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

Question

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?

Answer 1

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.}