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Let $H_0$ be diagonal and $H_1={\sum }_{j=1}^{N-1}{d}_j(E_{j,j+1}+E_{j+1,j})$, with every $d_j\ne 0$ and the adjacent absolute energy gaps distinct and nonzero. The squared drift adjoint action has distinct eigenvalues on these coupling components, so Lagrange interpolation isolates every adjacent symmetric coupling. One more commutator gives its antisymmetric partner. Their mutual brackets produce adjacent traceless diagonals, while brackets along the connected chain give all longer-range matrix units. Thus the special unitary Lie algebra is generated. Any nonzero-trace generator also supplies the central direction, giving all of $\mathfrak{u}(N)$. This is a sufficient condition in the finite-dimensional freely timed real-control model; connectivity alone does not remove frequency degeneracies.

In the complexification of a Lie algebra, defineThe relations in a becomeThe special unitary Lie algebra $\mathfrak{su}(2)$ has real basis $S_i=-i{\sigma }_i/2$, where the Pauli matrices satisfy $[{\sigma }_i,{\sigma }_j]=2i{ϵ}_{ijk}{\sigma }_k$. Its commutators are the same: $[S_i,S_j]={ϵ}_{ijk}{S}_k$. The complex-linear map $L_i\mapsto S_i$ therefore preserves the bracket and maps a basis to a basis. ConsequentlyThe real Lie algebras are not isomorphic: multiplying the boosts by $i$ changes their real form. For example, the Adjoint representation of a Lie algebra evaluated at $K_1$ has real eigenvalues $0,1,-1$, whereas the corresponding nonzero compact SU(2) generators have purely imaginary adjoint eigenvalues.

Every complex Lie algebra representation of the real algebra extends uniquely by complex linearity to its complexification of a Lie algebra. The finite-dimensional complex irreducible representations are therefore classified by a nonnegative integer highest weight $m$, just as for SU(2). There is one equivalence class for eachThese are statements about Lie algebra representations. They do not assert positive-definite unitarity for the noncompact real form, or automatic descent to the full Lorentz group. In particular, half-integer $j$ gives a sign under a $2\pi$ spatial rotation, so its integration needs a covering group.

The Von Neumann equation integrates to $\rho (t)=U(t)\rho (0)U(t{)}^{†}$, so the spectrum and purity of a density operator are invariant under any allowed open-loop control. These are the kinematic restrictions.

There is no additional orbit restriction here: the generators $i{\sigma }_z$ and $i{\sigma }_x$, together with their matrix commutator, span the special unitary Lie algebra:For freely chosen durations and piecewise continuous real control values, this gives density operator controllability. The connected reachable group is $SU(2)$; its action by conjugation is the same as that of $U(2)$, because a global phase cancels. Thus exactly the pairs ${\rho }_0\leftrightarrow {\rho }_2$ and ${\rho }_1\leftrightarrow {\rho }_3$ can be interconverted. No coherent control can turn either mixed state into either pure state, even though every orientation within each spectral orbit is reachable.

Assume piecewise-constant real controls with freely chosen duration, as in the usual finite-dimensional quantum controllability theorem. We can establish the strongest notion here: the Dynamical Lie algebra is $\mathfrak{u}(3)$, so the system has unitary operator controllability, and consequently both pure-state and fixed-spectrum density operator controllability.

To verify the hypotheses explicitly, use matrix units $E_{jk}=|j⟩⟨k|$ and put $K_0=-i{H}_0$, $X_{jk}=-i(E_{jk}+E_{kj})$, $Y_{jk}=E_{jk}-E_{kj}$. Then $K_1=-i{H}_1=d_1{X}_{12}+d_2{X}_{23}$ andSince $d_1{d}_2\ne 0$ and ${\omega }_1^2\ne {\omega }_2^2$, linear combinations of these two elements isolate $X_{12}$ and $X_{23}$. Their commutators with $K_0$ give ${\omega }_1{Y}_{12}$ and $-{\omega }_2{Y}_{23}$, respectively. Further commutators give two independent traceless diagonal generators, since $[X_{jk},Y_{jk}]=2i(E_{jj}-E_{kk})$, as well as the two $1$–$3$ generators. These eight generators span the special unitary Lie algebra $\mathfrak{su}(3)$.

Finally $\mathrm{Tr}{K}_0=i({\omega }_1+{\omega }_2)\ne 0$. Subtract its traceless component, already in the generated algebra, to obtain a nonzero multiple of $iI$. This proves the Lie algebra of the unitary group is all of $\mathfrak{u}(3)$, not merely $\mathfrak{su}(3)$. The compact-group controllability theorem then gives access to every unitary with a suitable control and some duration. This does not assert controllability at every predetermined short time or under additional bandwidth constraints. This proves that two adjacent transitions generate the qutrit unitary Lie algebra.

Neither monotonic decrease alone nor equality of the initial spectra guarantees convergence. A nonnegative Lyapunov function can approach a positive limiting value, and the set where $\dot{V}=0$ can contain trajectories other than the target. Different initial spectra already obstruct exact tracking: unitary time evolution preserves the eigenvalues of each density operator, while two matrices approaching one another in norm must have the same limiting eigenvalues.

Even the spectral obstruction can be absent while the Lyapunov quantum control stalls. Take a qubit with $H_0=Z/2$, $H_1=X$, ${\rho }_d(0)=|0⟩⟨0|$ and $\rho (0)=|1⟩⟨1|$. Both density operators commute with $H_0$, and the commutator of $X$ with $\rho$ has zero diagonal, so $f=\mathrm{Tr}({\rho }_d[-iX,\rho ])=0$. Both states remain constant. They have the identical spectrum $(1,0)$, but $\boxed{\Vert \rho (t)-{\rho }_d(t){\Vert }_{\mathrm{HS}}=\sqrt{2}\text{ for all }t}$. Moreover $iZ,iX$ generate the full special unitary Lie algebra $\mathfrak{su}(2)$, so the failure here is a limitation of this particular feedback law, not a lack of available state-transfer controls.

Assume the usual real Hamiltonian controls with freely chosen finite duration, rather than a fixed-time or amplitude-restricted task. Define the Dynamical Lie algebrathe real linear span closed under all nested commutators. Replacing each generator by $-i{H}_m$ gives the same algebra. The associated connected reachable group acts on propagators, on each unitary orbit of a density operator, and on pure states. For $N\ge 2$, the necessary and sufficient conditions are as follows.

For exact unitary operator controllability, $\boxed{\mathfrak{g}=\mathfrak{u}(N)}$. For operator control up to global phase, the possibilities are $\mathfrak{su}(N)$ or $\mathfrak{u}(N)$. If every Hamiltonian is traceless, $\det U(T)=\exp [-i{\int }_0^T\mathrm{Tr}H(t)dt]=1$, so the central phase direction is absent and exact access is to $SU(N)$. The full special unitary Lie algebra has dimension $N^2-1$ and the Lie algebra of the unitary group has dimension $N^2$.

For density operator controllability on every spectral orbit,Scalar phases do not affect conjugation. If only a particular density spectrum is being considered, a smaller group can suffice: for distinct eigenvalue multiplicities $n_j$ and centralizer ${\mathfrak{c}}_{\rho }=\{X\in \mathfrak{u}(N):[X,\rho ]=0\}$, the orbit criterion is $\dim \mathfrak{g}-\dim (\mathfrak{g}\cap {\mathfrak{c}}_{\rho })=N^2-{\sum }_j{n}_j^2$. A maximally mixed state has a one-point orbit and needs no controls. This restricted-orbit condition is weaker than controllability for every density spectrum.

For pure-state controllability, up to a fixed unitary change of basis,where the last two alternatives occur only for even $N$. Here $\mathfrak{sp}(n)$ is the standard compact symplectic Lie algebra acting on ${\mathbb{C}}^{2n}$, not the real noncompact symplectic algebra. These are precisely the unitary matrix algebras whose connected groups act transitively on normalized complex vectors, and also give the ray version of controllability. For $N=2$, $\mathfrak{sp}(1)=\mathfrak{su}(2)$. For even $N\ge 4$, the symplectic cases show why pure-state control can be weaker than density control: $\dim \mathfrak{sp}(n)=n(2n+1)$ is smaller than the nondegenerate density-orbit dimension $N^2-N$, and its scalar center acts trivially on that orbit. For $N=1$, the physical density and pure-state spaces are single points, while exact operator controllability still requires the phase algebra $\mathfrak{u}(1)$.

Use a real Lie algebra convention $[T_a,T_b]=f_{ab}{}^c{T}_c$. Its Adjoint representation of a Lie algebra acts on $\mathfrak{g}$ itself:Linearity is immediate. The Jacobi identity givesso $[{\mathrm{ad}}_X,{\mathrm{ad}}_Y]={\mathrm{ad}}_{[X,Y]}$, the defining Lie algebra representation condition. In the basis $T_b$, its matrix entries areThus the adjoint generators are the structure constants arranged as matrices. The representation has kernel equal to the center of a Lie algebra; it need not be faithful for an arbitrary algebra.

The Adjoint representation of a Lie group is ${\mathrm{Ad}}_g(Y)=gY{g}^{-1}$. It satisfies ${\mathrm{Ad}}_{gh}={\mathrm{Ad}}_g{\mathrm{Ad}}_h$, preserves the identity, and sends inverses to inverse matrices. Differentiating $e^{tX}Y{e}^{-tX}$ gives $[X,e^{tX}Y{e}^{-tX}]$, soConsequently the positive adjoint exponentials furnish the representation on elements $g=e^X$ and their products, and the intrinsic conjugation action defines it globally.

The inverse-conjugation formula in the PDF needs a minus adjoint exponent with this standard definition. Its first-order term is $Y-[X,Y]$, whereas $e^{+{\mathrm{ad}}_X}Y$ has first-order term $Y+[X,Y]$. For example $X=T_1,Y=T_2$ in a nonabelian algebra with $[T_1,T_2]\ne 0$ already distinguishes the two. The inverse conjugation and adjoint antirepresentations identity explains the group-order issue as well: $F(g)={\mathrm{Ad}}_{g^{-1}}$ satisfies $F(gh)=F(h)F(g)$. Retaining the printed inverse conjugation as a left action without this order reversal would not give an ordinary group representation.

The Killing form is the symmetric bilinear formTo prove degeneracy for a non-semisimple Lie algebra, take its nonzero solvable radical and the last nonzero member $\mathfrak{a}$ of its derived series of a Lie algebra. This is a nonzero abelian ideal of a Lie algebra. For $x\in \mathfrak{a}$, ${\mathrm{ad}}_x$ maps $\mathfrak{g}$ into $\mathfrak{a}$ and kills $\mathfrak{a}$. Every ${\mathrm{ad}}_y$ preserves $\mathfrak{a}$. Hence ${\mathrm{ad}}_x{\mathrm{ad}}_y$ maps the full space into $\mathfrak{a}$ and has zero restriction there, so its trace is zero. Thus $B(x,y)=0$ for every $y$. This is the abelian ideals lie in the radical of the Killing form argument, and establishes a nonzero kernel and vanishing determinant.

For a compact real semisimple Lie algebra, the adjoint action is unitary in an invariant positive inner product. Its infinitesimal generators are skew-Hermitian, givingStrictness follows because the adjoint kernel is the zero center. This explains the compactness criterion from the Killing form: “strictly negative” means negative-definite, not that every entry of its matrix is negative.

In the first three-generator example, take columns to be images of the basis $(X,Y,H)$. Direct use of the brackets givesTaking traces of products yields the Killing form for cyclic three-generator bracketsIt is nondegenerate, hence semisimple by the preceding degeneracy result, but it has mixed signature and is not compact. An explicit realization iswhich span the real traceless two-by-two matrices and satisfy exactly these brackets.

Changing the sign of $[X,H]$ changes the first and third adjoint matrices towhile ${\mathrm{ad}}_Y$ is unchanged. Nowthe negative-definite compact case. The basis $X=-i{\sigma }_1$, $Y=-i{\sigma }_2$, $H=-i{\sigma }_3$ realizes it as the real special unitary Lie algebra $\mathfrak{su}(2)$.

The complexification of a Lie algebra $L_0$ is $L=L_0{\otimes }_{\mathbb{R}}\mathbb{C}$, with the Lie bracket extended complex-bilinearly. Equivalently write $L=L_0\oplus i{L}_0$, whereComplex conjugation is an antilinear map and a Lie algebra automorphism whose fixed subalgebra is $L_0$.

If the solvable radical of $L_0$ is nonzero, its complexification is a nonzero solvable ideal of $L$. Conversely the solvable radical $R$ of $L$ is preserved by complex conjugation, because it is the unique largest solvable ideal. Consequentlyfor $z\in R$, both $(z+\overline{z})/2$ and $(z-\overline{z})/(2i)$ belong to $R_0$. If $R\ne 0$, $R_0\ne 0$ is a solvable ideal of $L_0$. This proves $L_0$ is semisimple if and only if $L_0{\otimes }_{\mathbb{R}}\mathbb{C}$ is semisimple. Consistently, the complex Killing form is just the complex-bilinear extension of the real one; its determinant in a real basis is unchanged by extending scalars.

Take the sl2R Lie algebra ${\mathfrak{sl}}_2(\mathbb{R})$ and the special unitary Lie algebra $\mathfrak{su}(2)$. Both complexify to ${\mathfrak{sl}}_2(\mathbb{C})$. This is immediate for the former; for the latter, the real basis $ih,e-f,i(e+f)$ consists of traceless matrices that are skew-Hermitian matrices and is also a complex basis of ${\mathfrak{sl}}_2(\mathbb{C})$.

They are not isomorphic as real Lie algebras. Their Killing forms are $B(X,Y)=4\mathrm{tr}(XY)$, but on $\mathfrak{su}(2)$ this is negative definite. On ${\mathfrak{sl}}_2(\mathbb{R})$ the basis $h,e+f,e-f$ has a diagonal Gram matrix with entries $8,8,-8$, so the signature of a quadratic form is $(2,1)$. A Lie algebra isomorphism preserves the Killing form and therefore its signature.

A real split semisimple Lie algebra has a Cartan subalgebra whose Adjoint representation of a Lie algebra is simultaneously diagonalizable over $\mathbb{R}$, so its root-space decomposition is defined over $\mathbb{R}$. The example ${\mathfrak{sl}}_2(\mathbb{R})$ is split: the diagonal Cartan subalgebra $\mathbb{R}h$ has eigenvalues $0,2,-2$ and real root spaces $\mathbb{R}e,\mathbb{R}f$. The example $\mathfrak{su}(2)$ is not split. Invariance makes every $\mathrm{ad}x$ skew-adjoint for the positive definite inner product $-B$, so its eigenvalues are purely imaginary. If it were diagonalizable over $\mathbb{R}$, all these eigenvalues would be zero and $\mathrm{ad}x=0$. The center is zero, so only $x=0$ has this property; no nonzero split Cartan subalgebra exists. Thus ${\mathfrak{sl}}_2(\mathbb{R})$ is split and $\mathfrak{su}(2)$ is not.

For a Lie bracket over $\mathbb{R}$ or $\mathbb{C}$, antisymmetry of a Lie bracket means $[x,y]=-[y,x]$. In particular $[x,x]=0$ in these characteristic-zero fields. The Jacobi identity isIt expresses compatibility of the bracket with its own adjoint action. Bilinearity must hold over the chosen base field, and the bracket must take its values in the same vector space.

For the matrix commutator, bilinearity and antisymmetry follow directly from distributivity. Associativity of matrix multiplication givesadding the two cyclic permutations cancels every monomial. Thus the Jacobi identity holds in the entire matrix algebra. For a specified linear subspace, the only additional bracket condition is closure: $AB-BA$ must belong to the subspace whenever $A,B$ do. It is unnecessary to require closure under the separate products $AB$ and $BA$.

To determine the special unitary Lie algebra, let $g(t)$ be a differentiable curve in the special unitary group with $g(0)=I$ and $A=g^{\prime }(0)$. Differentiating $g(t{)}^{†}g(t)=I$ gives $A^{†}+A=0$. Differentiating $\det g(t)=1$ at the identity gives $\mathrm{tr}A=0$. Conversely a traceless skew-Hermitian matrix has $e^{tA}$ unitary and $\det e^{tA}=e^{t\mathrm{tr}A}=1$, so it really is a tangent vector. ThereforeThis is a real Lie algebra of complex matrices: multiplication by $i$ generally leaves this real subspace. Its complexification is ${\mathfrak{sl}}_n(\mathbb{C})$, not the compact algebra itself. For $A,B\in \mathfrak{su}(n)$,Thus closure holds, and the already verified commutator identities establish all the Lie algebra axioms.

The cross-product Lie algebra on ${\mathbb{R}}^3$ is bilinear, antisymmetric and closed because the cross product has those properties. Its Jacobi identity follows from the vector triple-product identity:In the cyclic sum, the coefficients cancel by symmetry of the scalar product.

For the explicit relation to the SU(2) Lie algebra, take the three Pauli matrices and defineThese matrices are traceless and Skew-Hermitian, and the three images of the standard basis form a real basis of $\mathfrak{su}(2)$. Using the Pauli matrix commutator identity,Hence $F$ is a real Lie algebra isomorphism. The factor and sign $-i/2$ are essential for preserving the unscaled cross-product bracket. At the group level there is an Adjoint double cover from SU(2) to SO(3); the isomorphism of their tangent algebras does not identify the two global groups.

A semisimple Lie algebra over a field of characteristic zero is split if it has a Cartan subalgebra whose adjoint action is simultaneously diagonalizable over that field. Its root-space decomposition then needs no extension of scalars. The sl2R Lie algebra is split, with the real diagonal Cartan subalgebra. The real special unitary Lie algebra $\mathfrak{su}(2)$ is not: its adjoint operators are skew-adjoint for the positive definite inner product obtained by negating its Killing form. Thus a split adjoint operator would have to be zero.

Let $H_0=\mathrm{diag}(-{\omega }_1,0,-{\omega }_2)$ and $H_1=d_1(E_{12}+E_{21})+d_2(E_{23}+E_{32})$, with nonzero $d_j$, distinct positive ${\omega }_j$. The squared adjoint action of $-i{H}_0$ has different eigenvalues $-{\omega }_1^2,-{\omega }_2^2$ on the two adjacent skew-Hermitian couplings. Linear combinations of $-i{H}_1$ and its double commutator with $-i{H}_0$ isolate both couplings. Further commutators give their real antisymmetric partners, two traceless diagonal generators, and both $1$–$3$ generators, hence the special unitary Lie algebra $\mathfrak{su}(3)$. The drift has nonzero trace; subtracting its traceless part supplies $iI$, giving the Lie algebra of the unitary group $\mathfrak{u}(3)$ and unitary operator controllability under the usual freely timed real-control assumptions.