- #1

"Don't panic!"

- 601

- 8

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?