This is a follow on from my paper explaining renormalisation. A question was raised – why exactly do we need a cut-off. There is a deep reason to do with dimensional analysis, and the power series expansion used in perturbation theory. Along the way we will see renormalisation in a more general setting, and exactly why logarithms, like in the previous paper, so often crop up.
A More General Look At Renormalisation
Suppose we have a function G(x) that depends on some parameter λ ie G(x,λ). Then, so perturbation theory can be used, expand it in a power series about λ:
G(x) = G0 + G1(x)*λ + G2(x)*λ^2 + ……..
In perturbation theory, for theoretical convenience, it is usual to define a new function F(x) = (G(x) – G0)/G1 so:
F(x) = λ + F2(x) *λ^2 + ……..
It makes things like formal inversion etc easier. This seems a pretty innocuous thing to do, but from dimensional analysis, has a consequence that lies at the heart of where QFT infinities come from, and the need for a cut-off. To see this, suppose x has some kind of dimension such as, for example, momentum squared, and λ is dimensionless. A number of theories fall into this class; such as:
quantum electrodynamics where the fine structure constant is dimensionless, and only high energies are considered, so the electron mass is negligible;
the Weinberg-Salam model of electro-weak interactions;
the meson theory, again at high energies so the mass is negligible, used as an example in my previous paper. K^2 has dimensions of momentum squared and the coupling constant is dimensionless.
Suppose λ is small, then F(x) = λ, F has the dimensions of λ, so is dimensionless. This is also seen by its definition where G(x) – G0 is divided by G1(x). But lets expand F2(x) in a power series about x so F2(x) = F20 + F21*x + F22*x^2 + ……. = F20 + F21*x + O(x^2). Suppose x is small, so O(x^2) can be neglected, then F2(x) has the dimensions of x, hence to second order of λ, F(x) has the dimensions of x. Here we have a dimensional mismatch. This is the exact reason the equations blow up – in order for it to be dimensionless it cant depend on x. This can only happen if F2(x) is a constant or infinity. Either of course is death for our theory – but nature seemed to choose infinity – the reason for which will be examined later.
Now for the solution. The only way to avoid this is to divide x by some parameter, Λ, of the same units as x, so it becomes dimensionless.
The correct equation is:
F(x/Λ) = λ + F2(x/Λ) *λ^2 + F2(x/Λ) *λ^2 +………+ Fi(x/Λ) *λ^i + ……………
We see, due to dimensional analysis of the perturbation methods used, we have neglected a parameter in our theory, which can be interpreted as a cut-off. It is this oversight that has caused the trouble all along.
Consequence Of The Introduction Of Λ
To second order we have F(x/Λ) = λ + F2(x/Λ) *λ^2. (1)
The issue is while we know there is a Λ, we do not know its value so, as per the example in my previous paper, we want a formula without it. Similar to what was done before, we define the renormalised coupling constant:
λr = F(u/Λ) = λ + F2(u/Λ) *λ^2. (2)
Here u is some arbitrarily chosen value of x that yields a value of λr that can be measured.
Subtracting (2) from (1), and noting that to second order λr^2 = λ^2, we get:
F(x/Λ) = λr + (F2(x/Λ) – F2(u/Λ))*λr^2.
We want this to not depend on Λ, so F2(x/Λ) – F2(u/Λ) = f(x,u), where f(x,u) depends on x and u, but not Λ. Theories where this works to eliminate Λ are called renormalisable. Not all theories are renormalisable – but as we will see, if it is, this imposes restrictions on the equations.
Let g(x) = f(x,1) = F2(x/Λ) – F2(1/Λ) ⇒f(x,u) = g(x) – g(u). Let K(x) = F2(x/Λ) – g(x) ⇒ K(x) – K(u) = 0 ⇒ K(x) = K(u). But since x and u are independent, K can’t depend on x or u, so must only depend on Λ ie K = K(Λ). Thus:
F2(x/Λ) = g(x) + K(Λ).
We see the renormalisation condition, which is basically we want to get rid of the unknown Λ, determines the form of F2(x/Λ), namely it is the sum of a function of x and a function of Λ. The reason renormalisation works is when you subtract (2) from (1) the Λ dependant term cancels.
Why You Get Logarithms
An interesting consequence of this is it must involve logarithms. That the meson/meson scattering formula in the previous paper contained them is no accident.
Taking the derivative wrt to x in F2(x/Λ) = g(x) + K(Λ) ⇒ F2′(x/Λ)/Λ = g'(x). Let x =1. F2′(1/Λ)/Λ = g'(1) which will be called -α, where its conventional to use a minus sign because that’s what tends to occur in equations, such as the C in the previous paper. Let 1/Λ = y ⇒ F2′(y) = -α/y whose solution is F2(y) = -α*log(y) + C.
Hence we have:
F2(x/Λ) = -α*log (x/Λ) + C = α*log (Λ/x) + C = α*log (Λ) – α*log (x) + C.
As promised we see that α occurs in α*log (Λ) like the meson/meson scattering equation; justifying the negative sign.
This has exactly the same form as the equation for meson/meson scattering in the previous paper. However it can be simplified further to eliminate C. This is done by subtracting C*λ^2 from F(x/Λ) to give F(x/Λ) – C*λ^2. Using this new F we have:
F2(x/Λ) = α*log (Λ/x) = α*log (Λ) – α*log (x).
Why Did This Take So Long To Sort Out
We have seen the use of perturbation theory secretly requires another parameter to make sense. If you don’t include it, dimensional analysis shows you will get nonsense, with this nonsense manifesting in the infinities.
Even worse was an incorrect assumption about the coupling constant λ. Measurements showed it was much less than 1, so it looked reasonable to use in a perturbation expansion. But now we know there is a neglected parameter, Λ, in our equations, lets look at what happens to λ when that is taken into account.
To second order:
λ = λr + a*λr^2
F(x/Λ) = λ + α*log(Λ/x)*λ^2 = λr + a*λr^2 + α*log(Λ/x)*λr^2 = λr + (α*log(Λ/x) + a)*λr^2
But λr = F(u/Λ) = λr + (α*log(Λ/u) + a)*λr^2 ⇒ α*log(Λ/u) + a = 0 ⇒ a = -α*log(Λ/u) = α*log(u/Λ). Hence:
λ = λr + a*λr^2 = λr + α*log(u/Λ)*λr^2.
We see the coupling constant depends on this new parameter. Now, making the reasonable interpretation of Λ as a cut-off, lets remove it by taking the limit at infinity similar to the previous paper. When this is done, we see to first order, the coupling constant λ = λr, so, in our first order calculations, no problem arose. But at second order it blows up to -∞. Its also interesting to note the other reasonable choice to get rid of Λ, taking the limit to zero, also leads to it blowing up – this time to ∞.
In perturbation theory you want what we perturb about to be much less than one. But for it to actually be infinite – that’s really, really bad, and no wonder you get nonsense infinite answers.
Measurements gave small values of the coupling constant, which from the above equation, means Λ isn’t too large, or small. This is what fooled people all those years.
We have seen there is a secret parameter in our theories required by dimensional analysis. The inclusion of this parameter, and the renormalisation condition, leads to them having a certain form. For theories with a dimensional parameter, and a dimensionless coupling constant, to second order it is F(x/Λ) = λ + α*log(Λ/x)*λ^2.
Its very interesting that dimensional considerations show why there is a parameter missing. When it’s not introduced, you get nonsense. If its included, then requiring our equations to be renormalisable, constrains its form.
I posted the following paper before:
It extends these ideas a lot further by calculating higher order terms, and investigating the important renormalisation group. Trouble is it has a few (relatively minor) errors and isn’t 100% clear what’s going on in some areas.
I hope to do some further papers giving the third and higher order terms, plus the renormalisation group.