# Scale invariance in QFT?

About scale invariance in "beyond the standard model".

At the base of the analysis is the principle of scale invariance. So what is being said: what if there were another sector of the theory that interacts so weakly with the standard model that it hasn’t been noticed yet, and what if it were exactly scale-invariant?

It then mentioned: "A free massless particle is a simple example of scale invariant stuff because the zero mass is unaffected by rescaling. But quantum field theorists have long realized that there are more interesting possibilities — theories in which there are fields that get multiplied by fractional powers of the rescaling parameter.It is clear what scale invariance is in the quantum field theory. Fields can scale with fractional dimensions."

My question now is: What does he mean by that last sentence in bold? What is scale invariance in quantum field theory? Now I can say in QFT when electromagnetic field is quantized, there the photon has zero mass and is thus scale invariant. But what is being pointed out is to something else "more interesting" as said so what is that? And finally what does he mean by "fields can scale with fractional dimensions?"

• If you double the size of a square, its area increases by a factor of four. If you double the size of a cube, its volume increases by a factor of eight. These are examples of quantities that scale with integer dimension 2 and 3. Surprisingly there are other phenomena that scale with a non-integer dimensions, especially when self-similar and fractal objects are involved. This phenomenon can also be found near thermodynamic phase transitions which have close mathematical ties to quantum field theory. Maybe a theoretically inclined person can give an easy to understand QFT example. May 4, 2015 at 11:15
• Please provide proper reference (i.e. author, page, etc.) to quotes. Aug 4, 2015 at 11:36

What is meant by fractional scaling dimension is exactly what is says: Given a dilatation $x\mapsto\lambda x$, the field/operator $\mathcal{O}(x)$ behaves as $$\mathcal{O}(\lambda x) = \lambda^h\mathcal{O}(x)$$ with $h\in\mathbb{R}$ a possibly fractional or even irrational number.
The prime example of quantum field theories in which a fractional scaling dimension appears is for conformal field theories, which are always scale invariant because the scaling is just on of the conformal transformations. It is a bit unusual for an interesting theory to be scale and not conformally invariant, actually. What the author means with "scaling by fractional dimensions" is simply that quantum theories need not have integer $h$. Here's a "simple" example:
Consider a 2D theory of a Majorana fermion $\psi$ on the cylinder $\Sigma = S^1 \times \mathbb{R}$. The action is $$S[\psi] = \int_\Sigma \bar\Psi\gamma^\mu\partial_\mu\Psi$$ with $\Psi = (\psi \;\bar \psi)^T$. There is a conformal map to the complex plane such that $$S[\psi] = \int_\mathbb{C} \psi\bar\partial\psi + \bar\psi\partial\bar\psi$$ where the integration measure is in both cases already chosen invariant under conformal (and other) transformations. This theory is scale invariant iff under $z\mapsto\lambda z$,$\bar z \mapsto \bar\lambda\bar z$1 the fields behave as $$\psi(\lambda z,\bar\lambda\bar z) = \lambda^{1/2}\psi(z,\bar z)\text{ and } \bar\psi(\lambda z,\bar\lambda\bar z) = \bar\lambda^{1/2}\psi(z,\bar z)$$ where $\frac{1}{2}$ is clearly a fractional "scaling dimension". In the full quantum analysis, however, it turns out that there is a third independent state (which, by the state-field correspondence of CFTs, means there is a third independent field) that has scaling dimension $\frac{1}{16}$. This is essentially due to the possibility to choose anti-periodic boundary conditions for the spinor fields.
1It is an annoying convention to write $\bar\lambda$ for the factor of the second dilatation although it is not the complex conjugate of $\lambda$, just as $\bar h$ is not the complex conjugate of $h$ in the following
• Thank you for your nice answer @ACuriousMind. I have a query though, in the first equation $O(\lambda x) = \lambda ^h O(x)$ how can one know what $h$ should represent here (for example $h=1/2$ in your example). How can one know that? May 4, 2015 at 18:38
• Computing exponents like $h$ is very hard in general. In two-dimensional conformal QFT one can do it see for instance the original article by Friedan, Qiu and shenker journals.aps.org/prl/abstract/10.1103/PhysRevLett.52.1575 May 4, 2015 at 18:56