micromass said:
What does that even mean? ##L^2##-spaces show up plenty enough in classical probability. If you want to talk about variance, then you'll need something in ##L^2##.
Ok, That was not my focus. Sorry for not being precise. I was trying to help
Number Nine said:
What does "open a new channel mean"? If six is an option, then the probability is 3/6. If it isn't, then the
probability is 2/6. I don't see what QM adds in this case. There's nothing probabilistically groundbreaking about QM -- you square amplitudes and you get probabilities; there's nothing deep or unusual about it (at least, not in a probabilistic sense).
I fear you will have to rethink this.
My event was an even number, right?
The throwing of a dice was the general stochastic process in this context.
Without the 6 (let´s change to a five sides dice) there were two channels for my event to appear. Sides 2 and 4. -> probabilty 2/5, right?
By creating a new side, the 6, I have opened a new channel for my event to happen. Probability 3/6, larger.
My point is that, every time in classical probability when there is a change in the structure of the stochastic process, that increases the number of ways (channels) that conduce to the results which corresponds to "my event", its probability suffers an increase as well.
In QM, new channels for a given result may cause destructive interference when you take the square of a sum of complex numbers.
let ##z_1 ##, ##z_ 2 ## and ## z_3 ## be three complex numbers such that ##0 \le z_1 z_1^* + z_2 z_2^* + z_3 z_3^* < 1 ##
then, my point in this post is connected to the following mathematical fact:
For certain choices of complex numbers ##z_1 ##, ##z_ 2 ## and ## z_3 ##:
##(z_1 + z_2) (z_1 + z_2)^* > (z_1 + z_2 + z_3) (z_1 + z_2 + z_3)^* ## .
Best wishes,
DaTario