Normalization of ground state "1/2hw"

In summary, normalization of the ground state "1/2hw" refers to the process of adjusting the wave function of a quantum system to ensure that its total probability is equal to 1. This normalization is necessary for accurately describing the behavior of a quantum system, as it allows for the calculation of probabilities for different states of the system. The normalization of the ground state is particularly important in the study of quantum mechanics, as it forms the basis for understanding the behavior of particles and their interactions at the microscopic level. By normalizing the ground state, researchers are able to accurately predict the behavior of quantum systems and make meaningful interpretations of their results.
  • #1
Jeffrey Yang
39
0
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
 
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  • #2
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 [itex]10^{120}[/itex] 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
http://link.aps.org/abstract/RMP/V61/P1
 
Last edited:
  • #3
vanhees71 said:
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 [itex]10^{120}[/itex] times larger 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
http://link.aps.org/abstract/RMP/V61/P1

Thanks for your reply. Very advanced insight. :-).

So, there is no simple way to normalize out the 1/2hw in free space right?
 
  • #4
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
vanhees71 said:
The question thus is not, why there is "dark energy" but why is it [itex]10^{120}[/itex] times larger than expected.

This is backwards. The cosmological constant is 120 orders of magnitude smaller than the naive QFT calculation predicts.
 
  • #6
LastOneStanding said:
This is backwards. The cosmological constant is 120 orders of magnitude smaller than the naive QFT calculation predicts.
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.
 

Related to Normalization of ground state "1/2hw"

What is "Normalization of ground state 1/2hw"?

Normalization of ground state 1/2hw refers to the process of adjusting a wave function to ensure that the probability of finding the particle at any point in space is equal to 1. This is important in quantum mechanics as it allows for accurate calculations and predictions.

Why is normalization of ground state 1/2hw necessary?

Normalization of ground state 1/2hw is necessary because it ensures that the wave function conforms to the laws of quantum mechanics. Without normalization, the wave function would violate the principle of conservation of probability, making it unreliable for predicting the behavior of particles.

How is normalization of ground state 1/2hw calculated?

The normalization of ground state 1/2hw is calculated by finding the square of the wave function, integrating it over all space, and then dividing the original wave function by the square root of the result. This ensures that the probability of finding the particle at any point in space is equal to 1.

What is the significance of "1/2hw" in normalization of ground state 1/2hw?

The term "1/2hw" represents the energy of the ground state of a particle. This value is used in the normalization process to ensure that the resulting wave function is properly scaled and represents the most probable state of the particle.

Are there any limitations to normalization of ground state 1/2hw?

While normalization of ground state 1/2hw is an important concept in quantum mechanics, it does have some limitations. It assumes that the particle is in a stationary state, and it is not applicable to systems with time-dependent potentials or non-stationary states.

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