If you have read about quantum computing before, you have definitely stumbled across expressions like $|\psi\rangle$ or $\langle 0|1\rangle$. This is Dirac notation — also called bra-ket notation — and it is the universal language of quantum mechanics and quantum computing.

For a CS student it is just linear algebra in a clever costume. Once you see through the costume, everything clicks.

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The Core Idea: States Are Vectors

In quantum computing, the state of a qubit is a vector in a 2-dimensional complex vector space $\mathbb{C}^2$.

The two basis states — the quantum equivalents of 0 and 1 — are written as column vectors:

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

These are called kets. A ket $|\psi\rangle$ is just any column vector representing a quantum state.

A general single-qubit state is a linear combination (superposition) of these basis vectors:

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

where $\alpha, \beta \in \mathbb{C}$ and $|\alpha|^2 + |\beta|^2 = 1$.

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Bras: The Row Vector Counterpart

Every ket has a matching bra, written $\langle\psi|$. It is the conjugate transpose (Hermitian adjoint) of the ket:

$\langle\psi| = |\psi\rangle^\dagger = \begin{pmatrix} \alpha^* & \beta^* \end{pmatrix}$

where $\alpha^*$ is the complex conjugate of $\alpha$.

So for the basis states:

$\langle 0| = \begin{pmatrix} 1 & 0 \end{pmatrix}, \qquad \langle 1| = \begin{pmatrix} 0 & 1 \end{pmatrix}$

The notation is designed so that combining a bra and a ket gives you a scalar — an inner product.

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Inner Products: The Bracket

The inner product of two states $\langle\phi|$ and $|\psi\rangle$ is written $\langle\phi|\psi\rangle$ — a “bra-ket”, which is where the name comes from.

For the basis states:

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

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

$\langle 1|0\rangle = 0, \qquad \langle 1|1\rangle = 1$

This is exactly orthonormality — $|0\rangle$ and $|1\rangle$ are orthogonal unit vectors, just like the standard basis $\hat{e}_1$ and $\hat{e}_2$ you know from linear algebra.

The inner product $\langle\phi|\psi\rangle$ gives the probability amplitude for measuring state $|\phi\rangle$ given that the system is in state $|\psi\rangle$. The probability itself is $|\langle\phi|\psi\rangle|^2$.

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Outer Products: Building Matrices

The outer product $|\psi\rangle\langle\phi|$ produces a matrix instead of a scalar. This is the key to constructing quantum gates.

For example:

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

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

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

Notice that $|0\rangle\langle 0| + |1\rangle\langle 1| = I$ — the identity matrix. This is the completeness relation, and it shows up constantly when decomposing gates.

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Quantum Gates Are Unitary Matrices

A quantum gate transforms one qubit state into another. Since quantum evolution must be reversible and norm-preserving (probabilities must still sum to 1), every gate is represented by a unitary matrix $U$, where $UU^\dagger = I$.

Applying a gate to a state is ordinary matrix-vector multiplication:

$|\psi'\rangle = U|\psi\rangle$

Let’s go through the most important single-qubit gates.

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The Pauli-X Gate (Quantum NOT)

$X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = |0\rangle\langle 1| + |1\rangle\langle 0|$

It flips $|0\rangle \to |1\rangle$ and $|1\rangle \to |0\rangle$. Verify:

$X|0\rangle = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} = |1\rangle \checkmark$

$X|1\rangle = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} = |0\rangle \checkmark$

In the outer product form: $X|\psi\rangle = (|0\rangle\langle 1| + |1\rangle\langle 0|)(\alpha|0\rangle + \beta|1\rangle) = \alpha|1\rangle + \beta|0\rangle$. The amplitudes are swapped — exactly what NOT does.

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The Pauli-Z Gate (Phase Flip)

$Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = |0\rangle\langle 0| - |1\rangle\langle 1|$

$Z$ leaves $|0\rangle$ unchanged and flips the sign (phase) of $|1\rangle$:

$Z|0\rangle = |0\rangle, \qquad Z|1\rangle = -|1\rangle$

This does not change the measurement probabilities ($|-1|^2 = 1$), but phase matters for interference — it is the kind of gate that makes quantum algorithms work.

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The Pauli-Y Gate

$Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} = i(|1\rangle\langle 0| - |0\rangle\langle 1|)$

$Y$ combines a bit flip and a phase flip:

$Y|0\rangle = i|1\rangle, \qquad Y|1\rangle = -i|0\rangle$

Together $X$, $Y$, $Z$ are the Pauli matrices, the fundamental building blocks of single-qubit operations. You will see them everywhere.

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The Hadamard Gate

$H = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = \frac{1}{\sqrt{2}}(|0\rangle\langle 0| + |0\rangle\langle 1| + |1\rangle\langle 0| - |1\rangle\langle 1|)$

The Hadamard gate creates superposition. It maps the basis states to the diagonal basis $|+\rangle$ and $|-\rangle$:

$H|0\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \frac{|0\rangle + |1\rangle}{\sqrt{2}} = |+\rangle$

$H|1\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 0 \\ 1 \end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \end{pmatrix} = \frac{|0\rangle - |1\rangle}{\sqrt{2}} = |-\rangle$

$|+\rangle$ is a 50/50 superposition with both amplitudes positive. $|-\rangle$ is a 50/50 superposition with a phase difference — they look identical when measured, but behave differently under further gates.

Apply $H$ twice and you get back to where you started: $H^2 = I$.

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The S and T Phase Gates

$S = \begin{pmatrix} 1 & 0 \\ 0 & i \end{pmatrix}, \qquad T = \begin{pmatrix} 1 & 0 \\ 0 & e^{i\pi/4} \end{pmatrix}$

Both leave $|0\rangle$ unchanged and rotate the phase of $|1\rangle$ — by $90°$ for $S$ and $45°$ for $T$. Note $S = T^2$ and $Z = S^2 = T^4$.

These gates are essential for universal quantum computing. The set $\{H, T, \text{CNOT}\}$ is universal — any unitary operation can be approximated to arbitrary precision using only these three gates (Solovay-Kitaev theorem).

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Two Qubits: Tensor Products

For a 2-qubit system, states live in a $4$-dimensional space: $\mathbb{C}^2 \otimes \mathbb{C}^2$. The basis states are $|00\rangle$, $|01\rangle$, $|10\rangle$, $|11\rangle$, written as column vectors of length 4.

The tensor product $|a\rangle \otimes |b\rangle = |ab\rangle$ combines two single-qubit states:

$|00\rangle = |0\rangle \otimes |0\rangle = \begin{pmatrix}1\\0\end{pmatrix} \otimes \begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}1\\0\\0\\0\end{pmatrix}, \qquad |01\rangle = \begin{pmatrix}0\\1\\0\\0\end{pmatrix}, \qquad |10\rangle = \begin{pmatrix}0\\0\\1\\0\end{pmatrix}, \qquad |11\rangle = \begin{pmatrix}0\\0\\0\\1\end{pmatrix}$

A 2-qubit gate is a $4 \times 4$ unitary matrix.

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The CNOT Gate

The Controlled-NOT gate is the canonical 2-qubit gate:

$\text{CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix} = |00\rangle\langle 00| + |01\rangle\langle 01| + |11\rangle\langle 10| + |10\rangle\langle 11|$

The first qubit is the control, the second is the target. The target is flipped if and only if the control is $|1\rangle$:

Input	Output
$	00\rangle$
$	01\rangle$
$	10\rangle$
$	11\rangle$

Verify the last row:

$\text{CNOT}|11\rangle = \begin{pmatrix} 1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0 \end{pmatrix}\begin{pmatrix}0\\0\\0\\1\end{pmatrix} = \begin{pmatrix}0\\0\\1\\0\end{pmatrix} = |10\rangle \checkmark$

Combine CNOT with Hadamard to create entanglement. Start with $|00\rangle$, apply $H$ to the first qubit, then CNOT:

$|00\rangle \xrightarrow{H \otimes I} \frac{|0\rangle+|1\rangle}{\sqrt{2}} \otimes |0\rangle = \frac{|00\rangle+|10\rangle}{\sqrt{2}} \xrightarrow{\text{CNOT}} \frac{|00\rangle+|11\rangle}{\sqrt{2}}$

That final state is a Bell state — the qubits are now entangled.

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Putting It All Together

Here is the full notation reference at a glance:

Notation	Type	Meaning
$\|\psi\rangle$	ket	column vector — a quantum state
$\langle\psi\|$	bra	row vector — conjugate transpose of the ket
$\langle\phi\|\psi\rangle$	bracket	inner product — a scalar (probability amplitude)
$\|\psi\rangle\langle\phi\|$	outer product	a matrix (used to build gates)
$U\|\psi\rangle$	matrix-vector product	applying gate $U$ to state $\|\psi\rangle$
$\|ab\rangle$	tensor product	joint state of two qubits

Every quantum circuit you will ever read is just a sequence of these matrix multiplications. The notation looks exotic, but the machinery is linear algebra — eigenvectors, inner products, unitary matrices. If you are comfortable with those, you already have the tools to read and reason about quantum circuits.

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Further Reading

To practice computing with these by hand, the Qiskit Textbook has interactive circuit exercises. For a rigorous linear algebra treatment of the formalism, chapters 2–4 of Nielsen & Chuang are the standard reference. If you want a faster ramp, Quantum Computation: An Applied Approach by Hidary covers the same ground more concisely.