Postulate 1: Normalization and Alternate Bases

Recall our beloved qubit in superposition:

$$\ket{\psi} = \alpha\ket{0}+\beta\ket{1}.$$

The symbol $|\psi\rangle$ represents the quantum state of the qubit, describing its complete physical state in the state space that is a two-dimensional Hilbert space. 

The right-hand side of this equation simply means that the qubit is in some sort of "blend" of the $\ket{0}$ and $\ket{1}$ states, rather than just one or the other. Whether we call it a linear combination, a superposition, a blend, what's important is that it leads to uniquely quantum phenomenon!

There is a mathematical constraint on the complex numbers $\alpha$ and $\beta$ known as the normalization condition:

$$|\alpha|^2 + |\beta|^2 = 1.$$

This is to ensure $\ket{\psi}$ satisfies the first postulate of quantum mechanics, i.e. that individual quantum systems are described by unit vectors in a Hilbert space.

Let's show why this mathematical constraint guarantees this.

Justification for the normalization condition

Recall the norm of a complex vector $\vec{v}$:

$$\|\vec{v}\| = \sqrt{\vec{v} \cdot \vec{v}}.$$

Equivalently, using bra-ket notation, for a quantum state $|\psi\rangle$, the norm is given by:

$$\|\psi\| = \sqrt{\langle \psi | \psi \rangle}.$$

Since we require quantum states to be unit vectors, this means:

$$\|\psi\| = 1.$$

Rather than writing $\|\psi\| = 1$ explicitly, it''s more common in quantum mechanics to check the squared norm directly:

$$\|\psi\|^2 = \langle \psi | \psi \rangle = 1.$$

This avoids dealing with square roots unnecessarily.

Computing:

\begin{align*} \|\psi\|^2 &= \langle \psi | \psi \rangle \\ &= (\alpha^* \bra{0} + \beta^* \bra{1})(\alpha |0\rangle + \beta |1\rangle) \\ ...