Quantizing the Dirac field using commutation relations leads to an unbounded Hamiltonian?

Question

If we were to quantize the Dirac field using commutation relations instead of anticommutation relations we would end up with the Hamiltonian

$$ H = \int\frac{d^3p}{(2\pi)^3}E_p \sum_{s=1}^2 \Big( a^{s\dagger}_\textbf{p}a^s_\textbf{p} -b^{s\dagger}_{\textbf{p}}b^s_{\textbf{p}} \Big). \tag{3.90} $$

In Peskin and Schröeder they write that this is an unbounded Hamiltonian. And that by creating more and more particles with $b^\dagger$, we could lower the energy indefinitely.

My question: What do they mean when they say that we could lower the energy indefinitely by creating more and more particles with $b^\dagger$? Do they mean that we can lower the energy to negative infinity or do they just mean that we can lower the energy indefinitely to get to the ground state of the system? Also what does it mean for the Hamiltonian to be unbounded?

Answer 1

After this calculation, you now realize that the quantum field theory you just derive makes no sense, indeed, your hamiltonian is not bounded** below! But what does that mean? It means that we could tumble to states of lower and lower energy by continually producing $b^\dagger$ particles (to negative infinity). But in fact, there is no way to deal with this minus sign. So our theory must be wrong. Lets take a deeper look at the physic here.

The key piece of physics we missed is that Dirac equation is for spin $1/2$ particles, which are fermions, meaning they must obey Fermi-Dirac statistics, with the quantum state picking up a minus sign upon the interchange of any two particles. This is embedded into the structure of relativistic quantum field theory, and any attempt at doing otherwise will lead to inconsistency like the one above. So now we replace the commutators by anti-commutators, such that we get: $$ \mathcal{H}{=\int\frac{\mathrm{d}^3\vec p}{(2\pi)^3}E_{\vec p}\sum_{s=1}^2\left[{a^s_{\vec p}}^\dagger{a^s_{\vec p}}-{b^s_{\vec p}}{b^s_{\vec p}}^\dagger\right]\\ =\int\frac{\mathrm{d}^3\vec p}{(2\pi)^3}E_{\vec p}\sum_{s=1}^2\left[{a^s_{\vec p}}^\dagger{a^s_{\vec p}}+{b^s_{\vec p}}^\dagger{b^s_{\vec p}}-(2\pi)^3\delta^{(3)}(0)\right]} $$ Where we get rid of the constant as usual. (Note here that the constant is negative, which could in principle cancel the positive contribution from boson field.)

** An operator $\mathcal{O}$ is said to be unbounded if there does not exist a finite constant $B\geq0$ such that $$ \langle\psi|\mathcal{O}|\psi\rangle\leq B\langle\psi|\psi\rangle, $$ for all states in the domain of $\mathcal{O}$.