01 — Dirac Notation
Ket, Bra, and Braket — what they are and what they do
Why does this notation exist?

A qubit state is just a vector. You could write it as \(\begin{pmatrix}\alpha \\ \beta\end{pmatrix}\) every time, but that gets tedious. Dirac notation is shorthand that makes quantum math readable.

Ket: \(|\psi\rangle\) = column vector

A ket is just a column vector with a name inside the \(|\ \rangle\) symbol.

The basis states:

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

A general qubit:

$$|\psi\rangle = \alpha|0\rangle + \beta|1\rangle = \begin{pmatrix} \alpha \\ \beta \end{pmatrix}$$

That's it. \(|0\rangle\) is just a name for the column vector \(\begin{pmatrix}1\\0\end{pmatrix}\). Nothing deeper.

Bra: \(\langle\psi|\) = conjugate transpose of the ket

Take the ket, flip it to a row vector, and complex-conjugate each entry.

Step by step:

$$|\psi\rangle = \begin{pmatrix} \alpha \\ \beta \end{pmatrix} \quad\xrightarrow{\text{flip to row}}\quad (\alpha, \;\beta) \quad\xrightarrow{\text{conjugate}}\quad \langle\psi| = (\alpha^*, \;\beta^*)$$

Complex conjugate means: flip the sign of the imaginary part. So if \(\alpha = 3 + 2i\), then \(\alpha^* = 3 - 2i\). If \(\alpha\) is already real, nothing changes.

Examples:

$$|0\rangle = \begin{pmatrix}1\\0\end{pmatrix} \quad\Rightarrow\quad \langle 0| = (1, \; 0)$$

$$|1\rangle = \begin{pmatrix}0\\1\end{pmatrix} \quad\Rightarrow\quad \langle 1| = (0, \; 1)$$

$$|\psi\rangle = \begin{pmatrix}\frac{1+i}{2}\\ \frac{1-i}{2}\end{pmatrix} \quad\Rightarrow\quad \langle\psi| = \left(\frac{1-i}{2}, \; \frac{1+i}{2}\right)$$

Braket: \(\langle x | y \rangle\) = inner product (a single number)

Put a bra and a ket together. It's just row × column matrix multiplication:

$$\langle x | y \rangle = \text{(row vector)} \times \text{(column vector)} = \text{one number}$$

Example: \(\langle 0 | 1 \rangle\)

$$\langle 0|1\rangle = (1, \; 0) \cdot \begin{pmatrix}0\\1\end{pmatrix} = 1 \cdot 0 + 0 \cdot 1 = 0$$

Zero → orthogonal → measuring \(|1\rangle\) never gives outcome \(|0\rangle\).

Example: \(\langle 0 | 0 \rangle\)

$$\langle 0|0\rangle = (1, \; 0) \cdot \begin{pmatrix}1\\0\end{pmatrix} = 1 \cdot 1 + 0 \cdot 0 = 1$$

One → identical states → measuring \(|0\rangle\) always gives outcome \(|0\rangle\).

Example with a superposition: \(\langle 0 | \psi \rangle\)

If \(|\psi\rangle = \alpha|0\rangle + \beta|1\rangle\):

$$\langle 0|\psi\rangle = (1, \; 0) \cdot \begin{pmatrix}\alpha\\\beta\end{pmatrix} = \alpha$$

The inner product just picks out the amplitude for that basis state. Then:

$$|\langle 0|\psi\rangle|^2 = |\alpha|^2 = \text{probability of measuring } |0\rangle$$

The punchline

What each one is:

Ket \(|\psi\rangle\) — column vector. The qubit state.

Bra \(\langle\psi|\) — row vector, conjugated. The "mirror" of the ket.

Braket \(\langle x|y\rangle\) — single number. Multiply row × column. Measures overlap.

For probability: \(|\langle x|y\rangle|^2\) = probability of getting outcome \(|x\rangle\) when measuring state \(|y\rangle\).

Quick reference:

\(\langle 0|0\rangle = 1\) — same state, certain outcome

\(\langle 1|1\rangle = 1\) — same state, certain outcome

\(\langle 0|1\rangle = 0\) — orthogonal, impossible outcome

\(\langle 1|0\rangle = 0\) — orthogonal, impossible outcome

\(\langle 0|\psi\rangle = \alpha\) — picks out the \(|0\rangle\) amplitude

\(\langle 1|\psi\rangle = \beta\) — picks out the \(|1\rangle\) amplitude