|q⟩ Bad Qubits

intermediate · Physics · Hilbert Spaces & Dirac Formalism

Orthonormal Bases and Completeness

An orthonormal basis turns abstract vectors into concrete lists of numbers, and the completeness relation built from such a basis is one of the most-used identities in all of quantum mechanics. This lesson introduces both and shows how to wield the completeness relation as a calculational tool.

Orthonormal bases

A set {∣e1⟩,∣e2⟩,…}\{|e_1\rangle, |e_2\rangle, \ldots\} is orthonormal if its members are mutually orthogonal and individually normalized:

⟨ei∣ej⟩=δij={1i=j,0i≠j.\langle e_i | e_j \rangle = \delta_{ij} = \begin{cases} 1 & i = j, \\ 0 & i \ne j. \end{cases}

If, in addition, the set spans the whole Hilbert space, it is an orthonormal basis. The computational basis {∣0⟩,∣1⟩}\{|0\rangle, |1\rangle\} for a qubit is the standard example.

Expansion coefficients are inner products

In an orthonormal basis, any state expands as ∣ψ⟩=∑kck∣ek⟩|\psi\rangle = \sum_k c_k |e_k\rangle, and each coefficient is recovered by a single inner product. Taking the bra-ket with ∣ej⟩|e_j\rangle,

⟨ej∣ψ⟩=∑kck⟨ej∣ek⟩=∑kck δjk=cj.\langle e_j | \psi \rangle = \sum_k c_k \langle e_j | e_k \rangle = \sum_k c_k \,\delta_{jk} = c_j.

So ck=⟨ek∣ψ⟩c_k = \langle e_k | \psi \rangle. The orthonormality is doing all the work: it collapses the sum to a single surviving term.

The completeness relation (resolution of the identity)

Substitute ck=⟨ek∣ψ⟩c_k = \langle e_k | \psi \rangle back into the expansion:

∣ψ⟩=∑k⟨ek∣ψ⟩ ∣ek⟩=∑k∣ek⟩⟨ek∣ψ⟩=(∑k∣ek⟩⟨ek∣)∣ψ⟩.|\psi\rangle = \sum_k \langle e_k | \psi \rangle \, |e_k\rangle = \sum_k |e_k\rangle \langle e_k | \psi \rangle = \left( \sum_k |e_k\rangle\langle e_k| \right) |\psi\rangle.

Since this holds for every ∣ψ⟩|\psi\rangle, the operator in parentheses must be the identity:

  ∑k∣ek⟩⟨ek∣=I  \boxed{\;\sum_k |e_k\rangle\langle e_k| = I\;}

This is the completeness relation, also called the resolution of the identity. Each term ∣ek⟩⟨ek∣|e_k\rangle\langle e_k| is an outer product (a projector, covered in the next lesson), and they sum to the identity precisely when the basis is complete.

Parseval's identity and normalization

Sandwich the completeness relation inside ⟨ψ∣ ⋅ ∣ψ⟩\langle\psi|\,\cdot\,|\psi\rangle:

⟨ψ∣ψ⟩=∑k⟨ψ∣ek⟩⟨ek∣ψ⟩=∑k∣⟨ek∣ψ⟩∣2=∑k∣ck∣2.\langle\psi|\psi\rangle = \sum_k \langle\psi|e_k\rangle\langle e_k|\psi\rangle = \sum_k |\langle e_k|\psi\rangle|^2 = \sum_k |c_k|^2.

This is Parseval's identity. For a normalized physical state, ⟨ψ∣ψ⟩=1\langle\psi|\psi\rangle = 1, so

∑k∣ck∣2=1.\sum_k |c_k|^2 = 1.

The squared coefficients are exactly the Born-rule probabilities of measuring each basis outcome, and completeness guarantees they sum to 1 — total probability is conserved.

Worked example

Take the normalized state

∣ψ⟩=15((1+i)∣0⟩+3 ∣1⟩),|\psi\rangle = \frac{1}{\sqrt{5}}\Big( (1+i)|0\rangle + \sqrt{3}\,|1\rangle \Big),

so c0=(1+i)/5c_0 = (1+i)/\sqrt{5} and c1=3/5c_1 = \sqrt{3}/\sqrt{5}. Then ∣c0∣2=∣1+i∣2/5=2/5|c_0|^2 = |1+i|^2/5 = 2/5 and ∣c1∣2=3/5|c_1|^2 = 3/5, giving

∑k∣ck∣2=25+35=1,\sum_k |c_k|^2 = \frac{2}{5} + \frac{3}{5} = 1,

confirming both normalization and the completeness relation in this basis.

Try it

This is a numerical exercise — your code should return a number, not a circuit. For the state above, compute ∑k∣ck∣2\sum_k |c_k|^2 and confirm it equals 1.

Run your code to see the quantum state.

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