# Pointer state

Eigenstates 3

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In quantum Darwinism and similar theories, pointer states are quantum states that are less perturbed by decoherence than other states, and are the quantum equivalents of the classical states of the system after decoherence has occurred through interaction with the environment.

— Wikipedia on Pointer state

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In calculation, if a quantum state is in a superposition, that superposition is a superposition of eigenstates.

However, real superposition does not just includes states that make macroscopic senses.

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That is the major mistake of the many-worlds interpretation of quantum mechanics.

— Me@2017-12-30 10:24 AM

— Me@2018-07-03 07:24 PM

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# Block spacetime, 9

motohagiography 42 days ago [-]

I once saw a fridge magnet that said “time is natures way of making sure everything doesn’t happen all at once,” and it’s stuck with me.

The concept of time not being “real,” can be useful as an exercise for modelling problems where to fully explore the problem space, you need to decouple your solutions from needing them to occur in an order or sequence.

From an engineering perspective, “removing” time means you can model problems abstractly by stepping back from a problem and asking, what are all possible states of the mechanism, then which ones are we implementing, and finally, in what order. This is different from the relatively stochastic approach most people take of “given X, what is the necessary next step to get to desired endstate.”

More simply, as a tool, time helps us apprehend the states of a system by reducing the scope of our perception of them to sets of serial, ordered phenomena.

Whether it is “real,” or an artifact of our perception is sort of immaterial when you can choose to reason about things with it, or without it. A friend once joked that math is what you get when you remove time from physics.

I look forward to the author’s new book.

— Gödel and the unreality of time

— Hacker News

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2018.06.26 Tuesday ACHK

# Eigenstates 2.3.2

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eigenstates

~ classical states

~ definite states

— Me@2012-04-15 11:42:10 PM

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The concept of eigenstate is relative.

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First, you have to specify the eigenstate is of which physical observable.

A physical system can be at an eigenstate of one observable but at a superposition state of another observable.

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Second, you have to specify the state of that observable is eigen with respect to which observer.

— Me@2018-06-16 7:27 AM

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eigenstates

~ of which observable?

~ with respect to which observer?

— Me@2018-06-19 10:54:54 AM

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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

# Van der Waals equation 1.2

Whether $X_{\text{measured}}$ is bigger or smaller than $X_{\text{ideal}}$ ultimately depends on the assumptions and definitions used in the derivation of the ideal gas equation itself.

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In the ideal gas equation derivation, the volume used in the equation refers to the volume that the gas molecules can move within. So

$V_{\text{ideal}} = V_{\text{available for a real gas' molecules to move within}}$

Then, when deriving the pressure, it is assumed that there are no intermolecular forces among gas molecules. So

$P_{\text{ideal}} = P_{\text{assuming no intermolecular forces}}$

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These are the reasons that

$V_{\text{ideal}} < V_{\text{measured}}$

$P_{\text{ideal}} > P_{\text{measured}}$

$P_{\text{ideal}} V_{\text{ideal}} = nRT$

$\left(P_\text{measured} + a\left(\frac{n}{V}\right)^2\right) \left(V_\text{measured}-nb\right) = nRT$

— Me@2018-05-16 07:12:51 PM

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… the thing to keep in mind is that the “pressure we use in the ideal gas law” is not the pressure of the gas itself. The pressure of the gas itself is too low: to relate that pressure to “pressure for the ideal gas law” we have to add a number. While the volume occupied by the real gas is too large – the “ideal volume” is less than that. – Floris Sep 30 ’16 at 17:34

— Physics Stackexchange

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# Van der Waals equation 1.1

Why do we add, and not subtract, the correction term for pressure in [Van der Waals] equation?

Since the pressure of real gases is lesser than the pressure exerted by (imaginary) ideal gases, shouldn’t we subtract some correction term to account for the decrease in pressure?

I mean, that’s what we have done for the volume correction: Subtracted a correction term from the volume of the container V since the total volume available for movement is reduced.

asked Sep 30 ’16 at 15:20

— Physics Stackexchange

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Ideal gas law:

$P_{\text{ideal}} V_{\text{ideal}} = nRT$

However, since in a real gas, there are attractions between molecules, so the measured value of pressure P is smaller than that in an ideal gas:

$P_{\text{measured}} = P_{\text{real}}$

$P_{\text{measured}} < P_{\text{ideal gas}}$

Also, since the gas molecules themselves occupy some space, the measured value of the volume V is bigger that the real gas really has:

$V_{\text{measured}} > V_{\text{real}}$

$P_{\text{ideal}} V_{\text{ideal}} = nRT$

If we substitute $P_{\text{measured}}$ onto the LHS, since $P_{\text{measured}} < P_{\text{ideal}}$, the LHS will be smaller than the RHS:

$P_{\text{measured}} V_{\text{ideal}} < nRT$

So in order to maintain the equality, a correction term to the pressure must be added:

$\left(P_\text{measured} + a\left(\frac{n}{V}\right)^2\right) V_{\text{ideal}} = nRT$

$P_{\text{ideal}} V_{\text{ideal}} = nRT$

If we substitute $V_{\text{measured}}$ onto the LHS, since that volume is bigger that actual volume available for the gas molecules to move, the LHS will be bigger than the RHS:

$P_{\text{ideal}} V_{\text{measured}} > nRT$

So in order to maintain the equality, a correction term to the pressure must be subtracted:

$P_{\text{ideal}} \left(V_\text{measured}-nb\right) = nRT$

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In other words,

$V_{\text{measured}} > V_{\text{real}}$

$V_{\text{ideal}} = V_{\text{real}}$

$V_{\text{measured}} > V_{\text{ideal}}$

— Me@2018-05-13 03:37:18 PM

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Why? I still do not understand.

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How come

$P_{\text{measured}} = P_{\text{real}}$

but

$V_{\text{measured}} \ne V_{\text{real}}$?

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How come

$V_{\text{real}} = V_{\text{ideal}}$

but

$P_{\text{real}} \ne P_{\text{ideal}}$?

— Me@2018-05-13 03:22:54 PM

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The above is wrong.

The “real volume” $V_{\text{real}}$ has 2 possible different meanings.

One is “the volume occupied by a real gas”. In other words, it is the volume of the gas container.

Another is “the volume available for a real gas’ molecules to move”.

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To avoid confusion, we should define

$V_{\text{real}} \equiv V_{\text{measured}}$

$P_{\text{real}} \equiv P_{\text{measured}}$

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Or even better, avoid the terms $P_{\text{real}}$ and $V_{\text{real}}$ altogether. Instead, just consider the relationship between $(P_{\text{ideal}}, P_{\text{measured}})$ and that between $(V_{\text{ideal}}, V_{\text{measured}})$.

Whether $X_{\text{measured}}$ is bigger or smaller than $X_{\text{ideal}}$ ultimately depends on the assumptions and definitions used in the derivation of the ideal gas equation itself.

— Me@2018-05-13 04:15:34 PM

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# 時空兌換率

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$E = m c^2$

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$E = c^2 m$

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$1 \text{USD} \approx 8 \times 1 \text{HKD}$

1 美元 $\approx 8 \times$ 1 港元

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（而光速 c，則是時間和空間的兌換率。）

— Me@2018-05-11 09:10:00 PM

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# The Sixth Sense, 3

Mirror selves, 2 | Anatta 3.2 | 無我 3.2

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You cannot feel your own existence or non-existence. You can feel the existence or non-existence of (such as) your hair, your hands, etc.

But you cannot feel the existence or non-existence of _you_.

— Me@2018-03-17 5:12 PM

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Only OTHER people or beings can feel your existence or non-existence.

— Me@2018-04-30 11:29:08 AM

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# Quantum decoherence 8

12. On the other hand, consistent histories are just a particular convenient framework to formulate physical questions in a certain way; the only completely invariant consequence of this formalism is the Copenhagen school’s postulate that physics can only calculate the probabilities, they follow the laws of quantum mechanics, and when decoherence is taken into account, to find both the quantum/classical boundary as well as the embedding of the classical limit within the full quantum theory, some questions about quantum systems follow the laws of classical probability theory (and may be legitimately asked) while others don’t (and can’t be asked)[.]

— Decoherence is a settled subject

— Lubos Motl

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2018.04.24 Tuesday ACHK

# Digital physics, 8

Check whether this world is a Matrix:

Some physical results (such as Lorentz symmetry) in this universe cannot be simulated by a classical digital computer.

— Me@2011.08.21

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# Logical arrow of time, 6.3

“Time’s arrow” is only meaningful when considering with respect to an observer.

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c.f. the second law of thermodynamics

The direction of time is direction of losing microscopic information… by whom?

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“Time’s arrow” is only meaningful when considering with respect to an observer.

— Me@2018-01-01 6:14 PM

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# Quantum observer 1.3.2

Principle of Least Action, 7.2

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Without quantum superposition, there would be no principle of least action and thus we would not be able to see the classical macroscopic world.

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Our mind or perception is a superposition of eigenstates.

— Me@2012.04.14

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# Density matrix, 4

Consider a system that is in a mixed state. The system has 0.3 of probability in a pure state $|\psi_1 \rangle$ and 0.7 of probability in another pure state $|\psi_2 \rangle$. Then the density matrix $\rho$ is

$0.3 | \psi_1 \rangle \langle \psi_1 | + 0.7 | \psi_2 \rangle \langle \psi_2 |$

In the most general cases, neither $|\psi_1 \rangle$ nor $|\psi_2 \rangle$ is an eigenstate. So we cannot expect that $\rho$ is diagonal.

For example, if each of the pure state $|\psi_1 \rangle$ and $|\psi_2 \rangle$ is a superposition of two eigenstates $(|\phi_1\rangle, |\phi_2\rangle)$, then

$| \psi_1 \rangle = \frac{1}{\sqrt 2} |\phi_1 \rangle + \frac{1}{\sqrt 2} |\phi_2 \rangle$

$| \psi_2 \rangle = \frac{1}{\sqrt 3} |\phi_1 \rangle + \sqrt{\frac{2}{3}} |\phi_2 \rangle$

and

$\rho$

$= 0.3 | \psi_1 \rangle \langle \psi_1 | + 0.7 | \psi_2 \rangle \langle \psi_2 |$

$= 0.3 \left( \frac{1}{\sqrt 2} |\phi_1 \rangle + \frac{1}{\sqrt 2} |\phi_2 \rangle \right) \left( \frac{1}{\sqrt 2} \langle \phi_1 | + \frac{1}{\sqrt 2} \langle \phi_2 | \right)$

$+ 0.7 \left( \frac{1}{\sqrt 3} |\phi_1 \rangle + \sqrt{\frac{2}{3}} |\phi_2 \rangle \right) \left( \frac{1}{\sqrt 3} \langle \phi_1 | + \sqrt{\frac{2}{3}} \langle \phi_2 | \right)$

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For simplicity, assume that the eigenstates $\{ |\phi_1\rangle, |\phi_2\rangle \}$ form a complete orthonormal set.

If we use $\{ | \phi_1 \rangle, |\phi_2 \rangle \}$ as basis,

$\rho$

$= 0.3 \begin{bmatrix} \frac{1}{\sqrt 2} \\ \frac{1}{\sqrt 2} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt 2} & \frac{1}{\sqrt 2} \end{bmatrix} + 0.7 \begin{bmatrix} \frac{1}{\sqrt 3} \\ \sqrt{\frac{2}{3}} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt 3} & \sqrt{\frac{2}{3}} \end{bmatrix}$

$= \frac{0.3}{2} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} + \frac{0.7}{3} \begin{bmatrix} 1 & \sqrt{2} \\ \sqrt{2} & 2 \end{bmatrix}$

$=\cdots$

— Me@2018.03.12 11:51 AM

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# 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

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

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

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

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

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

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

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

— 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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# 機遇再生論 1.6

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（而這個意思，亦在「機遇再生論」的原文中，用作其理據。）

$P(A) = \frac{1}{N}$

$P(\text{not} A) = 1 - \frac{1}{N}$

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$P(A) = \frac{1}{N}$

$P(\text{not} A) = 1 - \frac{1}{N}$

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(問：那樣，為什麼要問多一次呢？）

「如果洗牌兩次，起碼一次洗到原本排列 A 的機會率是多少？」

$A_2$ = 兩次洗牌的結果，起碼一次洗到原本排列 A

$A_2$ 的互補事件為「不是 $A_2$」：

= 兩次洗牌的結果，不是起碼一次洗到原本排列 A

= 兩次洗牌的結果，都不是排列 A

$P(\text{not} A_2) = (1 - \frac{1}{N})^2$

$P(A_2)$
$= 1 - P(\text{not} A_2)$
$= 1 - (1 - \frac{1}{N})^2$

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$P(A_m)= 1 - (1 - \frac{1}{N})^m$

$P(A_m)$
$= 1 - (1 - \frac{1}{N})^m$
$= 1 - (1 - \frac{1}{52!})^{10,000,000}$

$1.239799930857148592 \times 10^{-61}$

— Me@2018-01-25 12:38:39 PM

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# Logical arrow of time, 6.2

Source of time asymmetry in macroscopic physical systems

Second law of thermodynamics

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— Bohr

— paraphrased

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Physics should deduce what an observer would observe,

not what it really is, for that would be impossible.

— Me@2018-02-02 12:15:38 AM

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2. Whatever an observer can observe is a consistent history.

observer ~ a consistent story

observing ~ gathering a consistent story from the quantum reality

3. Physics [relativity and quantum mechanics] is also about the consistency of results of any two observers _when_, but not before, they compare those results, observational or experimental.

4. That consistency is guaranteed because the comparison of results itself can be regarded as a physical event, which can be observed by a third observer, aka a meta observer.

Since whenever an observer can observe is consistent, the meta-observer would see that the two observers have consistent observational results.

5. Either original observers is one of the possible meta-observers, since it certainly would be witnessing the comparison process of the observation data.

— Me@2018-02-02 10:25:05 PM

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# phe-nomenon | 本體現象 || an observer | a causal diamond

My categories “phe-nomenon | 本體現象” and “an observer | a causal diamond” are equivalent, except that the latter focuses on physics.

— Me@2018-01-22 10:37:50 AM

# 穿梭時空 探索宇宙

The Hubble Ultra-Deep Field (Public Domain Image)

— NASA and the European Space Agency

— Me@2017-12-06 08:09:20 PM

2017.12.06 Wednesday ACHK

# 機遇再生論 1.5

（請參閱本網誌，有關「重言句」、「經驗句」和「印證原則」的文章。）

「同情地理解」的意思是，有些理論，雖然在第一層次的分析之後，有明顯的漏洞，但是，我們可以試試，代入作者發表該理論時的，心理狀態和時空情境；研究作者發表該理論的，緣起和動機；從而看看，該理論不行的原因，會不會只是因為，作者的語文或思考不夠清晰，表達不佳而已？

（而這個意思，亦在「機遇再生論」的原文中，用作其理據。）

$P(A) = \frac{1}{N}$

$P($not $A) = 1 - \frac{1}{N}$

— Me@2017-12-18 02:51:11 PM