- $Y$ operator. It is purely imaginary — return its imaginary part as a plain number.

Run your code to see the quantum state.