On Local Unitary Equivalence of Two and Three-qubit States

We study the local unitary equivalence for two and three-qubit mixed states by investigating the invariants under local unitary transformations. For two-qubit system, we prove that the determination of the local unitary equivalence of 2-qubits states only needs 14 or less invariants for arbitrary two-qubit states. Using the same method, we construct invariants for three-qubit mixed states. We prove that these invariants are sufficient to guarantee the LU equivalence of certain kind of three-qubit states. Also, we make a comparison with earlier works.

Nonlocality is one of the astonishing phenomena in quantum mechanics. It is not only important in philosophical considerations of the nature of quantum theory, but also the key ingredient in quantum computation and communications such as cryptography 1 . From the point of view of nonlocality, two states are completely equivalent if one can be transformed into the other by means of local unitary (LU) transformations. Many crucial properties such as the degree of entanglement 2, 3 , maximal violations of Bell inequalities [4][5][6][7] and the teleportation fidelity 8,9 remain invariant under LU transformations. For this reason, it has been a key problem to determine whether or not two states are LU equivalent.
There have been a plenty of results on invariants under LU transformations [10][11][12][13][14][15][16][17][18][19][20][21][22][23][24][25] . However, one still does not have a complete set of such LU invariants which can operationally determine the LU equivalence of any two states both necessarily and sufficiently, except for 2-qubit states and some special 3-qubit states. For the 2-qubit state case, Makhlin presented a set of 18 polynomial LU invariants in ref. 10. In ref. 20 the authors constructed a set of very simple invariants which are less than the ones constructed in ref. 10. Nevertheless, the conclusions are valid only for special (generic) two-qubit states and an error occurred in the proof. In this paper, we corrected the error in ref. 20 by adding some missed invariants, and prove that the determination of the local unitary equivalence of 2-qubits states only needs 14 or less invariants for arbitrary two-qubit states. Moreover, we prove that the invariants in ref. 20 plus some invariants from triple scalar products of certain vectors are complete for a kind of 3-qubit states.

Results
A general 2-qubit state can be expressed as: I  I  T  I  T I  T  1  4 , where I is the 2 × 2 identity matrix, σ i , i = 1, 2, 3, are Pauli matrices and  . Therefore there are only 9 linearly independent invariants: 〈μ i , μ j 〉, 〈v i , v i 〉, i = 1, 2, 3, and 〈μ 1 , μ j 〉, j = 2, 4, 6. We denote them as = L     are missed. By adding these missed invariants, we have remedied the error in ref. 20 and, moreover, generalized the method to the case of dim〈S i 〉 = 3 for i = 1 or 2.
The expression of a complete set of LU invariants depends on the form of the invariants. Different constructions of LU invariants may give different numbers of the invariants in the complete set, and may have different advantages. Obviously the eigenvalues of a density matrix are LU invariants. Based on the eigenstate decompositions of density matrices, in ref. 12  make the corresponding theorems incorrect even for generic cases studied in ref. 20. By adding these invariants, our set of invariants work for arbitrary 2-qubit states. In fact, a set of complete LU invariants characterizes completely the LU orbits in the quantum state space. Generally such orbits are not manifolds, but varieties. For example, the set of pure states is a symplectic variety 26 . For general mixed states, the situation is much more complicated 27 . Our results would highlight the analysis on the structures of LU orbits.
Now we come to discuss the case of three-qubit system. A three-qubit state ρ can be written as:  I  I  I  T  I  I  TI  I  T I  I   T  I  T  I  T I   T   1 T  T  T  T   S  T  T  T  T   S  T  T  T  T   ,  ,  ,  , are all invariants under LU transformations. Using the method in ref. 20, we now prove that these invariants together with the additional ones in theorem 3 are sufficient to guarantee the LU equivalence of certain kind of three-qubit states with at least two of dim〈S i 〉 = 3 for i = 1, 2, 3.

Theorem 3 Given two 3-qubit states ρ and ρˆ
, , ω =  and i j k , , 1, 2, , and = = S S dim dim 3 i i for at least two i ∈ 1, 2, 3, then ρ and ρˆ are LU equivalent. See Methods for the proof of Theorem 3. If at most one of dim〈S i 〉 is 3, things become more complicated. Now we give a comparison with the results in ref. 11. For 3-qubit states ρ and ρˆ, if Here we only need that two of the dim〈S i 〉 are 3. So we give the sufficient conditions for local unitary equivalence of more states than the ones given in ref. 11.

Conclusion
We study the local unitary equivalence for two and three-qubit mixed states by investigating the invariants under local unitary transformations. We corrected the error in ref. 20 by adding some missed invariants, and prove that the determination of the local unitary equivalence of 2-qubits states only needs 14 or less invariants for arbitrary two-qubit states. Moreover, we prove that the invariants in ref. 20 plus some invariants from triple scalar products of certain vectors are complete for a kind of 3-qubit states. Comparing with the results in ref. 11, it has been shown that we judge the LU equivalence for a larger class of 3-qubit states.

Proof of Theorem 1 Suppose
. From the construction of S 1 and S 2 , we have that ν µ ν 12 12 , α = 1, 2 and =T T det det 12 12 , one has that T 12 and T 12 have the same singular values. According to the singular value decomposition, there are  .

Proof of Theorem 2
We only need to prove the "only if " part, i.e., to find    1. If t t t , , 1 2 3 are all not equal, from (6) and (7) we can conclude that α α = ±î i for α = a b c , , and = i 1, 2. are linearly dependent. Then there is at least one α ∈ a b c { , , } i i i i 0 that is zero. Hence if P 1 T 1 and D 2 P 1 T 1 are linearly independent, we have that DP 2 T 2 can be linearly represented by P 1 T 1 and D 2 P 1 T 2 . Using ≠ t t t 0 1 2 3 and supposing a 1 = 0, we get that a 2 is also zero. Now e 1 in R can be chosen to be 1 or −1 freely. We can choose e 1 to assure that det R = 1. Similarly, for the case that P 2 T 2 and DP 2 T 2 are linear independent, we can also find R which has determinate one. Lastly, if P i T i and D 2 P i T i are linear dependent, then there are at least two members are zero in a b c { , , } i i i , i = 1, 2. Therefore, there is an α ∈ a b c { , , } satisfying α 1 = α 2 = 0, such that det R = 1. (ii) If there exists a t i = 0, say, t 3 = 0, then we have α α αα =ˆ1 2 1 2 for α = a, b from (8). And the invariant I can assure that =ˆĉ c cc 1 2 1 2 . From the discussion above, we have the conclusion.