Is the Klein-Gordon Hamiltonian unbounded below?

Question

This question is about the Klein-Gordon Hamiltonian for simplicity, but the problem seems to remain when dealing with other fields (e.g. Dirac, photon...).

One usually writes the Hamiltonian (density) as $\mathcal{H}=\frac{1}{2}\Pi^2+\frac{1}{2}(\nabla \phi)^2+\frac{1}{2}m^2\phi^2$, which is supposed to be bounded below (and, in this case, positive, as it is the sum of squares). From this fact, we can deduce the existence of a ground state, it being the state with minimum eigenvalue (and such a minimum is guaranteed to exist because $H$ is bounded, as previously stated).

So far, so good. When integrating $\mathcal{H}$ over $\mathrm d^3\vec x$ to get the total Hamiltonian, we arrive at the (schematic) result $H=\frac{1}{2}\int \mathrm d^3 \vec p\ \omega(p) n(p)+\text{const}$, where $\omega=+\sqrt{\vec p^2+m^2}$, and $n$ is the number operator. The thing is, the $\text{const.}$ term is actually infinite (and related to the zero-point energy), which we ignore by redefining the Hamiltonian to subtract off this constant term.

If I am getting this right, this means that we "switch" $H\to H-\infty$, which is obviously unbounded below. If this is the case, what happened to the ground state? How can we guarantee it still exists?

We encounter this very same problem in other theories, which seems to be related to the fact that we use canonical commutation relations (because these always lead to "harmonic oscillator relations", which give rise to the zero-point energy). We always ignore this infinity, by different methods (such as redefining the Hamiltonian, or imposing normal ordering), but I suspect that all of them make $H$ unbounded below. Is this unbound-ness really important? I feel it is, but if we can prove $\exists |0\rangle$ regardless of $H$ being unbounded, then it is not really an issue.

So, my question is, how can we sensibly get rid of these zero-point infinities without running into problems with the ground state? Or, can we prove the existence of the ground state for an unbounded Hamiltonian?

Answer 1

$\Pi(x)$ and $\Phi(x)$ are not operators. They are operator valued distributions (distributions that when acting on the one-particle functions become operators). Therefore, it does not make sense to do their square since it is like doing the square of the $\delta$ function. The normal ordering is a prescription on how to cure this ill definition, and obtain an operator that is bounded from below. The boundedness from below can be easily proved, and in addition $\int\omega(k)a^*(k)a(k)dk$ is a well-defined self-adjoint operator.

To be more precise: the flaw in your reasoning is in the statement "$\mathscr{H}$ is bounded from below". $\mathscr{H}$ does not make sense as an operator or a distribution if you do not manipulate suitably $\Phi$ and $\Pi$, therefore you cannot say that it is "positive". In order for it to make sense we do indeed a renormalization procedure, that involves introducing cut offs, manipulating the then well defined operator, subtracting some constants (infinite in the limit) and finally removing the cut offs obtaining a well-defined (and a posteriori positive) operator.

This simple renormalization procedure can be effectively performed by the rule "normal order the object", i.e. formally putting all the creation operator valued distributions to the left of annihilation operator valued distributions.