Werner versus Isotropic States: Physical Significance Can someone please explain the difference between Werner and Isotropic states based on physical significance rather than simply ($U\otimes U$) vs. ($U\otimes U^*$) invariant? (Where the $U$'s are unitary operators.)
Edited to add (based on Norbery Schuch's suggestion):
Werner states are mixed entangled states that are constructed with combinations of the projectors onto symmetric and antisymmetric subspaces, and are  ($U\otimes U$) invariant:
$\rho^W=p P_{-}/N_{-} + (1-p)P_{+}/N_{+}$
where $P_{-(+)}$ are the projectors onto the antisymmetric (symmetric) subspaces ( $P_{+}=0.5(1+P)$ and $P_{-}=0.5(1-P)$, where P is the permutation operator that exchanges the two subsystems) and $N_{\pm}$ are the dimensions of each ($\frac{d^2\pm d}{2}$).  $p$ is a parameter.
Isotropic states are combinations of the Identity and maximally entangled state P_+ and are ($U\otimes U^*$) invariant:
$\rho^I=\frac{(1-F)}{d^2-1}I+FP_+$
where $F$ is a parameter called Fidelity.
I don't understand what they represent physically in the lab.  Why are isotropic states called isotropic?  What's the difference between these two? How can we look at a state and see which class it falls under?  What do each parts of the construction represent? (Eg, I think the identity represents white noise.)  The only difference I've seen anywhere is that they're ($U\otimes U$) vs. ($U\otimes U^*$) invariant.  
Eg. there's a state that looks like $\rho=x|singlet><singlet| + \frac{1-x}{4}1$.  This looks like an isotropic state (though the singlet is antisymmetric) but I've seen it called a Werner state.
 A: As Norbert mentioned, for two qubits, isotropic and Werner States are equal up to local unitaries. What follows below applies for two qubits.
In general, you can define states that are invariant under $U\otimes f(U)$, where $f$ is some function of $U$. Since Unitaries are basically rotations, each one is a function of a rotation axis and angle, $U(\hat{a},\alpha)=\cos(\frac{\alpha}{2})I-i\sin(\frac{\alpha}{2})\hat{a}\cdot\vec{\sigma}$. So the above is equivalent to asking what states are invariant under $U(\hat{a},\alpha)\otimes U(\hat{b},\beta)$, where $\hat{b},\beta$ are some unknown functions of $\hat{a},\alpha$. 
It turns out the only solutions [1] are $\beta=\alpha$ and $\hat{a}=-O\hat{b}$, for some orthogonal (rotoreflection) matrix $O$ that satisfies $O^\dagger O=I, \det O=-1$. In this case the states invariant under $U(\hat{a},\alpha)\otimes U(-O^\dagger\hat{a},\alpha)$ for all $\alpha, \hat{a}$) are 
\begin{equation}
\rho(x)=\frac{1-x}{4}I+x\sum_{ij}O_{ij}\sigma_i\otimes\sigma_j.
\end{equation}
One can show that, independent of $O$, $\rho(x)$ is positive (i.e. a valid density matrix) for $-\frac{1}{3}\le x \le 1$, and entangled if and only if $\frac{1}{3}< x \le 1$. 
Note you can flip the signs of $O$ to get rid of the negative determinant, but then $x$ has to switch sign too, and you reverse the standard representation used in QM. Also, you can always transfer a sign from $\hat{a}$ to $\alpha$.
For a Werner state $O=-I$, and for an isotropic state $O=\text{diag}(1,-1,1)$, in which cases $U\otimes U(-O^\dagger\hat{a},\alpha)$ reduces to $U\otimes U$ and $U \otimes U^*$ respectively.
I would say the isotropic state is a misnomer. The Werner state is the one that is isotropic; its $O$ is proportional to the identity, which is invariant under change of basis rotations. This is not the case for what is called the isotopic state.
[1] O. Gamel, Phys. Rev. A 93, 062320, (2016) - Sec. VIII.C
