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beginner · Physics · Math Foundations: Linear Algebra

Norm and Normalization

What is a norm?

Given a vector with nn components v=(v1,v2,,vn)v = (v_1, v_2, \ldots, v_n), its Euclidean norm (also called the 2\ell^2-norm or simply "the length") is

v=k=1nvk2.\|v\| = \sqrt{\sum_{k=1}^{n} |v_k|^2}.

The vertical bars vk2|v_k|^2 are there for generality: when the components are complex numbers, vk2=vkvk|v_k|^2 = v_k^* v_k is the squared modulus. For real components the expression reduces to the familiar Pythagorean sum.

Example. For v=(1,2,2)v = (1, 2, 2):

v=12+22+22=1+4+4=9=3.\|v\| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3.

Normalization

A vector is called normalized (or a unit vector) if its norm equals 1. To normalize vv, divide every component by v\|v\|:

v^=vv=(v1v,v2v,,vnv).\hat{v} = \frac{v}{\|v\|} = \left(\frac{v_1}{\|v\|},\, \frac{v_2}{\|v\|},\, \ldots,\, \frac{v_n}{\|v\|}\right).

One check: v^=1\|\hat{v}\| = 1 always, because

v^=vv=1.\|\hat{v}\| = \frac{\|v\|}{\|v\|} = 1.

Why quantum mechanics insists on normalization

In quantum mechanics a state is represented by a state vector (or ket) ψ|\psi\rangle. The Born rule says the probability of finding the system in basis state k|k\rangle is kψ2|\langle k|\psi\rangle|^2 — the squared modulus of the corresponding amplitude. Probabilities must sum to 1:

kkψ2=1ψ=1.\sum_k |\langle k|\psi\rangle|^2 = 1 \quad\Longleftrightarrow\quad \|\,|\psi\rangle\,\| = 1.

This is the physical reason normalization is not optional: an un-normalized state vector gives probabilities that do not add up to 1, which is physically meaningless. Whenever you construct a state vector from scratch — say, by superposing several basis states — the first thing to do is normalize it.

Worked example

Suppose the initial state of a three-level system is proportional to (1,2,2)(1, 2, 2). The norm is 3 (computed above), so the normalized state is

ψ=(13,23,23).|\psi\rangle = \left(\frac{1}{3},\, \frac{2}{3},\, \frac{2}{3}\right).

Check: (13)2+(23)2+(23)2=19+49+49=99=1\left(\frac{1}{3}\right)^2 + \left(\frac{2}{3}\right)^2 + \left(\frac{2}{3}\right)^2 = \frac{1}{9} + \frac{4}{9} + \frac{4}{9} = \frac{9}{9} = 1. The probabilities sum to 1, as required.

Try it

This is a numerical exercise — your code should return a number, not a circuit. Start from the state vector (1,2,2)(1, 2, 2), normalize it, and return its first component (which equals 13\tfrac{1}{3}).

Run your code to see the quantum state.

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