Wave function measurement question

[Kevlwig]

In Born interpretation of the wave function it notes that the matter itself cannot be measured however the square of its absolute value is measurable. I am lost as to why the product can be measured but not the wave function itself. Can someone provide clarity?

 

[mikeyork]

Neither the wave-function nor its absolute square is measurable. The wave-function predicts -- through its absolute square -- the asymptotic relative frequency distribution of possible measurement results over an infinite number of repeatable experiments. Since such a frequency distribution is, by definition, non-negative, the wave-function itself, which may be negative, cannot serve this purpose.

 

[bhobba]

In Born interpretation of the wave function it notes that the matter itself cannot be measured however the square of its absolute value is measurable. I am lost as to why the product can be measured but not the wave function itself. Can someone provide clarity?

If you make the reasonable assumption that the average of the results of outcomes of an observation is linear ie if you take O1 and O2 as observations then the average of the observable O1 + O2 is the sum of those averages then with a bit of QM math you can actually show the Born Rule. Its quite intuitive and Von Neumann gave the proof in his book on QM - Mathematical Foundations of QM. I have also given the proof - see post 137 of the following:
https://www.physicsforums.com/threads/the-born-rule-in-many-worlds.763139/page-7

The last bit is what I was referrring to:

O = ∑ <bi|O|bj> |bi><bj| = ∑ Trace (O |bj><bi|) |bi><bj|

Now we use the linearity idea I mentioned (I said something slightly different in the above - the change better reflects this context - and fixed a minor error I made - I posted it years ago and was quite careful - yet all these years later spotted a minor error)

f(O) = ∑ Trace (O |bj><bi|) f(|bi><bj|) = Trace (O ∑ f(|bi><bj|)|bj><bi|) where f is the function that gives the average ie f(O) is the average of the observable O if you measure it.

Define P as ∑ f(|bi><bj|)|bj><bi| and we have f(O) = Trace (OP).

P, by definition, is called the state of the quantum system. The following are easily seen. Since f(I) = 1, Trace (P) = 1. Thus P has unit trace. f(|u><u|) is a positive number >= 0 since |u><u| is an effect. Thus Trace (|u><u| P) = <u|P|u> >= 0 so P is positive.

This is the Born Rule in its full generality and a little further math shows it leads to the squaring rule you mention without detailing the actual math. Its a good exercise doing it if you are studying QM. If you can't do it do another post and me and/or someone else will detail it.

But the problem is intuition can lead you astray and it did Von-Neumann. He used it as the basis of a proof there are no hidden variables in QM. Trouble is the rule may or may not apply to hidden variables. Such was Von-Neumans reputation it took many years before this was realized - except for a few that tried to point it out. It wasn't really until the 60's and Bell that it was widely known.

Still as a heuristic reason why the Born Rule is true its quite reasonable especially to tell beginners who say - that looks strange - where did it come from.

Thanks
Bill