Dirac Bra–Ket Notation & Outer Products

Paul Dirac introduced a notation in which the type of every object is visible at a glance. A ket $\lvert\psi\rangle$ is a vector of the state space, which for us is a column in $\mathbb{C}^n$. A bra $\langle\phi\rvert$ is the conjugate-transposed row. Putting a bra next to a ket closes the bracket and produces a number, the inner product $\langle\phi\vert\psi\rangle$ ("bra-c-ket"). Putting a ket next to a bra, $\lvert a\rangle\langle b\rvert$, leaves the bracket open and produces an operator. With these two rules, long calculations become mechanical: juxtapose symbols, and whenever a $\langle\cdot\rvert$ meets a $\lvert\cdot\rangle$ collapse them into a complex number.

The label inside the ket is just a name. For a qubit, $\lvert0\rangle=(1,0)^T$ and $\lvert1\rangle=(0,1)^T$ are the computational basis vectors; $\lvert+\rangle$, $\lvert-\rangle$, $\lvert{+i}\rangle$, $\lvert{-i}\rangle$ are the vectors met in the previous lessons, and for several qubits $\lvert q_0q_1\cdots q_{n-1}\rangle$ names the computational basis vector with index $k=\sum_jq_j2^{n-1-j}$.

Worked example. Let $\lvert\psi\rangle=\tfrac{1}{\sqrt3}\lvert0\rangle+\tfrac{1+i}{\sqrt3}\lvert1\rangle=\tfrac1{\sqrt3}(1,\ 1+i)^T$. It is normalised because $\tfrac13+\tfrac{\lvert1+i\rvert^2}{3}=\tfrac13+\tfrac23=1$. Its bra is

$\langle\psi\rvert=\tfrac{1}{\sqrt3}\langle0\rvert+\tfrac{1-i}{\sqrt3}\langle1\rvert=\tfrac{1}{\sqrt3}\bigl(1,\ 1-i\bigr).$

The amplitude of $\lvert1\rangle$ is $\langle1\vert\psi\rangle=\tfrac{1+i}{\sqrt3}$, while $\langle\psi\vert1\rangle=\tfrac{1-i}{\sqrt3}$ is its conjugate. The overlap with $\lvert+\rangle$ is $\langle+\vert\psi\rangle=\tfrac{1}{\sqrt2}\cdot\tfrac{1}{\sqrt3}\bigl(1+(1+i)\bigr)=\tfrac{2+i}{\sqrt6}$, so measuring $\lvert\psi\rangle$ in the $\pm$ basis gives $+$ with probability $\tfrac{\lvert2+i\rvert^2}{6}=\tfrac56$.

Worked examples. The four outer products of computational basis vectors are the matrix units:

$\lvert0\rangle\langle0\rvert=\begin{pmatrix}1&0\\0&0\end{pmatrix},\quad \lvert0\rangle\langle1\rvert=\begin{pmatrix}0&1\\0&0\end{pmatrix},\quad \lvert1\rangle\langle0\rvert=\begin{pmatrix}0&0\\1&0\end{pmatrix},\quad \lvert1\rangle\langle1\rvert=\begin{pmatrix}0&0\\0&1\end{pmatrix}.$

Adding them gives the Pauli operators in Dirac form: $X=\lvert0\rangle\langle1\rvert+\lvert1\rangle\langle0\rvert$ (it maps $\lvert1\rangle\mapsto\lvert0\rangle$ and $\lvert0\rangle\mapsto\lvert1\rangle$), $Z=\lvert0\rangle\langle0\rvert-\lvert1\rangle\langle1\rvert$ and $Y=-i\lvert0\rangle\langle1\rvert+i\lvert1\rangle\langle0\rvert$. For example $X\lvert0\rangle=\lvert0\rangle\langle1\vert0\rangle+\lvert1\rangle\langle0\vert0\rangle=0+\lvert1\rangle$.

A complex example: with $\lvert a\rangle=(1,\ i)^T$ and $\lvert b\rangle=(2,\ 1+i)^T$ the rule $a_i\overline{b_j}$ gives

$\lvert a\rangle\langle b\rvert=\begin{pmatrix}1\cdot2&1\cdot(1-i)\\ i\cdot2&i(1-i)\end{pmatrix}=\begin{pmatrix}2&1-i\\2i&1+i\end{pmatrix},$

and applying it to $\lvert c\rangle=(1,\ 0)^T$ yields the first column $(2,\ 2i)^T=2\lvert a\rangle=\langle b\vert c\rangle\lvert a\rangle$, since $\langle b\vert c\rangle=\bar2\cdot1=2$.

Worked example. For $\lvert+\rangle=\tfrac1{\sqrt2}(1,1)^T$ the rule $\psi_i\overline{\psi_j}$ gives $P_+=\lvert+\rangle\langle+\rvert=\tfrac12\begin{pmatrix}1&1\\1&1\end{pmatrix}$. Squaring, $P_+^2=\tfrac14\begin{pmatrix}2&2\\2&2\end{pmatrix}=P_+$, and $\mathrm{Tr}\,P_+=\tfrac12+\tfrac12=1$. Applied to $\lvert0\rangle$ it returns $\langle+\vert0\rangle\lvert+\rangle=\tfrac{1}{\sqrt2}\lvert+\rangle$, whose squared length $\tfrac12$ is the probability of the outcome $+$ when $\lvert0\rangle$ is measured in the $\pm$ basis. In general $\lVert P\lvert v\rangle\rVert^2=\lvert\langle\psi\vert v\rangle\rvert^2=\langle v\vert P\vert v\rangle$, which is how the Born rule will be written in the measurement postulate.

For the circular state $\lvert{+i}\rangle=\tfrac1{\sqrt2}(1,\ i)^T$ the projector has complex off-diagonal entries: $P_{01}=\psi_0\overline{\psi_1}=\tfrac1{\sqrt2}\cdot\tfrac{-i}{\sqrt2}=-\tfrac i2$ and $P_{10}=\tfrac{i}{2}$, so the matrix is Hermitian but not symmetric. Writing $+\tfrac i2$ in the top-right corner is the classic symptom of forgetting to conjugate the bra.

Worked example. In the computational basis, $\lvert0\rangle\langle0\rvert+\lvert1\rangle\langle1\rvert=\mathrm{diag}(1,0)+\mathrm{diag}(0,1)=I$. In the $\pm$ basis, $\lvert-\rangle\langle-\rvert=\tfrac12\begin{pmatrix}1&-1\\-1&1\end{pmatrix}$ and

$\lvert+\rangle\langle+\rvert+\lvert-\rangle\langle-\rvert=\frac12\begin{pmatrix}1&1\\1&1\end{pmatrix}+\frac12\begin{pmatrix}1&-1\\-1&1\end{pmatrix}=\frac12\begin{pmatrix}2&0\\0&2\end{pmatrix}=I .$

The completeness relation is the single most useful trick in Dirac notation: you may insert $I=\sum_i\lvert i\rangle\langle i\rvert$ anywhere in an expression. Inserting it between a bra and a ket gives $\langle\phi\vert\psi\rangle=\sum_i\langle\phi\vert i\rangle\langle i\vert\psi\rangle=\sum_i\overline{\phi_i}\psi_i$, recovering the component formula for the inner product.

Worked example (matrix elements in another basis). Compute the matrix elements of $X=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ in the $\pm$ basis. Since $X\lvert+\rangle=\tfrac1{\sqrt2}(1,1)^T=\lvert+\rangle$ and $X\lvert-\rangle=\tfrac1{\sqrt2}(-1,1)^T=-\lvert-\rangle$, we get $\langle+\vert X\vert+\rangle=1$, $\langle-\vert X\vert-\rangle=-1$, and $\langle+\vert X\vert-\rangle=-\langle+\vert-\rangle=0=\langle-\vert X\vert+\rangle$. The resolution formula therefore gives

$X=\lvert+\rangle\langle+\rvert-\lvert-\rangle\langle-\rvert=\frac12\begin{pmatrix}1&1\\1&1\end{pmatrix}-\frac12\begin{pmatrix}1&-1\\-1&1\end{pmatrix}=\begin{pmatrix}0&1\\1&0\end{pmatrix}.$

In the $\pm$ basis $X$ is diagonal, exactly as $Z=\lvert0\rangle\langle0\rvert-\lvert1\rangle\langle1\rvert$ is diagonal in the computational basis. This is a first glimpse of the spectral theorem studied in the next module: an operator written as a weighted sum of projectors onto an orthonormal basis.

Worked example (an expectation value, symbolically). Take $\lvert\psi\rangle=\tfrac1{\sqrt3}\lvert0\rangle+\tfrac{1+i}{\sqrt3}\lvert1\rangle$ from the first section and $Z=\lvert0\rangle\langle0\rvert-\lvert1\rangle\langle1\rvert$. Sandwiching $Z$ between the bra and the ket and closing every bracket,

$\langle\psi\vert Z\vert\psi\rangle=\langle\psi\vert0\rangle\langle0\vert\psi\rangle-\langle\psi\vert1\rangle\langle1\vert\psi\rangle=\lvert\langle0\vert\psi\rangle\rvert^2-\lvert\langle1\vert\psi\rangle\rvert^2=\frac13-\frac23=-\frac13 .$

We used $\langle\psi\vert i\rangle=\overline{\langle i\vert\psi\rangle}$, so each term is a modulus squared. No matrix was ever written down: the calculation is pure symbol manipulation, and it would read the same in $\mathbb{C}^{1024}$. This is the practical reason for Dirac notation. Matrices are best for small numeric checks and for code; bra–ket algebra is best for proofs and for reasoning about structure. Throughout the curriculum you should be able to move freely between the two, and the translator widget above exists precisely to train that reflex.

import { c, cconj, cmul, type Complex } from '@/lib/quantum-simulator/complex';

/** Outer product |a><b| as a dense matrix: entry (i, j) = a_i * conj(b_j). */
function outer(a: Complex[], b: Complex[]): Complex[][] {
  return a.map((ai) => b.map((bj) => cmul(ai, cconj(bj))));
}

const s = Math.SQRT1_2;
const plusI = [c(s), c(0, s)]; // |+i> = (|0> + i|1>)/sqrt(2)
const P = outer(plusI, plusI);
// P = [[0.5, -0.5i], [0.5i, 0.5]] : Hermitian, idempotent, trace 1
console.log(P.map((row) => row.map((z) => z.re.toFixed(2) + (z.im < 0 ? '-' : '+') + Math.abs(z.im).toFixed(2) + 'i').join('  ')).join('; '));

0.50+0.00i  0.00-0.50i; 0.00+0.50i  0.50+0.00i