If we look at two maximally entangled qubits, what is the symmetry
group of all possible states? It seems to me that it is U(1) x SO(4),
but does anyone know a reference to it?
We have the 4 maximally entangled Bell states (see
>) and we
can use a phase convention as follows:
Phi+ = ( |11> + |00> ) * sqrt(1/2)
Phi- = ( |11> - |00> ) * i * sqrt(1/2)
Psi+ = ( |10> + |01> ) * i * sqrt(1/2)
Psi- = ( |10> - |01> ) * sqrt(1/2)
In that way they span a 4-dimensional space of maximally entangled
states in R^4. Of course not in C^4, because Phi+ and Phi- could then
combine e.g. to |11>, which is a totally non-entangled separable
2-particle state. But as written, every combination with real-valued
coefficients is still maximally entangled, as the reader can easily
verify using reduced density matrices.
Obviously the symmetry group at this point is SO(4), but we still can
give each state an arbitrary phase, since the total phase does not alter
the entanglement. (We just can't give the Bell states an arbitrary phase
before combining them, but after a combination has been chosen we can!)
So with this freedom of phase choice added, I find U(1) x SO(4).
It would be tempting to try doing something with SU(2) x SU(2), which
is the double cover of SO(4), because it turns out that the SU(2)
transforms that Alice[*] can do with her qubit and the SU(2) that Bob
can do with his, are indeed precisely giving the SO(4) group that
describes the allowed combinations. This is just the left-isoclinic /
right-isoclinic rotation decomposition of SO(4)
But clearly this gives the non-uniqueness of the double cover: if Alice
and Bob both transform their qubit with the negative identity
[ 0 -1]
then the total 2-particle state is unchanged. So the group isn't
U(1)xSU(2)xSU(2), however physically appealing that would be, but
just U(1) x SO(4), I think..
Any other thoughts?
[*] Alice and Bob are just the usual names for the two single-qubit