# Rotationally invariant maximally entangled states in higher dimensions

Is there a straightforward generalization of the $$\mathbb{C}^2$$ Bell basis to $$N$$ dimensions? Is there a rotational invariant Bell state in higher dimensions? If yes, then what is the form of that state (how does it look like)? And, by rotational invariance, I mean that the state is invariant under applying the same unitary transformation $$U$$ to each qubit separately.

For example, $$|\psi^-\rangle = \frac{|0\rangle|1\rangle - |1\rangle|0\rangle}{\sqrt{2}} = \frac{|\gamma\rangle|\gamma_\perp\rangle - |\gamma_\perp\rangle|\gamma\rangle}{\sqrt{2}}$$, where $$|\gamma\rangle$$ is some quantum state in $$\mathbb{C}^2$$, and $$|\gamma_\perp\rangle$$ is orthogonal to $$|\gamma\rangle$$.

It would be helpful if I could see an example of the same, in say $$\mathbb{C}^4$$ space, perhaps in the computational basis {$$|0\rangle, |1\rangle, |2\rangle, |3\rangle$$} itself.

## 1 Answer

An orthonormal maxinammly entangled basis for two quNits is easily defined: $$|\Psi_{xy}\rangle=\frac{1}{\sqrt{N}}\sum_{i=0}^{N-1}\omega^{iy}|i,i+x\rangle,$$ where $$\omega$$ is an $$N$$-th root of unity, and $$x,y=0,1,\ldots,N-1$$.

I don't believe that there is a rotationally invariant maximally entangled state, except in the case $$N=2$$. You may want to look up 'twirling', which almost does the calculation you need (they find states invariant under $$U\otimes U^\star$$ instead of $$U\otimes U$$), however the way that I convinced myself is the following:

• Any state invariant under $$U\otimes U$$ must be invariant under particular examples.

• Let's start with the $$Z$$-equivalent, $$\tilde Z=\sum_{n=1}^N\omega^n|n\rangle\langle n|$$ The class of maximally entangled states that are invariant under this operation are $$\frac{1}{\sqrt{N}}\sum_{n=0}^{N-1}|n\rangle|N-1-n\rangle e^{i\phi_n},$$ where we have free choice over the phases $$\phi_n$$.

• Next consider the action of the permutation operation $$P=\sum_{n=0}^{N-2}|n+1\rangle\langle n|+|0\rangle\langle N-1|$$ For $$N=2$$, this maps back to the original state provided $$e^{2i(\phi_2-\phi_1)}=1$$. For all other $$N$$, we cannot map back to the original state. Terms like $$|1\rangle|N-2\rangle$$ become $$|2\rangle|N-1\rangle$$ which are not of the form $$|m\rangle|N-1-m\rangle$$, and are hence not in the original state.

• The claim that there are no rotationally invariant maximally entangled states when $N\geq 3$ is correct. Density operators that are invariant under the action $\rho \mapsto (U\otimes U)\rho (U \otimes U)^{\dagger}$ are called Werner states, and they are never pure when $N\geq 3$. Oct 8 '18 at 12:53