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From what I know the reason we have infinite vacuum energy is because according to Quantum Field Theory at every point in space we have something analogous to a harmonic oscillator but since the Zero Point Energy of the quantum harmonic oscillator is non-zero, vacuum energy becomes infinite to due infinite points in any finite region of space.

But if we quantized space shouldn't we get rid of this problem since there would finitely many points in a finite volume of space thus the vacuum energy would be very large in any finite volume of space but still finite because the sum of the points is finite.

Or would we still end up with infinite but countable points that are quantized. i.e would we still end up with a bijection but one that is now a bijection to the rationals instead of the reals from any finite volume in space.

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Yes, the vacuum energy of a spacetime lattice with finite spacing and periodic boundary conditions within a box of finite size is finite. One would not call this "quantizing", though, rather discretizing because we are not carrying out any "quantization procedure" in the sense of going from a classical to a quantum system.

In this approach, the finite size of the box takes care of infrared divergences since the largest allowed wavelength is that of the total size of the box, and the finite lattice spacing takes care of the ultraviolet divergences since the shortest allowed wavelength is of the order of the lattice spacing.

In fact, this is actually underlying QFT in most approaches - the path integral is defined by a limiting procedure on such a lattice (and physicists typically don't care whether that limit actually exists).

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As far as I know, the number of points to not have any influence on divergent behaviour. The infinite vacuum energy comes from the fact that we allow arbitrary frequencies for our quantum fields. There is no difference if we sum or integrate a constant from zero to infinity, the result is still infinite.

$$\sum_{k=0}^\infty {1\over 2} \sqrt{k^2+m^2} \rightarrow \infty$$

or

$$({L\over 2\pi})^3 \int^\infty_0 \sqrt{k^2+m^2} d^3k \rightarrow \infty $$

See here What are the calculations for Vacuum Energy?

The vacuum energy would become finite if there were a good reason to forbid frequencies above some value.

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  • $\begingroup$ Thanks but just to clarify even if there were finite number of quantized points in space this would still diverge? Since it is is dependent on the frequencies and not number of points in space itself. I wanted to know why we have to sum to infinity. $\endgroup$
    – FireFistAce
    Commented Dec 2, 2014 at 7:34
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    $\begingroup$ Yes I think so. I think the reason is that there is no good reason to stop the sum earlier. In the derivation of quantum field theory the general solutions of the equations of motion e.q. the Dirac Equation are used. This general solution is a superposition of all possible solutions, which includes all possible $k$ values. When we derive the Energy=the Hamiltonian from the corresponding Lagrangian and put the general solution in there we still have this sum over all possible solutions in there. This is were this infinity comes from. $\endgroup$
    – Tim
    Commented Dec 2, 2014 at 7:46
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    $\begingroup$ The reason discrete spacetime has been suggested is that it sets a minimum length, and therefore a minimum wavelength/maximum frequency and the integral is no longer to infinity. $\endgroup$
    – John Rennie
    Commented Dec 2, 2014 at 7:58
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    $\begingroup$ @JohnRennie That makes perfect sense. My answer is therefore wrong! $\endgroup$
    – Tim
    Commented Dec 2, 2014 at 8:07

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