global-phase
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| + | # Global phase | ||
| + | **Global phase** is a property of a quantum state that emerges from the mathematical description but is physically completely unobservable. Multiplying an entire quantum state by a complex number of unit magnitude $e^{i\phi}$ produces a state that is physically identical to the original in every possible measurement. | ||
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| + | To gain intuition, start from a single qubit with probability amplitudes $a = Ae^{i\alpha}$ and $b = Be^{i\beta}$ where $A, B \in \mathbb{R}$ and $\alpha, \beta \in (-\pi, \pi]$. | ||
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| + | $$\lvert\psi\rangle = Ae^{i\alpha}\lvert 0\rangle + Be^{i\beta}\lvert 1\rangle$$ | ||
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| + | If we create a new qubit $\lvert\psi' | ||
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| + | $$\lvert\psi' | ||
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| + | The states $\lvert\psi\rangle$ and $\lvert\psi' | ||
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| + | ## No change in probabilities | ||
| + | The [[born-rule]] gives measurement probabilities $|a|^2$ and $|b|^2$. Since $|e^{i\phi}| = 1$ for any real $\phi$, multiplying by a global phase leaves all probabilities unchanged. | ||
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| + | $$P_0 = |Ae^{i(\alpha+\phi)}|^2 = A^2|e^{i(\alpha+\phi)}|^2 = A^2$$ | ||
| + | $$P_1 = |Be^{i(\beta+\phi)}|^2 = B^2|e^{i(\beta+\phi)}|^2 = B^2$$ | ||
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| + | These are identical to the probabilities for $\lvert\psi\rangle$. No experiment can detect the global phase $\phi$. | ||
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| + | ## Relative phase | ||
| + | Because global phase is unobservable, | ||
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| + | $$\lvert\psi\rangle = e^{i\alpha}\!\left(A\lvert 0\rangle + Be^{i(\beta-\alpha)}\lvert 1\rangle\right) \sim A\lvert 0\rangle + Be^{i\varphi}\lvert 1\rangle \qquad \varphi = \beta - \alpha$$ | ||
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| + | The quantity $\varphi$ is the **relative phase** between the two probability amplitudes. Unlike global phase, relative phase is physically observable — it determines interference patterns and measurement outcomes in rotated bases. For example, $\lvert +\rangle = (\lvert 0\rangle + \lvert 1\rangle)/ | ||
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| + | ## Removing global phase | ||
| + | Using the equivalence $\lvert\psi\rangle \sim A\lvert 0\rangle + Be^{i\varphi}\lvert 1\rangle$, we can write the most general qubit state using only three real numbers: the magnitudes $A, B \geq 0$ (with $A^2 + B^2 = 1$) and the relative phase $\varphi \in (-\pi, \pi]$. Substituting $A = \cos(\theta/ | ||
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| + | $$\lvert\psi\rangle = \cos\frac{\theta}{2}\lvert 0\rangle + e^{i\varphi}\sin\frac{\theta}{2}\lvert 1\rangle$$ | ||
