# Negative of a qubit state

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

• 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! – sebhofer Mar 13 at 18:46
• @sebhofer does your concern boil down to a distinction between "indistinguishable" and "equal?" Either way can you offer an answer to highlight your concern? – Mark S Mar 13 at 18:55
• @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. – sebhofer Mar 13 at 19:32

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 $$\mathrm{e}^{i\pi} = -1$$.

• Indistinguishability does not necessarily mean that they are equal. Yes, we cannot distinguish them from each other, but they are different. – nippon Mar 7 at 17:23
• @nippon Aren't states rays in a Hilbert space (and thus $-\psi$ and $+\psi$ two representing elements of the same guy in the quotient)? – c.p. Mar 7 at 18:16
• @nippon: Thanks for pointing that out. Yes, states are not equal but for quantum computing they are equivalent. – Martin Vesely Mar 7 at 23:24
• 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! – sebhofer Mar 13 at 18:42
• @MartinVesely I know, but not everyone reads all the comments... – sebhofer Mar 13 at 18:52

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.

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

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.