How are two-qubit states like $|00\rangle+|01\rangle+|10\rangle-|11\rangle$ related to entanglement?

Question

I've been going through different rotations of two-qubit states and noticed the states

$$ |00\rangle+|01\rangle+|10\rangle-|11\rangle\\ |00\rangle+|01\rangle-|10\rangle+|11\rangle\\ |00\rangle-|01\rangle+|10\rangle+|11\rangle\\ |00\rangle-|01\rangle-|10\rangle-|11\rangle. $$

They appear instead of the four Bell states for particular initial conditions and rotations. I've written down reduced density matrices for the single qubit for those states - they are identities, just like Bell states and they seem like unfactorizable states. I couldn't find much info on these states. Perhaps someone could suggest resources or papers on these states or tell how they relate to Bell states and to entanglement in general. For example, can we perform some interesting things like teleportation or use them as some resource?

Answer 1

The four states are equivalent to Bell states under local unitaries. Consequently, they exhibit many properties of the Bell states. In particular, they can be used for many quantum protocols, such as quantum teleportation, that use entanglement as a resource.

We can see this by doing the following calculation. Let's denote the four states in the question as $$ |\alpha_{ab}\rangle = \frac12(|00\rangle+(-1)^a|01\rangle+(-1)^b|10\rangle+(-1)^{1+a+b}|11\rangle)\tag1 $$ and the Bell states as $$ |\beta_{ab}\rangle = \frac{1}{\sqrt2}(|0a\rangle+(-1)^b|1\overline{a}\rangle)\tag2 $$ where $a,b\in\{0,1\}$ and $\overline{a}=1-a$. Then $$ \begin{align} I\otimes H|\beta_{ab}\rangle&=\frac{1}{\sqrt2}(|0\rangle H|a\rangle+(-1)^b|1\rangle H|\overline{a}\rangle)\tag3\\ &=\frac12\left(|0\rangle[|0\rangle+(-1)^a|1\rangle]+(-1)^b|1\rangle[|0\rangle+(-1)^{1-a}|1\rangle]\right)\tag4\\ &=\frac12\left(|00\rangle+(-1)^a|01\rangle+(-1)^b|10\rangle+(-1)^{1-a+b}|11\rangle\right)\tag5\\ &=|\alpha_{ab}\rangle\tag6 \end{align} $$ where we used $(-1)^{-a}=(-1)^a$ in the last step before substituting from $(1)$.