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Two finite-dimensional Hermitian matrices are related by unitary conjugation if and only if they have the same multiset of eigenvalues. Necessity follows because conjugation preserves the characteristic polynomial; sufficiency follows by mapping orthonormal eigenbases in the spectral theorem. Consequentlyand no member of the first pair is equivalent to a member of the second. For an explicit example, the Hadamard gate $H$ gives $H{\rho }_0{H}^{†}={\rho }_2$ and $H{\rho }_3{H}^{†}={\rho }_1$. Multiplying $H$ by a scalar phase does not change either conjugation.

Write the unitary conjugation as $\rho (t)=U(t)\rho (0)U(t{)}^{†}$. The identity component is unchanged, and the generalized Bloch vector transforms byThese coefficients are real, by the same trace argument as in part (a). Unitary conjugation preserves the Hilbert-Schmidt inner product, so the conjugated traceless generators form another orthonormal basis. It follows that $O(U{)}^{T}O(U)=I$. The unitary group is connected, $O(I)=I$, and the determinant of an orthogonal matrix is $\pm 1$; hence $\det O(U)=1$. Thus Hamiltonian evolution rotates the generalized Bloch vector by an element of $SO(N^2-1)$.

Equivalently, in units $\hslash =1$, the Von Neumann equation givesCyclicity of the trace shows $A_{k\ell }=-A_{\ell k}$, so $A$ is a real skew-symmetric matrix. This is the generator form of Hamiltonian rotations of Bloch vectors, valid also for a time-dependent Quantum Hamiltonian. For $N>2$, Hamiltonians generally realize only a proper subgroup of all rotations of this real space; they preserve the whole density-operator spectrum, not merely its purity of a density operator.

The two density operators have the same eigenvalues $a,a,b,b$, so a permutation of basis vectors makes them unitarily equivalent. A dynamical equivalence must additionally respect the symplectic dynamical symmetry in quantum control.

Define the symplectic dual $\tilde{\rho }=J{\rho }^T{J}^{†}$. For a unitary $U$ preserving $J$, both $U^{T}JU=J$ and $UJ{U}^T=J$ hold. In particular $J{U}^{\ast }=UJ$, and thereforeIt follows by cyclicity of the trace that the symplectic invariant of a density operatoris unchanged by every reachable unitary conjugation. The given $J$ reverses the diagonal order, soConsequentlyHence the states are not dynamically equivalent, despite their identical ordinary spectra. This works for all the allowed values, including $a=0$, and for any subgroup satisfying the symmetry in part (c).

Use the Fourier transform convention $\hat{w}(E)={\int }_{\mathbb{R}}w(t)e^{itE}dt$. Since $w\ge 0$, it is real, so $\hat{w}(-E)=\hat{w}(E{)}^{\ast }$. The positive-frequency cutoff therefore also eliminates frequencies at or below $-\Delta$. Normalization gives $\int w=\hat{w}(0)=1$.

For spectral filtering of Hamiltonian terms, chooseIn an energy eigenbasis, its matrix elements areAt $i=j=0$ the frequency is zero, so the normalization preserves the ground-state expectation:The spectral filter is a positive weighted average of unitary conjugations; in particular the integral is bounded in operator norm by $\Vert h_Z\Vert$. We can choose it even without an extra assumed tail bound: the evenization of a nonnegative bandlimited filter proved in part (e) produces another admissible spectral filter with the same type of positive-time decay. Make that choice consistently in all the filtered terms and shell definitions below.

Make a bipartite spin rotation: rotate every odd site's spin through $\pi$ about the $x$ axis, using $U={\prod }_{n\mathrm{odd}}{e}^{-i\pi S_n^x}$. This unitary conjugation preserves the spin commutation relations. At an odd site it sends $(S^x,S^y,S^z)$ to $(S^x,-S^y,-S^z)$ and exchanges the spin raising operator and spin lowering operator.

Each bond joins one rotated and one unrotated site. In the transformed operators,ThusThe selected Néel state becomes an all-up reference state, so a single Holstein–Primakoff transformation convention works on both sublattices.

For part (a), the adjoint operator isUsing the canonical commutation relation $[X,P]=i\hslash I$,Thus $A$ and $A^{†}$ are the annihilation and creation operators of the quantum harmonic oscillator.

For part (b), writeSince $[A,K]=A^{†}$ and $[A^{†},K]=A$, differentiation of the unitary conjugation givesThe initial conditions are $A(0)=A$ and $A^{†}(0)=A^{†}$. Solving this pair of linear ordinary differential equations gives the Bogoliubov transformationThis is the transformation implemented by the single-mode squeeze operator.

For part (c), invert the definition of $A$:The transformation above impliesThe ground state has zero position and momentum means and variances $\hslash /(2m\omega )$ and $m\hslash \omega /2$. Hence the squeezed vacuum state $|\gamma ⟩=S(\gamma )|0⟩$ hasThe reciprocal squeezing factors cancel, soThe state therefore continues to saturate the Heisenberg uncertainty relation.

Set $|\Psi ⟩=U|0^m⟩$. Unitary conjugation givesThe operator negates $|\Psi ⟩$ and fixes every vector in $\mathcal{S}$ orthogonal to it. It is therefore the reflection in a hyperplane whose normal is $|\Psi ⟩$.