ABSTRACT (original article): Despite their importance in quantum theory, joint quantum measurements remain poorly understood. An intriguing conceptual and practical question is whether joint quantum measurements on separated systems can be performed without bringing them together. Remarkably, by using shared entanglement, this can be achieved perfectly when disregarding the postmeasurement state. However, existing localization protocols typically require unbounded entanglement. In this work, we address the fundamental question: “Which joint measurements can be localized with a finite amount of entanglement?” We develop finite-resource versions of teleportation-based schemes and analytically classify all two-qubit measurements that can be localized in the first levels of the resulting hierarchies. These levels include several measurements with exceptional properties and symmetries, such as the Bell state measurement and the elegant joint measurement. This leads us to propose a systematic classification of joint measurements based on entanglement cost, which we argue directly connects with the complexity of implementing those measurements. We illustrate how to numerically explore higher levels and construct generalizations to higher dimensions and multipartite settings. CITATION (original article): Jef Pauwels, Alejandro Pozas-Kerstjens, Flavio Del Santo, and Nicolas Gisin (2025), Classification of Joint Quantum Measurements Based on Entanglement Cost of Localization, Phys. Rev. X 15, 021013. https://doi.org/10.1103/PhysRevX.15.021013
This combined notebook contains the computational appendices for the two-qubit measurements localized using one shared ebit and three shared ebits. The two parts are self-contained and may be evaluated independently from the beginning of the corresponding section.
One shared ebit solutions
One shared ebit solutions
In this section we give all the two-qubit measurements that can be localized using one shared ebit between Alice and Bob. This is, we solve Eq. (4) in the manuscript. We do so by following the procedure in Appendix B.1. This involves i) finding representations of the Pauli group in terms of permutation matrices with phases, and ii) finding the intertwiners between such representations and the 1⊗Pauli representation. The fact that the left-hand side of Eq. (4) in the manuscript is Hermitian and its spectrum is {1,1,-1,-1} implies that we can restrict the search of representations of the Pauli group to tuples of the following 16 matrices (also shown in Appendix B.1):
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P1[a_,b_]:={{0,0,0,Exp[a]},{0,0,Exp[b],0},{0,Exp[-b],0,0},{Exp[-a],0,0,0}}P2a[a_,b_]:={{0,0,0,Exp[a]},{0,1,0,0},{0,0,-1,0},{Exp[-a],0,0,0}}P2b[a_,b_]:={{0,0,0,Exp[a]},{0,-1,0,0},{0,0,1,0},{Exp[-a],0,0,0}}P3[a_,b_]:={{0,0,Exp[a],0},{0,0,0,Exp[b]},{Exp[-a],0,0,0},{0,Exp[-b],0,0}}P4a[a_,b_]:={{0,0,Exp[a],0},{0,1,0,0},{Exp[-a],0,0,0},{0,0,0,-1}}P4b[a_,b_]:={{0,0,Exp[a],0},{0,-1,0,0},{Exp[-a],0,0,0},{0,0,0,1}}P5[a_,b_]:={{0,Exp[-a],0,0},{Exp[a],0,0,0},{0,0,0,Exp[-b]},{0,0,Exp[b],0}}P6a[a_,b_]:={{0,Exp[-a],0,0},{Exp[a],0,0,0},{0,0,1,0},{0,0,0,-1}}P6b[a_,b_]:={{0,Exp[-a],0,0},{Exp[a],0,0,0},{0,0,-1,0},{0,0,0,1}}P7a[a_,b_]:={{1,0,0,0},{0,0,0,Exp[a]},{0,0,-1,0},{0,Exp[-a],0,0}}P7b[a_,b_]:={{-1,0,0,0},{0,0,0,Exp[a]},{0,0,1,0},{0,Exp[-a],0,0}}P8a[a_,b_]:={{1,0,0,0},{0,0,Exp[a],0},{0,Exp[-a],0,0},{0,0,0,-1}}P8b[a_,b_]:={{-1,0,0,0},{0,0,Exp[a],0},{0,Exp[-a],0,0},{0,0,0,1}}P9a[a_,b_]:={{1,0,0,0},{0,-1,0,0},{0,0,0,Exp[a]},{0,0,Exp[-a],0}}P9b[a_,b_]:={{-1,0,0,0},{0,1,0,0},{0,0,0,Exp[a]},{0,0,Exp[-a],0}}P10a[a_,b_]:=DiagonalMatrix[{1,1,-1,-1}]P10b[a_,b_]:=DiagonalMatrix[{1,-1,1,-1}]P10c[a_,b_]:=DiagonalMatrix[{-1,1,1,-1}]P10d[a_,b_]:=DiagonalMatrix[{1,-1,-1,1}]P10e[a_,b_]:=DiagonalMatrix[{-1,1,-1,1}]P10f[a_,b_]:=DiagonalMatrix[{-1,-1,1,1}]Ps={P1,P2a,P2b,P3,P4a,P4b,P5,P6a,P6b,P7a,P7b,P8a,P8b,P9a,P9b,P10a,P10b,P10c,P10d,P10e,P10f};
Step 1: Find representations of SU(2)
Step 1: Find representations of SU(2)
We create all possible tuples, and discard those that do not satisfy the SU(2) algebra for any values of the phases.
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subsets=Subsets[Ps,{3}];Do[If[ii!=jj,AppendTo[subsets,{Ps[[ii]],Ps[[ii]],Ps[[jj]]}]],{ii,Length[Ps]},{jj,Length[Ps]}]Do[AppendTo[subsets,{Ps[[ii]],Ps[[ii]],Ps[[ii]]}],{ii,Length[Ps]}]
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constraintsXYZ=Assuming[a1∈Reals&&a2∈Reals&&a3∈Reals&&b1∈Reals&&b2∈Reals&&b3∈Reals,Table[Assuming[a1∈Reals&&a2∈Reals&&a3∈Reals&&b1∈Reals&&b2∈Reals&&b3∈Reals,Reduce[P[[1]][a1,b1].P[[2]][a2,b2]-P[[2]][a2,b2].P[[1]][a1,b1]==2P[[3]][a3,b3]&&P[[2]][a2,b2].P[[3]][a3,b3]-P[[3]][a3,b3].P[[2]][a2,b2]==2P[[1]][a1,b1]&&P[[3]][a3,b3].P[[1]][a1,b1]-P[[1]][a1,b1].P[[3]][a3,b3]==2P[[2]][a2,b2]&&P[[1]][a1,b1].P[[2]][a2,b2].P[[3]][a3,b3]==IdentityMatrix[4]]//FullSimplify],{P,subsets}]/.->0];(*SinceSubsetsgivesorderedsubsets,wemustconsideralternativeorderingsofthematrices*)constraintsXZY=Assuming[a1∈Reals&&a2∈Reals&&a3∈Reals&&b1∈Reals&&b2∈Reals&&b3∈Reals,Table[Assuming[a1∈Reals&&a2∈Reals&&a3∈Reals&&b1∈Reals&&b2∈Reals&&b3∈Reals,Reduce[P[[1]][a1,b1].P[[3]][a3,b3]-P[[3]][a3,b3].P[[1]][a1,b1]==2P[[2]][a2,b2]&&P[[3]][a3,b3].P[[2]][a2,b2]-P[[2]][a2,b2].P[[3]][a3,b3]==2P[[1]][a1,b1]&&P[[2]][a2,b2].P[[1]][a1,b1]-P[[1]][a1,b1].P[[2]][a2,b2]==2P[[3]][a3,b3]&&P[[1]][a1,b1].P[[3]][a3,b3].P[[2]][a2,b2]==IdentityMatrix[4]]//FullSimplify],{P,subsets}]/.->0];(*Keepthetuplesthat,forsomevalueofthephases,theysatisfytheSU(2)algebra*)setposXYZ=Complement[Range[Length[subsets]],Flatten[Position[constraintsXYZ,False]]];setposXZY=Complement[Range[Length[subsets]],Flatten[Position[constraintsXZY,False]]];relevant=subsets[[DeleteDuplicates[Join[setposXYZ,setposXZY]]]]
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{{P1,P2a,P8a},{P1,P2a,P8b},{P1,P2b,P8a},{P1,P2b,P8b},{P1,P3,P5},{P3,P4a,P7a},{P3,P4a,P7b},{P3,P4b,P7a},{P3,P4b,P7b},{P5,P6a,P9a},{P5,P6a,P9b},{P5,P6b,P9a},{P5,P6b,P9b},{P1,P1,P10a},{P1,P1,P10b},{P1,P1,P10e},{P1,P1,P10f},{P3,P3,P10a},{P3,P3,P10c},{P3,P3,P10d},{P3,P3,P10f},{P5,P5,P10b},{P5,P5,P10c},{P5,P5,P10d},{P5,P5,P10e}}
Now we explicitly calculate which are the conditions for which the matrices form a representation of the SU(2) algebra. Note that there may be different solutions for a single equation.
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conditionsXYZ=(First[Assuming[a1∈Reals&&b1∈Reals&&a2∈Reals&&b2∈Reals&&a3∈Reals&&b3∈Reals,Solve[constraintsXYZ[[#]]//FullSimplify]//FullSimplify]/.{->0,->0,->0}])&/@setposXYZ;conditionsXZY=(First[Assuming[a1∈Reals&&b1∈Reals&&a2∈Reals&&b2∈Reals&&a3∈Reals&&b3∈Reals,Solve[constraintsXZY[[#]]//FullSimplify]//FullSimplify]/.{->0,->0,->0}])&/@setposXZY;
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For every solution of the type XYZ there is an analogous solution of the type XZY that, in fact, produces the same triple of matrices. Thus, from now onwards, we only use the XYZ-type triples. We have also checked that all the solutions to a same triple lead to an equivalent set of matrices, and thus we use First to keep only one of the solutions.
Step 2: Compute the intertwiners
Step 2: Compute the intertwiners
Now that we have all possible right-hand sides of Eq. (4), we solve it for each of them. Equation (4) is a set of Sylvester-type equations. In order to solve them, we vectorize them per as Eq. (B7). For multiple equations, the simultaneous solution to all of them lies in the joint null space, given in Eq. (B8). Below we solve for this null space and find among all possible solutions one that is Hermitian. The program below produces instead of directly.
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solutionsXYZ={};DoLx=KroneckerProduct[IdentityMatrix[4],KroneckerProduct[IdentityMatrix[2],PauliMatrix[1]]]-KroneckerProduct[Transpose[relevant[[ii]][[1]][a1,b1]/.conditionsXYZ[[ii]]],IdentityMatrix[4]];Ly=KroneckerProduct[IdentityMatrix[4],KroneckerProduct[IdentityMatrix[2],PauliMatrix[2]]]-KroneckerProduct[Transpose[relevant[[ii]][[2]][a2,b2]/.conditionsXYZ[[ii]]],IdentityMatrix[4]];Lz=KroneckerProduct[IdentityMatrix[4],KroneckerProduct[IdentityMatrix[2],PauliMatrix[3]]]-KroneckerProduct[Transpose[relevant[[ii]][[3]][a3,b3]/.conditionsXYZ[[ii]]],IdentityMatrix[4]];vecs=NullSpace[Join[Lx,Ly,Lz,1]];ms=ArrayReshape[#,{4,4}]&/@vecs;(*De-vectorizethesolution*)M=Sum[Exp[]ms[[i]],{i,1,4}]//Simplify;(*Parametrizationof*)(*SolveforwhichparametersmakeitHermitian*)AppendTo[solutionsXYZ,{relevant[[ii]],M/.First[Solve[First[Assuming[∈Reals&&∈Reals&&∈Reals&&∈Reals&&∈Reals&&∈Reals&&∈Reals&&∈Reals&&>=0&&>=0&&>=0&&>=0&&a1∈Reals&&b1∈Reals&&a2∈Reals&&b2∈Reals&&a3∈Reals&&b3∈Reals,Reduce[M.ConjugateTranspose[M]==IdentityMatrix[4]]//Simplify]]]]}],{ii,Range[Length[relevant]]}
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The case number 5 seems to give issues when solving, so we deal with it separately and by hand
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ii=5;Lx=KroneckerProduct[IdentityMatrix[4],KroneckerProduct[IdentityMatrix[2],PauliMatrix[1]]]-KroneckerProduct[Transpose[relevant[[ii]][[1]][a1,b1]/.conditionsXYZ[[ii]]],IdentityMatrix[4]];Ly=KroneckerProduct[IdentityMatrix[4],KroneckerProduct[IdentityMatrix[2],PauliMatrix[2]]]-KroneckerProduct[Transpose[relevant[[ii]][[2]][a2,b2]/.conditionsXYZ[[ii]]],IdentityMatrix[4]];Lz=KroneckerProduct[IdentityMatrix[4],KroneckerProduct[IdentityMatrix[2],PauliMatrix[3]]]-KroneckerProduct[Transpose[relevant[[ii]][[3]][a3,b3]/.conditionsXYZ[[ii]]],IdentityMatrix[4]];vecs=NullSpace[Join[Lx,Ly,Lz,1]];ms=ArrayReshape[#,{4,4}]&/@vecs;MatrixForm[M=Sum[Exp[]ms[[i]],{i,1,4}]//Simplify]
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Out[]//MatrixForm=
-(a1- ϕ 3 r 3 | -(a1- ϕ 4 r 4 | -(a1- ϕ 1 r 1 | -(a1- ϕ 2 r 2 |
-(a1+a3- ϕ 3 r 3 | - -(a1+a3- ϕ 4 r 4 | -(a1+a3- ϕ 1 r 1 | - -(a1+a3- ϕ 2 r 2 |
- -(a1-a2- ϕ 4 r 4 | -(a1-a2- ϕ 3 r 3 | - -(a1-a2- ϕ 2 r 2 | -(a1-a2- ϕ 1 r 1 |
ϕ 4 r 4 | ϕ 3 r 3 | ϕ 2 r 2 | ϕ 1 r 1 |
We now find, by hand, a solution that makes Hermitian. Per as Lemma 2 in the manuscript, all other solutions will be achievable from this one via local unitary rotations.
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Assuming∈Reals&&∈Reals&&∈Reals&&∈Reals&&∈Reals&&∈Reals&&∈Reals&&∈Reals&&>=0&&>=0&&>=0&&a1∈Reals&&b1∈Reals&&a2∈Reals&&b2∈Reals&&a3∈Reals&&b3∈Reals,MatrixFormM.ConjugateTranspose[M]/.->0,->0,a1->0,->,->//Simplify
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1 | 0 | 0 | 0 |
0 | 1 | 0 | 0 |
0 | 0 | 1 | 0 |
0 | 0 | 0 | 1 |
Indeed, we see that there is an assignment that makes Hermitian. As a final double check, we look at whether M transforms the 1⊗Pauli matrices into permutation matrices with phases.
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t=M/.->0,->0,a1->0,->,->;
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Assuming[∈Reals&&∈Reals&&∈Reals&&∈Reals&&∈Reals&&∈Reals&&∈Reals&&∈Reals&&>=0&&>=0&&>=0&&a1∈Reals&&b1∈Reals&&a2∈Reals&&b2∈Reals&&a3∈Reals&&b3∈Reals,ConjugateTranspose[Transpose[t]].KroneckerProduct[IdentityMatrix[2],PauliMatrix[1]].Transpose[t]//MatrixForm//Simplify]Assuming[∈Reals&&∈Reals&&∈Reals&&∈Reals&&∈Reals&&∈Reals&&∈Reals&&∈Reals&&>=0&&>=0&&>=0&&a1∈Reals&&b1∈Reals&&a2∈Reals&&b2∈Reals&&a3∈Reals&&b3∈Reals,ConjugateTranspose[Transpose[t]].KroneckerProduct[IdentityMatrix[2],PauliMatrix[2]].Transpose[t]//MatrixForm//Simplify]Assuming[∈Reals&&∈Reals&&∈Reals&&∈Reals&&∈Reals&&∈Reals&&∈Reals&&∈Reals&&>=0&&>=0&&>=0&&a1∈Reals&&b1∈Reals&&a2∈Reals&&b2∈Reals&&a3∈Reals&&b3∈Reals,ConjugateTranspose[Transpose[t]].KroneckerProduct[IdentityMatrix[2],PauliMatrix[3]].Transpose[t]//MatrixForm//Simplify]
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0 | 0 | 0 | 1 |
0 | 0 | (a2+a3) | 0 |
0 | - -(a2+a3) | 0 | 0 |
1 | 0 | 0 | 0 |
It does, so we add the solution to the list.
Finally we transpose so that the columns represent the states.
And we normalize by the first phase we see in each column in order to make the solutions clearer to analyze.
We identify three types of solutions: the measurement in the computational basis (solutions 14-25), the measurement in the twisted basis (solutions 1-4, 6-13), and the Bell state measurement (solution 5).
Three shared ebits solutions
Three shared ebits solutions
In this section we give all the two-qubit measurements that can be localized using three shared ebits between Alice and Bob. This is, we solve Eq. (5) in the manuscript. We do so by following the procedure in Appendix B.2. This involves i) finding representations of the SU(2)⊗SU(2) in terms of permutation matrices with phases, ii) computing the intertwiners between such representations and that consisting of two copies of the Pauli matrices, iii) finding, among all such intertwiners, triples that form a representation of SU(2), and iv) computing the intertwiners between such representations an that of the form 1⊗Pauli. We begin by defining the permutation matrices with phases that are relevant, and some auxiliary functions
Step 1: Compute triples of permutations that reproduce SU(2)SU(2)
Step 1: Compute triples of permutations that reproduce SU(2)SU(2)
This part is equivalent to that of finding representations of SU(2) when solving Eq. (4) in the manuscript.
First we compute the triples and conditions that lead to a single representation of SU(2)
Second, we find the conditions that lead to two SU(2) representations to commute (so they build SU(2)SU(2))
Sanity check: conditionsR/L should have the same length than relevantTriples because we essentially have one dimension more due to the fact that Solve gives a list of solutions, even if the list has length 1.
Some of the triples cannot be made to commute. We thus identify which can, because these are those we care about
We thus have the list of sets of matrices and associated conditions that produce representations of SU(2)⊗SU(2).
Step 2: Compute the Mb that correspond to the representations
Step 2: Compute the Mb that correspond to the representations
We solve for the intertwiners between the Pauli⊗Pauli representation and the representations solved above. These are Sylvester-type equations, which are solved by vectorizing the matrices and computing the null space
Note that there is a call to First in the last line of the cell. This is fine because no null space contains more than one solution.
Finally, in order to have unitary matrices, we need to have them divided by their norm.
Finally, in order to have unitary matrices, we need to have them divided by their norm.
Step 3: Use the Mbs found above as the right-hand side of the one-ebit equation
Step 3: Use the Mbs found above as the right-hand side of the one-ebit equation
Delete the cases where anticommutation conditions fail to be satisfied or a potentially suitable My can not be found.
Now we have a list of lists of nullspaces + pairs {Mx, Mz}. Some of these can be Null, so let us eliminate those as well
This leads to some empty lists. Let us eliminate them
There are four cases in which there is an “incomplete” null space. We have been able to track these down to branch cuts. In order to identify which two are these, we need to go back to the original form of nullSpaces
And we proceed to solve each of them by hand, with the code that works best and suitable choices of constants
And now we can just have the full list of null spaces
The final step is combining the null spaces and solving for the rs and ϕs that give a unitary matrix
There are some errors, that stem from the issues we had with 2-dimensional null spaces. Let us fix them by hand
Now we simplify and normalize the states
Finally, we can set all the free parameters to zero, leading to the final set of solutions.
Several solutions correspond to the same bases. In the end, there will be only 8 unique bases.
CITE THIS NOTEBOOK
CITE THIS NOTEBOOK
Classification of joint quantum measurements based on entanglement cost of localization
by Alejandro Pozas-Kerstjens and Jef Pauwels
Wolfram Community, STAFF PICKS, June 10, 2026
https://community.wolfram.com/groups/-/m/t/3730885
by Alejandro Pozas-Kerstjens and Jef Pauwels
Wolfram Community, STAFF PICKS, June 10, 2026
https://community.wolfram.com/groups/-/m/t/3730885