# Locality, unitarity & vacuum energy

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1. Jul 9, 2015

### "Don't panic!"

I've read in a set of lecture notes that the requirement of locality and unitarity in QFT imply that the vacuum must have a non-zero energy associated with it (http://arxiv.org/pdf/1502.05296v1.pdf , top of page 3 under heading "What is the problem?").

My question is, why does the locality and unitarity of QFT imply this?

My understanding of locality and unitarity are as follows:

Locality requires that the quantum fields of a given QFT must satisfy equal-time commutation relations of the form $$[\phi (t,\mathbf{x}),\psi(t,\mathbf{y})]=i\delta^{(3)}(\mathbf{x}-\mathbf{y})$$ which naturally introduces an infinite constant term $\sim \delta^{(3)}(0)$ upon calculation (if $\phi$ and $\psi$ are expressed in terms of their Fourier modes).

Unitarity requires that the time evolution of any quantum system is such that the sum of the probabilities of all possible outcomes always sums to 1 (intuitively this ensures that if one considers all possible outcomes at a given instant, one of them will definitely occur). This implies that operators that evolve quantum systems, which in the case of QFT is the scattering matrix $S$, are themselves unitary, i.e. they satisfy $S^{\dagger}S=1$.

If I've understood these two concepts correctly, then I can't quite see at the moment why it follows that the vacuum energy is non-zero?

2. Jul 9, 2015

### fzero

I think the logic there is that locality and universality imply that the energy spectrum is bounded from below and that there is a state carrying zero momentum and the minimum energy, which we would identify with the vacuum state. These properties do not specify that the minimum energy is nonzero, which is why it is necessary to compute it given the field content and interactions of a particular QFT.

3. Jul 9, 2015

### "Don't panic!"

Specifically, is this because locality requires that fields satisfy equal-time commutation relations and upon calculating this explicitly (using creation and annihilation operators) we see that a constant energy term is introduced, and then unitarity requires that all particle states have positive-definite norms which further implies that the energy spectrum should be bounded from below (as if this were not the case this would lead to the existence of negative energies (associated with negative probabilities), and the the vacuum would be unstable)?

Is the point then, that locality and unitarity imply that there should be a state of minimum energy which is identified with the vacuum, but this does not imply anything about whether this minimum energy is zero or non-zero?

4. Jul 9, 2015

### fzero

Yex, those are part of the arguments that one would use in QFT. I would worry a bit that they rely strongly on the oscillator approach to the free particle and might miss important details about nontrivial interactions. Maybe Weinberg's book or similar fills in the argument.

Yes. Note that the calculation is not just the addition of the zero point energies for oscillator states, but also involves the interaction of the fields $\phi_i$ with the VEVs $\langle \phi_i \rangle$.

5. Jul 9, 2015

### "Don't panic!"

I've tried Weinberg's book and I couldn't really find anything particularly useful in there on the topic. Are there any other books you'd suggest?

6. Jul 9, 2015

### fzero

Weinberg seems to basically define the vacuum state as the one where there are no oscillator excitations. I'm not familiar with any texts that do anything differently I don't want to mislead you by overthinking this and I was probably being overcautious in casting any doubt. It's fairly clear that it's the right thing to do when you treat the interactions perturbatively. As you include interactions at each order you will have to renormalize and the full set of corrections will dictate the properties of the vacuum state in the interacting theory.