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:


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.

  • $\begingroup$ What do you mean by "physical significance"? Do you mean "find it in the lab" or do you mean "useful to prove something which might be physically relevant"? Also, it might help in answering the question if you give some details on Werner and isotropic states (otherwise the one answering has to do it, thereby increasing the threshold to answer). $\endgroup$ Nov 23, 2016 at 10:13
  • $\begingroup$ I don't think these states are relevant because they naturally appear in the lab. They are relevant because they form small (1-parameter) families of states, any state can be transformed into one of them, and as it happens these transformations can preserve certain interesting properties (entanglement, Bell violation, ...), and it is therefore sufficient to analyze them on Werner or isotropic states. $\endgroup$ Nov 23, 2016 at 12:07
  • $\begingroup$ @NorbertSchuch That doesn't answer the question. I know they're 1-parameter families, as is evident from the equations. I want to know the difference between them. Taking your example of being able to transform a state to one of them, how do you know which to pick? Would you call the example I gave in my edited post a Werner or an Isotropic state? Why is it called 'isotropic'? What's the difference between the two? $\endgroup$ Nov 23, 2016 at 16:17
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    $\begingroup$ That's why it's a comment ;-) But you start mixing up questions ... which makes it more tricky to answer. My point above is that I do not think there is a physical significance in the way you mean it. And the way you ask it now you ask for a wide-range review of the topic. Not sure there are people here who could give that without putting considerable work into it. $\endgroup$ Nov 23, 2016 at 17:02
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    $\begingroup$ As to your question: For qubits, isotropic = Werner, up to local unitaries. In fact, the state you give is exactly the Werner state (not the isotropic state). For $d>2$, I would call the state you give isotropic, because it is equivalent to it up to local unitaries (while it is not LU-equivalent to the Werner state), and usually the whole point of these states is to study entanglement properties. $\endgroup$ Nov 23, 2016 at 17:03

1 Answer 1


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


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