Dirac Notation -- Bras and Kets

A ket $|\psi\rangle$ denotes a vector in the Hilbert space $\mathcal{H}$ of a quantum system. The symbol $\psi$ is a label; it can be a wave function, a spin state, or any other descriptor. Kets encode the complete physical state of the system at a given time.

Kets can be added and multiplied by complex scalars following the usual vector-space rules. A superposition such as $|\psi\rangle = c_1 |\psi_1\rangle + c_2 |\psi_2\rangle$ is itself a valid state, as long as it is nonzero. This linearity is the mathematical expression of quantum superposition.

A basis $\{|j\rangle\}_{j=1}^n$ provides a concrete representation. Expanding a ket as $|\psi\rangle = \sum_j \psi_j |j\rangle$ gives a column of amplitudes $\psi_j = \langle j|\psi\rangle$. The choice of basis is arbitrary but often dictated by the observable being measured.

Dirac notation separates the abstract vector from its coordinates. This makes formulas cleaner and emphasizes physical content. For example, the Schrödinger equation can be written $i\hbar \frac{d}{dt}|\psi(t)\rangle = \hat{H}|\psi(t)\rangle$ without committing to a representation.

For every Hilbert space $\mathcal{H}$, there is a dual space $\mathcal{H}^*$ consisting of all continuous linear functionals on $\mathcal{H}$. The Riesz representation theorem guarantees that every such functional can be written as an inner product with a unique ket. Dirac denotes this functional by $\langle\psi|$, called a bra.

Given a ket $|\psi\rangle$, the corresponding bra $\langle\psi|$ is obtained by Hermitian conjugation: it is the row vector whose entries are the complex conjugates of the entries of $|\psi\rangle$. In equations, $\langle\psi| = (|\psi\rangle)^\dagger$.

Bras act on kets to produce complex numbers. The expression $\langle\phi|\psi\rangle$ is the inner product of $|\phi\rangle$ and $|\psi\rangle$. It is linear in the ket $|\psi\rangle$ and conjugate-linear in the bra $\langle\phi|$. This asymmetry reflects the complex structure of $\mathcal{H}$.

The bra-ket machinery makes it easy to talk about dual bases. If $\{|j\rangle\}$ is an orthonormal basis, the dual basis is $\{\langle j|\}$. A vector $|\psi\rangle$ can be reconstructed from its bra by conjugation, and vice versa.

The inner product is a map $\mathcal{H} \times \mathcal{H} \to \mathbb{C}$ that is conjugate-linear in the first argument and linear in the second. It satisfies $\langle\phi|\psi\rangle = \langle\psi|\phi\rangle^*$, is positive definite, and obeys the Cauchy-Schwarz inequality.

In a finite orthonormal basis, the inner product becomes the sum $\langle\phi|\psi\rangle = \sum_j \phi_j^* \psi_j$. This is the complex dot product with conjugation on the first vector. In function spaces it becomes an integral, $\langle\phi|\psi\rangle = \int \phi^*(x)\psi(x)\,dx$.

The inner product measures the overlap between two states. If $\langle\phi|\psi\rangle = 0$, the states are orthogonal. If $|\langle\phi|\psi\rangle| = 1$ and both states are normalized, they represent the same physical state up to a global phase.

Many quantum formulas are inner products in disguise. The probability of measuring outcome j is $|\langle j|\psi\rangle|^2$, the expectation value of an operator is $\langle\psi|A|\psi\rangle$, and transition amplitudes are $\langle\phi|U|\psi\rangle$. Dirac notation makes these structures transparent.

The norm of a ket is defined by $\|\psi\| = \sqrt{\langle\psi|\psi\rangle}$. A state with $\|\psi\| = 1$ is called normalized, or a unit vector. Normalization ensures that probabilities sum to 1 when the Born rule is applied.

Any nonzero ket can be normalized by dividing by its norm: $|\hat{\psi}\rangle = |\psi\rangle / \|\psi\|$. This rescaling does not change the physical state because a global phase or positive rescaling is unobservable. Quantum states are therefore rays in Hilbert space, not individual vectors.

The set of normalized states in $\mathcal{H} = \mathbb{C}^2$ forms the Bloch sphere. A general pure qubit state can be written $|\psi\rangle = \cos\frac{\theta}{2}|0\rangle + e^{i\phi}\sin\frac{\theta}{2}|1\rangle$, with $\theta \in [0,\pi]$ and $\phi \in [0,2\pi)$. Points on the sphere correspond to distinct physical states.

A state vector that is not normalized still carries the same directional information but must be normalized before computing probabilities. Many intermediate calculations are easier if normalization is imposed only at the end.

Two kets $|\phi\rangle$ and $|\psi\rangle$ are orthogonal if $\langle\phi|\psi\rangle = 0$. Orthogonal states are maximally distinct: if the system is definitely in one of them, it has zero probability of being found in the other.

A set of vectors is orthonormal if every vector has norm 1 and distinct vectors are orthogonal. In symbols, $\langle i|j\rangle = \delta_{ij}$, where $\delta_{ij}$ is the Kronecker delta. Orthonormal bases are the most convenient bases for calculations because inner products reduce to single terms.

The Gram-Schmidt process converts any linearly independent set into an orthonormal set with the same span. This guarantees that every finite-dimensional Hilbert space has an orthonormal basis. In infinite dimensions, separability ensures the existence of a countable orthonormal basis.

Orthonormal bases simplify expansions. The coefficient of $|j\rangle$ in $|\psi\rangle$ is simply $\psi_j = \langle j|\psi\rangle$, and the norm squared is $\|\psi\|^2 = \sum_j |\psi_j|^2$. These formulas are used constantly in quantum mechanics.

Let $\mathcal{H} = \mathbb{C}^3$ with orthonormal basis $\{|1\rangle, |2\rangle, |3\rangle\}$. Define $|\psi\rangle = i|1\rangle + 2|2\rangle - |3\rangle$ and $|\phi\rangle = |1\rangle - i|2\rangle + 2|3\rangle$. Compute $\langle\phi|\psi\rangle$, $\|\psi\|$, and check whether the states are orthogonal.

Step 1 -- Inner product. Using $\langle\phi| = \langle 1| + i\langle 2| + 2\langle 3|$, we get $\langle\phi|\psi\rangle = (1)(i) + (i)(2) + (2)(-1) = i + 2i - 2 = -2 + 3i$. Conjugation on the first vector is essential.

Step 2 -- Norm of $|\psi\rangle$. $\|\psi\|^2 = \langle\psi|\psi\rangle = |i|^2 + |2|^2 + |-1|^2 = 1 + 4 + 1 = 6$, so $\|\psi\| = \sqrt{6}$.

Step 3 -- Orthogonality. Since $\langle\phi|\psi\rangle = -2 + 3i \neq 0$, the states are not orthogonal. Their squared overlap is $|\langle\phi|\psi\rangle|^2 = (-2)^2 + 3^2 = 13$.