Scale in mathematics vs in quantum physics

[RockyMarciano]

What is the difference between the concept of scale in the SM QFTs and the scale concept in geometry?
In other words, QFT is a scale-dependent theory that is living on a scale-invariant space(either R4 or M4, related by Wick rotations). Mathematically scale-invariance and scale-dependence appear as incompatible, are they congenial in QFT? and if so how is that achieved?

 

[vanhees71]

I don't know how to answer you concrete question, because I've no clue what you mean by "difference of concept" here. A scale is a scale in geometry, and spacetime is a special kind of (in this case pseudo-Euclidean) geometry.

The reason why QFTs are very unlikely to be scale invariant is that scale invariance usually is anomalously broken, i.e., while the classical field theory may be scale invariant (e.g., pure Yang-Mills theory or QCD in the massless-quark limit), the quantized theory isn't.

 

[Demystifier]

Realistic QFT's are not scale invariant. Some toy QFT's are scale invariant, and they are used as an approximation to real physics because this symmetry simplifies calculations.

 

[RockyMarciano]

I don't know how to answer you concrete question, because I've no clue what you mean by "difference of concept" here. A scale is a scale in geometry, and spacetime is a special kind of (in this case pseudo-Euclidean) geometry.

The reason why QFTs are very unlikely to be scale invariant is that scale invariance usually is anomalously broken, i.e., while the classical field theory may be scale invariant (e.g., pure Yang-Mills theory or QCD in the massless-quark limit), the quantized theory isn't.

Right, so my question is how does the scale dependence of the QFTs that sustain the Standard model fit mathematically in a scale invariant spacetime?

 

[vanhees71]

Spacetime is not scale invariant to begin with, because in order to measure space and time intervals you need a unit of time (or space). In the SI units time is defined by a hyperfine transition in Cs, and that sets the scale "second" and defines also a length scale "light second" (in natural units you measure space and time intervals in seconds).

 

[RockyMarciano]

Realistic QFT's are not scale invariant.

That's what I wrote, and it follows naturally to wonder about the insertion of such theories in a scale-invariant Minkowski specetime. I'm aware that there are important difficulties in making certain physics theories mathematically sound, I'm just trying to corner some apparently evident from the mathematical point of view possible sources of those difficulties.

 

[RockyMarciano]

Spacetime is not scale invariant to begin with, because in order to measure space and time intervals you need a unit of time (or space). In the SI units time is defined by a hyperfine transition in Cs, and that sets the scale "second" and defines also a length scale "light second" (in natural units you measure space and time intervals in seconds).

When I say spacetime I'm referring to Minkowski spacetime or R4, I'm only referring to the mathematical definition of the mathematical spaces used in definitions of spacetime in QFT, sorry if that was a confusing shorthand.

 

[Demystifier]

That's what I wrote, and it follows naturally to wonder about the insertion of such theories in a scale-invariant Minkowski specetime. I'm aware that there are important difficulties in making certain physics theories mathematically sound, I'm just trying to corner some apparently evident from the mathematical point of view possible sources of those difficulties.

Usually the scale dependence in QFT is due to a mass term in the action. The mass does not depend on the spacetime metric, so the insertion is kind of trivial.

 

[RockyMarciano]

Usually the scale dependence in QFT is due to a mass term in the action. The mass does not depend on the spacetime metric, so the insertion is kind of trivial.

I can see that its appearance as a term in the action doesn't depend on the spacetime but it seems reasonable that it must be compatible with it if we are going to claim that the local gauge and the global symmetries that belong to that metric are perfectly congenial.

 

[Demystifier]

I can see that its appearance as a term in the action doesn't depend on the spacetime but it seems reasonable that it must be compatible with it if we are going to claim that the local gauge and the global symmetries that belong to that metric are perfectly congenial.

Well, the mass term is compatible in the sense that it transforms as a spacetime scalar.

 

[RockyMarciano]

Well, the mass term is compatible in the sense that it transforms as a spacetime scalar.

Yes, but you surely are aware that scale in QFT involves also the energy scale and also vector bosons.

 

[Demystifier]

Yes, but you surely are aware that scale in QFT involves also the energy scale and also vector bosons.

How are vector bosons relevant to the scale?

 

[RockyMarciano]

It is the interaction that introduces the scale in the 4-dimensional QFTs, no?.

 

[vanhees71]

The point is that even in theories without any dimensionful quantity as pure Yang Mills or massless QED/QCD, the scale invariance of the classical theory is broken when quantizing the theory. You can understand this formally in two ways:

(a) perturbatively: Evaluating quantum corrections, i.e., loop diagrams, you encounter divergences and thus have to renormalize the theory. E.g., in massless QED you have to renormalize the fermion self energy, which is linearly divergent. Now you cannot subtract the divergency at zero four-momentum, because there is an essential singularity in the self-energy function, because the threshold is at 0, because both the photon and the fermion are massless. So you have to subtract at a space-like four-momentum which introduces the renormalization scale $\Lambda$, and your running coupling and wave-function normalization becomes scale dependent, and scale-invariance is broken. Also the fermion doesn't stay massless, but you generate a mass dynamically, which also solves the problem with IR and collinear divergences (there's a famous paper by Eric Weinberg and Coleman on this, getting famous under the buzzword "mass without mass"). It's clear that through the introduction of a scale the scale invariance is broken.

(c) In the path integral the scale invariance transformation leads to a non-trivial change in the path-integral measure and thus the quantum action doesn't fulfill the naive Ward-Takahashi identity of the classical theory (which is the tracelessness of the energy-momentum tensor). Thus the symmetry is broken also in this "non-perturbative" sense. That's known as "trace anomaly".

http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.44.1733
http://journals.aps.org/prd/abstract/10.1103/PhysRevD.23.2262

 

[RockyMarciano]

Thanks for the informative post with which I obviously agree. It helps understanding how the scale dependence arises in renormalized QFTs.
My question was seeking a different thing though. As I said I'm interested in spelling out what's mainly behind the difficulties to make the renormalizable QFTs mathematically sound.
And one candidate that is obvious to me (I am trying to find out if for anyone else here) is the fact that when trying to reconcile the global symmetries of the mathematical space our theory lives in with the structure of the local gauge renormalizable QFTs that produce accurate predicitions, one finds this contrast between the scale invariance of the former and the scale dependence of the latter, irrespective of how this dependency comes about.

 

[PeterDonis]

a scale-invariant space(either R4 or M4, related by Wick rotations).

Under what definition of "scale invariant" are R4 and M4 (by which I assume you mean $R^4$ with a Euclidean and Minkowski metric, respectively) scale invariant? Any metric space has a length scale and so is not "scale invariant" under the obvious interpretation of that term.

 

[RockyMarciano]

Under what definition of "scale invariant" are R4 and M4 (by which I assume you mean $R^4$ with a Euclidean and Minkowski metric, respectively) scale invariant? Any metric space has a length scale and so is not "scale invariant" under the obvious interpretation of that term.

One precision first, R4 and M4 were meant to refer to affine Euclidean and Minkowskian spaces in 4 dimensions respectivley. But anyway the Euclidean geometry as a metric space is too scale-invariant. Also Minkowski space is not a metric space according to the mathematical definition of metric space, which is related but independent of whether it has a metric tensor or not, it is better not to conflate the two,
The concept of scaling transformation or dila(ta)tion and its correspondent symmetry leading to the concept of similarity in euclidean geometry is such a basic one that I didn't think it needed further explanation in an thread labeled advanced. Look up https://en.wikipedia.org/wiki/Similarity_(geometry) or https://en.wikipedia.org/wiki/Scaling_(geometry) or https://en.wikipedia.org/wiki/Dilation_(metric_space).
It is directly related to the notion of unitarity in the case we are discussing here, QFT, in the sense that without it you don't have the possibility of normalizing vectors in the presence of a gauge connection.
By the way, what specific length scale were you referring to? AFAIK neither Euclidean nor Minkowski geometries have an absolute scale length.

 

[PeterDonis]

the Euclidean geometry as a metric space is too scale-invariant

Again, under what definition of "scale invariant" is this true? Under the definition that's most obvious to me, that distances should be invariant under scale transformations (dilations), it is obviously false. The Wikipedia link you gave to the article on Dilations gives the definition of a dilation transformation which supports what I just said.

Minkowski space is not a metric space according to the mathematical definition of metric space

Yes, that's correct if we insist that a metric be positive definite; under this terminology Minkowski space is only a pseudometric space. But I don't see what this has to do with scale invariance; you can still define dilation transformations in a pseudometric space.

what specific length scale were you referring to?

The length scale defined by the metric; given a metric, there is an invariant length between any two points in the space. A dilation transformation changes this length.

 

[RockyMarciano]

Again, under what definition of "scale invariant" is this true? Under the definition that's most obvious to me, that distances should be invariant under scale transformations (dilations), it is obviously false.

What is false? You just wrote the definition everyone uses:"that distances should be invariant under scale transformations", this is scale-invariance and of course the definition I'm using. I honestly don't know what you are up to. But I would respectfully suggest for you to discuss these very basic geometric questions in a different thread as to not get very offtopic here and not distract from my OP question. Perhaps this migh help you too: http://math.stackexchange.com/questions/23129/why-is-euclidean-geometry-scale-invariant

The length scale defined by the metric; given a metric, there is an invariant length between any two points in the space. A dilation transformation changes this length.

And here you seem to be mixing up the physical notion of dimensionful length referred to some physical standard and the mathematical concept of distance that is a dimensionless quantity in the physical sense.

 

[PeterDonis]

You just wrote the definition everyone uses:"that distances should be invariant under scale transformations", this is scale-invariance and of course the definition I'm using.

I assume that by "scale transformations" you mean dilations? If so, then Euclidean N-space is obviously not scale invariant. A dilation transformation is a map from Euclidean N-space to itself such that the distance $d(x, y)$ between two points gets mapped to $r d(x, y)$, where $r$ is some nonzero real number. Any such transformation where $r \neq 1$ obviously changes distances, and therefore Euclidean N-space is not scale invariant by your own definition.

I would respectfully suggest for you to discuss these very basic geometric questions in a different thread as to not get very offtopic here and not distract from my OP question.

It seems to me that the definition of "scale invariant" that you are using, which is what I am trying to clarify, is crucial for understanding your question. If you mean "invariant under dilations", then that's fine, we can discuss what quantities in a QFT are invariant under dilations. But if that's the definition you are using, then your claim that Euclidean space is scale invariant is incorrect; so I'm trying to figure out whether you are just mistaken about that, or whether I'm not correctly understanding the definition of "scale invariant" that you are using.

Perhaps this migh help you too: http://math.stackexchange.com/questions/23129/why-is-euclidean-geometry-scale-invariant

This only shows that some geometric quantities in Euclidean geometry are scale invariant. It does not show that all geometric quantities in Euclidean geometry are scale invariant, which is what you are claiming.

you seem to be mixing up the physical notion of dimensionful length referred to some physical standard and the mathematical concept of distance that is a dimensionless quantity in the physical sense.

I haven't talked about "physical lengths" at all. The metric is a mathematical function; it takes two points and gives you a number. "Scale invariant" means that that number should not change under scale transformations. But dilation transformations with $r \neq 1$, as above, change the number that you get when you plug in the same two points.

 

[RockyMarciano]

I assume that by "scale transformations" you mean dilations? If so, then Euclidean N-space is obviously not scale invariant. A dilation transformation is a map from Euclidean N-space to itself such that the distance $d(x, y)$ between two points gets mapped to $r d(x, y)$, where $r$ is some nonzero real number. Any such transformation where $r \neq 1$ obviously changes distances, and therefore Euclidean N-space is not scale invariant by your own definition.
It seems to me that the definition of "scale invariant" that you are using, which is what I am trying to clarify, is crucial for understanding your question. If you mean "invariant under dilations", then that's fine, we can discuss what quantities in a QFT are invariant under dilations. But if that's the definition you are using, then your claim that Euclidean space is scale invariant is incorrect; so I'm trying to figure out whether you are just mistaken about that, or whether I'm not correctly understanding the definition of "scale invariant" that you are using.
This only shows that some geometric quantities in Euclidean geometry are scale invariant. It does not show that all geometric quantities in Euclidean geometry are scale invariant, which is what you are claiming.
I haven't talked about "physical lengths" at all. The metric is a mathematical function; it takes two points and gives you a number. "Scale invariant" means that that number should not change under scale transformations. But dilation transformations with $r \neq 1$, as above, change the number that you get when you plug in the same two points.

Oh, now I see where your confusion comes from. Scale invariance is certainly not the same as particular scalars not changing, even if in the end the outcome is equivalent to that, scale invariance is about preserving ratios. It is obviously the distance function that is preserved not some particular scalar or real number you obtain from a particular distance function, i.e. in mathematics, in the context of flat spaces that number is always re-scalable or normalizable in a translationally and rotationally invariant (pseudo)euclidean space. The special feature of having uniform isotropic scaling transformations producing similarities in the euclidean sense is what everybody understand by scale invariance(again not scalar invariance), you don't find this feature say in the other two constant curvature classical geometries:spherical and hyperbolic.

 

[PeterDonis]

Scale invariance is certainly not the same as particular scalars not changing, even if in the end the outcome is equivalent to that, scale invariance is about preserving ratios.

Ok, so now you're changing your definition. Before, you said scale invariance was about preserving distances; now it's ratios. Which definition do you want to use?

 

[RockyMarciano]

Ok, so now you're changing your definition. Before, you said scale invariance was about preserving distances; now it's ratios. Which definition do you want to use?

I paraphrased your own wording back when I still thought you meant the form of distance functions when you said distances. In euclidean geometry if you want to compare specific distances at one scale you obtain the same results of the comparison in any other scale, that is scale invariance.There is not one scale that is favored. This is not the case in renormalizable QFTs as commented by Demystifier and vanhees71 above, and this discrepancy between the physical theory and the Minkoski space with Poincare invariance in which is set in might not be trivial IMO.

 

[vanhees71]

I don't understand the confusion. The transformation describing scalings is clearly defined the space-time vectors obey $x \rightarrow x'= \lambda x$, $\lambda \neq 0$ real, and any field $\varphi(x) \rightarrow \varphi'(x')=\lambda^{-\delta_{\varphi}} \varphi(x)=\lambda^{-\delta_{\varphi}} \varphi(x/\lambda')$. For a scalar field, e.g., in (1+3) space-time dimensions the scaling dimension is $\delta_{\varphi}=1$.

For a massless free Klein-Gordon field you have
$$\mathcal{L}=\frac{1}{2} \partial_{\mu} \varphi \partial^{\mu} \varphi.$$
Now we have
$$\varphi'(x')=\frac{1}{\lambda} \varphi(x)$$
and
$$\partial_{\mu}' \varphi'(x')=\frac{\partial x^{\nu}}{\partial x_{\mu}'} \frac{1}{\lambda} \partial_{\nu} \varphi(x) = \frac{1}{\lambda^2} \partial_{\mu} \varphi(x).$$
Further
$$\mathrm{d}^4 x' = \lambda^4 \mathrm{d}^4 x.$$
Thus the action is scaling invariant. The Noether theorem tells you that the trace of the energy-momentum tensor vanishes.

As an interaction the only scaling-invariant possibility is with a dimensionless coupling, i.e.,
$$\mathcal{L}_{\text{int}}=-\frac{\lambda}{4!} \varphi^4.$$
Thus you have massless $\varphi^4$ theory. When renormalizing it you must inevitably introduce a space-like momentum as the subtraction point, which introduces a momentum scale and thus breaks inevitable dilation invariance, i.e., the symmetry of the classical model is broken when quantizing the model, leading to the trace anomaly.

 

[RockyMarciano]

[...] When renormalizing it you must inevitably introduce a space-like momentum as the subtraction point, which introduces a momentum scale and thus breaks inevitable dilation invariance, i.e., the symmetry of the classical model is broken when quantizing the model, leading to the trace anomaly.

Exactly, this is what I'm describing, the trace anomaly(aka conformal anomaly, scale anomaly...). Surely you know this scale anomaly appearance is related to whether the space of the theory is odd, or even dimensional, for instance the three dimensional(2+1) conformal QFT remains conformal after quantization: no scale anomaly; so it is not inevitably linked to quantization. Two spatial dimensions are always conformally flat so that might help explain the absence of scale anomaly when introducing momentum scale in 2+1 conformal QFT.
So my point is, don't you consider that at least in the physically relevant case of 3+1 dimensions, the scale anomaly is one obvious candidate for the source of mathematical difficulties when trying to make 4 dimensional renormalizable QFT rigorous?

 

[RockyMarciano]

To phrase it differently, my doubt is why is not the scale anomaly seen as a hint that the global symmetry is not phenomenolgically adequate as guiding principle?

 

[vanhees71]

I don't know what you mean. Since the scale invariance in many models is broken, it's simply not a symmetry in these models.

 

[RockyMarciano]

Since the scale invariance in many models is broken, it's simply not a symmetry in these models.

Well, yes. But unless I'm missing something It so happens that a space with Poincare symmetry must have scale invariance as symmetry,i.e. it must have a symmetry preserving relative distances, which as I clarified in a previous post is not the same as having dilations as an extra isometry(except for the identity), the group of Poincare doesn't include dilations. This is related to the extension of the Lorentz algebra with the generator of translations $[P_\mu, P_\nu] = 0$ in the Poincare algebra. Breaking scale-invariance breaks this extension.

So it seems inevitable to naively ask, given this, how is Poincare symmetry kept?

 

[RockyMarciano]

Actually my question is kind of trivial in a very obvious way(so I wonder if the lack of answers means nobody is understanding what I'm talking about, if so please ask), the Poincare symmetry relevant to QFT is concerned with infinitesimal transformations and therefore is not affected by the breaking of the scale invariance.
But the not totally trivial part I was trying to put forth is that, even if the trace anomaly, being a global anomaly, is not considered relevant for the physical relevance and consistency of QFT(i.e. only the gauge anomalies are essential for consistency so it's a good thing they cancel in the SM), the purely mathematical problem of trying to obtain a valid(in the sense of rigorous as used by mathematicians) mathematical model of the SM gauge that exists on Minkowski 4-space seems to be connected to the global anomalies and more specifically the trace anomaly.
In other words I can see how there is a sense in which the scale anomaly (or global anomalies in general) remaining uncancelled is not a direct problem for the physical consistency of renormalizable 4-dimensional QFT the consistency of which hinges on the cancellation of the gauge anomalies, but isn't it an obstruction mathematically to formally obtain the rigorous theory on Minkowski space?

 

[Demystifier]

So it seems inevitable to naively ask, given this, how is Poincare symmetry kept?

In general, one should distinguish symmetry of general laws from symmetry of special solutions.

Take, for instance, one free photon with 4-momentum $k$. It is not a Lorentz invariant object (and hence not Poincare invariant object) because in another Lorenz frame it has momentum $k'\neq k$. Nevertheless, this photon is described by Lorentz invariant laws. But one photon with one particular momentum is a particular solution of these laws, and it is not a Lorentz invariant solution. In free QED, which is the theory of free photons and electrons, the only Lorentz invariant solution is the vacuum.

Of course, free particles are not very interesting from a physical point of view, but the example above serves to understand the difference between symmetry of laws and symmetry of solutions. Equipped with this conceptual understanding, now we can try to understand something less trivial.

Now consider interacting QED. Take, for instance, scattering of two electrons with given initial momenta. In calculation of the scattering amplitude one encounters loop diagrams which appear divergent. To make them convergent one takes some cut-off, and this violates scale invariance and Poincare invariance. Physically, the effective (i.e. renormalized) coupling constant, which initially was a true constant, now depends on energy. Energy is not scale and Lorentz invariant, so naively one would say that scale and Poincare invariance are violated. But one should recall that it is obtained during a study of a special case: scattering of two electrons with given initial momenta. So this "violation" is merely the absence of symmetry in the special solution. The general laws are still Poincare and scale invariant.