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I know this it's perhaps, a little "stupid" question, but... What's the purpose of phase in Quantum?

Per example, in quantum Computing, the phase don't seems to be very important in the amplitudes and probability of the outcomes, before a qubit be measured. But I know that phase it's important for something, I just don't know exactly for what (quantum entanglement, perhaps?)

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    $\begingroup$ Do the answers below clarify what you had in mind? $\endgroup$ – keisuke.akira Jul 4 at 22:20
  • $\begingroup$ If you're happy with the answers then please accept one of them. $\endgroup$ – keisuke.akira Jul 6 at 8:16
  • $\begingroup$ I'm still struggling a little with the Phase and Quantum Entanglement, but I need to look more carefully to some details of them, but it's not your fault, I'm a newbie in Quantum, I've been studying Quantum Information Science, since a year ago... :/ $\endgroup$ – Rúben André Barreiro Jul 6 at 8:44
  • $\begingroup$ I know per example, that the phase it's very useful in Grover's Algorithm, inverting the phase and amplitudes, in order to find the correct solution, but I want to understand very well the phase and the entanglement, it's where I'm struggling more, right now... $\endgroup$ – Rúben André Barreiro Jul 6 at 8:51
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This is simply not true. It is the global phase that is not physically relevant -- which can be immediately seen from the linearity of the Schrodinger equation: two states $| \psi \rangle, e^{i \phi} | \psi \rangle$ are both solutions to the same equation.

However, consider a qubit state, $| \psi \rangle \in \mathbb{C}^{2}$, with two different choices of local (a.k.a. relative) phases: $| + \rangle \equiv \frac{1}{\sqrt{2}}(| 0 \rangle + | 1 \rangle) \text{ and } | - \rangle \equiv \frac{1}{\sqrt{2}}(| 0 \rangle - | 1 \rangle)$, which are orthogonal to each other and hence completely distinguishable (for example, see this answer). For an explanation of the difference(s) between global and local phases, see this answer.

One way to deal with the ambiguity of global phases is to work in a complex projective space (or with density matrices) -- which is what one does, for example in quantum computing, where we represent qubit states on the Bloch sphere.

Also, phases are relevant for both amplitudes and probabilities of outcomes (again, it is the global phase that is not physically relevant). For example, see this answer, where two states that have same probability of outcomes when measured in the $\sigma_z$ basis yield different probabilities when measured in a different basis (say $\sigma_x$). Apart from this, phases are clearly very important for several quantum computing protocols like quantum phase estimation, in cyclic evolutions i.e., Berry phase, etc.

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Phase is really one of the things that makes quantum computing what it is! In fact, I think there is a quote by Aaronson like how quantum is "probability theory with negative numbers."

The place it comes in is when looking at the interference of different amplitudes. We don't see the phase directly when measuring states like $|+\rangle = |0\rangle + |1\rangle$ and $|-\rangle = |0\rangle - |1\rangle$ in the Z basis. Both states will give a 50/50 distribution. But if you apply a Hadamard transform to them, $$H|+\rangle = H|0\rangle + H|1\rangle = (|0\rangle + |1\rangle) + (|0\rangle - |1\rangle) = |0\rangle,$$ $$H|-\rangle = H|0\rangle - H|1\rangle = (|0\rangle + |1\rangle) - (|0\rangle - |1\rangle) = |1\rangle.$$

The minus signs cause different terms to cancel out. So when changing basis or combining different states, the phase controls which amplitudes constructively and destructively interfere.

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To give one example. A phase is very important in Grover algorithm used for searching in an unordered database. A record you are looking for (i.e. having specific features) is "marked" by negative amplitude (i.e. phase is $\pi$). Based on different phase in comparison with other records in the database the one you are searching for is picked and its probability is amplified, or in other words the record is found.

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