Fermionic vacuum energy from Hamiltonian

Question

I have some questions about some commentsthat Zee makes treating this problem in Sec II.2 of his QFT book. The Hamiltonian density of a spin-1/2 field is

$$ \mathcal{H}=\bar\psi(i\vec\gamma\cdot\vec\partial+m)\psi. $$

It follows that the Hamiltonian is

$$ H=\int\!d^3x\mathcal{H}=\int\!d^3p\,\sum_s p_0\left[ \hat b^\dagger(p,s)\hat b(p,s)-\hat d(p,s)\hat d^\dagger(p,s) \right]. $$

The anticommutation relation $\{\hat d(p_1,s_1),\hat d^\dagger(p_2,s_2)\}=\delta^{(3)}(\vec p_1-\vec p_2)\delta_{s_1s_2}$ yields

$$ H=\int\!d^3p\,\sum_s p_0\left[ \hat b^\dagger(p,s)\hat b(p,s)+ \hat d^\dagger(p,s)\hat d(p,s) \right] - \int\!d^3p\,\delta^{(3)}(\vec0) \sum_sp_0. $$

Zee says that the term with the $\delta^{(3)}(\vec0) $ in it should "fill us with fear." Why should it? It looks perfectly fine to me.

Answer 1

Zee says it should "fill us with fear and loathing" originally, because it implies that the Hamiltonian contains a term proportional to $\delta^3(0)$, which is, in a sense, an infinite c-number (how can an operator yield an infinite observable?). However, you can either notice that only energy differences are observable, so the fact that the second term is infinite is irrelevant, or you can employ the 'regularisation' that Zee does and show that it is the analogous to the sum over all modes of the zero-point energies, as in the scalar field, albeit negative.