# Normalization of ground state "1/2hw"

Tags:
1. Nov 26, 2014

### Jeffrey Yang

Hello everyone:

I didn't have a complete view of the quantum field theory and cannot understand this question. We now there will always be fluctuation field in the universe which corresponds to the ground state energy 1/2hw of harmonic oscillator.

In the free space, we will use box normalization to plane wave. If this field is fluctuation field, the integrated energy should be 1/2hw. How to get this result?

Thanks

2. Nov 28, 2014

### vanhees71

Even in the box-regularized case you get an infinite ground-state energy when doing the calculation in a naive way. The reason for this failure is that the field operators in quantum field theory are operator valued distributions, which you cannot multiply in a mathematically rigorous way. That's why already at this stage you have to renormalize by subtracting the infinite zero-point energy. As long as you neglect gravity, there's no way to observe the absolute value of total energy. Only energy differences between different states of the system are observable quantities, and thus there's no problem in relativistic QFT with the infinite zero-point energy arising from the sloppy calculation of the operator of the total field energy (the Hamiltonian of the free fields in this case).

In cosmology, where General Relativity becomes important, there is a really big problem with this, because even when renormalizing the zero-point energy you need to adjust the corresponding parameters of the Standard model to an extreme accuracy. The reason is that the Higgs boson is a spin-0 field and it's mass is quadratically divergent. This is known as the fine-tuning problem. The question thus is not, why there is "dark energy" but why is it $10^{120}$ times smaller than expected. It's one of the least understood problems in contemporary physics. A famous review on this issue is

Weinberg, Steven: The cosmological constant problem, Rev. Mod. Phys. 61, 1, 1989

Last edited: Dec 1, 2014
3. Dec 1, 2014

### Jeffrey Yang

So, there is no simple way to normalize out the 1/2hw in free space right?

4. Dec 1, 2014

### vanhees71

In free space (as long as you consider only special relativity and thus no gravitation) you simply subtract the zero-point energy without any observable effect.

5. Dec 1, 2014

### VantagePoint72

This is backwards. The cosmological constant is 120 orders of magnitude smaller than the naive QFT calculation predicts.

6. Dec 1, 2014

### vanhees71

Argh :-(. You are, of course, right. I corrected my sentence in the original posting. It must of course read:

The question thus is not, why there is "dark energy" but why is it 10120 times smaller than expected.