# Adjoint representation in Liouville-von Neumann equation

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

• The parameterization is derived for su(N) in Hioe and Eberly, 1981. For N=2 the $s_j$ are the Pauli-Matrices, for N=3 the Gell-Mann-Matrices are used. A general set of generators for N is given in the reference as well. However, if it serves simplicity, we can restrict ourselves to the N=2 case. – carlosvalderrama Mar 6 '18 at 11:23
• Just to clarify, my remarks in the questions have not been checked at all. They are rather claims and guesses, and unfortunately I am unable to (dis)prove them. – carlosvalderrama Mar 6 '18 at 11:29

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).

• 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.

• First of all, thanks a lot for your answer! The problem is that I have not understood yet the abstract concepts involved. For example, I am not sure what the adjoint representation actually is. Is it a matrix? Which dimensions are involved? I have been looking for an easy example that does not invoke additional abstract concepts (automorphisms, endomorphisms, homomorphisms, tangent space at the origin, etc.) without any luck and hope that somebody could provide me with a practical explanation based on the problem above. – carlosvalderrama Mar 6 '18 at 18:47
• Your answer is definitely helpful to complete the puzzle. However, would you be so kind and comment on the case where the Liouvillian $\mathcal{L}(\rho) = -\mathrm{i}\hbar^{-1}[ H, \rho ] + \mathcal{G}(\rho)$ does not only consist of the commutator but also includes an additional dissipation term? How does the adjoint representation look like if there is no commutator? – carlosvalderrama Mar 6 '18 at 18:55
• Thanks for your update. What I still fail to understand is how the transformation from matrix elements to vector elements can be justified, since Lie groups, algebras, and their representations seem to consider only one underlying vector space (see Q0, which I added to the question). Is the adjoint representation responsible for this transformation from $N \times N$ Hermitian matrices to $N^2-1$ real vectors, or is there another concept involved? – carlosvalderrama Mar 8 '18 at 12:32
• No, no further concepts: the hermitean matrix or real vector bases are two equivalent bases of the same, adjoint, representation. Normally, in SU(N), the dimensionality completely specifies the irrep if it is real, as here. You are probably reading too much of depth and consequence here... It is covered in the leading chapters of any decent book on Lie Groups in physics.... – Cosmas Zachos Mar 8 '18 at 13:47
• So the adjoint representation does not define which basis is used, and e.g. by using $\mathrm{Tr}\{\cdot s_j\}$ we perform a change of basis. I think this was my major block. – carlosvalderrama Mar 8 '18 at 19:05

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

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., $L_{ij} = \mathrm{Tr}\{\mathrm{ad}_{\hbar^{-1}H}(s_j)s_i\}.$
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.