I am trying to transform the ODE $\partial_t \rho = \mathcal{L}(\rho)$ for a $N\times N$ matrix $\rho$ $$\rho = \frac{1}{N}I_N + \sum_{j} d_j s_j$$ to a more suitable basis $s_j$ (the generalized Pauli matrices) using the Hilbert-Schmidt product. The transformed ODE reads $\partial_t \vec d = M \vec d + \vec d_\infty$, where $\vec d$ is a vector now and the matrix $M$ has the entries $$ M_{ij} = \mathrm{Tr}\{\mathcal{L}(s_j) s_i\}.$$ Ignoring the vector $\vec d_\infty$, the solution of this ODE is $\vec d(t) = \exp (M t) \vec d(0)$. On the other hand, the solution of the original ODE would be $\rho (t) = \exp (\mathcal{L} t) \rho (0)$, which can be transformed to $$d_i(t) = \mathrm{Tr}\{ \exp (\mathcal{L} t) \rho (0)s_i\}.$$ As a crosscheck, I would like to confirm that both solutions are equivalent, i.e., $$\mathrm{Tr}\{ \exp (\mathcal{L} t) \rho (0)s_i\} = \sum_j \exp (Mt)_{ij} \mathrm{Tr}\{ \rho (0) s_j\}.$$

Any ideas how to accomplish this?

For details about the application in particular see my question in physics.SE: https://physics.stackexchange.com/questions/390458/adjoint-representation-in-liouville-von-neumann-equation


1 Answer 1


I found that an intermediate step via the Liouville space was the key to understand the problem. The matrix $\rho$ is stacked in column-major ordering into a vector $\vec r$. Then, the superoperator $\mathcal{L}(\rho)$ can be written as $L\vec r$, where $L$ is a $N^2 \times N^2$ matrix. Subsequently, the ODE can be solved $$\vec r(t) = \exp\left(Lt\right) \vec r(0).$$

Then, the trace operations can be replaced with matrix-vector products so that $\vec d = S \vec r$ and $$\vec r = \vec r_0 + \frac{1}{2}S^\dagger \vec d,$$ where $S$ is a $(N^2 - 1) \times N^2$ matrix containing the basis matrices, $\dagger$ denotes the conjugate transpose, and $\vec r_0$ is the term $N^{-1}I_N$ stacked into a vector. Consequently, the matrix $$M = \frac{1}{2}SLS^\dagger.$$

Now, the solutions $$S\exp(Lt)\vec r(0)=\exp(Mt)S\vec r(0)$$ must be equal. Using $\frac{1}{2}S S^\dagger = I_{N^2-1}$, we can compare the expressions $$\frac{1}{2}S\exp\left(Lt\right)S^\dagger = \exp\left(\frac{1}{2}S L S^\dagger t\right)$$ and check whether they are equal with the Taylor expansion (ignoring the scalar $t$) $$ \frac{1}{2}S \left( I_{N^2} + L + \frac{1}{2} L^2 + \frac{1}{6} L^3 + \dots\right) S^\dagger = I_{N^2 - 1} + \frac{1}{2}S L S^\dagger + \frac{1}{2}\left(\frac{1}{2}S L S^\dagger\right)^2 + \frac{1}{6} \left( \frac{1}{2}S L S^\dagger\right)^3 + \dots$$

The equality of the first two terms is trivial to proof. Generally, the condition $$\frac{1}{2 i!} S L^i S^\dagger = \frac{1}{i!}\left(\frac{1}{2}S L S^\dagger\right)^i = \frac{1}{i!}\left(\frac{1}{2}S L S^\dagger\frac{1}{2}S L S^\dagger\frac{1}{2}S L S^\dagger\dots\right)$$ must hold. Now it becomes apparent, that if $$L = \frac{1}{2}S^\dagger S L,$$ then the two solutions are equal. Indeed, after left-multiplication with $S$ the expressions on both sides equal $SL$.

It would seem that the number of equation is reduced and the system is underdetermined, however, due to the nature of the density matrix $\rho$ the original equation system contains a redundant equation. In summary, there is no general relation but for this special case the crosscheck succeeds.


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