# Are $(|00\rangle-|11\rangle)/\sqrt2$ and $(|11\rangle-|00\rangle)/\sqrt2$ the same quantum state?

The state $$(|00\rangle-|11\rangle)/\sqrt2$$ is an entangled state. If we think about the state $$(|11\rangle-|00\rangle)/\sqrt2$$, is this also entangled, but with maybe a phase change? The above two can not be similar? Maybe, the second state is entangled but not maximally?

• – glS
Nov 26 '21 at 9:07

The kets $$|\psi\rangle$$ and $$e^{i\theta}|\psi\rangle$$ represent the same quantum state for any $$|\psi\rangle$$ and any phase factor $$e^{i\theta}$$ with $$\theta\in[0,2\pi)$$. No observable quantity depends on the global$$^1$$ phase $$e^{i\theta}$$ of a state.

In particular,

$$|\psi\rangle=\frac{|00\rangle-|11\rangle}{\sqrt2}$$

and

$$e^{i\pi}|\psi\rangle=\frac{|11\rangle-|00\rangle}{\sqrt2}$$

represent the same state. The state is maximally entangled.

$$^1$$ Relative phase, i.e. phase on a proper subset of kets in a sum, does have observable consequences.