Global Phase Is Unobservable

Every quantum state has an overall complex factor that you can multiply it by without changing any measurement outcome. This factor is called the global phase, and understanding why it cannot be detected is one of the first steps toward reasoning correctly about quantum states.

What global phase is

A qubit state $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ is said to carry a global phase $e^{i\phi}$ when the entire state is multiplied by that factor:

Both amplitudes are scaled by the same complex number. The magnitude of any unit complex number is exactly 1: $|e^{i\phi}| = 1$ for all real $\phi$.

Why the Born rule kills global phase

The Born rule says the probability of outcome $b$ is $|\langle b|\psi\rangle|^2$. For a computational-basis measurement on the state above, the probabilities are

The factor $e^{i\phi}$ disappears because squaring the magnitude of a complex number of unit magnitude always gives 1. This argument works for every measurement basis, not just the computational one: you can rotate the measurement axis however you like, and the global phase still cancels. Two states related by a global phase are therefore physically identical — no experiment can distinguish them.

A concrete example

Take the simplest case: $|0\rangle$ and $e^{i\pi}|0\rangle = -|0\rangle$.

For $|0\rangle$: $\alpha = 1$, so $P(0) = |1|^2 = 1$ and $P(1) = 0$.

For $-|0\rangle$: $\alpha = -1$, so $P(0) = |-1|^2 = (-1)^2 = 1$ and $P(1) = 0$.

The two predictions are identical. No matter how many times you run the experiment, you cannot tell these states apart.

Try it

Apply the Born rule to the state $e^{i\pi}|0\rangle = -|0\rangle$ and return the probability of measuring $|0\rangle$. You should find the same answer as for plain $|0\rangle$ — confirming that the global phase $-1$ is unobservable.

Sign in on the full site to ask questions and join the discussion.