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For a complex antisymmetric tensor $A\in {\wedge }^2{\mathbb{C}}^4$, the SU(4) action is $A\mapsto UA{U}^T$. The unitary skew-diagonalization of an antisymmetric matrix gives $A=\mathrm{diag}(aϵ,bϵ)$ in suitable coordinates, where $ϵ=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$. If $|a|\ne |b|$, preserving $A{A}^{†}$ forces a stabilizing unitary matrix to preserve the two coordinate planes. The identity $VϵV^T=(\det V)ϵ$ then forces both two-by-two blocks to have determinant one, using $\det U=1$ also when one of $a,b$ vanishes. The exact stabilizer is $SU(2)\times SU(2)$. If $|a|=|b|\ne 0$, the stabilizer is the larger compact symplectic group $USp(4)$. If $a=b=0$, it is all of $SU(4)$. The real structure of the SU4 exterior square imposes $b=\bar{a}$ in this form, so a nonzero real six-dimensional field has the larger symplectic stabilizer, whereas unequal paired singular values are generic for a complex six-dimensional field.

The Lie algebra of the compact symplectic group consists of $X^{†}=-X$ and $XJ+J{X}^T=0$. Equivalently $X=\left(\begin{array}{cc}A& B\\ -\overline{B}& \overline{A}\end{array}\right)$ with $A^{†}=-A$ and $B^T=B$. Its real dimension is $n(2n+1)$, and its complexification is the symplectic Lie algebra.

The fundamental case. Set $⟨\varphi ⟩=v{e}_4$, where $v\ne 0$. Fixing this vector forces the fourth column of a unitary matrix to be $e_4$, and hence also fixes its fourth row. The remaining three-by-three block has determinant one. ThereforeEvery nonzero complex vector can be brought to this form by an SU(4) transformation: complete its normalized direction to a unitary basis and adjust the phase of a complementary basis vector to obtain determinant one. Thus a generic nonzero fundamental vacuum expectation value has a conjugate of $H_3$ as its exact unbroken subgroup. The particular embedded $H_3$ of the question is fixed only when the vacuum expectation value lies along $e_4$.

The antisymmetric case. With $ϵ=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, takeSince $VϵV^T=ϵ$ for $V\in SU(2)$, this antisymmetric tensor is fixed by $H_2$ for every $a,b$. Its transformation is $A\mapsto UA{U}^T$, so any stabilizing $U$ also preservesIf $|a|\ne |b|$, its two eigenspaces force $U=\mathrm{diag}(V,\tilde{V})$ with unitary two-by-two blocks. For each nonzero block, $VϵV^T=(\det V)ϵ$ forces the block determinant to be one. If one block is zero, the SU(4) determinant condition forces the other block determinant to be one as well. HenceThe unitary skew-diagonalization of an antisymmetric matrix shows that a generic complex $6$ has two distinct paired singular values; choosing an SU(4) rather than a $U(4)$ basis merely leaves an allowed overall phase in $a,b$. Thus its generic stabilizer is conjugate to $H_2$.

If instead $|a|=|b|\ne 0$, the form is proportional, up to unitary coordinates and phases, to a nondegenerate standard symplectic form. Its stabilizer is the compact symplectic group $USp(4)$, of dimension ten, containing $H_2$ properly. The explicit choice $a=b=1$ is a concrete counterexample to inferring an exact $H_2$ stabilizer from the existence of two singlets alone. For a real $6$, the real structure of the SU4 exterior square imposes $b=\bar{a}$ in this form, so these equal magnitudes are compulsory. Every nonzero real $6$ therefore has the larger compact symplectic group stabilizer. The zero vacuum expectation value preserves all of $SU(4)$.

The adjoint cases. The $H_2$-fixed and $H_3$-fixed traceless matrices have respectively the formsFor a Hermitian adjoint scalar field, choose $a$ real. More generally a complex $a$ gives the same stabilizers. The repeated eigenvalues implyHere $S$ means that the product of the block determinants is one. These groups contain respectively $H_2$ and $H_3$, together with an additional continuous $U(1)$. Explicit extra transformations are $\mathrm{diag}(e^{it}{I}_2,e^{-it}{I}_2)$ and $\mathrm{diag}(e^{it}{I}_3,e^{-3it})$. They commute with $X_2$ and $X_3$ for every $t$, proving that neither exact endpoint $H_2$ or $H_3$ occurs for a single adjoint vacuum expectation value.

Nor do these larger subgroups survive for a generic Hermitian traceless matrix: four distinct eigenvalues leave only the maximal torus $U(1{)}^3$. For a generic complex adjoint matrix, there need not even be a continuous stabilizer; generically its unitary centralizer is just the centre of a group ${\mathbb{Z}}_4$. Thus the displayed degenerate adjoint directions are special. All these statements concern the stabilizer of a possible vacuum expectation value; no unspecified scalar potential can be assumed to select a particular direction or a particular reality condition.

The stated relation places every generator, and therefore the whole Dynamical Lie algebra, inside the standard compact symplectic Lie algebra $\mathfrak{sp}(\ell )\subset \mathfrak{su}(2\ell )$. Indeed the relation is closed under linear combinations and commutators, and differentiating evolution givesThus it supplies a symplectic dynamical symmetry in quantum control, rather than a claim that $J$ commutes with every Quantum Hamiltonian.

For $\ell \ge 2$, b1 is false: even the full compact symplectic group does not act transitively on every mixed-state spectral orbit. Its algebra has dimension $\ell (2\ell +1)<(2\ell {)}^2-1$. An explicit obstruction is the symplectic invariant of a density operatorwhich is unchanged by a symplectic conjugation. For $N=4$ and the printed $J$, let $a,b>0$, $a\ne b$, $2a+2b=1$. The isospectral operators $\mathrm{diag}(a,a,b,b)$ and $\mathrm{diag}(a,b,a,b)$ have invariants $4ab$ and $2(a^2+b^2)$, respectively, and cannot be interconverted. For the exceptional case $\ell =1$, $\mathfrak{sp}(1)=\mathfrak{su}(2)$, so b1 cannot be decided from inclusion alone; equality gives density-state controllability.

b2 is not implied by the displayed relation: more information is needed. It permits the full symplectic algebra but also a single commuting generator, or even zero generators. The full $Sp(\ell )$ is pure-state controllable: identify ${\mathbb{C}}^{2\ell }$ with ${\mathbb{H}}^{\ell }$, extend any unit vector to a quaternionic orthonormal basis, and map one such basis to another. A one-axis diagonal subgroup, in contrast, cannot change level populations.

The required extra information is the generated algebra, obtained by computing its real span of iterated commutators. Within this specified symplectic inclusion, equality $\mathfrak{g}=\mathfrak{sp}(\ell )$ gives pure-state controllability; its dimension is $\ell (2\ell +1)$. The mere inclusion does not establish equality. For $N=2$, equality also answers b1 affirmatively. None of these strictly symplectic generators supplies arbitrary global phase.

Set $X_j=i{H}_j$. The symplectic Lie condition is $X_j^{T}J+J{X}_j=0$, equivalently $H_j^{T}J+J{H}_j=0$. For the supplied diagonal $H_0$, the nonzero entries of $J$ pair levels $1,4$ and $2,3$; the corresponding energy sums are $-\alpha +\alpha =0$ and $-\beta +\beta =0$. Thus $H_{0}J+J{H}_0=0$ for all real $\alpha ,\beta$.

For the supplied real symmetric control matrix, direct multiplication givesBoth generators therefore satisfy the same alternating-form condition. That condition is closed under real linear combinations, and for two such generatorsas follows by substituting $X^{T}J=-JX$ and $Y^{T}J=-JY$ into $(XY{)}^{T}J-(YX{)}^{T}J$. Hence the whole Dynamical Lie algebra obeys it.

If $\dot{U}=X(t)U$ and $U(0)=I$, differentiating givesThus every reachable propagator satisfies $U^{T}JU=J$. Also $J^T=-J$, $J^{†}J=I$, so this is a symplectic dynamical symmetry in quantum control: the generated group lies in the corresponding compact symplectic group, unitarily equivalent to $Sp(2)$. It is not necessary to claim equality with $Sp(2)$, and special choices of $\alpha ,\beta$ can give a smaller group.

Let $P_0,P_1$ be any two rank-one orthogonal projections. Choose $0<q<1/N$ and $p=1-(N-1)q$, so $p>q$, and form the mixed density operatorsThey have the same spectrum. If density operator controllability holds, a reachable $U$ obeys $U{\rho }_0{U}^{†}={\rho }_1$, and subtraction of $qI$ gives $U{P}_0{U}^{†}=P_1$. Thus density operator controllability implies pure-state controllability, even if its definition is phrased using only mixed density operators.

The converse is false in general. For $N=4$, the full compact symplectic group $Sp(2)$ is transitive on the complex unit sphere: identify ${\mathbb{C}}^4$ with ${\mathbb{H}}^2$, extend any quaternionic unit vector to an orthonormal quaternionic basis, and map one such basis to another. This gives pure-state controllability. Yet the invariant in part (d) prevents it from connecting the indicated isospectral mixed density operators, so it fails density operator controllability. This counterexample uses the full symplectic group, not an assertion that every parameter choice in part (c) generates it.

For $N=2$, the converse is true. Every density operator has the form $\rho =qI+(p-q)P$ for a rank-one $P$, with $p+q=1$. If $p\ne q$, pure-state control of $P$ controls its entire fixed-spectrum unitary orbit of a density operator. If $p=q=1/2$, the orbit contains only $I/2$. Therefore the two controllability notions coincide for a qubit. The Lie classification gives the same conclusion, since $Sp(1)=SU(2)$.

A Lie group is a finite-dimensional smooth manifold with a group structure for which multiplication and inversion are smooth. The group defined here is the compact symplectic group $Sp(n)=U(2n)\cap Sp(2n,\mathbb{C})$, rather than the full complex symplectic group. Also, the displayed matrix expression requires $B$ to be $2n\times 2n$; the printed $n\times n$ size is incompatible with $J$.

Since $J^2=-I$, we have $J^{-1}=-J$. The matrix exponential commutes with conjugation, as follows term by term from its absolutely convergent power series. ConsequentlyTo construct logarithm charts for the compact symplectic group, consider the real vector spaceDifferentiating the defining identities at $I$ gives precisely these conditions. Conversely if $X\in \mathfrak{k}$, $e^X$ is unitary andso $e^X\in Sp(n)$. These are the infinitesimal conditions of the compact symplectic Lie algebra.

Near $I$, the convergent matrix logarithm series $\log (I+C)={\sum }_{m\ge 1}(-1{)}^{m+1}{C}^m/m$ is smooth and inverse to the matrix exponential near zero. These local inverses respect transpose, conjugate transpose, and conjugation; also $\log (A^{-1})=-\log A$ when both matrices are sufficiently close to $I$. Shrink their neighborhoods accordingly. If $A\in Sp(n)$ there, unitarity givesThe symplectic identity is equivalent to $A^t=J^{-1}{A}^{-1}J$, so, putting $X=\log A$, it gives $X^t=-J^{-1}XJ=JXJ$, equivalently $XJ+J{X}^t=0$. Thus this local logarithm restricts to a bijection between a neighborhood of $I$ in $Sp(n)$ and an open neighborhood of zero in $\mathfrak{k}$. Its inverse is the restricted matrix exponential. These restrictions are manifold charts; left multiplication translates them to every $A_0\in Sp(n)$ via $A\mapsto \log (A_0^{-1}A)$. The chart overlaps are smooth compositions of multiplication, exponential and logarithm. The subspace topology is Hausdorff and second countable, inherited from the finite-dimensional matrix space, so these charts give a smooth manifold.

Closure under products and inverses follows from $AJ{A}^t=J$ and unitarity. Matrix multiplication is polynomial in real and imaginary entries, and inversion on the unitary group is $A\mapsto A^{\ast }$, a real linear operation. Their restrictions are smooth in the charts just constructed. Hence $Sp(n)$ is a Lie group without needing a closed-subgroup theorem.

Write $X$ in $n\times n$ blocks. The two infinitesimal conditions giveThe skew-Hermitian matrix $P$ has $n^2$ real parameters; the complex symmetric matrix $Q$ has $n(n+1)/2$ complex parameters, hence $n(n+1)$ real parameters. ThereforeFor $n=1$, any two-by-two matrix satisfies $AJ{A}^t=(\det A)J$, so the group is $SU(2)$. Explicitly,The rows are orthonormal and the determinant is one; conversely unitarity and determinant one force this form. This is the SU(2) as the three-sphere parametrization. The map and its inverse, extraction of the first row, are smooth. Thus $Sp(1)$ is diffeomorphic to $S^3$.

Over $\mathbb{R}$ or $\mathbb{C}$, a Lie algebra is a vector space with a bilinear map $[,]$ which is alternating and obeys the Jacobi identity:Bilinearity and alternation imply $[X,Y]=-[Y,X]$.

For a Matrix Lie group, identify the tangent space $\mathfrak{g}=T_{I}G$ with derivatives ${\gamma }^{\prime }(0)$ of smooth curves through $I$. Product curves show that the sum of two such derivatives is again tangent, and reparametrization supplies scalar multiples. In particular this is a real vector space, even when the matrices have complex entries. For $X,Y\in \mathfrak{g}$, choose a curve $\gamma (s)$ with ${\gamma }^{\prime }(0)=X$. Conjugating a curve with derivative $Y$ shows thatDifferentiate this curve in the finite-dimensional vector space $\mathfrak{g}$. The result isThe matrix commutator is bilinear and alternating, and expanding the six terms proves its Jacobi identity. Thus this construction gives the Lie algebra of a matrix Lie group, with the appropriate bracket, using actual group curves rather than an assumed commutator closure.

For the unitary group, differentiating $\gamma (t{)}^{†}\gamma (t)=I$ gives $X^{†}+X=0$. Conversely, if $X^{†}=-X$, then $e^{tX}$ is unitary and is a curve with derivative $X$. ConsequentlyThis is the unitary Lie algebra. The diagonal entries are purely imaginary, contributing $N$ real parameters, and each upper off-diagonal entry contributes two real parameters. A matrix-unit basis of the unitary Lie algebra isThese $N+2\left(\genfrac{}{}{0ex}{}{N}{2}\right)=N^2$ anti-Hermitian matrix units and their combinations are linearly independent over $\mathbb{R}$ and span every allowed entry.

The symplectic stabilizer is a subgroup: the identity preserves $J$, and if $MJ{M}^T=J$ and $NJ{N}^T=J$, thenMultiplying $MJ{M}^T=J$ on the left by $M^{-1}$ and on the right by $M^{-T}$ gives $M^{-1}J{M}^{-T}=J$. Products and inverses remain unitary. This is the compact symplectic group, often denoted $USp(2n)$ or $Sp(n)$.

Differentiating the stabilizer equation gives $XJ+J{X}^T=0$. For $X=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$, this saysTogether with $X^{†}=-X$, these are equivalentlyThese conditions are also sufficient: $e^{tX}$ is unitary, andHence it stays in the subgroup. They characterize its compact symplectic Lie algebra without adding any trace condition. In fact the trace automatically vanishes. The free anti-Hermitian block $A$ has $n^2$ real parameters, and the complex symmetric matrix $B$ has $n(n+1)$ real parameters. Thus

Here is a complete matrix-unit basis of the compact symplectic Lie algebra. The generators supplying $A$ areThe two families supplying the real and imaginary parts of $B$ are, for $1\le i\le j\le n$,The denominator merely avoids double-counting diagonal entries. Each displayed generator obeys both defining tangent conditions. The $H,P,Q$ generators form a real basis of the allowed $A$ blocks, and the $U,V$ generators form a real basis of the complex symmetric $B$ blocks. Thus they are independent and their total number is $n(2n+1)$. They give all generators required by the real compact algebra, including the case $n=1$.

For a nonsingular skew-symmetric unitary $J$, if each skew-Hermitian control generator obeys $X^{T}J+JX=0$, then so do their real linear combinations and commutators. Differentiating $U^{T}JU$ shows that it stays equal to $J$. The reachable group is therefore contained in a unitary conjugate of the compact symplectic group. This is preservation of an alternating bilinear form, not commutation with $J$.