 Solution

Read the unsuperscripted $a_{\mathbf{p}}$ in the printed Hamiltonian as $a_{\mathbf{p}}^s$: the repeated spin label is otherwise missing. Relabel $\mathbf{p}\mapsto -\mathbf{p}$ in its antiparticle term, and insert both projections from c. Keeping the original operator ordering givesThe difference of the two orthogonal projectors in a is $h(\mathbf{p})/E$, explaining the cancellation of the factors $E$. Under inverse Fourier transformation, $\mathbf{p}$ becomes $-i\nabla$ acting on $\psi$. HenceSpatial indices run from $1$ to $3$. This is the local Hamiltonian density of the free Dirac field. It also follows from $\mathcal{H}=\pi \dot{\psi }-\mathcal{L}$ with $\pi =i{\psi }^{†}$.

The original antiparticle term is $-b{b}^{†}$, so the result retains its infinite additive vacuum energy. If the Hamiltonian is normal-ordered, its occupation-number expression becomes $\int d^{3}p/(2\pi {)}^{3}E{\sum }_s(a^{s†}{a}^s+b^{s†}{b}^s)$ and its local expression is $:\mathcal{H}:$, with the same subtraction. The displayed unsymmetrized Hamiltonian density is Hermitian after spatial integration under vanishing boundary terms; a manifestly Hermitian local form replaces the derivative term by $-\frac{i}{2}{\psi }^{†}\mathbf{\alpha }\cdot \overleftrightarrow{\nabla }\psi$ and differs by a spatial divergence.