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

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?

• Related: physics.stackexchange.com/q/17893/2451 and links therein. Oct 31, 2017 at 18:33
• They are only trying to give a physical explanation of why one can neglect this infinite constant. One way of viewing this is saying that you can fill an infinite well with $b^\dagger$, and then, on top of it, you set your zero of energy where you can define particles created with $a^\dagger$. Oct 31, 2017 at 18:44
• @Soap312 Where do you get an infinite constant from? Oct 31, 2017 at 19:06
• Sorry, I have misread your equation I will post a complete answer in a few minutes. Oct 31, 2017 at 19:41

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 $\mathcal{O}$.
• the last formula should have a $\ge$ instead of a $\le$. Oct 31, 2017 at 20:28
• Thank you for your answer @Soap312. How can we apply ** to our case where the $a$'s and the $b$'s are our operators? Btw for the Hamiltonian to be bounded from below shouldn't we have $$⟨ψ|\mathcal{0}|ψ⟩\geq B⟨ψ|ψ⟩$$ instead? Nov 1, 2017 at 7:18
• @Turbotanten, yes as you can see, I gave a general definition, for $B\geq0$. So in your case $B\leq0$ such that indeed we should flip the inequality. Nov 1, 2017 at 14:12