What does it mean for the Hamiltonian to not be bounded?

[Turbotanten]

If we were to quantize the Dirac field using commutation relations instead of anticommutation relations we would end up with the Hamiltonian, see Peskin and Schroeder

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

They write that this Hamiltonian is not bounded below. And that by creating more and more particles with $b^\dagger$ we could lower the energy indefinitely.

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 not be bounded below?

 

[vanhees71]

A Hamiltonian is called "bounded from below", if there exists $E_0$ such that for all vectors in Hilbert space
$$\langle \psi|\hat{H}|\psi \rangle \geq E_0.$$

Now you can show easily that (after putting everything in a large box, so that the momentum-spin Fock states become true Hilbert-space states) the Fock states with $N_b(\vec{p},\sigma)$ are eigenstates of the Hamiltonian with eigenvalues $-\sum_{\vec{p},\sigma} E_p N(\vec{p},\sigma)$, which you can make $\rightarrow -\infty$ by just occupying more and more single-b-particle states.

To get a finite Hamiltonian (via normal ordering) bounded from below you have to quantize Dirac fields as fermions, which holds for any fields with an half-integer spin number. This is the famous spin-statistics theorem.

For details about the general case, see Weinberg, Quantum Theory of Fields, Vol. 1.

 

[Turbotanten]

Great answer but I kinda got lost in your second paragraph.

What does $N_b(\vec{p},\sigma)$ represent? I understand that $\vec{p}$ is the momentum but what is $\sigma$ and $N_b$ in your case?

When you write that $N_b(\vec{p},\sigma)$ are eigenstates of the Hamiltonian do you mean that
$H N_b(\vec{p},\sigma) = -E_pN_b(\vec{p},\sigma)$

 

[andrien]

$b^{s\dagger}b^s$ is some sort of occupation number and counts the number of particles. More number of particles created with $b^{\dagger}$ will increase this occupation number and will give negative contribution to Hamiltonian.

 

[vanhees71]

The Fock states are defined as the common eigenstates of the number operators $\hat{N}_{a}(\vec{p},\sigma)=\hat{a}^{\dagger}(\vec{p},\sigma) \hat{a}(\vec{p},\sigma)$ and $\hat{N}_{b}(\vec{p},\sigma)=\hat{b}^{\dagger}(\vec{p},\sigma) \hat{b}(\vec{p},\sigma)$. These states are eigenstates of the Hamiltonian since (for particles in a finite box)
$$\hat{H}=\sum_{\vec{p},\sigma} E_(p) [\hat{N}_a(\vec{p},\sigma)-\hat{N}_b(\vec{p},\sigma)].$$
If you, e.g., choose a state with all $N_a(\vec{p},\sigma)=0$ you can make the eigenvalue of $\hat{H}$ arbitrarily small, i.e., make it $\rightarrow \infty$ by occupying more and more $b$-particle states.

 

[Turbotanten]

The Fock states are defined as the common eigenstates of the number operators $\hat{N}_{a}(\vec{p},\sigma)=\hat{a}^{\dagger}(\vec{p},\sigma) \hat{a}(\vec{p},\sigma)$ and $\hat{N}_{b}(\vec{p},\sigma)=\hat{b}^{\dagger}(\vec{p},\sigma) \hat{b}(\vec{p},\sigma)$. These states are eigenstates of the Hamiltonian since (for particles in a finite box)
$$\hat{H}=\sum_{\vec{p},\sigma} E_(p) [\hat{N}_a(\vec{p},\sigma)-\hat{N}_b(\vec{p},\sigma)].$$
If you, e.g., choose a state with all $N_a(\vec{p},\sigma)=0$ you can make the eigenvalue of $\hat{H}$ arbitrarily small, i.e., make it $\rightarrow \infty$ by occupying more and more $b$-particle states.

Thanks! Now I finally get it!