|q⟩ Bad Qubits

intermediate · Physics · Hilbert Spaces & Dirac Formalism

Representing States in a Basis

A vector is an abstract object, but to compute with it we pick a basis and write down its components. This lesson makes the bridge between the coordinate-free ket ∣ψ⟩|\psi\rangle and the column of numbers we actually manipulate.

Components are basis-dependent

Fix an orthonormal basis {∣ek⟩}\{|e_k\rangle\}. Any state expands as

∣ψ⟩=∑kck∣ek⟩,ck=⟨ek∣ψ⟩.|\psi\rangle = \sum_k c_k |e_k\rangle, \qquad c_k = \langle e_k | \psi \rangle.

The numbers ckc_k are the components (or coordinates) of ∣ψ⟩|\psi\rangle in that basis. The vector itself does not change when you choose a different basis — only its list of components does. This is the same idea as describing the same arrow by different (x,y)(x, y) pairs after rotating your axes.

The column-vector representation

Collecting the components into a column gives the familiar matrix representation. For a qubit in the computational basis,

∣ψ⟩=c0∣0⟩+c1∣1⟩  ⟷  (c0c1),⟨ψ∣  ⟷  (c0‾c1‾).|\psi\rangle = c_0 |0\rangle + c_1 |1\rangle \;\longleftrightarrow\; \begin{pmatrix} c_0 \\ c_1 \end{pmatrix}, \qquad \langle\psi| \;\longleftrightarrow\; \begin{pmatrix} \overline{c_0} & \overline{c_1} \end{pmatrix}.

Inner products then become ordinary matrix multiplication of a row by a column, and operators become matrices. Choosing a basis is what makes Dirac notation computable.

Changing basis

Suppose we also have a second orthonormal basis {∣fk⟩}\{|f_k\rangle\}. The components in the new basis are again given by inner products, ck′=⟨fk∣ψ⟩c'_k = \langle f_k | \psi \rangle. Inserting the completeness relation of the old basis,

ck′=⟨fk∣ψ⟩=∑j⟨fk∣ej⟩⟨ej∣ψ⟩=∑jUkj cj,c'_k = \langle f_k | \psi \rangle = \sum_j \langle f_k | e_j \rangle \langle e_j | \psi \rangle = \sum_j U_{kj}\, c_j,

where Ukj=⟨fk∣ej⟩U_{kj} = \langle f_k | e_j \rangle are the entries of a unitary change-of-basis matrix. The state is fixed; the unitary just rewrites its coordinates. The Hadamard gate, for instance, is exactly the unitary that converts between the computational {∣0⟩,∣1⟩}\{|0\rangle, |1\rangle\} basis and the {∣+⟩,∣−⟩}\{|+\rangle, |-\rangle\} basis.

Real-amplitude states and the RYRY rotation

A useful family of single-qubit states has real, non-negative amplitudes:

∣ψ(θ)⟩=cos⁡ ⁣θ2 ∣0⟩+sin⁡ ⁣θ2 ∣1⟩.|\psi(\theta)\rangle = \cos\!\frac{\theta}{2}\,|0\rangle + \sin\!\frac{\theta}{2}\,|1\rangle.

These are produced from ∣0⟩|0\rangle by the rotation gate RY(θ)RY(\theta), whose matrix is

RY(θ)=(cos⁡θ2−sin⁡θ2sin⁡θ2cos⁡θ2),RY(\theta) = \begin{pmatrix} \cos\frac{\theta}{2} & -\sin\frac{\theta}{2} \\ \sin\frac{\theta}{2} & \cos\frac{\theta}{2} \end{pmatrix},

so that RY(θ)∣0⟩=∣ψ(θ)⟩RY(\theta)|0\rangle = |\psi(\theta)\rangle. The components depend on the single angle θ\theta, and normalization cos⁡2+sin⁡2=1\cos^2 + \sin^2 = 1 holds automatically.

Worked example

To represent

∣ψ⟩=12∣0⟩+32∣1⟩|\psi\rangle = \tfrac{1}{2}|0\rangle + \tfrac{\sqrt{3}}{2}|1\rangle

we need cos⁡(θ/2)=1/2\cos(\theta/2) = 1/2, i.e. θ/2=π/3\theta/2 = \pi/3, so θ=2π/3\theta = 2\pi/3. Then sin⁡(θ/2)=sin⁡(π/3)=3/2\sin(\theta/2) = \sin(\pi/3) = \sqrt{3}/2, matching the ∣1⟩|1\rangle component. The probabilities are ∣c0∣2=1/4|c_0|^2 = 1/4 and ∣c1∣2=3/4|c_1|^2 = 3/4, summing to 1.

Try it

Prepare the state ∣ψ⟩=12∣0⟩+32∣1⟩|\psi\rangle = \tfrac{1}{2}|0\rangle + \tfrac{\sqrt{3}}{2}|1\rangle on one qubit using an RYRY rotation, and return the circuit. The grader checks the full statevector.

Run your code to see the quantum state.

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