The problem of induction 3.2

The meaning of induction is that

we regard, for example, that

“AAAAA –> the sixth is also A”

is more likely than

“AA –> the second is also A”


We use induction to find “patterns”. However, the induced results might not be true. Then, why do we use induction at all?

There is everything to win but nothing to lose.

— Hans Reichenbach

If the universe has some patterns, we can use induction to find those patterns.

But if the universe has no patterns at all, then we cannot use any methods, induction or else, to find any patterns.


However, to find patterns, besides induction, what are the other methods?

What is meaning of “pattern-finding methods other than induction”?

— Me@2012.11.05

— Me@2018.12.10



2018.12.10 Monday (c) All rights reserved by ACHK

The problem of induction 3


In a sense (of the word “pattern”), there is always a pattern.


Where if there are no patterns, everything is random?

Then we have a meta-pattern; we can use probability laws:

In that case, every (microscopic) case is equally probable. Then by counting the possible number of microstates of each macrostate, we can deduce that which macrostate is the most probable.


Where if not all microstates are equally probable?

Then it has patterns directly.

For example, we can deduce that which microstate is the most probable.

— Me@2012.11.05



2018.11.19 Monday (c) All rights reserved by ACHK

Monty Hall problem 1.6

Sasha Volokh (2015) wrote that “any explanation that says something like ‘the probability of door 1 was 1/3, and nothing can change that…’ is automatically fishy: probabilities are expressions of our ignorance about the world, and new information can change the extent of our ignorance.”

— Wikipedia on Monty Hall problem



2018.11.02 Friday ACHK

The square root of the probability

Probability amplitude in Layman’s Terms

What I understood is that probability amplitude is the square root of the probability … but the square root of the probability does not mean anything in the physical sense.

Can any please explain the physical significance of the probability amplitude in quantum mechanics?

edited Mar 1 at 16:31

asked Mar 21 ’13 at 15:36


Part of you problem is

“Probability amplitude is the square root of the probability […]”

The amplitude is a complex number whose amplitude is the probability. That is \psi^* \psi = P where the asterisk superscript means the complex conjugate.{}^{[1]} It may seem a little pedantic to make this distinction because so far the “complex phase” of the amplitudes has no effect on the observables at all: we could always rotate any given amplitude onto the positive real line and then “the square root” would be fine.

But we can’t guarantee to be able to rotate more than one amplitude that way at the same time.

More over, there are two ways to combine amplitudes to find probabilities for observation of combined events.


When the final states are distinguishable you add probabilities:

P_{dis} = P_1 + P_2 = \psi_1^* \psi_1 + \psi_2^* \psi_2


When the final state are indistinguishable,{}^{[2]} you add amplitudes:

\Psi_{1,2} = \psi_1 + \psi_2


P_{ind} = \Psi_{1,2}^*\Psi_{1,2} = \psi_1^*\psi_1 + \psi_1^*\psi_2 + \psi_2^*\psi_1 + \psi_2^* \psi_2


The terms that mix the amplitudes labeled 1 and 2 are the “interference terms”. The interference terms are why we can’t ignore the complex nature of the amplitudes and they cause many kinds of quantum weirdness.

{}^1 Here I’m using a notation reminiscent of a Schrödinger-like formulation, but that interpretation is not required. Just accept \psi as a complex number representing the amplitude for some observation.

{}^2 This is not precise, the states need to be “coherent”, but you don’t want to hear about that today.

edited Mar 21 ’13 at 17:04
answered Mar 21 ’13 at 16:58


— Physics Stack Exchange



2018.08.19 Sunday (c) All rights reserved by ACHK

Quantum Computing, 2

stcredzero 3 months ago

A note for the savvy: A quantum computer is not a magic bit-string that mysteriously flips to the correct answer. A n-qubit quantum computer is not like 2^n phantom computers running at the same time in some quantum superposition phantom-zone. That’s the popular misconception, but it’s effectively ignorant techno-woo.

Here’s what really happens. If you have a string of n-qubits, when you measure them, they might end up randomly in [one] of the 2^n possible configurations. However, if you apply some operations to your string of n-qubits using quantum gates, you can usefully bias their wave equations, such that the probabilities of certain configurations are much more likely to appear. (You can’t have too many of these operations, however, as that runs the risk of decoherence.) Hopefully, you can do this in such a way, that the biased configurations are the answer to a problem you want to solve.

So then, if you have a quantum computer in such a setup, you can run it a bunch of times, and if everything goes well after enough iterations, you will be able to notice a bias towards certain configurations of the string of bits. If you can do this often enough to get statistical significance, then you can be pretty confident you’ve found your answers.

— An Argument Against Quantum Computers

— Hacker News



2018.05.17 Thursday ACHK

Mixed states

To me the claim that mixed states are states of knowledge while pure states are not is a little puzzling because of the fact that it is not possible to uniquely recover what aspects of the mixed state are subjective and what aspects are objective.

The simple case is this:

Let’s work with a spin-1/2 particle, so there are states:

|0 \rangle
|1 \rangle
|+ \rangle = \frac{1}{\sqrt{2}} \left( |0 \rangle + |1 \rangle \right)
|- \rangle = \frac{1}{\sqrt{2}} \left( |0 \rangle - |1 \rangle \right)

The mixed state corresponding to 50% |0> + 50% |1> is the SAME as the mixed state corresponding to 50% |+> + 50% |->.

— Daryl McCullough

— Comment #13 November 19th, 2011 at 2:00 pm

— The quantum state cannot be interpreted as something other than a quantum state


\frac{1}{2}_c | + \rangle \langle + | + \frac{1}{2}_c | - \rangle \langle - |

=\frac{1}{2}_c \left( \frac{1}{\sqrt{2}}_q | 0 \rangle + \frac{1}{\sqrt{2}}_q | 1 \rangle \right) \left( \frac{1}{\sqrt{2}}_q \langle 0 | + \frac{1}{\sqrt{2}}_q \langle 1 | \right)+ \frac{1}{2}_c \left( \frac{1}{\sqrt{2}}_q | 0 \rangle - \frac{1}{\sqrt{2}}_q | 1 \rangle \right) \left( \frac{1}{\sqrt{2}}_q \langle 0 | - \frac{1}{\sqrt{2}}_q \langle 1 | \right)

=\frac{1}{2}_c \frac{1}{\sqrt{2}}_q \frac{1}{\sqrt{2}}_q \left( | 0 \rangle + | 1 \rangle \right) \left( \langle 0 | + \langle 1 | \right)+ \frac{1}{2}_c \frac{1}{\sqrt{2}}_q \frac{1}{\sqrt{2}}_q \left( | 0 \rangle - | 1 \rangle \right) \left( \langle 0 | - \langle 1 | \right)

=\frac{1}{2}_c \frac{1}{2}_q \left( | 0 \rangle + | 1 \rangle \right) \left( \langle 0 | + \langle 1 | \right) + \frac{1}{2}_c \frac{1}{2}_q \left( | 0 \rangle - | 1 \rangle \right) \left( \langle 0 | - \langle 1 | \right)

=\frac{1}{2}_c \frac{1}{2}_q \left( | 0 \rangle \langle 0 | + | 1 \rangle \langle 1 | + | 0 \rangle \langle 0 | + | 1 \rangle \langle 1 | \right)

=\frac{1}{2}_c \frac{1}{2}_q \left( 2_c | 0 \rangle \langle 0 | + 2_c | 1 \rangle \langle 1 | \right)

= \frac{1}{2}_q | 0 \rangle \langle 0 | + \frac{1}{2}_q | 1 \rangle \langle 1 |

— Me@2018-03-11 03:14:57 PM


How come the classical probabilities \frac{1}{2}_c of a density matrix in one representation can become quantum probabilities \frac{1}{2}_q in another?

\frac{1}{2}_c | + \rangle \langle + | + \frac{1}{2}_c | - \rangle \langle - | = \frac{1}{2}_q | 0 \rangle \langle 0 | + \frac{1}{2}_q | 1 \rangle \langle 1 |

1. Physically, whether we label the coefficients as “classical probabilities” or “quantum probabilities” gives no real consequences. The conflict lies only in the interpretations.

2. The interpretation conflict might be resolved by realizing that probabilities, especially classical probabilities, is meaningful only when being with respect to an observer.

For example,

\frac{1}{2}_c | + \rangle \langle + | + \frac{1}{2}_c | - \rangle \langle - | = \frac{1}{2}_q | 0 \rangle \langle 0 | + \frac{1}{2}_q | 1 \rangle \langle 1 |

represents the fact that the observer knows that the system is either in state |+\rangle \langle+| or |-\rangle \langle-|, but not |0 \rangle \langle 0| nor |1 \rangle \langle 1|.


\frac{1}{2}_c | 0 \rangle \langle 0 | + \frac{1}{2}_c | 1 \rangle \langle 1 |

represents the fact that the observer knows that the system is either in state |0 \rangle \langle 0| or |1 \rangle \langle 1|, but not |+\rangle \langle+| nor |-\rangle \langle-|.

— Me@2018-03-13 08:10:46 PM



2018.03.14 Wednesday (c) All rights reserved by ACHK

Quantum Indeterminacy

注定外外傳 1

Quantum indeterminacy is the apparent necessary incompleteness in the description of a physical system, that has become one of the characteristics of the standard description of quantum physics.

Indeterminacy in measurement was not an innovation of quantum mechanics, since it had been established early on by experimentalists that errors in measurement may lead to indeterminate outcomes. However, by the later half of the eighteenth century, measurement errors were well understood and it was known that they could either be reduced by better equipment or accounted for by statistical error models. In quantum mechanics, however, indeterminacy is of a much more fundamental nature, having nothing to do with errors or disturbance.

— Wikipedia on Quantum indeterminacy

Quantum indeterminacy is the inability to predict the behaviour of the system with 100% accuracy, even in principle.

If everything is connected

, quantum indeterminacy is due to the logical fact that, by definition, a “part” cannot contain (all the information of) the “whole”.

An observer (A) cannot separate itself from the system (B) that it wants to observe, because an observation is an interaction between the observer and the observed


In order to get a perfect prediction of a measurement result, observer (A) must have all the information of the present state of the whole system (A+B). However, there are two logical difficulties.

First, observer A cannot have all the information about (A+B).

Second, observer A cannot observe itself to get (all of) its present state information, since an observation is an interaction between two entities. Logically, it is impossible for something to interact with itself directly. Just as logically, it is impossible for your right hand to hold your right hand itself. 

So the information observer A can get (to the greatest extent) is all the information about B, which is only part of the system (A+B) it (A) needs to know in order to get a prefect prediction for the evolution of the system B.

— Me@2015-09-14 08:12:32 PM

2015.09.15 Tuesday (c) All rights reserved by ACHK

假名定律 1.2

反白論前傳:冠名篇 2.2

Jesus, Buddha, Einstein 3.2

這段改編自 2010 年 4 月 10 日的對話。

(安:經濟學家張五常先生提過,有一篇經濟論文指出,凱恩斯(John Maynard Keynes)的經濟理論,和「凱恩斯學派」的經濟理論,不盡相同,雖然整個學派是以「凱恩斯」來命名。)




又例如,「機會率」有兩大學派,「頻率學派」和「貝葉斯學派」。「貝葉斯學派」雖然以數學家貝葉斯(Thomas Bayes)來命名,但是,貝葉斯並不算是,「貝葉斯學派的成員」(Bayesian),因為,「貝葉斯學派」中有很多理論,例如,「貝葉斯學派」對「機會率」的詮釋,也不是貝葉斯本人的意見。

Bayes himself might not have embraced the broad interpretation now called Bayesian. It is difficult to assess Bayes’s philosophical views on probability, since his essay does not go into questions of interpretation.

— Wikipedia on Thomas Bayes

— Me@2014.04.11

I like your Christ. I do not like your Christians. Your Christians are so unlike your Christ. The materialism of affluent Christian countries appears to contradict the claims of Jesus Christ that says it’s not possible to worship both Mammon and God at the same time.

– Mohandas K. Gandhi

2014.04.11 Friday (c) All rights reserved by ACHK

測不準原理 1.11

這段改編自 2010 年 4 月 10 日的對話。



方便起見,我把「甲」稱為 A,而「甲+乙」,則簡稱為 B。

是「影響」還是「更換」,要視乎你把 A 和 B 標籤為,「同一個系統的兩個不同(時間)版本」,還是「兩個不同的系統」。










— Me@2014.02.05

2014.02.05 Wednesday (c) All rights reserved by ACHK

測不準原理 1.10

這段改編自 2010 年 4 月 10 日的對話。







方便起見,我把「甲」稱為 A,而「甲+乙」,則簡稱為 B。

是「影響」還是「更換」,要視乎你把 A 和 B 標籤為,「同一個系統的兩個不同(時間)版本」,還是「兩個不同的系統」。


— Me@2014.02.01

2014.02.02 Sunday (c) All rights reserved by ACHK

測不準原理 1.9

這段改編自 2010 年 4 月 10 日的對話。











留意,以上並不是「測不準原理」的真身,而只是「測不準原理」的其中一個例子 —— 應用「測不準原理」,來解釋「觀察者效應」的由來。



— Me@2014.01.29

2014.01.29 Wednesday (c) All rights reserved by ACHK

測不準原理 1.8

這段改編自 2010 年 4 月 10 日的對話。


1. 有些物理量的配對,是 incompatible observables(不相容觀察量)。

如果一個物量系統的物理量,「甲」和「乙」並不相容,該系統就沒有可能,同時處於「甲」的 eigenstate(本徵態)和「乙」的 eigenstate。換句話說,該系統不可能有一個狀態,同時是甲乙的「本徵態」。

2. 兩件事不可以同時發生,不代表不可以同時不發生。



3. 如果甲乙這兩個物理量互不相容,甲的標準差( \sigma_a )和乙的標準差( \sigma_b ),相乘之積一定不小於 \frac{\hbar}{2},而 \hbar 是「約化普朗克常數」(reduced Planck constant)。

\sigma_{a} \sigma_{b} \geq \frac{\hbar}{2}


4. 這數式背後想帶出的物理意義是,對於互不相容的兩個物理量「甲」和「乙」,

雖然你可以刻意建構一個量子物理系統,令到其對應的「物理量甲」,所對應的「標準差」極之細小,而「極之細小」在這裡的意思是,任意細小 —— 細小到你指定的程度;但是,你要付出的代價是,該個物理系統的「物理量乙」,所對應的「標準差」,就會相應變大。







— Me@2014.01.26

2014.01.26 Sunday (c) All rights reserved by ACHK

測不準原理 1.7

這段改編自 2010 年 4 月 10 日的對話。



\sqrt{\frac{1}{3}} | A \rangle + \sqrt{\frac{2}{3}} | B \rangle

這個『能量疊加狀態』的一個物理系統,如果複製成很多個相同的系統,然後各自量度能量數值的話,那堆能量數據的分佈,所對應的『標準差』,將等於  0.9428J。  


\sqrt{\frac{1}{3}} | A \rangle + \sqrt{\frac{2}{3}} | B \rangle

的『標準差』是 0.9428J。


你想講的是,每個量子態,都有對應的「標準差」(standard deviation)。而「標準差」就反映了,一堆數據的分散程度。)






\sqrt{\frac{1}{10}} | A \rangle + \sqrt{\frac{9}{10}} | B \rangle

的『標準差』是 0.6J。


\sqrt{\frac{1}{2}} | A \rangle + \sqrt{\frac{1}{2}} | B \rangle

的『標準差』,則是 1J。


乙有 1/2 的機會,會被量度出,帶有 1J 的能量(「本徵態 A」的對應能量數值);而亦有 1/2 的機會,會被量度出,帶有 3J 的能量(「本徵態 B」的對應能量數值)。兩個可能數值,出現的機會率相同或者相若時,我們就「無從估計」,系統會出現兩個數值中的哪一個。

但是,甲卻有 9/10,即是有九成的機會率,會被量度出,帶有 3J 的能量(「本徵態 B」的對應能量數值)。那樣,我們就可以說,我們「相對確定」,系統帶有 3J 的能量。

— Me@2014.01.23

2014.01.23 Thursday (c) All rights reserved by ACHK

測不準原理 1.6

這段改編自 2010 年 4 月 10 日的對話。

那樣,如果該物理系統,並不處於 A、B、C 狀態,即是不處於任何一個,「能量本徵態」的話,情況又會如何呢?


\sqrt{\frac{1}{3}} | A \rangle + \sqrt{\frac{2}{3}} | B \rangle


那樣,你在量度之前,並不會知道,你得到的能量數值是 1J (「本徵態 A」的對應數值),還是 3J (「本徵態 B」的對應數值)。但是,你會知道,你有 1/3 的機會率,會得到 1J; 而亦有 2/3 的機會率,會得到 3J。

換句話說,如果將該物理系統複製成,120 萬個相同系統,然後量度它們各自的能量數值的話,你會發現,將有大概 1/3 的成員,即是 40 萬個,帶有 1J 的能量;另外有大概 2/3 的成員,即是 80 萬個,帶有 3J 的能量。

因為現在不只有一點數據,而是有一大堆的數據,所以我們可以討論,這堆數據的「標準差」(standard deviation)。「標準差」是一個統計學的測量,用來反映一堆數據的分散程度。「標準差」越大,就代表一堆數據越分散;「標準差」越小,就代表一堆數據越集中。

例如,在剛才的例子中,總共有 120 萬個能量數值。當中大概 40 萬個是 1J; 而大概 80 萬個是 3J。如果要找到這堆數據的「標準差」,你就要先運算出它們的「平均值」:

\frac{400000(1J) + 800000(3J)}{1200000}

= 2.333J


\sqrt{\frac{400000(1-2.333)^2 + 800000(3-2.333)^2}{1200000}}

= 0.9428J



\sqrt{\frac{1}{3}} | A \rangle + \sqrt{\frac{2}{3}} | B \rangle

這個『能量疊加狀態』的一個物理系統,如果複製成很多個相同的系統,然後各自量度能量數值的話,那堆能量數據的分佈,所對應的『標準差』,將等於 0.9428J。  


\sqrt{\frac{1}{3}} | A \rangle + \sqrt{\frac{2}{3}} | B \rangle

的『標準差』是 0.9428J。

— Me@2014.01.14

2014.01.14 Tuesday (c) All rights reserved by ACHK

測不準原理 1.5

這段改編自 2010 年 4 月 10 日的對話。


如果一個物理系統的狀態,有可能同時是物理量「甲」和物理量「乙」的 eigenstate(本徵態),「甲」和「乙」就為之 compatible observables(相容觀察量)。不可能的話,「甲」和「乙」就為之「不相容觀察量」。

你首先記住這一點。然後,我要跳去另一個問題 —— 如果一個物理系統,並不是處於(例如)能量的本徵態,我們會量度到什麼能量數值呢?

其實,你都會度到其中一個本徵態,所對應的數值,簡稱 eigenvalues(本徵值/特徵值)。





例如,假設一個物理系統,有「能量本徵態」 A、B 和 C,而順序對應的「能量本徵值」是 1J、3J 和 5J。

如果該物理系統正處於「本徵態 A」,你就一定會量度到能量數值 1J。換句話說,只要知道系統正處於「本徵態 A」,即使不用量度,你也知道系統當時,所帶的能量值是 1 焦耳。同理,如果該物理系統正處於「本徵態 B」,你就一定會量度到能量數值 3J;如果該物理系統正處於「本徵態 C」,你則一定會量度到能量數值 5J。

那樣,如果該物理系統,並不處於 A、B、C 狀態,即是不處於任何一個,「能量本徵態」的話,情況又會如何呢?


\sqrt{\frac{1}{3}} | A \rangle + \sqrt{\frac{2}{3}} | B \rangle


那樣,你在量度之前,並不會知道,你得到的能量數值是 1J(「本徵態 A」的對應數值),還是 3J(「本徵態 B」的對應數值)。但是,你會知道,你有 1/3 的機會率會得到 1J,有 2/3 的機會率會得到 3J。

— Me@2014.01.07

2014.01.07 Tuesday (c) All rights reserved by ACHK

測不準原理 1.4

這段改編自 2010 年 4 月 10 日的對話。




量子力學中,有一個術語,叫做 eigenstate(本徵態),意思是「本身帶有特徵的狀態」,簡稱「特別態」。例如,如果你正在考慮的物理系統,是一粒粒子,而該粒子正處於一個「位置的本徵態」,那樣,原則上,在量度那粒子之前,你就可以百分百準確地,預測到它在下一刻的位置。或者說,你毋須量度,也可以準確知道,那粒子在下一刻的位置。



— Me@2013.12.25

2013.12.25 Wednesday (c) All rights reserved by ACHK

量子力學 1.17

因果律 1.22 | 語意互相推卸責任論 1.22 | Verification principle, 5.22 | 西瓜 9.22 | Make a difference, 3.3

這段改編自 2010 年 4 月 3 日的對話。


如果,就連在原則上,你都講不出,如何分辨它們誰是誰非 —— 所有可能的實驗結果,「量子自由版本」和「量子決定版本」,都必定一模一樣的話,「量子自由論」和「量子決定論」就根本是「同義句」。







— Me@2013.10.14

2013.10.14 Monday (c) All rights reserved by ACHK