# 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! Mar 13, 2020 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? Mar 13, 2020 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. Mar 13, 2020 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. Mar 7, 2020 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, 2020 at 18:16
• @nippon: Thanks for pointing that out. Yes, states are not equal but for quantum computing they are equivalent. Mar 7, 2020 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! Mar 13, 2020 at 18:42
• @MartinVesely I know, but not everyone reads all the comments... Mar 13, 2020 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 $(\mathrm{e^{i\varphi}}|\psi\rangle) (\mathrm{e^{-i\varphi}}\langle\psi|) = |\psi\rangle \langle\psi|$ Mar 9, 2020 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.