# Fermionic vacuum energy from Hamiltonian

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.

• @NiharKarve Nihar, how could it be wrong to have it inside the integral? If it should multiply the integral, then that is equally written with it multiplying the integrand. Please explain yourself, Nihar. You also asked why I think it's fine. I think it's fine, Nihar, because I don't see a problem with it and doesn't "fill me with fear." Why should it be outside the integral, Nihar? Why do you tell me it's wrong, Nihar, and then ask me to explain to you when I made this thread seeking an explanation which you have not provided? Nihar, just saying it's wrong is not helpful. – hodop smith Nov 18 '20 at 15:37
• Nihar means that when $\delta(x)$ is under a $\int dx$ you get a finite answer of unity, but this is not case here. It matters not whether you write the $\delta(0)$ inside or outside the integral --- both are correct as you say --- but it is infinite wherever you put it, and being infinite is "scary". – mike stone Nov 18 '20 at 17:17
• Yes, that was a mistake on my part – Nihar Karve Nov 19 '20 at 2:08

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.