Single-world interpretation, 7

One consequence is that every observation can be thought of as causing the combined observer-object’s wavefunction to change into a quantum superposition of two or more non-interacting branches, or split into many “worlds”.

— Wikipedia on Many-worlds interpretation

That is incorrect.

Let’s consider the double-slit experiment. For simplicity, we regard the event “a person reads the device reading” as a classical event.

Before installing the measuring device, we do not know which slit a photon goes through. The photon state is in a superposition of eigenstates: 

| photon > = a | left > + b | right >

(According to the meaning of probability, |a|^2 + |b|^2 = 1.) In other words, if we send enough such kind of photons through the double-slit apparatus, we get the interference pattern. 

After installing the measuring device, we know which slit a photon goes through. According to the Copenhagen interpretation, when the photon passes through the double-slit apparatus, the photon-state “collapses” to one of the two eigenstates, such as | left >. However, a more accurate point of view is that, according to the quantum decoherence interpretation, the photon-and-device state becomes a superposition of a lot of eigenstates. Most of such eigenstates are corresponding to the macrostate of passing-through-the-left-slit, |left>_macro_state. 

The above many-worlds-interpretation statement assumes that there is a |right>_macro_state.

It is true in a sense that, since the photon-and-device involves a lot of particles, there are so many eigen-microstates. Some are certainly corresponding to the |right>_macro_state.

It is false in a sense that the weighting of the |right>_macro_state is so small that such macrostate is not meaningful in a macroscopic context, for example:

| photon-and-device > = 10^23 |left>_macro_state + 0.001 |right>_macro_state + other possible macrostates

— Me@2012-04-07 11:03:12 AM

2012.04.09 Monday (c) All rights reserved by ACHK

Probability 3.1

這段改編自 2010 年 6 月 2 日的對話。

15 個錢幣之中,有 3 個是金的。假設你要由那 15 個錢幣之中,抽 4 個出來,抽到 3 個金幣的機會率是多少?

已知整個過程是隨機的,即是抽中各個錢幣的機會均等。

這一題有兩種運算方法。

第一種運算方法,我稱之為「機會率方法」,簡稱「P 」。

P 方法:

首先,假想有四個空格:

(_)(_)(_)(_)

然後,你要抽四個錢幣出來,逐一填滿那四個空格。為簡便起見,我們暫時先把題目改為:抽到「頭三個是金」和「尾一個非金」的機會率是多少?

(金)(金)(金)(X)

接著,你可以考慮,第一個抽到金幣的機會是多少。因為原本有 15 個錢幣給你抽,所以第一個分數的分母是 15。那 15 個錢幣中,有 3 個是你所要的金幣,所以第一個分數的分子是 3。結論是,第一個分數是 3/15。

(3/15)(金)(金)(X)

到你抽第二個錢幣時,只剩下 14 個,所以第二個分母是 14。那 14 個錢幣中,還有兩個是金的,所以第二個分子是 2。

(3/15)(2/14)(金)(X)

如此類推,你亦可以得到餘下的兩個分數。

(3/15)(2/14)(1/13)(12/12)

但是,原本的題目並沒有規定 4 個錢幣之中,哪 3 個是金的,所以,我們要乘 4C3 於剛才的那個中途答案之上,因為,4 個之中放 3 個金幣,共有 4C3 種放法。4C3 即是 「4 選 3」,等於 4。

(3/15)(2/14)(1/13)(12/12)4C3 = 0.008791

這個方法,是透過幾個「機會率分數」的相乘來獲得答案,所以,我稱之為「機會率方法 P」。

— Me@2012.04.09

2012.04.09 Monday (c) All rights reserved by ACHK