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I have a quick question: Is the qubit state $|\psi\rangle$ the same as $-|\psi\rangle$?

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  • $\begingroup$ Be careful with the answers below, the two are certainly not equal! Personally I would not call them "the same". Also note that if a phase is not global (that is, if you have a state such as $|\psi_1\rangle + e^{i\phi}|\psi_2\rangle$) the phase matters... a lot! $\endgroup$ – sebhofer Mar 13 at 18:46
  • $\begingroup$ @sebhofer does your concern boil down to a distinction between "indistinguishable" and "equal?" Either way can you offer an answer to highlight your concern? $\endgroup$ – Mark S Mar 13 at 18:55
  • $\begingroup$ @MarkS In part yes. After all, we would never call the vectors (1,0) and (-1,0) (in the Euclidean space) equal. Also, as I pointed out above, I think it's important to stress that this is only true for a global phase, and not for the phases in a superposition. This might not be clear if you call them equal. $\endgroup$ – sebhofer Mar 13 at 19:32
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States $|\psi\rangle$ and $-|\psi\rangle$ differ in global phase only and thus they are indistinguishable. So, the answer is: state $-|\psi\rangle$ is the equivalent to $|\psi\rangle$.

The global phase in this case is $\pi$ because $\mathrm{e}^{i\pi} = -1$.

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    $\begingroup$ Indistinguishability does not necessarily mean that they are equal. Yes, we cannot distinguish them from each other, but they are different. $\endgroup$ – nippon Mar 7 at 17:23
  • $\begingroup$ @nippon Aren't states rays in a Hilbert space (and thus $-\psi$ and $+\psi$ two representing elements of the same guy in the quotient)? $\endgroup$ – c.p. Mar 7 at 18:16
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    $\begingroup$ @nippon: Thanks for pointing that out. Yes, states are not equal but for quantum computing they are equivalent. $\endgroup$ – Martin Vesely Mar 7 at 23:24
  • $\begingroup$ I think you should edit this answer! Claiming that they are "the same" seems pretty missleading to me. For me "the same" means "equal", which they are clearly not! $\endgroup$ – sebhofer Mar 13 at 18:42
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    $\begingroup$ @MartinVesely I know, but not everyone reads all the comments... $\endgroup$ – sebhofer Mar 13 at 18:52
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They are physically indistinguishable, also their density matrices are the same because $$\big(-|\psi\rangle\big)\big(-\langle\psi|\big) = |\psi\rangle \langle\psi|$$ But mathematically they are two different vectors. And it's better to not forget about this when doing calculations.

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    $\begingroup$ This can be generalized, having $\mathrm{e^{i\varphi}}|\psi\rangle$, a density matrix is $(\mathrm{e^{i\varphi}}|\psi\rangle) (\mathrm{e^{-i\varphi}}\langle\psi|) = |\psi\rangle \langle\psi|$ $\endgroup$ – Martin Vesely Mar 9 at 22:43
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Yes.

The factor $e^{i\phi}$ of the state $e^{i\phi}|\psi\rangle$ (which in this case is $-1$) is called a "global phase". It does not have physically observable consequences (i.e., you can not come up with an experiment to figure out what the global phase of a state is) and can be safely ignored.

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