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Nov 26, 2020, 4:46:02 PM11/26/20

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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?

My reasoning:

We have the 4 maximally entangled Bell states (see

<https://en.wikipedia.org/wiki/Bell_state#Bell_basis>) 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)

<https://en.wikipedia.org/wiki/Rotations_in_4-dimensional_Euclidean_space#Isoclinic_decomposition>

But clearly this gives the non-uniqueness of the double cover: if Alice

and Bob both transform their qubit with the negative identity

[-1 0]

[ 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

sub-systems..

--

Jos

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?

My reasoning:

We have the 4 maximally entangled Bell states (see

<https://en.wikipedia.org/wiki/Bell_state#Bell_basis>) 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)

<https://en.wikipedia.org/wiki/Rotations_in_4-dimensional_Euclidean_space#Isoclinic_decomposition>

But clearly this gives the non-uniqueness of the double cover: if Alice

and Bob both transform their qubit with the negative identity

[-1 0]

[ 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

sub-systems..

--

Jos

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