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In a standard QFT class, you're being indoctrinated that there is the "infinite vacuum energy density problem". (This is sometimes paraphrased as the "cosmological constant problem", which is in my opinion a misnomer since the calculation that calculates a finite value for the vacuum density and relates it to the cosmological constant is ill-founded and dubious.)

The argument goes as follows: When defining the Hamiltonian $H$ from the Lagrangian density $\mathcal{L}$, we first define the Lagrangian function $L := \int \operatorname{d}^3 x \mathcal{L}$ and the momenta $\pi(x) := \frac{\partial L}{\partial \dot \phi(x)}$ and finally the Hamiltonian $H := \int \operatorname{d}^3 x \pi \dot \phi - L$.

We then promote all the quantities that occured to operators and calculate the decomposition of $H$ into creation and annihilation operators $a^\dagger$, $a$.

Then, the lecturer makes a big fuss (prototypical example) about the Hamiltonian being of the form $\int \operatorname{d}^3 p E_p \frac{1}{2} \left( a^\dagger a + a a^\dagger \right)$ = $\int \operatorname{d}^3 p E_p \left( a^\dagger a + \delta(0) \right)$ and the $\delta(0)$ being an infinite contribution to the vacuum energy density. (I always found the argument fishy because we can't know beforehand if some quantisation strategy will succeed or not, so we should rather start with a quantised Hamiltonian and ask whether its classical limit is the theory we started with, which can be answered with yes.)

My question now is: Does a similar problem also occur in the Feynman path integral approach, or in any approach from algebraic quantum field theory?

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Minor comment to v2: The formula $\pi(x) := \frac{\partial L}{\partial \dot \phi(x)}$ is meaningless because one should distinguish between functional and function, cf. e.g. this post. –  Qmechanic Jun 5 '12 at 20:38
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1 Answer 1

up vote 4 down vote accepted

In algebraic QFT, the Hamiltonian is always well-defined, and the vacuum state is an eigenstate of zero energy. Thus no such terms arise - they are eliminated by careful renormalizations before a continuum limit is taken.

In the path integral approach, the problem is still present, hidden in the renormalization prescriptions for making perturbative sense of the contributions to the path integral. Note that infinities must be cancelled to get finite results, and as $\infty-\infty$ is undetermined there remain finite renormalizations that can be fixed by a number of different recipes that (should) give equivalent parameterizations of the same manifold of renormalized field theories.

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Makes me think that the problem is not a physical problem (and thus doesn't have to be resolved, so there is definitely no cosmological constant problem coming from QFT) but just a problem of finding a good definition. –  Turion Jun 10 '12 at 15:15
@Turion: In a fixed background metric, the cosmological constant gives just a shift in the energy, and hence immaterial. But if the metric is an operator, the cosmological constant gives not a constant energy shift hence represents a nontrivial metric interaction changing the theory. Thus the problem of the cosmological constant is indeed independent of QFT renormalization questions. –  Arnold Neumaier Jun 10 '12 at 15:26
How can the problem be hidden in the renormalization prescriptions for making perturbative sense of the contributions to the path integral in a free theory? –  fqq May 3 at 15:34
@fqq: I had explained how. Which constants one gets is found out only after successful renormalization, as only then the formal infinities are converted into finite quantities. –  Arnold Neumaier May 5 at 10:36
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