I am pretty new to Quantum Computing and am not exactly sure what the phase is. Could you please explain in terms of the Bloch sphere and point out how to mathematically calculate and represent the phase?


1 Answer 1


Quantum phase does not have an analog in classical world, it is simply one of quantum phenomenon (feature,...). To put it simply mathematicaly, you can write any qubit as $$ |q\rangle = \alpha|0\rangle + \beta|1\rangle, $$ i.e. a superposition of states $|0\rangle$ and $|1\rangle$ (you can choose any other two orthogonal state but I think that quantum analogs to 0 and 1 bit are more illustrative).

Parameters $\alpha$ and $\beta$ are complex numbers, so you can write them in exponential form, i.e. $\alpha = |\alpha|\mathrm{e}^{i\varphi_\alpha}$, similarly $\beta= |\beta|\mathrm{e}^{i\varphi_\beta}$. Now, put this into our definition of qubit: $$ |q\rangle = |\alpha|\mathrm{e}^{i\varphi_\alpha}|0\rangle + |\beta|\mathrm{e}^{i\varphi_\beta}|1\rangle, $$ and factor out $\mathrm{e}^{i\varphi_\alpha}$: $$ |q\rangle = \mathrm{e}^{i\varphi_\alpha}[|\alpha||0\rangle + |\beta|\mathrm{e}^{i(\varphi_\beta-\varphi_\alpha)}|1\rangle]. $$

$\varphi_\alpha$ is called global phase. Whole term $\mathrm{e}^{i\varphi_\alpha}$ can be neglected as two qubits which differ in global phase only are physically undistinguishable. So, after substituting $\varphi = \varphi_\beta-\varphi_\alpha$ we are left with $$ |q\rangle = |\alpha||0\rangle + |\beta|\mathrm{e}^{i\varphi}|1\rangle. $$ A parameter $\varphi$ is called relative phase.

  • 2
    $\begingroup$ To add to this - $\lvert\alpha\rvert$ and $\lvert\beta\rvert$ represent the lengths of $\alpha$ and $\beta$ and are therefore real. We also know that $\lvert\alpha\rvert^2 + \lvert\beta\rvert^2 = 1$, and so it is convenient to represent $\lvert\alpha\rvert$ and $\lvert\beta\rvert$ as $\cos\frac{\theta}{2}$ and $\sin\frac{\theta}{2}$ respectively, where $0 \leq \theta \leq \pi$. $\theta$ and $\psi$ then give us the 2 angles we need to plot a point on the Bloch Sphere. $\endgroup$
    – giri
    Sep 11, 2020 at 7:14
  • $\begingroup$ @AdityaGiridharan: Thanks for expanding my answer. $\endgroup$ Sep 11, 2020 at 7:16

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