I am reading the QFT book by Peskin and Schroeder. They compute the Hamiltonian for a free quantum scalar field and find \begin{equation} H = \int \frac{d^3p}{(2\pi)^3}\omega_p (a_p^\dagger a_p + \frac{1}{2}\delta(0)). \end{equation} Then they say that the second term cannot be detected experimentally, so it will be ignored. The same refers to spatial momenta \begin{equation} P^i = \int \frac{d^3p}{(2\pi)^3}p^i (a_p^\dagger a_p + \frac{1}{2}\delta(0)). \end{equation} The second term in the formula above in the book by P-S is ignored straightaway.

I wonder to what extent it is accurate. It seems to me, that for many things to work properly, the singular term just should not be there. For example, one normally expects that single particle states $|p\rangle$ should realize a unitary irreducible representation of the Poincare group. To this end, one needs that \begin{equation} H |p \rangle = \omega_p |p\rangle, \qquad P^i |p\rangle = p^i |p\rangle. \end{equation} These formulas are spoiled if the vacuum energy is added.

Besides that, in some standard computations one actually uses that the vacuum has the vanishing energy. For example, one often uses that \begin{equation} e^{iHt}|0\rangle = |0\rangle \end{equation} when computes transition amplitudes. This is equivalent to $H|0\rangle = 0$ or, equivalently, that the vacuum is invariant under translations in time. The same one usually says about spatial translations.

In other words, it seems to me that the singular term is not just immaterial and one can do whatever one wants with it. Instead, it should actually be absent. Is that right? If it is zero, is there any clear way to explain how to get rid of it? For example, one may postulate that before replacing fields with operators, $a$ and $a^*$ should be properly ordered. After all, ordering ambiguity is, indeed, there and it should be fixed somehow (By the way, is there any systematic discussion of what is the right way to fix the ordering ambiguity before one promotes fields to operators when quantizing things? Or one just picks the most natural ordering and hopes for the best?). If it is non-zero, does that mean that the vacuum is not Poincare invariant? Is that ok?

At the same time, Wikipedia says that the vacuum energy not only does not have to be zero, but it is also measurable. For example, in the Casimir effect. This seems to contradict to the statement by Peskin and Schroeder.

Yet another thing is that people seem to compute the vacuum energy as ${\rm log}\; {\rm det} (\Box + m^2)$ . How can one see that it does compute the vacuum energy? Actually, it looks a bit surprising to me that it does, as $H$ is the time component of a vector, while ${\rm log}\; {\rm det} (\Box + m^2)$ is Lorentz invariant. And finally, how is it related to the vacuum one-loop diagram?

I would appreciate if you help clarifying these matters or give a reference where it is written in the accessible manner.

  • 1
    $\begingroup$ possible duplicates: physics.stackexchange.com/q/364240/50583, physics.stackexchange.com/q/709135/50583; for the claim about the Casimir effect see: physics.stackexchange.com/a/746546/50583 $\endgroup$
    – ACuriousMind
    Commented Feb 25, 2023 at 20:01
  • $\begingroup$ @ACuriousMind yes, thanks. It indeed overlaps significantly, though, I have a somewhat expanded version. Anyway, I do not find that your comment on renormalisation there fully answers my questions. In particular, I am wondering whether that term is there at first place: by choosing a different ordering one can get rid of it. Then, assume that you keep it and then regularize it somehow. If the result is not zero, one finds that the vacuum is not Poincare invariant. Is it ok? Moreover, its presence would spoil many standard computations $\endgroup$
    – Dr.Yoma
    Commented Feb 25, 2023 at 20:31
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    $\begingroup$ Existence of the Casimir force is not a proof of non-zero value of zero point energy or the Casimir boundary effect (force due to modification of zero point energy between the bodies caused by the boundary condition). There are good reasons to think that the brief calculations of Casimir forces using zero point energy defect are "analytical tricks" that provide a plausible and even correct result for the force that is actually due to the EM interaction between polarizable bodies (no zero point radiation needed) and should properly be analyzed that way. $\endgroup$ Commented Feb 26, 2023 at 1:10
  • $\begingroup$ This is known as the Lifshitz theory of the Casimir force. Cf. Jaffe's paper arxiv.org/pdf/hep-th/0503158.pdf $\endgroup$ Commented Feb 26, 2023 at 1:10


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