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.{}^{[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.

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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,{}^{[2]} 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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2018.08.19 Sunday (c) All rights reserved by ACHK

Universal wave function, 20

The physical (synthetic) universal wave function logically cannot be found by any local observers.

The definition of “universe” is “all the things”. So there is no outside.

A global observer has to be outside the universe.

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However, a mathematical (analytic) universal function is possible.

It applies to theoretical/model universe, which can be used to develop interpretations of quantum mechanics and successively approximate the physical universe.

— Me@2012-04-16

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2018.07.21 Saturday (c) All rights reserved by ACHK

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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2018.07.03 Tuesday (c) All rights reserved by ACHK

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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2018.06.19 Tuesday (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

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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

~~~

… 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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2018.05.16 Wednesday (c) All rights reserved by ACHK

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
Ram Bharadwaj

— 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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2018.05.13 Sunday (c) All rights reserved by ACHK

時空兌換率

這段改編自 2015 年的對話。

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我的相對論教授說,所謂

E = m c^2

在某些意思之下,沒有那麼特別,因為,你可以把它看成,貨幣的兌換。

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E = c^2 m

能量 =(光速二次方)\times 質量

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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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公式中的 c^2(光速平方),角色其實正正就是,能量 E 和質量 m 之間的「貨幣兌換率」。

(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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2018.05.11 Friday (c) All rights reserved by ACHK

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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2018.04.30 Monday (c) All rights reserved by ACHK

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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2018.04.16 Monday (c) All rights reserved by ACHK

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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2018.04.09 Monday (c) All rights reserved by ACHK

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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2018.03.14 Wednesday (c) All rights reserved by 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

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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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2018.03.14 Wednesday (c) All rights reserved by ACHK

機遇再生論 1.6

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所以,「同情地理解」,亦可稱為「意念淘金術」。

機遇再生論,可以同情地理解為,有以下的意思:

(而這個意思,亦在「機遇再生論」的原文中,用作其理據。)

假設,你現在手中,有一副樸克牌,存在於某一個排列 A 。洗牌一次之後,排列仍然是 A 的機會極微。

一副完整的撲克牌,共有 N = 52! \approx 8.07 \times 10^{67} 個,可能的排列。亦即是話,洗牌後仍然是排列 A 的機會率,只有 \frac{1}{N}

由於分母 N 太大(相當於 8 之後,還有 67 個位),洗牌後,理應變成另外一個排列 B 。

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

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

洗了一次牌後,發覺排列是 B 不是 A 後,我們可以再問,如果再洗一次牌,「是 A」和「不是 A」的機會,分別是多少?

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由於,機會率只是與未知的事情有關,或者說,已知的事件,發生的機會率必為 1;所以,如果發生了第一次洗牌,而你又知道其結果的情況下,問「如果再洗一次牌,『是 A』和『不是 A』的機會,分別是多少」,第二次洗牌各個可能結果,發生的機會率,與第一次洗牌的結果無關。

第二次洗牌結果為組合 A 的機會率,仍然是

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

不是組合 A 的機會率,仍然是

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

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(問:那樣,為什麼要問多一次呢?)

我是想釐清,我真正想問的是,並不是這個問題,而是另一個:

如果在第一次洗牌之前,亦即是話,一次牌都未洗的話,問:

「如果洗牌兩次,起碼一次洗到原本排列 A 的機會率是多少?」

把該事件標示為 A_2

A_2 = 兩次洗牌的結果,起碼一次洗到原本排列 A

再把該事件的機會率,標示為 P(A_2)

由於 P(A_2) 相對麻煩,我們可以先行運算其「互補事件」的機會率。

A_2 的互補事件為「不是 A_2」:

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

其機會率為

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

那樣,我們就可推斷,

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

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同理,在一次牌都未洗的時候,問:

如果洗牌 m 次,起碼一次洗到原本排列 A 的機會率是多少?

答案將會是

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

留意,N = 52! \approx 8.07 \times 10^{67},非常之大,導致 (1 - \frac{1}{N}) 極端接近 1。在一般情況,m 的數值還是正常時, P(A_m) 會仍然極端接近 0。

例如,你將會連續洗一千萬次牌(m = 10,000,000),起碼有一次,回到原本排列 A 的機會是:

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

你用一般手提計算機的話,它會給你 0。你用電腦的話,它會給你

1.239799930857148592 \times 10^{-61}

— Me@2018-01-25 12:38:39 PM

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2018.02.13 Tuesday (c) All rights reserved by ACHK

Logical arrow of time, 6.2

Source of time asymmetry in macroscopic physical systems

Second law of thermodynamics

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Physics is not about reality, but about what one can say about reality.

— 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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1. Physics is about what an observer can observe about reality.

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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2018.02.03 Saturday (c) All rights reserved by ACHK