Is the Cosmological Constant Problem a Misunderstanding of Zero-Point Energy?

[PeterDonis]

There is no proper framework of quantum mechanics that includes GR.

Yes, there is: quantum field theory in curved spacetime. "Energy gravitates" is true in that framework, so it is a valid physical principle for me to use.

What we do not have is a complete theory of quantum gravity, i.e., a theory in which spacetime is quantized (or emerges from some more fundamental entity that is quantized). Whether such a theory will be necessary in order to resolve the cosmological constant problem is an open question. But that does not undermine what I have been saying, since I am not proposing a solution to the cosmological constant problem. I am simply making a physical argument for what the Hamiltonian of the quantum harmonic oscillator should be given the fact that energy gravitates. A treatment of the QHO in curved spacetime along the lines of QFT in curved spacetime should be perfectly fine as a basis for that.

the part of your physical argument that is supposed to show how you uniquely determine the Hamiltonian does not involve any gravitational effects.

You are clearly not even reading what I post so I don't see any point in responding other than to say that you are wrong here. I have already given the details several times. I'm not going to repeat them again.

If there are arguments other than the wave function and the consequences that support your viewpoint over others then please talk about them and not things related to the wave function.

I have not made any argument at all based on the wave function. My argument has been based on the Hamiltonian. The Hamiltonian is not the wave function. I have already explained this.

Again, you are clearly not even reading what I post.

You also effectively added an arbitrary constant to your Hamiltonian because you have not shown that your choice is better than any other.

Once more, you are clearly not even reading what I post. I have made a physical argument for why the quantum Hamiltonian should have the same form as the classical Hamiltonian, $p^2 / 2 + x^2 / 2$, period, without any other constant added. You clearly disagree with that argument, but that does not mean I haven't made it or that it involves adding an arbitrary constant to the Hamiltonian. The whole point is that $p^2 / 2 + x^2 / 2$ is not arbitrary.

 

[strangerep]

[...] if you look at the most famous and earliest example of these virtual QFT effects... the Lamb shift of the hydrogen atom. It arises from the same sort of sums over vacuum diagrams as the above.
And this has a physical effect, it contributes to a shift in energy levels. So, by the equivalence principle it must gravitate, and that means you can couple the theory in the way we just did. So there is in fact some sort of reality to this afterall (and indeed nontrivial computations in this formalism have been done and verified in famous neutron interferometry experiments in the presence of gravitational fields).

I'm aware of the original neutron interferometry experiments with gravity, but those results could be accounted for by just a Schrödinger equation with Newtonian potential.

If you would please give me reference(s) to more recent experiments that probe this with greater precision, I'd be most grateful.

 

[DrClaude]

This thread has run its course. Thanks to all that contributed.

Thread closed.