Divergent vacuum uncertainty of fields in QFT, how to resolve?

In summary: No, that would not work. ##\phi(x)## is an operator-valued distribution, and squaring it is not even well-defined.Just to iterate what @A. Neumaier said, ##\phi^{2}(x)## is not a well defined quantity due to ##\phi(x)## being an operator-valued distribution, it doesn't represent the actual uncertainty in ##\phi(x)##.
  • #1
HomogenousCow
737
213
If you calculate the uncertainty of a scalar field in the vacuum state, i.e. ##\langle0\left| \phi^2\right|0\rangle##, you get a divergent integral that comes out to something like

$$\frac{1}{4\pi^2}\int_0^\Lambda \frac{k^2 dk}{\sqrt{{m^2}+{k^2}}}$$

Where ##\Lambda## is some momentum cutoff. How does one make sense of this divergent quantity? Can it be made finite by some "zero-th order" renormalization?
 
Last edited:
Physics news on Phys.org
  • #2
It means the uncertainty in the value of the field at a point is infinite. Since it's not possible to measure the value of a field at an exact mathematical point, this is not a significant issue.

If you multiply together fields at slightly different points, you can express the result as a sum of other operators (at one of the points, or at their average), times coefficient functions that blow up as the points come together. This is called the "operator product expansion".
 
  • Like
Likes vanhees71
  • #3
So would something like ##\int \phi(x)\phi(x+\epsilon)d^3 \epsilon## give a finite vacuum expectation value? Where the integral runs over some finite volume around the origin.
 
  • #4
Yes, I think so. Just by dimensional analysis, ##\langle 0|\phi(x)\phi(x+\epsilon)|0\rangle\sim 1/\epsilon^2##, so the integral should be finite (all this is in 4 spacetime dimensions).
 
  • #5
Okay I see. I guess if we took the volume to infinity and also divide by it we might get some kind of macroscopic quantity. Will try this later.

I don't quite understand the physical intuition behind this quantity though. Say we have some kind of instrument that measures the field, shouldn't it be the volume average of ##\phi^2## where both factors are measured at the same point? (##\langle0\left|\int \phi^2 g(x) dx \right|0\rangle## is divergent, ##g(x)## being some weight function)
 
  • #6
Avodyne said:
Since it's not possible to measure the value of a field at an exact mathematical point, this is not a significant issue.
But energy, given by the field Hamiltonian, involves products of fields at the exact same point. And energy is measurable. So I wouldn't say that it isn't a significant issue.

In my opinion, the resolution of the problem is that continuous QFT is just an effective theory, while the fundamental theory is discrete. In other words, there is a fundamental finite UV cutoff ##\Lambda## which makes things finite.
 
  • Like
Likes bhobba
  • #7
HomogenousCow said:
If you calculate the uncertainty of a scalar field in the vacuum state, i.e. ##\langle0\left| \phi^2\right|0\rangle##, you get a divergent integral
The standard formula for uncertainties makes sense only for operators, not for distributions.
##\phi(x)## is not an operator but a distribution, and squaring it is not even well-defined.
 
  • Like
Likes dextercioby, vanhees71 and DarMM
  • #8
Just to iterate what @A. Neumaier said, ##\phi^{2}(x)## is not a well defined quantity due to ##\phi(x)## being an operator-valued distribution, it doesn't represent the actual uncertainty in ##\phi(x)##.

Well defined operators are quantities like
$$\phi(f) = \int{\phi(x)f(x)d^{4}x}$$
where you then work out the uncertainty by computing ##\langle \phi(f)^{2} \rangle## .
The need to integrate ##\phi(x)## against a function is what makes it an operator-valued distribution, it maps from (tempered Schwartz) functions to operators, but is not an operator itself.

Quantities like ##\phi^{2}(x)## need to be "repaired" as such in order to consitute legitimate operator-valued distributions. Unfortunately there is currently no properly developed mathematical theory of doing so, just a collection of ad-hoc methods known under the term "renormalization".
 
  • Like
Likes dextercioby, weirdoguy and bhobba
  • #9
Demystifier said:
In my opinion, the resolution of the problem is that continuous QFT is just an effective theory, while the fundamental theory is discrete. In other words, there is a fundamental finite UV cutoff ##\Lambda## which makes things finite.

That is a common modern view - and mine is similar except that perhaps the deeper theory is not necessarily discreet.

Wilson's view is interesting here:
https://quantumfrontiers.com/2013/06/18/we-are-all-wilsonians-now/

Thanks
Bill
 
  • Like
Likes Demystifier
  • #10
DarMM said:
it doesn't represent the actual uncertainty in ϕ(x)ϕ(x)\phi(x).

So what does? Can we multiply fields at different points in spacetime?
 
  • #11
DarMM said:
Unfortunately there is currently no properly developed mathematical theory of doing so, just a collection of ad-hoc methods known under the term "renormalization".
Renormalization looks ad hoc when one thinks of the renormalized theory as a fundamental theory. However, when one changes perspective and thinks of renormalized theories as effective theories, then renormalization does not longer look ad hoc. Such a view has been developed by Wilson and Weinberg, among others.
 
  • Like
Likes bhobba
  • #12
HomogenousCow said:
So what does?
Nothing. ##\phi(x)## isn't an operator, so there is no uncertainty for it. ##\phi(f)## is an operator and has corresponding uncertainty ##\langle \phi(f)^{2} \rangle##.

HomogenousCow said:
Can we multiply fields at different points in spacetime?
Fields can be multiplied at different spacetime points to give quantities like ##\phi(x)\phi(y)##, but once again these are not operators, but still only operator valued distributions.
 
  • #13
DarMM said:
Nothing. ##\phi(x)## isn't an operator, so there is no uncertainty for it. ##\phi(f)## is an operator and has corresponding uncertainty ##\langle \phi(f)^{2} \rangle##.Fields can be multiplied at different spacetime points to give quantities like ##\phi(x)\phi(y)##, but once again these are not operators, but still only operator valued distributions.

So would ##\int \phi(x)\phi(x+r)g(r)d^3r## work? With g being square normalized and localized around the origin.
 
  • #14
bhobba said:
the deeper theory is not necessarily discreet.
Yes, it looks at all the details you may wish to hide...
 
  • Like
Likes bhobba
  • #15
Demystifier said:
Renormalization looks ad hoc when one thinks of the renormalized theory as a fundamental theory. However, when one changes perspective and thinks of renormalized theories as effective theories, then renormalization does not longer look ad hoc. Such a view has been developed by Wilson and Weinberg, among others.
This is not quite what I'm talking about. I'm saying that there is no mathematical theory of taking an ill-defined distributional multiple like ##\phi^{2}(x)## to a well-defined distribution. What physicists have developed in renormalization theory constitutes a formal adhoc early form of such a theory.

This is essentially what the Wilson group with ##\Lambda \rightarrow \infty## is doing. You can see it as a flow across distributional space that terminates on well defined versions of a continuum distributional product and since these distributional products appear in the action it gives a well-defined action.

It's still ad-hoc as it involves unproven bounds and hasn't been formalised.
 
  • #16
HomogenousCow said:
So would ##\int \phi(x)\phi(x+r)g(r)d^3r## work? With g being square normalized and localized around the origin.
No as you still have ##x## not integrated against. The quantity you are looking for is ##\langle \phi(f)^{2} \rangle##.
 
  • #17
DarMM said:
This is essentially what the Wilson group with Λ→∞ is doing.
But from an effective theory point of view, this limit is unphysical and hence irrelevant.
 
  • #18
Demystifier said:
But from an effective theory point of view, this limit is unphysical and hence irrelevant.
Let's be clear, the procedure still results in a well-defined distributional product which is what I am discussing. You obtain a well-defined version of continuum limit products.

Regardless of how physical or relevant one views the resultant continuum theory it still constitutes a construction of a distributional product.
 
  • #19
Demystifier said:
Renormalization looks ad hoc when one thinks of the renormalized theory as a fundamental theory.
Just to say I don't think this is true either. Even if the continuum theory is viewed as fundamental the renormalized theory has a clear motivation. Renormalization of products in the action is simply Wick-ordering with respect to the interacting vacuum. This of course has a singular appearance when one expands about the free vacuum/path integral measure due to them being relatively singular.

The problem is we don't have proper (not ad-hoc) theory of constructing non-Gaussian measures over functional spaces.
 
  • Like
Likes dextercioby and maline

Related to Divergent vacuum uncertainty of fields in QFT, how to resolve?

1. What is divergent vacuum uncertainty in QFT?

Divergent vacuum uncertainty refers to the issue in quantum field theory (QFT) where calculations of physical quantities, such as energy or momentum, result in infinite values. This is due to the fact that QFT takes into account the fluctuation of virtual particles in the vacuum, which can lead to infinite values in certain calculations.

2. Why is divergent vacuum uncertainty a problem in QFT?

Divergent vacuum uncertainty is a problem because it leads to inconsistencies and inaccuracies in calculations, making it difficult to make precise predictions about physical phenomena. In order to make meaningful predictions, these divergences must be resolved.

3. How is divergent vacuum uncertainty resolved in QFT?

There are various techniques used to resolve divergent vacuum uncertainty in QFT. One common method is renormalization, which involves redefining the physical quantities to account for the infinite values. Another approach is to introduce a cutoff or regulator, which limits the values of the calculations to a finite range.

4. Are there any limitations to resolving divergent vacuum uncertainty in QFT?

While renormalization and other techniques can effectively resolve divergent vacuum uncertainty, there are still some limitations. For example, these techniques may not work for all calculations and may introduce errors or inconsistencies in certain cases. Additionally, the concept of virtual particles and their effects on the vacuum is still not fully understood, leading to ongoing debates and research in the field.

5. How does resolving divergent vacuum uncertainty impact our understanding of the physical world?

Resolving divergent vacuum uncertainty is crucial for making accurate predictions and understanding the physical world at a fundamental level. It allows for the development of more precise and reliable theories, which can then be used to make advancements in various fields such as particle physics and cosmology. By addressing this issue, we can gain a deeper understanding of the fundamental nature of our universe.

Similar threads

Replies
1
Views
836
  • Quantum Physics
Replies
4
Views
976
  • Quantum Physics
Replies
8
Views
2K
Replies
0
Views
607
Replies
1
Views
572
Replies
5
Views
1K
  • Quantum Physics
Replies
1
Views
1K
Replies
8
Views
1K
Replies
3
Views
486
  • Quantum Physics
Replies
5
Views
963
Back
Top