# What is the "quantum phase" of a quantum state?

On this page IMBQ docs, until the sentence '..and since the global phase of a quantum state is not detectable..' I follow everything. However 'quantum phase' is introduced without any explaination? What is this?

• be sure not to confuse the term "global quantum phase" with "quantum phase" in the context of "phase transitions" these are not the same "phase". The "quantum phase" you are referring to is the former. Jul 19 at 14:37
• – glS
Jul 19 at 20:49

It is a term of the form $$e^{i\phi}$$. So, the point is that two states $$|\psi\rangle$$ and $$|\phi\rangle=e^{i\phi}|\psi\rangle$$ differ only by this phase. We call it a global phase because it's affecting the whole of the state rather than just part of it.
You can easily see that this has no observable consequences. For example, if we measure in the standard basis, what's the probability of getting the answer $$|0\rangle$$? $$p_0=|\langle0|\phi\rangle|^2=|e^{i\phi}\langle0|\psi\rangle|^2=|\langle0|\psi\rangle|^2$$ The probability is the same for both cases. As it will be for any measurement basis and outcome.
• Where does it come from? It 's just a common factor. In the same way that you can write $77+55=11(7+5)$. It's just that if you impose that state vectors have to have length 1, the only factor you could possibly pull out is of the form $e^{i\phi}$. For example, any single-qubit state is $\alpha e^{i\phi}|0\rangle+\beta e^{i\psi}|1\rangle$ where $\alpha$ and $\beta$ are real. It looks like we need 4 parameters to describe the state. But I could just write it as $e^{i\phi}(\alpha|0\rangle+e^{i(\psi-\phi)}\beta|1\rangle)$ at which point the claim is that you can neglect the $e^{i\phi}$ and only Jul 19 at 14:52
• use 3 parameters (which you reduce further by the constraint $\alpha^2+\beta^2=1$) Jul 19 at 14:52