(I’ve said this before, but that was many years ago and could do with restating).

Quantum mechanics deals with infinite numbers all the time. Particularly in the process known as regularisation within renormalisation.

At school we are taught that ∞≠∞+1. This type of infinity is called ‘shift invariant’ or ‘Archimedean’, and its use is largely due to its promotion by Cantor in his cardinal numbers. Cantor also came up with ‘ordinal infinity’. Let ω be the number of natural numbers (or the set of natural numbers), then 1+ω=ω≠ω+1. Cantor ordinals are interesting, but not very useful.

A more useful type of infinity was developed by Robinson in the 1950s. https://en.wikipedia.org/wiki/Hyperreal_number. Here, ω≠ω+1=1+ω. This is non-Archimedean, non-shift-invariant. Robinson’s hyperreal numbers are consistent with other systems of infinite numbers developed by mathematicians: Laurent, Veronese, du Bois-Reymond, Levi-Civita, Hahn, Hardy, Bessel, Dahn-Göring, Conway, Giodorno-Katz and Ehrlich. Robinson has developed these furthest. Robinson’s hyperreal numbers also include the infinitesimals; the most familiar infinitesimal in Quantum Mechanics is, of course, ‘dx’. Robinson has rigorously proved that every theorem that is true for the real numbers is also true for the hyperreal numbers. To the non-mathematician this is starkly incredible, because it includes the theorem that every infinite number has a unique factorisation.

The concept of ‘limit’ needs review. There are two different types of limit, the shift invariant limit eg. lim _x→∞ (1/x) and the non-shift-invariant limit lim _x→a (1/x). Robinson works with the shift invariant limit a lot. But here I use the non-shift-invariant limit. So far as I’ve been able to ascertain, the non-shift-invariant limit on the hyperreals allows us to banish the concept of “divergence”. Every sequence has a limit, every series has a limit, every integral has a unique solution. Some integrals that are potentially useful in quantum mechanics are in the attached image. For rigor, please use ω instead of ∞.

But there is a subtlety. Quantum entanglement. The factorisation of infinite numbers relies on the ability to choose. For example if we choose ω to be even then ω+1 is odd. We can choose (ie. observe) the quantum state of one particle in an entangled pair, and that gives us the quantum state of the other particle. We can choose the value of an integral to have a value other than the value in the attached image, and that then alters the value of the other integrals.

I’d like to see some more detailed working on that one, poindexter.

dv said:

I’d like to see some more detailed working on that one, poindexter.

Elementary, my dear Watson.

I’ll look it up.

mollwollfumble said:

dv said:

I’d like to see some more detailed working on that one, poindexter.

Elementary, my dear Watson.

I’ll look it up.

Oh dang it, my notes are hand written not on computer and I have to find which notebook I’ve written them in.

The procedure is simple enough. Use Riemann sums to convert the integral into a series.

Then use the methods in https://en.wikipedia.org/wiki/Divergent_series to evaluate the series.
Or, easier, a different method that yields the exact same results,
split the sequence into a smooth component and an oscillating component, discard the oscillating component and take the smooth component to infinity.

eg. 1-2+3-4+5-6+7-8+ …
=(1,-1,2,-2,3,-3,4,-4,…)
=1/4 + (3/4.-5/4,7/4,-9/4,11/4,-13/4,15/4.-17/4, …)
=1/4 + 1/4 (2n+1)(-1)^(n+1)
the function 2n+1 is smooth and (-1)^(n+1) is purely oscillatory, so the product can be discarded, leaving
1-2+3-4+5-6+7-8+ … = 1/4

And you can find this on https://en.wikipedia.org/wiki/1_%E2%88%92_2_%2B_3_%E2%88%92_4_%2B_%E2%8B%AF

The same principle gives the integral of x cos(x) from zero to infinity.

mollwollfumble said:

mollwollfumble said:

dv said:

I’d like to see some more detailed working on that one, poindexter.

Elementary, my dear Watson.

I’ll look it up.

Oh dang it, my notes are hand written not on computer and I have to find which notebook I’ve written them in.

The procedure is simple enough. Use Riemann sums to convert the integral into a series.

Then use the methods in https://en.wikipedia.org/wiki/Divergent_series to evaluate the series.
Or, easier, a different method that yields the exact same results,
split the sequence into a smooth component and an oscillating component, discard the oscillating component and take the smooth component to infinity.

eg. 1-2+3-4+5-6+7-8+ …
=(1,-1,2,-2,3,-3,4,-4,…)
=1/4 + (3/4.-5/4,7/4,-9/4,11/4,-13/4,15/4.-17/4, …)
=1/4 + 1/4 (2n+1)(-1)^(n+1)
the function 2n+1 is smooth and (-1)^(n+1) is purely oscillatory, so the product can be discarded, leaving
1-2+3-4+5-6+7-8+ … = 1/4

And you can find this on https://en.wikipedia.org/wiki/1_%E2%88%92_2_%2B_3_%E2%88%92_4_%2B_%E2%8B%AF

The same principle gives the integral of x cos(x) from zero to infinity.

Yeah but the thing is, it’s wrong? The oscillatory part of that integral continues to blow up as x increases. It doesn’t converge.

Oops, let’s try that again.

The non-shift-invariant limit at infinity doesn’t know which part of the cycle it’s in.
So it equates to the mean value.

That’s how renormalisation works in QM. Discard the oscillation and keep the mean.

Except when two particles are entangled. In which case the value can be anywhere in the range.

mollwollfumble said:

Oops, let’s try that again.

The non-shift-invariant limit at infinity doesn’t know which part of the cycle it’s in.
So it equates to the mean value.

That’s how renormalisation works in QM. Discard the oscillation and keep the mean.

Except when two particles are entangled. In which case the value can be anywhere in the range.

Perhaps I can put it this way. In travelling from Point A to Point B, light travels out to infinity in both time and space according to QM. But we take the mean value path, which is the straight line joining A and B.

All these different infinities are a mystery to me, as is the whole process of renormalisation, but in QM do the infinite integrals diverge, as in your examples?