Phase Revisited: Interference & Physical Meaning

Overview

Earlier we learned that a qubit state has complex amplitudes, and that global phase is unobservable while relative phase can matter.

This page answers the natural next question: how can phase matter if probabilities come from magnitudes like $|\alpha|^2$ and $|\beta|^2$?

The answer is interference: when multiple amplitude contributions combine, their phases can reinforce or cancel, redistributing probability across outcomes.

We will explain this without time evolution or gates—only by comparing different measurements and how the same state looks in different bases.

Intuition

Think about adding arrows (vectors) in the plane.

• Two arrows pointing the same way add to a longer arrow.
• Two arrows pointing opposite ways can cancel.

Quantum amplitudes behave like arrows too—except they live in the complex plane.

Probabilities are squares of magnitudes, but the magnitude you square can come from a sum of amplitudes. If the summed amplitudes partially cancel, the probability can decrease; if they reinforce, it can increase.

That is the core meaning of phase: it controls how amplitudes add.

Formal Description

Probability is the square of an overlap

For a pure state $|\psi\rangle$ and a measurement outcome state $|b\rangle$,

$$P(b)=|\langle b|\psi\rangle|^2.$$

Now suppose $|b\rangle$ itself can be written as a superposition of computational basis states, such as

$$|b\rangle = c_0|0\rangle + c_1|1\rangle,$$

with complex coefficients $c_0$ and $c_1$.

If

$$|\psi\rangle = \alpha|0\rangle + \beta|1\rangle,$$

then the overlap becomes

$$\langle b|\psi\rangle = c_0^*\alpha + c_1^*\beta.$$

This is a sum of complex numbers. The relative phases of $\alpha$ and $\beta$ matter because they affect whether this sum is large or small.

Relative phase can change probabilities in a different basis

In the computational basis, the outcomes correspond directly to $|0\rangle$ and $|1\rangle$, so probabilities depend only on $|\alpha|^2$ and $|\beta|^2$.

But in a different basis, each outcome overlap involves sums like $c_0^*\alpha + c_1^*\beta$, and phase can influence those sums.

Worked Example

Compare the two states

$$|\psi_+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle+|1\rangle),\quad |\psi_-\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle-|1\rangle).$$

Same probabilities in the computational basis

For both states:

• amplitude of $|0\rangle$ has magnitude $1/\sqrt{2}$
• amplitude of $|1\rangle$ has magnitude $1/\sqrt{2}$

So measuring in the computational basis gives $P(0)=P(1)=1/2$ for both.

Different probabilities in the $\{|+\rangle,|-\rangle\}$ basis

Define

$$|+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle+|1\rangle),\quad |-\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle-|1\rangle).$$

Notice that $|\psi_+\rangle$ is exactly $|+\rangle$, and $|\psi_-\rangle$ is exactly $|-\rangle$.

So:

• If the state is $|\psi_+\rangle$, measuring in the $\{|+\rangle,|-\rangle\}$ basis gives “$+$” with probability 1.
• If the state is $|\psi_-\rangle$, the same measurement gives “$-$” with probability 1.

This is interference in its simplest form: a relative minus sign changed nothing in one measurement basis, but made outcomes deterministic in a different basis.

Turtle Tip

Common Pitfalls

Quick Check

What’s Next

Interference is the mechanism by which quantum computations shape probabilities. Next we’ll consolidate the state picture by comparing pure and mixed states side-by-side, including how the Bloch sphere generalizes to the Bloch ball for mixed states.