How to systematically generate all 2-qubit density matrices up to local operations

How to systematically generate all 2-qubit density matrices up to local operations Ask Question Asked 12 days ago Modified 11 days ago Viewed 105 times 4 In order to test ideas about entanglement, I wish to generate, in a systematic way, all 2-qubit density matrices without unnecessary over-production. By 'systematic' I mean one iterates over a finite set of parameter values, and by 'all' I mean one would cover the space more and more densely as the number of values (for each parameter, in some finite range of values) increases. There are various ways to generate 'all' (in the above sense) 2-qubit density matrices; one would require 15 parameters so the set of samples will be large and unwieldy and slow to work with. We know that entanglement properties are unaffected by local operations such as local rotations. For my purposes it is sufficient to get one density matrix for each equivalence class, where matrices are equivalent if they are related by local rotations. That will reduce the number of parameters by 6 so now we have a 9-parameter problem which is a bit more manageable. It may be possible to reduce the number of parameters further still. My question is, what is the least number of parameters required if matrices related by LOCC are in the same equivalence class, and how can I parametrize so as to produce one matrix per equivalence class, such that all equivalence classes will be visited in the continuous limit of the parameter values? If the answer is unknown for LOCC, then is there some restricted set within LOCC for which a method is known? entanglementdensity-matrix Share Improve this question Follow edited Mar 4 at 14:54 asked Mar 4 at 14:42 Andrew Steane 1435 5 bronze badges Hi Andrew, I'm a bit unclear if you're asking about equivalence classes that are invariant under local unitaries only, or whether you are actually interested in LOCC (where there can be cooperative stochastic operations). You seem to mention both (but the answers are quite different). –  DaftWullie Commented Mar 4 at 16:34 Either or both! Under LOCC would be good (since presumably that's a smaller number of parameters remaining) but failing that then under local unitaries only. –  Andrew Steane Commented Mar 4 at 18:10 Add a comment 1 Answer Sorted by: Highest score (default) Date modified (newest first) Date created (oldest first) 1 I guess this depends on how you define equivalence class. In the case of LOCC, I think the answer is actually kind of boring - everything is in the same equivalence class. Let Alice & Bob share a maximally entangled state. Alice creates ρ 𝜌 . She teleports one half of it to Bob. So sharing a Bell pair is always enough. It's the representative of the equivalence class. For local unitary equivalence, then I agree with your counting. I expect that you could turn it into a rigorous argument that you can't do better (otherwise it would imply some sort of obviously impossible compression of data). What is interesting is how to parametrise an arbitrary two-qubit density matrix with just 9 real parameters. Thankfully, there's a good way to do this (formally, see here). Let σ ⃗  =(X,Y,Z) 𝜎 → = ( 𝑋 , 𝑌 , 𝑍 ) be the vector of Pauli matrices and Σ ⃗  =(X⊗X,Y⊗Y,Z⊗Z) Σ → = ( 𝑋 ⊗ 𝑋 , 𝑌 ⊗ 𝑌 , 𝑍 ⊗ 𝑍 ) be a vector of two-qubit operator. You have ρ= 1 4 (I⊗I+( n ⃗  1 ⋅ σ ⃗  )⊗I+I⊗( n ⃗  2 ⋅ σ ⃗  )+ m ⃗  ⋅ Σ ⃗  ) 𝜌 = 1 4 ( 𝐼 ⊗ 𝐼 + ( 𝑛 → 1 ⋅ 𝜎 → ) ⊗ 𝐼 + 𝐼 ⊗ ( 𝑛 → 2 ⋅ 𝜎 → ) + 𝑚 → ⋅ Σ → ) where n ⃗  1 , n ⃗  2 , m ⃗  ∈ R 3 𝑛 → 1 , 𝑛 → 2 , 𝑚 → ∈ 𝑅 3 all have length 1 or less. This doesn't entirely capture the positivity requirements of ρ 𝜌 , but I don't know how to build those in nicely. (To explain this representation, the main trick is that the general case has two-qubit correlations ∑ ij T ij σ i ⊗ σ j ∑ 𝑖 𝑗 𝑇 𝑖 𝑗 𝜎 𝑖 ⊗ 𝜎 𝑗 but the SU(2)/SO(3) correspondence lets you take the singular value decomposition of T and correspond the two 3×3 3 × 3 real matrices that are the left/right singular vectors to the two unitaries that you apply.) Share Improve this answer Follow edited Mar 5 at 10:00 answered Mar 5 at 8:05 DaftWullie 64.4k4 4 gold badges 59 59 silver badges 148 148 bronze badges The opening point about teleporting is interesting, but it is not what "only LOCC" means I think. If we allow A and B to use up an ebit and still call what they are doing "LOCC" then it means that by this kind of "LOCC" they can bring a pair of qubits from separable to fully entangled, which is contrary to the standard definitions. For second part thanks and I will look into it. I'm still not sure the counting is correct because some pairs of different operations can have the same effect on a given state. –  Andrew Steane Commented Mar 5 at 8:36 1 Nice answer! How is Σ ⃗  Σ → defined? You might already have the positivity requirements. I calculated the eigenvalues of ρ 𝜌 for m ⃗  =0 𝑚 → = 0 in Mathematica and I get eigenvalues 1 4 (1± n ⃗  2 1 + n ⃗  2 2 ±2 n ⃗  2 1 n ⃗  2 2 − − − − √ − − − − − − − − − − − − − − − √ )= 1 4 (1± 2±2 − − − − √ ) 1 4 ( 1 ± 𝑛 → 1 2 + 𝑛 → 2 2 ± 2 𝑛 → 1 2 𝑛 → 2 2 ) = 1 4 ( 1 ± 2 ± 2 ) , which can become -1/4. This can possibly be saved by the m ⃗  ⋅ Σ ⃗  𝑚 → ⋅ Σ → term? –  AccidentalTaylorExpansion Commented Mar 5 at 9:46 @AccidentalTaylorExpansion Yes, you definitely need the m ⃗  𝑚 → term for positivity. Just think about a simple case such as |00⟩⟨00| | 00 ⟩ ⟨ 00 | where you have n ⃗  1 = n ⃗  2 =(0,0,1) 𝑛 → 1 = 𝑛 → 2 = ( 0 , 0 , 1 ) . You need m ⃗  =(0,0,1) 𝑚 → = ( 0 , 0 , 1 ) as well. –  DaftWullie Commented Mar 5 at 10:02 @AndrewSteane this depends on how you're understanding equivalence class. If you require a two-way conversion between states, that's a lot more messy (I don't even want to think about it!). Perhaps I misunderstood what you intended by "generate" (I was thinking "take one state and produce all the different states you want from it" and that's the sense in which I answered the LOCC part). –  DaftWullie Commented Mar 5 at 10:08 OK. By "generate" I mean a numerical method which spits out density matrices. –  Andrew Steane Commented Mar 5 at 14:25 Show 1 more comment Your Answer Sign up or log in Sign up using Google Sign up using Email and Password Post as a guest Name Email Required, but never shown Post Your Answer By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. 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