What effect does phase flip error have in computation?

Question

Phase flip error changes $\alpha|0\rangle + \beta|1\rangle$ into $\alpha|0\rangle - \beta|1\rangle$, but outcome probabilities are still $|\alpha|^2$ and $|-\beta|^2=|\beta|^2$. So, what is the effect of phase flip error in computation and why it should be corrected?

Answer 1

You make a good point saying that a phase flip is not detectable is the $Z$-basis.

If you only perform measurements in the computational basis you cannot distinguish between a state $|{\psi}\rangle = \alpha|{0}\rangle +\beta|{1}\rangle$ and the phase-flipped state $|{\phi}\rangle = \alpha|{0}\rangle -\beta|{1}\rangle$ since the measurement probabilities are squared amplitudes and $|\beta|^2 = |-\beta|^2$.

However, if the state is part of a larger circuit, this phase will matter as sheezymcdeezy pointed out. Imagine a circuit of a single qubit and two $H$ gates. If the phase flip (effectively a $Z$ gate) occurs in between the two, you get the following: $$HZH|\psi\rangle = HZ(\alpha|+\rangle+ \beta|-\rangle) = H(\alpha|-\rangle+\beta|+\rangle) = \beta|{0}\rangle +\alpha|{1}\rangle,$$ where you can clearly see the impact on the final state and measurements.

You can also notice the phase flip if you are not restricted to the computational basis measurements and you can measure in the $X$-basis $\left(\{|+\rangle, |-\rangle\}\right)$ you will effectively notice the phase flip. To visualize this, you can rewrite $|\psi\rangle$ as: $$|{\psi}\rangle = \alpha|{0}\rangle +\beta|{1}\rangle = \alpha\frac{|+\rangle+|-\rangle}{\sqrt{2}} + \beta\frac{|+\rangle-|-\rangle}{\sqrt{2}} = \frac{\alpha+\beta}{\sqrt{2}}|+\rangle+ \frac{\alpha-\beta}{\sqrt{2}}|-\rangle = \alpha'|{+}\rangle +\beta'|{-}\rangle$$ Doing the same for $|\phi\rangle$ yields: $$|\phi\rangle = \frac{\alpha-\beta}{\sqrt{2}}|+\rangle+ \frac{\alpha+\beta}{\sqrt{2}}|-\rangle = \beta'|{+}\rangle +\alpha'|{-}\rangle$$ You can clearly see that in this case the $X$-measurement on the phase-flipped state will give a different probability. Effectively, the phase flip exchanged the probability of measuring the $|+\rangle$ and $|-\rangle$ states.

Note that this change of basis is equivalent to applying an $H$ gate before measuring in the computational basis.