= 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 <2/c>. Keeping the original operator ordering gives
$$
\begin{aligned}
H&=\int\frac{d^3p}{(2\pi)^3}\,
\widetilde\psi^\dagger(t,\mathbf p)
\left[\frac12\sum_s\left(u^s(\mathbf p)u^{s\dagger}(\mathbf p)
-v^s(-\mathbf p)v^{s\dagger}(-\mathbf p)\right)\right]
\widetilde\psi(t,\mathbf p)\\
&=\int\frac{d^3p}{(2\pi)^3}\,
\widetilde\psi^\dagger(t,\mathbf p)
(\boldsymbol\alpha\cdot\mathbf p+\beta m)
\widetilde\psi(t,\mathbf p).
\end{aligned}
$$
The difference of the two <orthogonal projectors> in <2/a> is $h(\mathbf p)/E$, explaining the cancellation of the factors $E$. Under inverse <Fourier transformation>, $\mathbf p$ becomes $-i\boldsymbol\nabla$ acting on $\psi$. Hence
$$
\boxed{H=\int d^3x\,\mathcal H,\qquad
\mathcal H=\psi^\dagger(-i\boldsymbol\alpha\cdot\boldsymbol\nabla+\beta m)\psi
=\bar\psi(-i\gamma^i\partial_i+m)\psi.}
$$
Spatial 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 $\pi=i\psi^\dagger$.
The original antiparticle term is $-bb^\dagger$, so the result retains its infinite additive <vacuum energy>. If the <Hamiltonian> is <normal-ordered>, its occupation-number expression becomes $\int d^3p/(2\pi)^3\,E\sum_s(a^{s\dagger}a^s+b^{s\dagger}b^s)$ 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 $-\tfrac i2\psi^\dagger\boldsymbol\alpha\cdot\overleftrightarrow{\boldsymbol\nabla}\psi$ and differs by a spatial <divergence>.
Back to article page
