Matrix representations

2026-10-19

# Operators as matrices

Using closure (inserting the identity) twice:

$$ \hat X = \hat X\cdot\sum_{a'}|a'\rangle\langle a'| = \sum_{a'}\hat X|a'\rangle\langle a'| $$

$$ = \sum_{a''}|a''\rangle\langle a''|\cdot\sum_{a'}\hat X|a'\rangle\langle a'| $$

Any arbitrary operator:

$$ \boxed{\hat X = \sum_{a''}\sum_{a'}|a''\rangle\langle a''|\hat X|a'\rangle\langle a'|} $$

If $N$ is the dim. of the ket space, there are $N^2$ no.s of the form $\langle a''|\hat X|a'\rangle$ $\implies$ we can arrange all $N^2$ no.s in an $N \times N$ matrix. In $\langle a''|\hat X|a'\rangle$, $a''$ is the row and $a'$ is the column:

$$ \hat X \doteq \begin{pmatrix} \langle a^{(1)}|\hat X|a^{(1)}\rangle & \langle a^{(1)}|\hat X|a^{(2)}\rangle & \cdots \\ \langle a^{(2)}|\hat X|a^{(1)}\rangle & \langle a^{(2)}|\hat X|a^{(2)}\rangle & \cdots \\ \vdots & \vdots & \ddots \end{pmatrix} $$

where $|a^{(i)}\rangle$ denotes the eigenket of $\hat A$ (Hermitian), and $\doteq$ means “is represented by”.

Note that:

$$ \langle a''|\hat X|a'\rangle = \langle a'|\hat X^\dagger|a''\rangle^{\ast} $$

↳ Hermitian adjoint.

# Multiplying operators

The arrangement of the $X$ matrix is deliberate, to follow a row $\times$ column multiplication.

$$ \langle a'|\hat X|a''\rangle = \left(\langle a'|\hat X\right)\cdot|a''\rangle = \langle a'|\left(\hat X|a''\rangle\right) \in \mathbb{C} $$

Say, $\hat Z = \hat X\hat Y$

$$ \begin{aligned} \langle a''|\hat Z|a'\rangle &= \langle a''|\hat X\hat Y|a'\rangle \\ &= \langle a''|\hat X\left(\sum_{a'''}|a'''\rangle\langle a'''|\right)\hat Y|a'\rangle \end{aligned} $$

$$ \langle a''|\hat Z|a'\rangle = \sum_{a'''}\langle a''|\hat X|a'''\rangle\langle a'''|\hat Y|a'\rangle $$

which is (matrix) $= \sum$ (matrix) $\cdot$ (matrix), exactly matrix multiplication.

# Kets as column vectors

How can we represent $|\gamma\rangle = \hat X|\alpha\rangle$ using base kets?

Recall that $|\gamma\rangle = \sum_{a'}c_{a'}|a'\rangle$ and $c_{a'} = \langle a'|\gamma\rangle$.

$\therefore$ finding the expansion coefficients first,

$$ \begin{aligned} \langle a'|\gamma\rangle &= \langle a'|\hat X|\alpha\rangle \\ &= \langle a'|\hat X\left(\sum_{a''}|a''\rangle\langle a''|\right)|\alpha\rangle \\ &= \sum_{a''}\underbrace{\langle a'|\hat X|a''\rangle}_{\text{matrix}}\cdot\underbrace{\langle a''|\alpha\rangle}_{\text{const. numbers}} \end{aligned} $$

To realize this, we can find the expansion coefficient $c_{a'}$ for $|\alpha\rangle$ as $\langle a'|\alpha\rangle$, and hence

$$ |\alpha\rangle = \sum_{a'}\langle a'|\alpha\rangle|a'\rangle \doteq \begin{pmatrix} \langle a^{(1)}|\alpha\rangle \\ \langle a^{(2)}|\alpha\rangle \\ \vdots \end{pmatrix} $$

Similarly,

$$ |\gamma\rangle \doteq \begin{pmatrix} \langle a^{(1)}|\gamma\rangle \\ \langle a^{(2)}|\gamma\rangle \\ \vdots \end{pmatrix} $$

Now we have:

$$ \begin{pmatrix} \langle a^{(1)}|\gamma\rangle \\ \langle a^{(2)}|\gamma\rangle \\ \vdots \end{pmatrix} = \underbrace{\begin{pmatrix} \langle a^{(1)}|\hat X|a^{(1)}\rangle & \langle a^{(1)}|\hat X|a^{(2)}\rangle & \cdots \\ \langle a^{(2)}|\hat X|a^{(1)}\rangle & \langle a^{(2)}|\hat X|a^{(2)}\rangle & \cdots \\ \vdots & \vdots & \ddots \end{pmatrix}}_{\text{operator matrix}} \begin{pmatrix} \langle a^{(1)}|\alpha\rangle \\ \langle a^{(2)}|\alpha\rangle \\ \vdots \end{pmatrix} $$

That’s so cool.

# Bras as row vectors

We can do the same stuff for bras: $\langle\gamma| = \langle\alpha|\hat X$

find $c_{a'}$
represent all as matrices.

We can write a general bra as:

$$ \langle\gamma| = \sum_{a'}c_{a'}\langle a'| \implies \langle\gamma|a'\rangle = c_{a'} $$

$$ \implies \langle\gamma| = \sum_{a'}\langle\gamma|a'\rangle\langle a'| = \sum_{a'}\langle\alpha|\hat X|a'\rangle\langle a'| $$

or, inserting the identity $\sum_{a'}\Lambda_{a'}$:

$$ \begin{aligned} \langle\gamma|a''\rangle &= \langle\alpha|\hat X|a''\rangle \\ &= \langle\alpha|\left(\sum_{a'}|a'\rangle\langle a'|\right)\hat X|a''\rangle \\ &= \sum_{a'}\langle\alpha|a'\rangle\langle a'|\hat X|a''\rangle \end{aligned} $$

These are the coefficients $c_{a''}$. Hence we can write the coefficients of a bra in a row:

$$ \boxed{ \begin{aligned} \langle\gamma| &\doteq \begin{pmatrix} \langle\gamma|a^{(1)}\rangle & \langle\gamma|a^{(2)}\rangle & \langle\gamma|a^{(3)}\rangle & \cdots \end{pmatrix} \\ &= \begin{pmatrix} \langle a^{(1)}|\gamma\rangle^{\ast} & \langle a^{(2)}|\gamma\rangle^{\ast} & \cdots \end{pmatrix} \end{aligned} } $$

Hence the inner product $\langle\beta|\alpha\rangle$ can be thought of as a row vector ($\langle\beta|$ bra) times a column vector ($|\alpha\rangle$ ket).

$$ \begin{aligned} \langle\beta|\alpha\rangle &= \sum_{a'}\langle\beta|a'\rangle\langle a'|\alpha\rangle \equiv \langle\beta|\left(\sum_{a'}\Lambda_{a'}\right)|\alpha\rangle \\ &= \begin{pmatrix} \langle a^{(1)}|\beta\rangle^{\ast} & \langle a^{(2)}|\beta\rangle^{\ast} & \cdots \end{pmatrix} \begin{pmatrix} \langle a^{(1)}|\alpha\rangle \\ \langle a^{(2)}|\alpha\rangle \\ \vdots \end{pmatrix} \in \mathbb{C} \end{aligned} $$

A complex no.

And the outer product can then be:

$$ |\beta\rangle\langle\alpha| \doteq \begin{pmatrix} \langle a^{(1)}|\beta\rangle\langle a^{(1)}|\alpha\rangle^{\ast} & \langle a^{(1)}|\beta\rangle\langle a^{(2)}|\alpha\rangle^{\ast} & \cdots \\ \langle a^{(2)}|\beta\rangle\langle a^{(1)}|\alpha\rangle^{\ast} & \langle a^{(2)}|\beta\rangle\langle a^{(2)}|\alpha\rangle^{\ast} & \cdots \\ \vdots & \vdots & \ddots \end{pmatrix} $$

$\to$ a matrix.

So what we’ve done is essentially that we are saying each bra or ket can be broken down into a sum of basis vectors, found the right coefficients and slapped them in for linear algebra!

# Observables in their own eigenbasis

For an observable (Hermitian) $\hat A$, its matrix representation becomes very simply its eigenvalue or 0, if its eigenkets are the base kets.

$$ \hat A = \sum_{a''}\sum_{a'}|a''\rangle\underbrace{\langle a''|\hat A|a'\rangle}_{\text{matrix}}\langle a'| $$

If the basis vectors are the $|a'\rangle$, then the matrix is diagonal:

$$ \langle a''|\hat A|a'\rangle = \langle a'|\hat A|a'\rangle\delta_{a'a''} $$

$$ \boxed{\langle a''|\hat A|a'\rangle = a'\delta_{a'a''}} $$

(a diagonal square matrix)

$$ \implies \hat A = \sum_{a''}\sum_{a'}|a''\rangle a'\delta_{a'a''}\langle a'| $$

Dead easy.

$$ \boxed{\hat A = \sum_{a'}a'|a'\rangle\langle a'| = \sum_{a'}a'\Lambda_{a'}} $$

note: $\Lambda_{a'} = |a'\rangle\langle a'|$