Global vs relative phase

Two states that look completely different in the math — yet are physically identical.

Consider two quantum states. The first is the familiar equal superposition:

$$|\psi\rangle = \frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle$$

The second looks quite different — every amplitude has been multiplied by $e^{i\pi/4}$ (a rotation of 45° in the complex plane):

$$e^{i\pi/4}|\psi\rangle = \frac{e^{i\pi/4}}{\sqrt{2}}|0\rangle + \frac{e^{i\pi/4}}{\sqrt{2}}|1\rangle$$

Both states look different on paper. But here's the question: do they behave differently when you run an experiment? Take a moment to guess before reading on.

But now change something more subtle. Instead of rotating all amplitudes by the same amount, rotate only the $|1\rangle$ amplitude:

$$|\psi'\rangle = \frac{1}{\sqrt{2}}|0\rangle + \frac{e^{i\pi/4}}{\sqrt{2}}|1\rangle$$

Now the two states are distinguishable. Measure in a different basis and you will see a different outcome distribution. The phase matters — but only when it is a difference between components.

Global phase rotates the whole state together. Relative phase changes the angle between components.

A quantum state with two amplitudes looks like this:

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

Both $\alpha$ and $\beta$ are complex numbers. They each live somewhere on the complex plane — each with a direction (angle) and a magnitude (distance from the origin). The phase of an amplitude is its angle on the complex plane.

Global phase — the whole state rotates together

Multiplying by $e^{i\varphi}$ rotates every amplitude by the same angle $\varphi$. Both $\alpha$ and $\beta$ rotate equally:

$$e^{i\varphi}|\psi\rangle = e^{i\varphi}\alpha|0\rangle + e^{i\varphi}\beta|1\rangle$$

What determines a measurement probability? The rule from M09: $P = |\alpha|^2 = \alpha \cdot \alpha^*$. Now apply this to $e^{i\varphi}\alpha$:

$$|e^{i\varphi}\alpha|^2 = (e^{i\varphi}\alpha)(e^{i\varphi}\alpha)^* = (e^{i\varphi}\alpha)(e^{-i\varphi}\alpha^*) = e^{0} \cdot \alpha\alpha^* = |\alpha|^2$$

The global phase $e^{i\varphi}$ multiplies by its own conjugate $e^{-i\varphi}$ and cancels to 1. The probability is unchanged. This is not a coincidence — it is a mathematical inevitability, visible in one step.

Relative phase — the angle between components changes

Now rotate only the $|1\rangle$ amplitude by $\theta$, leaving $|0\rangle$ untouched:

$$|\psi'\rangle = \alpha|0\rangle + e^{i\theta}\beta|1\rangle$$

The angle between the two components has changed. The individual probabilities $|\alpha|^2$ and $|e^{i\theta}\beta|^2 = |\beta|^2$ are the same as before — but if you measure in a different basis (one that can see the angle between them), you will get a different result. Interference depends on this angle.

Global phase cancels exactly in every probability calculation. Relative phase survives.

Formal state notation

Let $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ where $\alpha, \beta \in \mathbb{C}$ and $|\alpha|^2 + |\beta|^2 = 1$ (normalization). The probability of measuring $|0\rangle$ is $|\alpha|^2$; the probability of measuring $|1\rangle$ is $|\beta|^2$.

Why global phase is unobservable — the derivation

Apply a global phase $e^{i\varphi}$ to every amplitude. The new state is:

Relative phase survives

Now consider a state with a relative phase $\theta$ on the $|1\rangle$ component:

$$|\psi'\rangle = \alpha|0\rangle + e^{i\theta}\beta|1\rangle$$

Individual probabilities in the standard ($|0\rangle, |1\rangle$) basis:

Micro practice — three conceptual checks

Slide the global phase and watch probabilities stay fixed. Slide the relative phase and watch them shift.

The state starts as $|\psi\rangle = \frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle$ — the equal superposition you saw many times in Track 1. Use the two sliders to apply a global phase $\varphi$ and a relative phase $\theta$ independently. The probability bars and amplitude vectors update in real time, and the live formula at the bottom shows the exact state you're looking at.