# The square root of the probability, 4.2

Eigenstates 3.4.2

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The difference between quantum and classical is due to the indistinguishability of cases.

— Me@2020-12-26 01:25:03 PM

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Statistical effects of indistinguishability

The indistinguishability of particles has a profound effect on their statistical properties.

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The differences between the statistical behavior of fermions, bosons, and distinguishable particles can be illustrated using a system of two particles. The particles are designated A and B. Each particle can exist in two possible states, labelled $\displaystyle{ |0 \rangle }$ and $\displaystyle{|1\rangle}$, which have the same energy.

The composite system can evolve in time, interacting with a noisy environment. Because the $\displaystyle{|0\rangle}$ and $\displaystyle{|1\rangle}$ states are energetically equivalent, neither state is favored, so this process has the effect of randomizing the states. (This is discussed in the article on quantum entanglement.) After some time, the composite system will have an equal probability of occupying each of the states available to it. The particle states are then measured.

If A and B are distinguishable particles, then the composite system has four distinct states: $\displaystyle{|0\rangle |0\rangle}$, $\displaystyle{|1\rangle |1\rangle}$ , $\displaystyle{ |0\rangle |1\rangle}$, and $\displaystyle{|1\rangle |0\rangle }$. The probability of obtaining two particles in the $\displaystyle{|0\rangle}$ state is 0.25; the probability of obtaining two particles in the $\displaystyle{|1\rangle}$ state is 0.25; and the probability of obtaining one particle in the $\displaystyle{|0\rangle}$ state and the other in the $\displaystyle{|1\rangle}$ state is 0.5.

If A and B are identical bosons, then the composite system has only three distinct states: $\displaystyle{|0\rangle |0\rangle}$, $\displaystyle{ |1\rangle |1\rangle }$, and $\displaystyle{{\frac {1}{\sqrt {2}}}(|0\rangle |1\rangle +|1\rangle |0\rangle)}$. When the experiment is performed, the probability of obtaining two particles in the $\displaystyle{|0\rangle}$ is now 0.33; the probability of obtaining two particles in the $\displaystyle{|1\rangle}$ state is 0.33; and the probability of obtaining one particle in the $\displaystyle{|0\rangle}$ state and the other in the $\displaystyle{|1\rangle}$ state is 0.33. Note that the probability of finding particles in the same state is relatively larger than in the distinguishable case. This demonstrates the tendency of bosons to “clump.”

If A and B are identical fermions, there is only one state available to the composite system: the totally antisymmetric state $\displaystyle{{\frac {1}{\sqrt {2}}}(|0\rangle |1\rangle -|1\rangle |0\rangle)}$. When the experiment is performed, one particle is always in the $\displaystyle{|0\rangle}$ state and the other is in the $\displaystyle{|1\rangle}$ state.

The results are summarized in Table 1:

— Wikipedia on Identical particles

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# Cheer up?

What you encourage people, don’t say “don’t worry, be happy” or “cheer up”.

They are nonsense, because they are not actionable.

What I really need to know is HOW to be happy.

What I really need to know is what ACTION to take to solve my problem.

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When you see a doctor, what if, without giving any other advice, he just tells you to “cheer up”?

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— Me@2011.06.26

— Me@2020-12-30

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# 機遇創生論 1.7.2

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「自由意志」問題方面，如果要討論的話，要先釐清「人有沒有自由意志」的意思，因為，它有超過一個常用的詮釋：

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1. 思：

2. 因：

3. 果：

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1. 思：

2. 因：

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「自由論」和「決定論」，其實沒有實質上的分別。

— Me@2020-12-29 05:46:05 PM

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# 1994

This was my Art result. I had a great art teacher Mr Lo in that year. He taught us a lot of design concepts.

— Me@2020-12-29 10:52:04 AM

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# Ex 1.14 Lagrange equations for L’

Structure and Interpretation of Classical Mechanics

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Show by direct calculation that the Lagrange equations for $\displaystyle{L'}$ are satisfied if the Lagrange equations for $\displaystyle{L}$ are satisfied.

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Equation (1.69):

$\displaystyle{C \circ \Gamma[q'] = \Gamma[q]}$

Equation (1.70):

$\displaystyle{L' = L \circ C}$

Equation (1.71):

$\displaystyle{L' \circ \Gamma[q'] = L \circ C \circ \Gamma[q'] = L \circ \Gamma[q]}$

The Lagrange equation:

\displaystyle{ \begin{aligned} D ( \partial_2 L \circ \Gamma[q]) - (\partial_1 L \circ \Gamma[q]) &= 0 \\ \end{aligned}}

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\displaystyle{ \begin{aligned} &\partial_2 L \circ \Gamma[q] \\ &= \frac{\partial}{\partial v} L(t, x, v) \\ &= \frac{\partial}{\partial v} L'(t, x', v') \\ &= \frac{\partial}{\partial x'} L'(t, x', v') \frac{\partial x'}{\partial v} + \frac{\partial}{\partial v'} L'(t, x', v') \frac{\partial v'}{\partial v} \\ \end{aligned}}

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Since it is just a coordinate transformation $\displaystyle{x = F(t, x')}$, $\displaystyle{x}$ has no explicitly dependent on $\displaystyle{v'}$. Similarly, if we consider the coordinate transformation $\displaystyle{x' = G(t, x)}$, $\displaystyle{x'}$ has no explicitly dependent on $\displaystyle{v}$. So

\displaystyle{ \begin{aligned} \frac{\partial x'}{\partial v} &= 0 \\ \end{aligned}}

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\displaystyle{ \begin{aligned} &\partial_2 L \circ \Gamma[q] \\ &= \frac{\partial}{\partial v'} L'(t, x', v') \frac{\partial v'}{\partial v} \\ \end{aligned}}

— Me@2020-12-28 04:03:24 PM

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# The square root of the probability, 4

Eigenstates 3.4

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quantum ~ classical with the indistinguishability of cases

— Me@2020-12-23 06:19:00 PM

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In statistical mechanics, a semi-classical derivation of the entropy that does not take into account the indistinguishability of particles, yields an expression for the entropy which is not extensive (is not proportional to the amount of substance in question). This leads to a paradox known as the Gibbs paradox, after Josiah Willard Gibbs who proposed this thought experiment in 1874‒1875. The paradox allows for the entropy of closed systems to decrease, violating the second law of thermodynamics. A related paradox is the “mixing paradox”. If one takes the perspective that the definition of entropy must be changed so as to ignore particle permutation, the paradox is averted.

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# Beauty and the Beast, 2

Beauty 7

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Beauty is the goal, but may not be the mean.

— Me@2020-10-29 08:29:31 AM

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The path to beauty may not be itself beautiful.

— Me@2020-11-03 11:32:52 AM

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# 尋覓 1.3

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1. 甲A

2. 甲B

3. 甲C

4. 乙A

5. 乙B

6. 乙C

7. 丙A

8. 丙B

9. 丙C

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— Me@2010.06.01

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–- Me@2010.06.01

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— Me@2010.06.07

— Me@2020-12-23 10:39:26 PM

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# 1980s, 2

— Photomyne colorization

— Me@2020-12-22 04:39:33 PM

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# Problem 2.4

A First Course in String Theory

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2.4 Lorentz transformations as matrices

A matrix L that satisfies (2.46) is a Lorentz transformation. Show the following.

(b) If $\displaystyle{L}$ is a Lorentz transformation so is the inverse matrix $\displaystyle{L^{-1}}$.

(c) If $\displaystyle{L}$ is a Lorentz transformation so is the transpose matrix $\displaystyle{L^{T}}$.

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(b)

\displaystyle{ \begin{aligned} (\mathbf{A}^\mathrm{T})^{-1} &= (\mathbf{A}^{-1})^\mathrm{T} \\ L^T \eta L &= \eta \\ \eta &= [L^T]^{-1} \eta L^{-1} \\ [L^T]^{-1} \eta L^{-1} &= \eta \\ [L^{-1}]^T \eta L^{-1} &= \eta \\ \end{aligned}}

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(c)

\displaystyle{ \begin{aligned} L^T \eta L &= \eta \\ (L^T \eta L)^{-1} &= (\eta)^{-1} \\ L^{-1} \eta^{-1} (L^T)^{-1} &= \eta \\ L^{-1} \eta (L^T)^{-1} &= \eta \\ \eta &= L \eta L^T \\ L \eta L^T &= \eta \\ \end{aligned}}

— Me@2020-12-21 04:24:33 PM

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# Pointer state, 3

Eigenstates 3.3 | The square root of the probability, 3

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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 include eigenstates 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

— Me@2020-12-18 06:12 PM

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Mathematically, a quantum superposition is a superposition of eigenstates. An eigenstate is a quantum state that is corresponding to a macroscopic state. A superposition state is a quantum state that has no classical correspondence.

The macroscopic states are the only observable states. An observable state is one that can be measured directly or indirectly. For an unobservable state, we write it as a superposition of eigenstates. We always write a superposition state as a superposition of observable states; so in this sense, before measurement, we can almost say that the system is in a superposition of different (possible) classical macroscopic universes.

However, conceptually, especially when thinking in terms of Feynman’s summing over histories picture, a quantum state is more than a superposition of classical states. In other words, a system can have a quantum state which is a superposition of not only normal classical states, but also bizarre classical states and eigen-but-classically-impossible states.

A bizarre classical state is a state that follows classical physical laws, but is highly improbable that, in daily life language, we label such a state “impossible”, such as a human with five arms.

An eigen-but-classically-impossible state is a state that violates classical physical laws, such as a castle floating in the sky.

For a superposition, if we allow only normal classical states as the component eigenstates, a lot of the quantum phenomena, such as quantum tunnelling, cannot be explained.

If you want multiple universes, you have to include not only normal universes, but also the bizarre ones.

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Actually, even for the double-slit experiment, “superposition of classical states” is not able to explain the existence of the interference patterns.

The superposition of the electron-go-left universe and the electron-go-right universe does not form this universe, where the interference patterns exist.

— Me@2020-12-16 05:18:03 PM

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One of the reasons is that a quantum superposition is not a superposition of different possibilities/probabilities/worlds/universes, but a superposition of quantum eigenstates, which, in a sense, are square roots of probabilities.

— Me@2020-12-18 06:07:22 PM

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# 機遇創生論 1.7

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（問：「種子論」，其實就即是「自由決定論」？）

「種子論」重於討論，人生如何逹到成功。（留意，這裡的「成功」，是在你自己定義下的成功，而不是在世俗標準下。）

「自由決定論」則重於研究，宇宙既然依物理定律運行，那就代表，人的一舉一動，甚至思想意志，在宇宙創生那刻，就已經決定了？

（問：「自由決定論」即是問，世間上，有沒有「自由意志」？）

「自由決定論」的重點跟人（或者其他意識體）沒有直接的關係。

「自由決定論」的重點在於研究，「Laplace 因果律」是否正確。

「Laplace 因果律」就是：

「如果『因果律』是正確，人就沒有自由」只是「因果律」的一個例子。而這個例子因為直接和人相關，所以，人們特別重視。但是，即使那樣，那仍不是「因果律」的重點。

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（另一話題：）

「自由意志」問題方面，如果要討論的話，要先釐清「人有沒有自由意志」的意思，因為，它有超過一個常用的詮釋：

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1. 思：

2. 因：

3. 果：

— Me@2020-12-11 06:43:58 PM

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# 2006, 3

— Me@2020-12-07 04:06:39 PM

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# How to Find Lagrangians

Lagrange’s equations are a system of second-order differential equations. In order to use them to compute the evolution of a mechanical system, we must find a suitable Lagrangian for the system. There is no general way to construct a Lagrangian for every system, but there is an important class of systems for which we can identify Lagrangians in a straightforward way in terms of kinetic and potential energy. The key idea is to construct a Lagrangian L such that Lagrange’s equations are Newton’s equations $\displaystyle{\vec F = m \vec a}$.

— 1.6 How to Find Lagrangians

— Structure and Interpretation of Classical Mechanics

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2020.12.06 Sunday ACHK

# Logical arrow of time, 6.4.2

Logical arrow of time, 6.1.2

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The source of the macroscopic time asymmetry, aka the second law of thermodynamics, is the difference between prediction and retrodiction.

In a prediction, the deduction direction is the same as the physical/observer time direction.

In a retrodiction, the deduction direction is opposite to the physical/observer time direction.

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— guess —

If a retrodiction is done by a time-opposite observer, he will see the entropy increasing. For him, he is really making a prediction.

— guess —

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— Me@2013-10-25 3:33 AM

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A difference between deduction and observation is that in observation, the probability is updated in real time.

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each update time interval ~ infinitesimal

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In other words, when you observe a system, you get new information about that system in real time.

Since you gain new knowledge of the system in real time, the probability assigned to that system is also updated in real time.

— Me@2020-10-13 11:27:59 AM

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# 心動

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2020.12.02 Wednesday ACHK

Posted in OCD

# 尋覓 1.2

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— Me@2020-11-29 10:36:46 PM

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