I'm more of a quantum optics guy than a quantum info guy, and deal mainly in master equations. I'm interested in operator-sum form, and I'd like to derive the errors in this form for a small quantum system that I'm simulating.

The catch: The quantum system is driven by an external (classical) field modelled with a sinusoidal function, and the damping rates are low, so I can't make a rotating wave approximation to eliminate this time dependence. Given that I must solve the master equation numerically by integration, and the result of each integration at time $t$ is not sufficient information to figure out these errors, and I need to do some work to recover the superoperator matrix that has operated on a vectorised density matrix. i.e. I feed the master equation a vectorised density matrix with a single entry of 1 and the rest zero, and build the matrix like that for a particular time $\tau$. Am I on the right track here (sanity check)? More explicitly, if $\mathrm{vec}(\rho_{ij,t=\tau})$ is the vectorised (so it's a column vector) form of a density matrix with a single entry of 1 in position $i,j$, at $t=0$ that has been evolved to time $\tau$, then a matrix to take the vector form of the density matrix from $t=0$ to $t=\tau$ is given as $\mathbf{M}=\sum_{i,j}\mathrm{vec}(\rho_{ij,t=0})\mathrm{vec}(\rho_{ij,t=\tau})^\dagger$.

The question: Given this superoperator $\mathbf{M}$ that does $\mathbf{M}\,\mathrm{vec}(\rho_0)=\mathrm{vec}(\rho_\tau)$, how can I get Krauss operators for the operator-sum equivalent of $\mathbf{M}$ that are in a useful form? i.e. the system in question is a qubit or a qutrit and another qubit or qutrit. I'd like to be able to do the operator sum in the form of tensor products of spin matrices on each channel if possible.

Side question: Is $\mathbf{M}$ a Choi matrix?

Final note: I awarded the acceptance to Pinja, as I used the paper Pinja suggested. I have provided an answer myself below that fills in the details.

  • $\begingroup$ What do you mean by "system in question is a qubit or a qutrit and another qubit or qutrit." -- what is the "other system"? Are you talking about the ancilla needed to implement this channel using unitaries + tracing out? In that case, note that the dimension of the ancilla can be up to D^2, so qubits won't do. $\endgroup$ Commented Nov 11, 2011 at 18:24
  • $\begingroup$ No, at the moment it's just a toy model consisting of two small quantum systems that are coupled, and have different T1 and T2 times. The answer to this question is not of serious concern. It's more a point of interest, as it could be a handy to know more about how to do this in the future. $\endgroup$
    – qubyte
    Commented Nov 12, 2011 at 2:22
  • $\begingroup$ Can I get this question migrated to CS Theory rather than Physics please? $\endgroup$
    – qubyte
    Commented May 4, 2012 at 5:02
  • $\begingroup$ Well... I think this would have been fine here, but okay. $\endgroup$
    – David Z
    Commented May 8, 2012 at 15:33
  • $\begingroup$ Thanks. Sorry, just not a big fan of Physics.SE, and anyway, I think research oriented QI questions fit better here (after being convinced). $\endgroup$
    – qubyte
    Commented May 8, 2012 at 15:35

4 Answers 4


I worked on a very similar problem on my Masters thesis, in which I studied the non-Markovian dynamics of a driven qubit in a dissipative environment. My interest was in checking that the master equation I obtained was completely positive, but this is just one side of your problem. The question turned out to be very non-trivial if no RWA is made, but I was able to get some results using Ref. [J Mod. Opt. 54, 1695 (2007)] and exploiting the fact that the qubit is weakly coupled to the environment. I'll beat my drum and also give the Ref. to an article where I present some of these results, [P. Haikka and S. Maniscalco, Phys. Rev. A 81, 052103 (2010)], you may find it useful.

  • $\begingroup$ Ah! It turns out I've been looking at the Andersson paper one for a few days now. It does seem very promising, and gives the most concrete recipe. I like having a method to apply to problems. To be honest, I need to find a patch of time to really sit down and look at this. It's more of a personal project at the moment. $\endgroup$
    – qubyte
    Commented Nov 17, 2011 at 2:29

The references given in answer to Quantum mechanics as a Markov process — in particular Carlton Caves' on-line notes "Completely positive maps, positive maps, and the Lindblad form" — survey physical ideas and mathematical tools that are helpful in answering the question.

A key point is associated to the specific question asked "How can I get Kraus operators for the operator-sum equivalent of $M$ that are in a useful form?" For large quantum systems, a generic superoperator $M$ will not have an algorithmically compressible form. Moreover, Kraus representations are non-unique, and to the best of my (non-expert) knowledge there is no procedure that is both general and efficient for finding Kraus representations of a given $M$ that have a "useful form" (by whatever criteria are given for a form being "useful"). That deciding quantum separability is NP-hard suggests that no efficient, general representation-finding algorithm exists, even when $M$ is numerically given in its entirety.

To make progress, it may be helpful to ask heuristic questions: "What is special about my particular superoperator? Can I exhibit a set of Lindbladian generators for it that have useful symmetry properties and/or generate compatible compressive flows on the Hilbert state-space? Are these Lindbladian properties associated to a natural Hilbert basis in which $M$ has a sparse, factored, or otherwise algorithmically compressible representation?"

If questions like these could be efficiently answered by "turning an algorithmic crank", then quantum physics would be a far less interesting subject! :)

  • $\begingroup$ This is pretty much what I was hoping was not the case, but thought would be. Sadly the system only has exploitable symmetry in the case of only dephasing with no depopulation. There's a very appealing form of the Lindblad master equation that collects terms which aren't of Krauss form into a non-hermitian Hamiltoninan, which for the case of no time-dependence in the Hamiltonian can be used to choose a basis which naturally expresses decay as the remaining Krauss terms. Neat, but no help for me. $\endgroup$
    – qubyte
    Commented Nov 11, 2011 at 16:35
  • $\begingroup$ One of the references in Caves' notes is Wolf and Cirac Dividing quantum channels (arXiv:math-ph/0611057), which I recommend without the least warranty of having personally grasped the (many and subtle) quantum informatic issues that this article discusses! :) $\endgroup$
    – John Sidles
    Commented Nov 11, 2011 at 16:43
  • $\begingroup$ Nice, I'll take a look at that. One interesting thing that I possibly should have noted about the undriven version of the system above is that its time independence means that you can find $\mathbf{M}$ directly with matrix exponentiation (not efficient, but this system is small enough). You can also construct $\mathbf{M}$ using some guessed Krauss operators, and a few simultaneous equations later gives you the mapping between the two, allowing the extraction of error rates on the various channels. $\endgroup$
    – qubyte
    Commented Nov 11, 2011 at 16:53

I think what you might be looking for is this: The Real Density Matrix. It gives you a recipe for converting between various superoperator representations (including using a tensor product basis of Paulis). A detailed quantum process tomography experiment utilizing the results are here: Quantum Process Tomography of the Quantum Fourier Transform. More generally, Havel has also derived algorithms to convert to minimal Kraus representations here: Procedures for Converting among Lindblad, Kraus and Matrix Representations of Quantum Dynamical Semigroups.

Edit to answer the added question: unfortunately, this area is plagued by inconsistent notation and conventions. Nevertheless, I will give you the one that seems most natural to me. So let me take ${\rm vec}(\rho)$ to be the operation of taking the the rows of $\rho$ and stacking them on top of each other. That is ${\rm vec}(|i\rangle\langle j|)=|i\rangle\otimes|j\rangle$ rather than ${\rm vec}(|i\rangle\langle j|)=|j\rangle\otimes|i\rangle$, which would be "column stacking" (But! This of course depends on your convention for the Kronecker product---here I am taking the second index to "vary most rapidly"). Let's instead define the "column stacking" convention as ${\rm col}(\rho)$. Now, neither of the matrices which act as ${\rm \bf M}^{\rm row}{\rm vec}(\rho_0)={\rm vec}(\rho_t)$ or ${\rm \bf M}^{\rm col}{\rm col}(\rho_0)={\rm col}(\rho_t)$ are the "Choi" matrix. The Choi matrix is defined as $$ {\rm\bf C} = \sum_{i,j} ({\rm\bf 1}\otimes |i\rangle\langle j|) {\rm \bf M}^{\rm row} (|i\rangle\langle j|\otimes {\rm\bf 1}), $$ or, equivalently $$ {\rm\bf C} = \sum_{i,j} (|i\rangle\langle j|\otimes {\rm\bf 1}) {\rm \bf M}^{\rm col} ({\rm\bf 1}\otimes |i\rangle\langle j|). $$ Note that since all these representations are defined in terms of the basis $\{|i\rangle\langle j|\otimes |k\rangle\langle l|\}$, the transformation between them is just a permutation.

  • $\begingroup$ This is interesting, it could be exactly what I'm looking for... $\endgroup$
    – qubyte
    Commented Nov 12, 2011 at 2:28
  • $\begingroup$ I just saw your addition. Thank you, this is very useful. I originally took your version of vec, but now I use the columns stacked. Thank Wikipedia for that one. Perhaps I should adopt your notation for clarity. $\endgroup$
    – qubyte
    Commented Nov 16, 2011 at 10:36

As Pinja noted, a paper by Andersson et al. (arXiv)(DOI) has been especially useful. The paper goes into a great deal of detail, and I finally sat down today to take a proper look at it. As an example problem, I picked two qubits with an exchange interaction to check this which is a minimal version of what I'm considering. To begin, the master equation is given by

$$ \dot\rho = \Lambda(\rho). $$

The method requires that basis operators of the system are chosen. It is convenient to give these in terms of the Pauli matrices in the case of two qubits, but for a qutrit one would employ the Gell-Mann matrices. Defining $\sigma_i = \mathbf{1},\sigma_x,\sigma_y,\sigma_z$ for each qubit, this system has a basis built up of the tensor products of these with a factor of $1/2$ for normalisation, yielding 16 operators $G_i$ e.g. $G_5 = G_{xx} = (\sigma_x\otimes\sigma_x)/2$. Sticking with Hermitian operators keeps things neat as well, since some daggers can be neglected.

A special matrix is now composed called $L$, which is related to the master equation.

$$ L_{n,m} = \mathrm{Tr}[G_n\Lambda(G_m)]. $$

If we are dealing with the master equation as a matrix acting on a vectorised density operator as discussed in the question, then this can be expressed as

$$ L_{n,m} = \mathrm{vec}(G_n)^\dagger\,\Lambda\,\mathrm{vec}(G_m), $$

which allows L to be derived in a single matrix equation, but that's getting a little off topic.

In the sample case I considered, $L$ is does not contain time varying terms, so it may be exponentiated to get a new matrix $F$, which is related to the solution of the master equation $\phi$

$$ F(t) = \exp(Lt). $$

$F$ can be used to get a Choi matrix $S$, which is exactly what I need. At this point, a basis needs to be chosen for the future Krauss operators. I'm quite happy with the Pauli operators so I'll stick with those for this next equation,

$$ S_{a,b} = \sum_{n,m}F_{m,n}\mathrm{Tr}[G_nG_aG_sG_b]. $$

Finally, the wonderful part.

$$ \rho_t=\phi_{n,m}(\rho_0,t) = S_{n,m}(t)G_n\rho_0 G_m^\dagger $$

As you can see, $S$ is a matrix of weights for a sum of superoperators in a useful basis that I can select. This has been referred to as the process matrix (arXiv)(DOI) which is unique to a process in a given basis. In the sample case, in which the master equation has no time dependent terms on the RHS, the solution can be directly verified by representing $\Lambda$ in matrix form and exponentiating it to get $\phi(t)=\exp(\Lambda t)$.

This works in the time independent case for quits and qutrits as expected. I need to check that this works in the case of time dependence.


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