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If each local term is already minimized by the global ground state, subtract its minimum $a_Z$ to obtain a positive operator. Nonnegative normalized spectral filtering of Hamiltonian terms preserves positivity and leaves the ground eigenvalue $a_Z$ unchanged. Thus the same state minimizes every filtered term. This preserves frustration freeness but does not create it for an initially frustrated decomposition.

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.

Sum the spectral filtering of Hamiltonian terms over the original finite decomposition. The total Hamiltonian operator commutes with its own evolution, soThe absence of all ground-to-excited matrix elements, proved in part (a), makes each filtered term block diagonal relative to $P_0=|{\varphi }_0⟩⟨{\varphi }_0|$ and its orthogonal complement. ThereforeA filtered term may act on the whole system, since unitary time evolution spreads its original support. Neither commutation with $P_0$ nor preservation of its ground-state expectation asserts that the global ground state minimizes each individual term. In particular, this construction does not generally turn a frustrated decomposition into a frustration-free Hamiltonian.

Use the Fourier transform convention $\hat{w}(E)={\int }_{\mathbb{R}}w(t)e^{itE}dt$ and define the spectral filtering of Hamiltonian terms byThe normalization $\hat{w}(0)=1$ means $\int w=1$. Since $w$ is real and nonnegative, $\hat{w}(-E)=\overline{\hat{w}(E)}$. Thus the stated positive-frequency cutoff also implies $\hat{w}(E)=0$ for $E\le -\Delta$. The integral exists in operator norm, because its integrand has norm at most $w(t)\Vert h_Z\Vert$.

Let $H|{\varphi }_i⟩=E_i|{\varphi }_i⟩$, with $E_i-E_0\ge \Delta$ for $i>0$. In this eigenbasis,Every matrix element coupling the unique ground state to an excited state vanishes, in either direction. HenceMoreover, linearity and conservation of $H$ under its own Heisenberg picture evolution giveEach filtered term is Hermitian, but it need not remain supported on its original set $Z$. The tail assumption supplies stronger long-time control than the integrability used here; no additional tail estimate is needed for the requested identities.

The filtered decomposition is not necessarily frustration-free. Commuting with $P_0$ makes $|{\varphi }_0⟩$ an eigenvector of each term; it does not make that eigenvector a lowest-energy state of each term. This is the distinction expressed by commuting with a ground projector does not imply frustration freeness.

For an explicit positive, two-local counterexample on three qubits, let $P_0^{(i)}=|0⟩⟨0{|}_i$ and $P_1^{(i)}=|1⟩⟨1{|}_i$, and use the two distinct interaction sets $\{1,2\}$ and $\{2,3\}$:Both are positive semidefinite and commute with their sum. The total energy on a computational basis vector isThus $H=h_{12}+h_{23}$ has unique ground state $|000⟩$, energy $1$, and spectral gap $1$. Since $[H,h_{12}]=[H,h_{23}]=0$, spectral filtering of Hamiltonian terms leaves both terms unchanged for every normalized filter:But $h_{12}|000⟩=|000⟩$, whereas $h_{12}|010⟩=0$. The global ground state therefore does not minimize $g^{(12)}$, proving that the resulting decomposition is not a frustration-free Hamiltonian.

There is a useful positive result if the original decomposition already is a frustration-free Hamiltonian. If $a_Z={\lambda }_{\min }(h_Z)$ and $h_Z|{\varphi }_0⟩=a_Z|{\varphi }_0⟩$, then $h_Z-a_{Z}I\ge 0$. Averaging its unitary conjugates with nonnegative $w$ gives $g^{(Z)}-a_{Z}I\ge 0$, while $g^{(Z)}|{\varphi }_0⟩=a_Z|{\varphi }_0⟩$. Thus filtering preserves existing frustration freeness; it does not create it for an arbitrary gapped Hamiltonian.