Dirac Bra-Ket Notation

Paul Dirac introduced a notation so compact and suggestive that it has become the universal language of quantum mechanics. This lesson makes the bra, the ket, and their pairing precise.

Kets are vectors

A ket $|\psi\rangle$ is just a vector in a Hilbert space $\mathcal{H}$ . The symbol inside the angle bracket is a label — it can be a name ( $|\psi\rangle$ ), a number ( $|0\rangle$ , $|1\rangle$ ), or any identifier convenient for the problem. In $\mathbb{C}^n$ a ket is a column vector, for example

|0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \qquad |1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}.

A general single-qubit state is a complex linear combination $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ .

Bras are linear functionals

Every ket $|\psi\rangle$ has a partner called a bra, written $\langle\psi|$ . The bra is the conjugate transpose of the ket: where the ket is a column vector, the bra is the row vector of complex conjugates,

|\psi\rangle = \begin{pmatrix} \alpha \\ \beta \end{pmatrix} \quad\Longrightarrow\quad \langle\psi| = \begin{pmatrix} \overline{\alpha} & \overline{\beta} \end{pmatrix}.

A bra is a machine that eats a ket and outputs a complex number — a linear functional. We will say much more about this dual nature in the next lesson.

The bra-ket is the inner product

Placing a bra next to a ket and merging the two vertical bars gives a bra-ket, which is exactly the inner product:

\langle \phi | \psi \rangle = \langle\phi| \, |\psi\rangle = (\text{row vector}) \times (\text{column vector}).

For $|\phi\rangle = \big(\begin{smallmatrix} \gamma \\ \delta \end{smallmatrix}\big)$ and $|\psi\rangle = \big(\begin{smallmatrix} \alpha \\ \beta \end{smallmatrix}\big)$ this is

\langle \phi | \psi \rangle = \overline{\gamma}\,\alpha + \overline{\delta}\,\beta.

The notation makes the inner-product axioms visually obvious. Conjugate symmetry, $\langle \phi | \psi \rangle = \overline{\langle \psi | \phi \rangle}$ , is the statement that swapping row and column and conjugating recovers the same number's conjugate.

Orthonormality in bra-ket form

The computational basis is orthonormal:

\langle 0 | 0 \rangle = 1, \quad \langle 1 | 1 \rangle = 1, \quad \langle 0 | 1 \rangle = \langle 1 | 0 \rangle = 0,

compactly $\langle i | j \rangle = \delta_{ij}$ with the Kronecker delta. Using this, the amplitude of $|0\rangle$ in $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ is extracted by a single bra-ket:

\langle 0 | \psi \rangle = \alpha \langle 0|0\rangle + \beta \langle 0|1\rangle = \alpha.

The equal superposition $|+\rangle$

A state you will meet constantly is

|+\rangle = \frac{1}{\sqrt{2}}\big(|0\rangle + |1\rangle\big) = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}.

It has equal amplitude $1/\sqrt{2}$ on both basis kets, so $|\langle 0|+\rangle|^2 = |\langle 1|+\rangle|^2 = 1/2$ : measuring it yields 0 or 1 with equal probability. The Hadamard gate produces it from $|0\rangle$ , $H|0\rangle = |+\rangle$ , which is what you will build below.

Try it

Prepare the ket $|+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ on one qubit and return the circuit. The grader checks the full statevector, so both amplitudes must equal $1/\sqrt{2}$ .

Run your code to see the quantum state.