Orthogonality

Two vectors are orthogonal when they point in completely independent directions — knowing the component of a vector along one direction tells you nothing about its component along the other. This idea, simple in geometry, is fundamental in quantum mechanics: quantum states that are orthogonal are perfectly distinguishable by an ideal measurement.

The inner product

Given two column vectors in $\mathbb{R}^n$,

their inner product (real dot product) is

The test for orthogonality is then simply:

Geometric meaning

The inner product is related to the angle $\theta$ between the vectors by $\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos\theta$. Setting this to zero gives $\cos\theta = 0$, so $\theta = 90°$. Orthogonal vectors are perpendicular.

A concrete example

Take $\mathbf{u} = (3, 4)$ and $\mathbf{v} = (4, -3)$.

Because the inner product is zero, $\mathbf{u}$ and $\mathbf{v}$ are orthogonal. You can also verify this geometrically: $\mathbf{u}$ lies in the first quadrant pointing roughly "up-right", while $\mathbf{v}$ points roughly "right-down", and they meet at a right angle.

Why it matters in quantum mechanics

In Dirac notation a quantum state is written as a ket $|{\psi}\rangle$. Two states $|{\psi}\rangle$ and $|{\phi}\rangle$ are orthogonal when $\langle\phi|\psi\rangle = 0$, where $\langle\phi|$ is the corresponding bra (the conjugate transpose). The standard basis states $|{0}\rangle$ and $|{1}\rangle$ satisfy $\langle 0 | 1 \rangle = 0$: measuring a qubit prepared in $|{0}\rangle$ will never return the outcome associated with $|{1}\rangle$, and vice versa. That perfect distinguishability rests entirely on orthogonality.

Try it

This is a numerical exercise — your code should return a number. Given the vectors $\mathbf{u} = (3, 4)$ and $\mathbf{v} = (4, -3)$, compute their dot product and return 1 if they are orthogonal, or 0 if they are not.

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