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Symmetry Effects On Quantum Master Equations
This demonstration illustrates the effects of permutation symmetry on quantum master equations. Although we consider only permutation symmetry, the essential idea may apply to any symmetry.
Permutation operator | |
Permute the logical values in each Ket[…] or Ket[<|…|>] | |
Describes the quantum operations | |
The Choi matrix of a supermap | |
Describes the Lindblad equations | |
A basis of matrices or operators that are all traceless except for one. | |
Converts the Lindblad equation into a normal linear differential equation | |
Solves the Lindblad equation | |
Finds a numerical solution to a given Lindblad equation |
Functions useful to handle quantum operations.
In[75]:=
Needs["QuantumMob`Q3`"]
Throughout this document, symbol S will be used to denote qubits and Pauli operators on them.
In[76]:=
Let[Qubit,S]
Here are some shortcut utilities.
In[77]:=
toYoungForm[gg:{_?GelfandPatternQ,__}]:=Map[toYoungForm,gg];toYoungForm[gp_?GelfandPatternQ]:=YoungForm@ToYoungTableau@gp;toYoungForm[bs_Association]:=KeyMap[toYoungForm,toYoungForm/@bs];toYoungForm[other_]:=other
Three-Qubit Case
Consider a system of three (or more) qubits, which involves irreducible subspaces of larger than one dimensions.
In[81]:=
$n=3;kk=Range[$n];SS=S[kk,$];
Initial State
Initial State
Start with constructing the Schur basis.
The first four irreducible representations are one-dimensional and equivalent to one another. The last two irreducible representations are two-dimensional and equivalent to each other.
In[89]:=
bs=SchurBasis[SS];bs=KeyGroupBy[bs,Last,Values];toYoungForm[bs]bb=Catenate[bs];
Out[91]=
,
++,
++,
,
--,-,
+-,-
1 | 1 | 1 |
0
S
1
0
S
2
0
S
3
1 | 1 | 2 |
0
S
1
0
S
2
1
S
3
3
0
S
1
1
S
2
0
S
3
3
1
S
1
0
S
2
0
S
3
3
1 | 2 | 2 |
0
S
1
1
S
2
1
S
3
3
1
S
1
0
S
2
1
S
3
3
1
S
1
1
S
2
0
S
3
3
2 | 2 | 2 |
1
S
1
1
S
2
1
S
3
1 | 1 |
2 | |
2
3
0
S
1
0
S
2
1
S
3
0
S
1
1
S
2
0
S
3
6
1
S
1
0
S
2
0
S
3
6
0
S
1
1
S
2
0
S
3
2
1
S
1
0
S
2
0
S
3
2
1 | 2 |
2 | |
0
S
1
1
S
2
1
S
3
6
1
S
1
0
S
2
1
S
3
6
2
3
1
S
1
1
S
2
0
S
3
0
S
1
1
S
2
1
S
3
2
1
S
1
0
S
2
1
S
3
2
Take an initial state. This is the most general form of density matrix that is invariant under the symmetric group.
In[93]:=
mat=RandomMatrix[4]⊕(RandomMatrix[2]⊗One[2]);mat=Topple[mat].mat;mat=Chop[mat/Tr[mat]];mat//MatrixFormdd=Outer[Dyad[#1,#2,SS]&,bb,bb];in=Total@Flatten[dd*mat];
Out[96]//MatrixForm=
0.253062 | -0.0192915+0.00759828 | 0.00503343+0.0338309 | -0.0464404-0.00876152 | 0 | 0 | 0 | 0 |
-0.0192915-0.00759828 | 0.157611 | -0.0302057-0.00869276 | -0.0104722-0.0673122 | 0 | 0 | 0 | 0 |
0.00503343-0.0338309 | -0.0302057+0.00869276 | 0.121335 | 0.0973724-0.0398234 | 0 | 0 | 0 | 0 |
-0.0464404+0.00876152 | -0.0104722+0.0673122 | 0.0973724+0.0398234 | 0.190283 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0.0661748 | 0 | -0.0274016+0.0502753 | 0 |
0 | 0 | 0 | 0 | 0 | 0.0661748 | 0 | -0.0274016+0.0502753 |
0 | 0 | 0 | 0 | -0.0274016-0.0502753 | 0 | 0.0726803 | 0 |
0 | 0 | 0 | 0 | 0 | -0.0274016-0.0502753 | 0 | 0.0726803 |
Check that the matrix representation in the Schur basis is the same as the intended matrix.
In[99]:=
new=MatrixIn[in,bb];new-mat//Chop//MatrixForm
Out[100]//MatrixForm=
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
Check that the initial state is a density operator (a positive operator with unit trace).
In[101]:=
Eigenvalues[mat]Tr[mat]
Out[101]=
{0.310083,0.250664,0.147245,0.126778,0.126778,0.0142987,0.0120775,0.0120775}
Out[102]=
1.
Check also that the initial state is invariant under all symmetry transformations in the symmetric group. It suffices to test with transpositions, which are generators of the symmetric group.
In[103]:=
cyc=Permutation[#,SS]&/@Cycles/@List/@Subsets[kk,{2}]
Out[103]=
{Permutation[Cycles[{{1,2}}],{,,}],Permutation[Cycles[{{1,3}}],{,,}],Permutation[Cycles[{{2,3}}],{,,}]}
S
1
S
2
S
3
S
1
S
2
S
3
S
1
S
2
S
3
In[104]:=
new=cyc〚1〛**in**cyc〚1〛;new-in//Garner//Chopnew=cyc〚2〛**in**cyc〚2〛;new-in//Garner//Chopnew=cyc〚3〛**in**cyc〚3〛;new-in//Garner//Chop
Out[105]=
0
Out[107]=
0
Out[109]=
0
Lindblad Equation
Lindblad Equation
Let us now construct the Lindblad equation.
In[110]:=
Let[Real,B];opH=Total@Map[Total[S[First@#,All]**S[Last@#,All]]&,Subsets[kk,{2}]]+B*Total@S[kk,3]
Out[110]=
x
S
1
x
S
2
x
S
1
x
S
3
y
S
1
y
S
2
y
S
1
y
S
3
z
S
1
z
S
2
z
S
1
z
S
3
x
S
2
x
S
3
y
S
2
y
S
3
z
S
2
z
S
3
z
S
1
z
S
2
z
S
3
Check that the Hamiltonian is invariant under the symmetric group.
In[111]:=
cyc〚1〛**opH**cyc〚1〛-opH//Elaboratecyc〚2〛**opH**cyc〚2〛-opH//Elaboratecyc〚3〛**opH**cyc〚3〛-opH//Elaborate
Out[111]=
0
Out[112]=
0
Out[113]=
0
In[114]:=
Let[Real,Γ]opL=Join[Sqrt[Γ["+"]]S[kk,4],Sqrt[Γ["-"]]S[kk,5]]
Out[115]=
,,,,,
+
S
1
Γ
+
+
S
2
Γ
+
+
S
3
Γ
+
-
S
1
Γ
-
-
S
2
Γ
-
-
S
3
Γ
-
In[116]:=
ops=Prepend[opL,opH];MatrixForm/@ops
Out[117]=
+++++++++B++,,,,,,
x
S
1
x
S
2
x
S
1
x
S
3
y
S
1
y
S
2
y
S
1
y
S
3
z
S
1
z
S
2
z
S
1
z
S
3
x
S
2
x
S
3
y
S
2
y
S
3
z
S
2
z
S
3
z
S
1
z
S
2
z
S
3
+
S
1
Γ
+
+
S
2
Γ
+
+
S
3
Γ
+
-
S
1
Γ
-
-
S
2
Γ
-
-
S
3
Γ
-
Here is the solution to the Lindblad equation.
In[118]:=
Clear[rho]$tmax=5;rho[t_]=Block[{Γ,B},Γ["+"]=.3;Γ["-"]=.7;B=0.2;NLindbladSolve[ops,in,{t,0,$tmax}]];
Here is a typical output state.
Verifying the Solutions
Verifying the Solutions
Choose a output state at a particular time. Note that the time must be smaller than $tmax chosen when numerically solving the Lindblad equation.
Check that any output state is invariant under all transpositions (and hence under the whole symmetric group).
The invariance also implies that any output state must be block diagonal in the Schur basis grouped by the standard Young tableaux.
In this arrangement of the Schur basis, the initial as well as output state must be block diagonal.
Conclusion
Conclusion
So far, we have verified the numerical solution in several ways.
Now, let us get back to the point and examine whether the Lindblad equation may be solved separately in each irreducible subspaces specified by the Schur basis. The answer is no. You can see this by noting that the trace in each block is not conserved.
Note that the (overall) trace is preserved.