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 1


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.

  • 3
    $\begingroup$ 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 $N\geq 3$. $\endgroup$ Oct 8, 2018 at 12:53

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