Adjoint representation in Liouville-von Neumann equation

Question

I am having trouble understanding the adjoint representation of a Lie algebra in the scope of a very specific example, so I thought physics.SE would be the best place to ask.

Background: A $N \times N$ density matrix $\rho$ contains only $N^2 - 1$ non-redundant real quantities. Therefore, it is convenient to represent it as real vector $d$ (called coherence or pseudospin vector). One possible parameterization [1] of the density matrix is $$\rho = N^{-1}I + \frac{1}{2} \sum_{j=1}^{N^2-1} d_j s_j,$$ where $s_j$ are the generalized Pauli matrices and $I$ is the identity. This parameterization can be plugged into the Liouville-von Neumann equation $\partial_t \rho = \mathcal{L}(\rho)$ and after applying $\mathrm{Tr}\{ \cdot s_i\}$ on both sides, one receives a differential equation for the vector $d$: $\partial_t d = L d$, where $L$ is a real $(N^2-1)\times (N^2-1)$ matrix with the elements $L_{ij} = \mathrm{Tr}\{ \mathcal{L}(s_j) s_i \}$.

Q0: Can the adjoint representation represent elements of a vector space in a different vector space?
Basically, we transform the density matrix (Hermitian, $N \times N$) to a vector (real, $N^2-1$), which is an isomorphism between two vector spaces. However, if I am not mistaken, the adjoint representation considers only one vector space.

Q1: Is (and if yes, how) the matrix $L$ related to the adjoint representation of the $\mathfrak{su}(N)$ algebra?
As far as I understood, the $s_j$ (traceless and Hermitian) span the $\mathfrak{su}(N)$ algebra and can be used to compose both density matrix $\rho$ and Hamiltonian $H$, where the latter enters the Liouvillian $$\mathcal{L}(\rho) = -\mathrm{i}\hbar^{-1} [ H, \rho ].$$ Then, $L$ resembles $\mathrm{ad}_{H} (\rho)$, where the Lie bracket $-\mathrm{i} [\cdot, \cdot]$ is used (I think this bracket must be used with the Hermitian version of the generalized Pauli matrices).

Q2: Does the relation in Q1 change when considering a general Liouvillian?
Of course the commutator term will remain in the Liouvillian, but possibly a dissipation term $\mathcal{G}(\rho)$ (similar to the Lindblad master equation) will be added. The matrix $L$ can be derived for any Liouvillian, but can it still be called adjoint representation?

Q3: How does this all relate to the equation $$\mathrm{Ad}_{\exp(x)} = \exp(\mathrm{ad}_x)?$$
The solution for the vector ODE is $d(t) = \exp({Lt})d(0)$. If $L$ is the adjoint representation of the Liouvillian, then the solution for the original density matrix reads $\rho(t) = \exp(\mathcal{L}t) \rho(0) \exp(-\mathcal{L}t)$. This would make sense, but is it correct? And does it hold for general Liouvillians?

[1] Hioe, F. T., & Eberly, J. H. (1981). $N$-level coherence vector and higher conservation laws in quantum optics and quantum mechanics. Physical Review Letters, 47(12), 838.

Edit #1: Added Q0 for a more basic understanding.
Edit #2: I have asked a separate question on maths.SE regarding the change of basis https://math.stackexchange.com/questions/2682901/matrix-exponential-and-change-of-basis

Answer 1

It is unclear to me where your block is, but I suspect it is in the trivial routine translation to adjoint vectors from abstract Lie generators, here in the defining (fundamental) representation. Since the canonical paradigm for all physicists is the Pauli matrices and su(2), that's what suffices to illustrate here, before you complicate life for yourself with the trivial extension to general N.

So you wish to recognize how (15) relates to (1) for the 3 Pauli matrices. To spare you complication, let us just use adjoint 3-vectors $\vec d$ instead of $d_i$s, and recall the standard-normalization generators of su(2) in the fundamental are half the Pauli matrices. $$ \rho=\frac{1}{2} I+ \frac{1}{4} \vec{d}\cdot \vec{\sigma}. $$

Likewise, for the sake of argument, take the most general Hermitian Hamiltonian $$ H=\hbar \gamma \vec{B}\cdot \vec{\sigma}~/2, $$ where a piece proportional to the identity would not matter, and γ is real for Hermiticity.

The von Neumann equation then reduces to (1) of your reference, $$ i \hbar \partial_t \rho = [H,\rho] \qquad \Longrightarrow \qquad \partial_t \vec{d}= \gamma ~\vec{B}\times \vec{d} . $$

Your adjoint representation 3×3 matrix then, sending a 3-vector $\vec d$ to a velocity 3-vector by merely left-multiplication is $\mathbb{L}=\gamma \vec{B}\times$, which is to say $L_{ij}=\gamma \epsilon^{ikj} B^k$. This amounts to a familiar 3-space rotation around an axis parallel to $\vec B$, further amounting to you equation (15)--you should be able to generalize by inserting arbitrary su(N) structure constants in the commutator!

• Again, bifundamental 2×2 commutator relations were faithfully mapped to 3×3 left-multiplication relations on vectors. (You can see how to generalize to arbitrary N).

Indeed, the generic solution is thus your left-multiplication $$ \vec{d}(t)= \exp (t~\mathbb{L}) ~~\vec{d}(0). $$ The point of your reference is that all operations are now left-multiplication matrix operations on 3 vectors ($N^2-1$-vectors in the general case), instead of commutations, so, left and right multiplications on 2×2 (defining) matrices. In this language, the adjoint has reduced to just a routine representation whose generators are in the 3×3 space of the structure constants.

Your answer then is just $$ \rho(t) = \frac{1}{2} I+ \frac{1}{4} \vec{d}(t) \cdot \vec{\sigma}, $$ as you noted, and you need not have gone the full bifundamental Ad way, $$ \rho(t) = e^{-itH/\hbar} \rho(0)~ e^{it H/\hbar } $$ for the generic matrix solution of the von Neumann equation. (Differentiate both sides to see that).

• This expression can be re-written more abstractly as $$ \rho(t) = e^{\operatorname{ad}(-itH/\hbar)} \rho(0) = \operatorname{Ad}(\exp (-itH/\hbar)) ~\rho(0). $$

But for your very last in-text ($\cal L$) formula, your expressions appear sound.

As you transcribe your expressions, just remind yourself whether you are operating on fundamental rep matrices (via commutation) or else, equivalently, adjoint vectors (via left multiplication).

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• Note added in response to comments. Well, that's why I gave you the su(2) example. Instead of looking at commutators, you get rid of them as described, and look at mere rotations of 3-vectors. The set of three (one for each component of B) 3×3 matrices $\mathbb{L}$ are the generators of su(2) in the adjoint representation. So, the full rotations are the exponentials of linear combinations of these matrices, merely left-multiplying 3-vectors.

That's all you need to know about them, and you may virtually forget about commutators at this level, so, yes, even with a dissipation correction in the Liouvillian you produce the suitable $\mathbb{L}$, plug it into the answer, etc... That's the point: that you may forget about the $\cal L$ 2×2 matrices and stick to the $\mathbb{L}$ 3×3 matrices, where you do not (directly) evaluate commutators... (Any irreducible 3-d rep of su(2) is the adjoint, so isomorphic to this: you do not need to know much about commutators here.) It might be easier if you simply tried an explicit example.

The generalization to su(N) is straightforward. Again, you transition from N×N matrices and commutators to $(N^2-1)\times (N^2-1)$ matrices left-multiplying $(N^2-1)$-vectors, and forget about the adjoint representation being special.

Possibly looking at the Pauli matrix article linked, you could remind yourself how infinitesimal rotations in the bifundamental amount to left-rotations in the adjoint; it is sophomore angular momentum. There, you learned about the Pauli vector map of 3-vectors to 2x2 hermitean traceless matrices: they are both the triplet (adjoint) rep of su(2) in alternate realizations.

Answer 2

This answer provides a summary of the other answers/comments above for future reference. See the accepted answer for more detail.

Note that there is still work in progress and my conclusions may be wrong

Answer Q0:
No. The adjoint representation does not depend on the underlying vector space.

Answer Q1:
Indeed they are related. However, $\mathrm{ad}_{\hbar^{-1}H}$ has to be written in $3\times 3$ form in order to be equal to $L$, i.e., $L_{ij} = \mathrm{Tr}\{\mathrm{ad}_{\hbar^{-1}H}(s_j)s_i\}.$

Answer Q2:
The relation remains the same assuming the general Liouvillian still contains the commutator term. Of course, the extra term $\mathcal{G}$ has to be expressed in $3\times 3$ form as well.

Answer Q3:
TODO