## Monday, November 27, 2017 ... //

### Quantum mechanics invalidates naive dimensional analysis

Anomalous dimensions are counterintuitive for beginners, important, and omnipresent

I have mentioned this subtle surprising consequence of quantum mechanics in the blog post against the Lagrangians two days ago. Too many people think that the quantization only means to add elegant hats above all the degrees of freedom and increase the frequency at which they say words like "weird" or "entanglement". But they still think classically.

That's not what you have needed to do to think scientifically after 1925. Quantum mechanics is a fundamentally different theory whose framework is very different conceptually and whose quantitative predictions may sometimes be similar as in classical physics but sometimes they can be different or totally different. And even when quantum mechanics ends up with the same or similar conclusions as classical physics, the derivation or argumentation may be very different. One of the omnipresent changes that quantum mechanics forces upon us is a new, modified dimensional analysis in field theory. What's going on?

In classical field theory, you have the action $$S=\int d^D x\,{\mathcal L}$$ if one exists at all. Because the action is dimensionless (in SI units, it has the dimension of the action i.e. the same as $$\hbar$$ but I will set $$\hbar=1$$) – exponents e.g. in the Feynman integrand $$\exp(iS/\hbar)$$ have to be dimensionless – the Lagrangian density $${\mathcal L}$$ has the dimension of $$M^D$$, the mass to the power of the spacetime dimension. The mass has the dimensions of $$[M]=1/[L]$$ in the $$c=\hbar=1$$ units.

Because we often write the bosonic kinetic terms in the Lagrangian density as${\mathcal L}_{\rm kin} = \frac 12 \partial_\mu \Phi \partial^\mu \Phi$ i.e. without any adjustable coefficients – we may normalize $$\Phi$$ so that the coefficient isn't there or it is just $$1/2$$ etc. – we may see that in $$D$$ spacetime dimensions, $$\partial_\mu$$ has the dimension of mass so $$\Phi$$ has to have the dimension $$M^{(D-2)/2}$$ for the Lagrangian density to have units of $$M^D$$. For example, in the $$D=4$$ spacetime that many readers incorrectly believe to inhabit, $$\Phi$$ – bosonic fields and/or gauge potentials – have the dimension of mass $$M$$.

Naively, a beginner is very likely to believe that this derivation of the units of the Klein-Gordon field $$\Phi$$ has to be valid in quantum mechanics, too. After all, the hat doesn't change anything about the units, does it? Well, the hat doesn't change the units directly but the relationship between the degrees of freedom – which become formidable operators thanks to their elegant hats – profoundly transforms (they don't commute with each other, among other things) which is why the dimensional analysis above is ultimately corrected – and may be completely wrong – in quantum mechanics.

The actual units of the operator are $$M^\Delta$$ where $$\Delta = \Delta_0+\gamma(g)$$ where $$\gamma(g)$$ is the anomalous dimension. Whatever is the difference between the actual dimension $$\Delta$$ and the naively, classically derived dimension $$\Delta_0$$ is called the anomalous dimension. It may depend on the gauge couplings and similar parameters in the field theory.

What is the simplest, canonical example? Textbooks of quantum field theory may pick some complicated examples to start with. But I believe the exponential operator in $$D=2$$ is the best, solvable, pretty, introductory example. What is the theory and what is the operator? Just consider a Klein-Gordon field $$X$$ in the $$D=2$$ dimensional spacetime – well, let's call it the world sheet (the history of a propagating string in string theory). The Klein-Gordon action (the kinetic term) is$S = \frac 12 \int d^2\sigma\, \partial_\alpha X \partial^\alpha X$ It's a normal Klein-Gordon action. Because we want to call this realm "world sheet" and not "spacetime", I chose a new name for coordinates, $$\sigma^\alpha$$ instead of $$x^\mu$$, and the indices were taken from the beginning of the Greek alphabet. Great. On the world sheet, you may still expand the quantum field $$X(\sigma^0,\sigma^1)$$ in terms of creation and annihilation operators multiplied by plane waves, just like you do in the $$D=4$$ Klein-Gordon theory.

You're used to the calculation of correlation functions of operators that are polynomial in $$X$$, right? But in $$D=2$$, we have something new and interesting. What are the units of $$X$$? In $$D$$ dimensions, it's "naively" $$M^{(D-2)/2}$$, as I derived previously. As you can see, for $$D=2$$, it's zero: the field $$X$$ is dimensionless; the two derivatives $$\partial_\alpha$$ already have the right units of mass to cancel the two-dimensional integral over the world sheet coordinates. For this reason, you may place it in the exponent and consider the operator${\mathcal O} = \exp(ikX)$ where $$k$$ is some coefficient. Note that $$ikX$$ must be dimensionless and because $$X$$ has no power of "length on the world sheet", $$k$$ doesn't have one, either. Both of them have some units involving the "length in the spacetime" but the spacetime is a different space than the world sheet where this theory is defined – in effect, it's some internal "configuration space", so these spacetime units are dimensionless from the world sheet viewpoint. (To allow $$X$$ to be written in normal "spacetime units", some coefficients that depend on $$\alpha'$$, which has the units of squared length, should be added everywhere but I will omit them.)

Just like the exponent inside $$\exp(\dots)$$ has to be dimensionless – e.g. because the Taylor expansion for the exponential adds the exponent and its powers to $$1$$ so all of those terms have to have the same units and be dimensionless, just like $$1$$ – the exponential itself is dimensionless, too. We may also reverse this statement: the argument of a logarithm should better be dimensionless and so is the logarithm itself.

The key question is: Is the exponential operator $${\mathcal O}$$ really dimensionless?

Of course, the answer is No. It's shocking but the exponential isn't really dimensionless in two-dimensional quantum field theory. Similarly and more generally, operators in quantum field theory almost never have exactly the same dimensions that you would derive by the classical dimensional analysis above. How is it possible? And what is the dimension of $${\mathcal O}$$?

A funny and powerful tool in conformal field theory – and our $$D=2$$ Klein-Gordon theory is conformal because there is no dimensionful parameter anywhere – is the state-operator correspondence. You know that holomorphic functions map the complex plane to a part of another complex plane so that the angles are preserved, right? And the exponential and logarithm are (locally) holomorphic functions, too. The exponential may map a cylinder to the whole complex plane (without the origin at zero).

Note that $$\exp(\sigma^0+ i\sigma^1)$$, the exponential of a complex number, is periodic in $$\sigma^1$$ with the periodicity $$2\pi$$, OK? This periodicity means that $$(\sigma^0,\sigma^1)$$ naturally parameterize an infinitely long cylinder. The circle with a constant $$\sigma^0$$ that goes around the cylinder gets mapped to a circle around zero in the complex plane by the exponential, OK? Physics of the Klein-Gordon field theory on the cylinder and on the complex plane is basically equivalent.

And a funny thing is that the insertion of an operator at the point $$z=0$$ of the complex plane is equivalent to exciting some state at $$\sigma^0\to -\infty$$, at the "infinite past", of the cylinder. Because of the properties of $${\mathcal O} = \exp(ikX)$$ under the translations of $$X$$ in the "target spacetime" $$X\to X+\Delta X$$, we may see that this operator carries some "spacetime momentum", and the corresponding state has to carry the same momentum.

It means that the operator $${\mathcal O} = \exp(ikX)$$ is equivalent to a state with the spacetime momentum $$k$$. But if the periodicity of $$\sigma^1$$ is $$2\pi$$, the momentum is the same as velocity (sorry if there is a numerical prefactor) and the state on the cylinder that we created at $$\sigma^0\to -\infty$$ is "moving" as a function of the Minkowski world sheet time $$\sigma^0_M$$:$X(\sigma^0) = X(0) + k \sigma^0_M.$ It's this simple. We created some state that describes a string at some (delocalized) position but the position is moving with the velocity proportional to the momentum. The time we used is the Minkowski world sheet time $$\sigma_M^0$$. But in the true conformal field theory, we need to convert it to the Euclideanized time which means we need to add some $$-i$$ (to be sure about the right sign takes quite some effort, let me not go there):$X(\sigma^0) = X(0) - ik \sigma^0_E.$ That's great. Let's insert two copies of the operator $${\mathcal O}$$ at two places of the complex plane – one at the origin and another one elsewhere. In the cylinder variables, this corresponds to starting with a momentum $$k$$ state in the infinite past and probing it by another operator$\exp[-ikX(\sigma^0_E)]$ at some time $$\sigma^0_E$$. I had to assign the opposite momenta to them (i.e. use two operators that are Hermitian conjugates of each other) for the total momentum to vanish, otherwise the expectation value would be zero. But in the previous displayed equation, we saw the dependence of $$X$$ on $$\sigma^0_E$$: it was linear with the simple coefficient (velocity) $$-ik$$. So the insertion of the extra operator will scale like$\exp(ik\cdot ik \sigma^0_E) = \exp(-k^2 \sigma^0_E)$ This is how the expectation value$\langle {\mathcal O}(\sigma^0_E=-\infty) {\mathcal O}^\dagger(\sigma^0_E)\rangle$ depends on $$\sigma^0$$. The dependence is the exponential of $$\sigma^0_E$$ multiplied by a constant, namely $$k^2$$. But $$\sigma_E^0$$ is just the logarithm of the radial coordinate in the plane,$\exp(-k^2 \sigma^0_E) = \exp(-k^2 \ln r) = \frac{1}{r^{k^2}}$ and this exponential of a multiple of logarithm may be rewritten as a simple power of $$r$$. The exponent is $$-k^2$$ or $$k^2$$ if we write the power in the denominator. And because this power of $$r$$ has to be divided between two copies of the operator $${\mathcal O}$$, we see that the dimension of $${\mathcal O}$$ is $$k^2/2$$ – I really mean the units are those of $$M^{k^2/2}$$. Depending on the coefficient $$k$$ in the exponential, the "spacetime momentum", the exponential field operator may have any real non-negative dimension!

Recall that classically, the dimension of the exponential operator $${\mathcal O}$$ is zero. Equivalently, when we insert two operators near to one another, there is no singularity, nothing special happening when the loci of the two operators nearly coincide, no diverging prefactor that would be a power of $$r$$; the expectation value would still scale like $$1$$ as a function of the small distance between the two points $$r$$. Quantum mechanics adds $$\alpha' k^2/2$$ to the (left-moving) dimension [I restored the $$\alpha'$$ with units of area to allow for SI units in the spacetime] – and the corresponding power of $$r$$, a new short-distance singularity, to the two-point function. Note that this nonzero dimension was forced upon us by quantum mechanics because two nearby field operators naturally have a singularity that scales like a power law of $$r$$ – it's one of the manifestations of the short-distance i.e. ultraviolet divergences (which make it subtle to place operators too close to each other). However, these ultraviolet effects are rather fundamental and directly follow from quantum mechanics because, as you could have seen, the nonzero exponent directly followed from the ability of the exponential operator to create "something that moves in the target spacetime".

We could have also deduced the anomalous dimension from the normal ordering of the exponential operator which is needed if you go beyond heuristic derivations – but I omitted the normal ordering above. So the nonzero dimension of $$:\exp(ik\hat X):$$ doesn't come from the hats, it really comes from the colons. ;-) The same normal ordering that makes the normalization of the operator finite (the correlation functions with other operators are finite, not including infinite factors from short-distance divergences) also forces the anomalous dimension on the operator. These two are inseparable.

You couldn't calculate anything in perturbative string theory if you were ignorant of the fact that the exponential operator has a nonzero dimension. Because the dimension is $$k^2/2$$, there are very interesting new effects that happen when $$k^2/2$$ is an integer. You can have new states associated with the exponential operator that behave almost indistinguishably from operators that are just polynomial in $$X$$ and the world sheet derivatives. In fact, there can be $$SU(2)$$ symmetries mixing such operators, and so on.

Anomalous dimensions and logarithmic divergences

Let me mention a related master example whose lesson is a bit different. It's an example that is more likely to appear in the "outdated" textbooks of quantum field theory that are focused on perturbative quantum field theory in $$D=4$$. Return to $$D=4$$ and consider some quantum field theory, e.g. the Standard Model. Operators in such theories also have anomalous dimensions – quantum corrections to the dimensions you would naively calculate by the classical dimensional analysis. The dimensions of operators may still be determined by looking at the correlation function$\langle {\mathcal O}(x,y,z,t) {\mathcal O}^\dagger(x',y',z',t')\rangle$ which will have some extra factor of $$1/r^{2\gamma}$$, some power of the distance between the two spacetime points, which will modify the dimension $$\Delta$$ of the operator $${\mathcal O}$$ by the extra term $$\gamma$$. How do these generally fractional powers $$1/r^{2\gamma}$$ of the distance between the two operators emerge in perturbative quantum field theory?

The answer is interesting. It's all about the logarithmic divergences. Why? Well,$\eq{ \langle {\mathcal O}(x,y,z,t) {\mathcal O}(x',y',z',t')\rangle &= \frac{1}{r^{2\gamma}} G\\ &=G (1 -2\gamma \ln r+\dots ) }$ The funny thing is that if we assume that the anomalous dimension $$\gamma$$ is small, and it should better be small if the field theory is weakly coupled and therefore well approximated by the classical field theory (and its dimensional analysis), we may expand the power of $$r$$, $$1/r^{2\gamma} = \exp(-2\gamma \ln r)$$, using the Taylor series and only pick the leading and the first subleading term. And because the logarithm of the base appears in the general power if it is written using an exponential, we get a term that goes like $$\ln r$$.

Well, as I mentioned before, arguments of logarithms should really be dimensionless so it's not quite $$r$$ that appears there if you're careful. The argument is some product $$r\Lambda$$ where $$\Lambda$$ has the compensating units of mass. It is some cutoff scale or another mass scale associated with the procedure to subtract the short-distance divergences. At any rate, you see that the coefficient of $$\ln r$$, and therefore a coefficient of $$\ln \Lambda$$, is proportional to $$\gamma$$, the anomalous dimension.

So if you write the correlation function – including the logarithmically divergent, $$\ln \Lambda$$ quantum corrections – in the format that looks like the Ansatz above, the coefficient of the logarithmically divergent correction divided by the leading term is equal to $$-2\gamma$$. You may directly extract the anomalous dimension from the logaritmically divergent term (if you omit the $$\ln \Lambda$$ factor itself)!

This is an important lesson because the logarithmically divergent Feynman diagrams (loop diagrams) tell you how the "real dimensions" (dependence of the magnitude on chosen units of length) differ from the naive classical ones. These logarithmically divergent diagrams play the same role as $$k^2/2$$ in my first example from the world sheet. More generally, logarithmically divergent Feynman diagrams tell you "how coupling constants run".

Classically, coupling constants are dimensionless in $$D=4$$. However, quantum mechanically, they're not quite dimensionless – they also have the anomalous dimension which is opposite to the anomalous dimension of the operators that these coupling constants multiply in the Lagrangian. We say that "they run". And they only run slowly, logarithmically, which means $$1/g(E_1)^2 - 1/g(E_2)^2 \sim \beta \ln(E_1/E_2)$$, and the speed may be determined from the logarithmically divergent diagrams. The first ones already start at the one-loop level, so they're suppressed by $$g^2$$, two gauge vertices, relatively to the finite tree-level terms. It means that the scaling of the anomalous dimension is$\gamma(e) \sim e^2$ where $$e$$ is a gauge coupling. I chose the letter $$e$$ for the electromagnetic coupling, the electron's charge in natural units, as an example. You may roughly say that the operators in QED have dimensions that differ from the naive classical ones by a multiple of $$1/137.036$$ because the latter is proportional to $$e^2$$, the fine-structure constant! QED is rather weakly coupled but the classical analysis is wrong in many respects. In particular, even the dimensions of the field operators may differ roughly by 1% from your classical estimates.

Again, we see that the logarithmically divergent Feynman diagrams have physical consequences – they modify dimensions of operators and they make coupling constants run. On the other hand, the power law divergences – even though they look more dramatic because they go to infinity more quickly – are less important and there's some sense in which all the power law divergent terms may be simply forgotten or set to zero. The dimensional regularization makes this point rather self-evident. But the logarithmic divergences, while seemingly "milder", are "more real". They contain real nonzero terms, genuine physical effects, and every one-loop calculation that completely erases the log divergent terms is bound to make physically wrong predictions.

Spin field and the anomalous dimension $$1/16$$

In the blog post about the non-Lagrangian field theories, I also mentioned that the Ising model has the spin field $$\sigma$$ which isn't a polynomial function of the basic Majorana fermion field $$\psi$$ and its derivatives. It's clear why it's not one: its dimension is $$\Delta=1/16$$. Again, this strange fractional dimension – the units are mass to the $$1/16$$ power – entirely follows from quantum mechanics.

There are many ways to calculate this dimension but as I mentioned, a simple one is$\eq{ 2\Delta &= \zav{ \frac 12 + \frac 32 + \frac 52 + \dots }-(1+2+3+\dots) =\\ &= \frac{1}{24} + \frac{1}{12} = \frac{1}{8}. }$ The sum of positive integers is $$-1/12$$, the sum of positive half-integers is $$+1/24$$, and the difference between them is $$1/8$$ which must be twice the dimension of the spin field. Why? It's because the spin field is the operator in the state-operator correspondence that has to be included if you want to switch from the sector of states where $$\psi$$ is periodic to the sector where it's antiperiodic or vice versa. So the spin field $$\sigma$$ must have the action of mapping the ground state of the periodic sector to the ground state of the antiperiodic sector or vice versa. The scaling dimensions of the operators are linked to the energies of states on the cylinder. So the dimension of $$\sigma$$ is linked to the energy difference $$E_{0,P}-E_{0,A}$$ between the energies of ground states in the periodic and antiperiodic sectors, respectively. The ground state energies come from the zero-point quantum fluctuations and they're ultimately proportional to the sum of positive integers or half-integers, depending on the sector of the string.

If you really can't swallow that the sum of integers is $$-1/12$$ etc. (the "infinite part" we omit when we make this statement is a specific example of the "power law divergences" that may be generally ignored, as I mentioned above), I assure you that there are lots of calculations that avoid the sum of integers, formulate it in a more careful way, or obfuscate its being $$-1/12$$ in various ways that are not too insightful. But if you really know what you're doing and you feel it's more than just some mathematical masturbation, you will agree that it's the finite invariant part of the sum of integers that matters here and it's simply $$-1/12$$.

To summarize, quantum mechanics changes lots of things and invalidates lots of assumptions that were taken for granted in classical physics. One of them is the dimensional analysis that was easy to made and usually implied that all quantities we consider have units of $$M^\Delta$$ where $$\Delta$$ is either integer or half-integer. In quantum mechanics, $$\Delta$$ may be a fraction such as $$1/16$$ or it may grow with the spacetime momentum of the exponential operators of strings, $$k^2/2$$, or it may be proportional to logarithmically divergent diagrams in quantum field theory that scale like $$e^2$$ for some coupling constant $$e$$.

All these things are probably unavoidably counter-intuitive for a beginner. But they're omnipresent facts and basic knowledge for any professional theoretical (or phenomenological) particle physicist. Those are expected to "start to get it" some year or two before they complete their PhD – if some "genius who proposes a theory of everything" doesn't get anomalous dimensions, he's surely less promising a "genius" than a good grad student.

In particular, logarithmically divergent diagrams may be the most important divergent diagrams because they can't be thrown away. Instead, they tell us about a new and unavoidable, quantum source of the physical phenomena's dependence on the distance scale, quantum corrections to the operators' dimensions etc. These short-distance divergences aren't pathologies. They're very interesting physics in consistent quantum field theories and their existence ultimately follows from nothing else than the ordinary nonzero commutators between operators – something that is needed to guarantee that things move and quantities evolve with time (because $$[H,X]\neq 0$$).

The nonzero commutators – e.g. with the Hamiltonian – are needed in quantum mechanics for quantities to change with time (because of the Heisenberg equations of motion). By the conformal map between the plane around the origin and the cylinder, the dependence on time is basically the same as the dependence on the radial distance – i.e. the distance between two operators. This dependence is also linked to the commutators. Those are generally nonzero and interesting. And this dependence on the distance between two operators in a product modifies the units of the whole product. That's a way to see that the anomalous dimensions unavoidably follow from the defining novelty of quantum mechanics, the nonzero commutators. You should really accept that these phenomena are important, inseparable, and unavoidable.