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?


1 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.

  • $\begingroup$ Awesome answer; as any good answer, though, it brings new questions to my mind... For example, canonical commutation relations are writen as products of fields, so they are tricky as well (and normal ordering may be inadequate? I mean, $:[a,a^\dagger]:=0$, right?) I have to think about this for some time. On the other hand, products of distributions are well defined when the supports are disjoint, and we usually think of a field and its momentum as belonging to different spaces (right?), so the product might be well defined in the case of $\phi\ \pi$, so c.c.r's might be well defined after all $\endgroup$ Commented Jul 4, 2015 at 20:33
  • $\begingroup$ Well the last part of my comment makes no sense, please ignore it (I can't edit it, and I don't want to delete it, because of the first part :P) $\endgroup$ Commented Jul 4, 2015 at 20:50
  • $\begingroup$ @Taylor Well, a first point is that there actually is ongoing mathematical research on defining products of distributions, with applications in calculating the $n$-point functions of perturbative QFT (especially in curved spacetime, where the situation is less understood), see e.g. this paper. However this is not exactly what you were wondering (but it is not completely unrelated as well)... $\endgroup$
    – yuggib
    Commented Jul 4, 2015 at 20:56
  • $\begingroup$ @ Taylor The commutation relations present some difficulties: one is that to write them correctly as commutation of operators you should integrate the fields (or the creation/annihilations) with functions on the one-particle space. Then the commutation makes sense as a commutation of operators, but only on a sufficiently regular dense subspace of the whole Hilbert space. To actually avoid this problem, the exponentiated form (Weyl relations) is used as the correct form of the CCR (there are also other reasons for this, but I don't want to go too much into detail). $\endgroup$
    – yuggib
    Commented Jul 4, 2015 at 21:00
  • $\begingroup$ @Taylor Anyways, it is a "lucky" (or misleading) coincidence that the CCR for the operator valued distributions (fields or annihilators/creators) can be formally written in a way that makes sense in a distributional way: this is surely useful in (formal) calculations, but it has to be used with care and keeping in mind that the true operators are defined otherwise and it is necessary to always check that the formal manipulations actually make sense mathematically. $\endgroup$
    – yuggib
    Commented Jul 4, 2015 at 21:05

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