# Quantum Computing, 3

Instead of requiring deterministic calculation, you allow (quantum) probabilistic calculation. What you gain is the extra speed.

— Me@2018-02-08 01:50:06 PM

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# Classical probability, 7

Classical probability is macroscopic superposition.

— Me@2012.04.23

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That is not correct, except in some special senses.

— Me@2019-05-02

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That is not correct, if the “superposition” means quantum superposition.

— Me@2019-05-03 08:44:11 PM

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The difference of the classical probability and quantum probability is the difference of a mixed state and a pure superposition state.

In classical probability, the relationship between mutually exclusive possible measurement results, before measurement, is OR.

In quantum probability, if the quantum system is in quantum superposition, the relationship between mutually exclusive possible measurement results, before measurement, is neither OR nor AND.

— Me@2019-05-03 06:04:27 PM

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# Mixed states, 4

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How is quantum superposition different from mixed state?

The state $\displaystyle{|\Psi \rangle = \frac{1}{\sqrt{2}}\left(|\psi_1\rangle +|\psi_2\rangle \right)}$

is a pure state. Meaning, there’s not a 50% chance the system is in the state $\displaystyle{|\psi_1 \rangle }$ and a 50% it is in the state $\displaystyle{|\psi_2 \rangle}$. There is a 0% chance that the system is in either of those states, and a 100% chance the system is in the state $\displaystyle{|\Psi \rangle}$.

The point is that these statements are all made before I make any measurements.

— edited Jan 20 ’15 at 9:54

— answered Oct 12 ’13 at 1:42

— Andrew

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Given a state, mixed or pure, you can compute the probability distribution $\displaystyle{P(\lambda_n)}$ for measuring eigenvalues $\displaystyle{\lambda_n}$, for any observable you want. The difference is the way you combine probabilities, in a quantum superposition you have complex numbers that can interfere. In a classical probability distribution things only add positively.

— Andrew Oct 12 ’13 at 14:41

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— How is quantum superposition different from mixed state?

— Physics StackExchange

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2019.04.23 Tuesday ACHK

# The problem of induction 3.3

“Everything has no patterns” (or “there are no laws”) creates a paradox.

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If “there are 100% no first order laws”, then it is itself a second order law (the law of no first-order laws), allowing you to use probability theory.

In this sense, probability theory is a second order law: the law of “there are 100% no first order laws”.

In this sense, probability theory is not for a single event, but statistical, for a meta-event: a collection of events.

Using meta-event patterns to predict the next single event, that is induction.

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Induction is a kind of risk minimization.

— Me@2012-11-05 12:23:24 PM

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# The problem of induction 3.1.2

Square of opposition

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“everything has a pattern”?

“everything follows some pattern” –> no paradox

“everything follows no pattern” –> paradox

— Me@2012.11.05

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My above statements are meaningless, because they lack a precise meaning of the word “pattern”. In other words, whether each statement is correct or not, depends on the meaning of “pattern”.

In common usage, “pattern” has two possible meanings:

1. “X has a pattern” can mean that “X has repeated data“.

Since the data set X has repeated data, we can simplify X’s description.

For example, there is a die. You throw it a thousand times. The result is always 2. Then you do not have to record a thousand 2’s. Instead, you can just record “the result is always 2”.

2. “X has a pattern” can mean that “X’s are totally random, in the sense that individual result cannot be precisely predicted“.

Since the data set X is totally random, we can simplify the description using probabilistic terms.

For example, there is a die. You throw it a thousand times. The die lands on any of the 6 faces 1/6 of the times. Then you do not have to record those thousand results. Instead, you can just record “the result is random” or “the die is fair”.

— Me@2018-12-18 12:34:58 PM

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# 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.

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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

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# The problem of induction 3

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

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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.

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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

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# 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

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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
nbro

asked Mar 21 ’13 at 15:36
Deepu

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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. ${}^{}$ 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.

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When the final states are distinguishable you add probabilities: $P_{dis} = P_1 + P_2 = \psi_1^* \psi_1 + \psi_2^* \psi_2$

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When the final state are indistinguishable, ${}^{}$ you add amplitudes: $\Psi_{1,2} = \psi_1 + \psi_2$

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

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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

dmckee

— Physics Stack Exchange

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# 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

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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

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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 $|1 \rangle \langle 1|$.

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

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

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# Quantum Indeterminacy

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

# Probability 4

using probability

~ ignoring the details

~ caring about only the results

— Me@2014-05-21 09:28:35 PM

# 假名定律 1.2

Jesus, Buddha, Einstein 3.2

（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。）

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

# 測不準原理 1.11

（安：但是，根據我們之前，有關「時間定義」的討論，「影響」和「更換」，沒有絕對的分別。在某個意思之下，「影響」和「更換」，是同一樣東西。

『甲+乙』（甲加乙）是一個新的物理系統，所以，其眾多量度數據的統計模式，自然和『甲』的統計模式，有所不同。

— Me@2014.02.05

# 測不準原理 1.10

（安：你的意思是，「測不準原理」成立，是「觀察者效應」成立的原因，而不是相反。一般人之所以錯，是因為不小心地，把「測不準原理」和「觀察者效應」的因果關係倒轉了。）

（安：但是，根據我們之前，有關「時間定義」的討論，「影響」和「更換」沒有絕對的分別。在某個意思之下，「影響」和「更換」，是同一樣東西。

— Me@2014.02.01

# 測不準原理 1.9

「測不準原理」所處理的，是有關在建構一個物理系統時，所要作出的考慮和妥協；而不是處理，在量度一個已有物理系統時，對該個物理系統原本的演化，所做成的影響。

（安：那樣，為什麼一般人也錯誤以為，「測不準原理」和「觀察者效應」，有直接關係呢？）

（安：這個講法合理正確，為何你說它「難於正確」呢？）

『觀察者』或者『量度儀器』，一定會和原本的物理系統，有相互作用，導致互相影響。換句話說，『量度儀器』（乙）必然地加入了，它想量度的『原本物理系統』（甲）。換而言之，『甲』和『乙』在一起，形成了一個新的物理系統。

（安：你的意思是，「測不準原理」成立，是「觀察者效應」成立的原因，而不是相反。一般人之所以錯，是因為不小心地，把「測不準原理」和「觀察者效應」的因果關係倒轉了。）

— Me@2014.01.29

# 測不準原理 1.8

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

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」，

「標準差甲」和「標準差乙」，並不能同時「任意細小」。

「測不準原理」所處理的，是有關在建構一個物理系統時，所要作出的考慮和妥協；而不是處理，在量度一個已有物理系統時，對該個物理系統原本的演化，所做成的影響。

— Me@2014.01.26

# 測不準原理 1.7 \sqrt{\frac{1}{3}} | A \rangle + \sqrt{\frac{2}{3}} | B \rangle \sqrt{\frac{1}{3}} | A \rangle + \sqrt{\frac{2}{3}} | B \rangle

（安：等一等，讓我先整理一下。

（安：那又怎樣？那跟「測不準原理」，又有什麼關係呢？）

「標準差」和「確定性」有著密切的關係。具體而言，一個量子態（例如）能量的「標準差」越大，即代表了可能的能量數值越分散。那樣，在量度之前，能量的「不確定性」就越大。 \sqrt{\frac{1}{10}} | A \rangle + \sqrt{\frac{9}{10}} | B \rangle \sqrt{\frac{1}{2}} | A \rangle + \sqrt{\frac{1}{2}} | B \rangle

— Me@2014.01.23