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

神的旨意 2.3

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神:我是神,而你只是人。神的心思,凡人不一定能夠理解。所以,作為凡人的你,應該心存謙卑,完全信任你的神,奉行神的旨意。

甲:不合理呀!

如果要殺害無辜,才可上天堂,而不殺害無辜,會被罰落地獄的話,那代表著,你天堂的居民,都是壞人;而你地獄的居民,反而是善人。

那樣,你的所謂「天堂」,根本是我的地獄。而你雖然自封為「神」,卻根本是我的「魔」。

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魔:我是魔又如何?我的法力遠高於你。我可以令你求生不得,求死不能。

甲:會有真正善良的神,去保護我。

魔:難保我的法力高過善神。「魔」就不可以是全能的嗎?

甲:如果是全能,你怎會有動機,去虐待凡人呢?既是全能者,又怎會那麼無聊,損人不利己?

另外,如果你法力高強,乃至「全能」,你就不可能是「全惡」的。你必定有善部和惡部。你的善部,就是善神。

魔:為什麼「全能」者不可「全惡」?

— Me@2018-08-13 11:54:47 AM

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2018.08.13 Monday (c) All rights reserved by ACHK

The Jacobian of the inverse of a transformation

The Jacobian of the inverse of a transformation is the inverse of the Jacobian of that transformation

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In this post, we would like to illustrate the meaning of

the Jacobian of the inverse of a transformation = the inverse of the Jacobian of that transformation

by proving a special case.

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Consider a transformation \mathscr{T}: \bar{x}^i=\bar{x}^i (x^1,x^2), which is an one-to-one mapping from unbarred x^i‘s to barred \bar{x}^i coordinates, where i=1, 2.

By definition, the Jacobian matrix J of \mathscr{T} is

J= \begin{pmatrix} \displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}} \\ \displaystyle{\frac{\partial \bar{x}^2}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} \end{pmatrix}

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Now we consider the the inverse of the transformation \mathscr{T}:

\mathscr{T}^{-1}: x^i=x^i(\bar{x}^1,\bar{x}^2)

By definition, the Jacobian matrix \bar{J} of this inverse transformation, \mathscr{T}^{-1}, is

\bar{J}= \begin{pmatrix} \displaystyle{\frac{\partial x^1}{\partial \bar{x}^1}} & \displaystyle{\frac{\partial x^1}{\partial \bar{x}^2}} \\ \displaystyle{\frac{\partial x^2}{\partial \bar{x}^1}} & \displaystyle{\frac{\partial x^2}{\partial \bar{x}^2}} \end{pmatrix}

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On the other hand, the inverse of Jacobian J of the original transformation \mathscr{T} is

J^{-1}=\displaystyle{\frac{1}{ \begin{vmatrix} \displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}} \\ \displaystyle{\frac{\partial \bar{x}^2}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} \end{vmatrix} }} \begin{pmatrix} \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} & \displaystyle{-\frac{\partial \bar{x}^1}{\partial x^2}} \\ \displaystyle{-\frac{\partial \bar{x}^2}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}} \end{pmatrix}

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If \bar{J} = J^{-1}, their (1, 1)-elementd should be equation:

\displaystyle{\frac{\partial x^1}{\partial \bar{x}^1}}\stackrel{?}{=}\displaystyle{\frac{1}{\displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}}\displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}}-\displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}}\displaystyle{\frac{\partial \bar{x}^2}{\partial x^1}} }} \bigg( \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} \bigg)

Let’s try to prove that.

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Consider equations

\bar{x}^1 = \bar{x}^1(x^1,x^2)

\bar{x}^2 = \bar{x}^2(x^1,x^2)

Differentiate both sides of each equation with respect to \bar{x}^1, we have:

A := 1=\displaystyle{\frac{\partial \bar{x}^1}{\partial \bar{x}^1}=\frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}+\frac{\partial \bar{x}^1}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}}

B := 0 = \displaystyle{\frac{\partial \bar{x}^2}{\partial \bar{x}^1}=\frac{\partial \bar{x}^2}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}+\frac{\partial \bar{x}^2}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}}

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A \times \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}}:~~~~~C := \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}=\frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}\frac{\partial \bar{x}^2}{\partial x^2}+\frac{\partial \bar{x}^1}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}\frac{\partial \bar{x}^2}{\partial x^2}}

B \times \displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}}:~~~~~D := \displaystyle{0=\frac{\partial \bar{x}^2}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}\frac{\partial \bar{x}^1}{\partial x^2}+\frac{\partial \bar{x}^2}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}\frac{\partial \bar{x}^1}{\partial x^2}}

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D-C:

\displaystyle{ \frac{\partial \bar{x}^2}{\partial x^2}= \bigg( \frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial \bar{x}^2}{\partial x^2} - \frac{\partial \bar{x}^2}{\partial x^1}\frac{\partial \bar{x}^1}{\partial x^2}\bigg) \frac{\partial x^1}{\partial \bar{x}^1}},

results

\displaystyle{ \frac{\partial x^1}{\partial \bar{x}^1}}=\frac{\displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}}}{\displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial \bar{x}^2}{\partial x^2} - \frac{\partial \bar{x}^1}{\partial x^2}\frac{\partial \bar{x}^2}{\partial x^1}}}

— Me@2018-08-09 09:49:51 PM

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2018.08.09 Thursday (c) All rights reserved by ACHK

Problem 14.5a1

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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(a) Consider the left NS’ sector. Write the precise mass-squared formula with normal-ordered oscillators and the appropriate normal-ordering constant.

~~~

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\displaystyle{\alpha' M_L^2 = \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}

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What is normal-ordering?

Put all the creation operators on the left.

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What for?

p.251 “It is useful to work with normal-ordered operators since they act in a simple manner on the vacuum state. We cannot use operators that do not have a well defined action on the vacuum state.”

“The vacuum expectation value of a normal ordered product of creation and annihilation operators is zero. This is because, denoting the vacuum state by |0\rangle, the creation and annihilation operators satisfy”

\displaystyle{\langle 0 | \hat{a}^\dagger = 0 \qquad \textrm{and} \qquad \hat{a} |0\rangle = 0}

— Wikipedia on Normal order

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— This answer is my guess. —

\displaystyle{\sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I}

\displaystyle{= \sum_{n \in \mathbf{Z}^-} \bar \alpha_{-n}^I \bar \alpha_n^I + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_n^I}

\displaystyle{= \sum_{n \in \mathbf{Z}^+} \bar \alpha_{n}^I \bar \alpha_{-n}^I + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_n^I}

\displaystyle{= \sum_{n \in \mathbf{Z}^+} \left[ \bar \alpha_{n}^I \bar \alpha_{-n}^I - \bar \alpha_{-n}^I \bar \alpha_{n}^I + \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_n^I}

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\displaystyle{= \sum_{n \in \mathbf{Z}^+} \left[ \bar \alpha_{n}^I, \bar \alpha_{-n}^I \right] + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_n^I}

= \displaystyle{\sum_{n \in \mathbf{Z}^+} n \eta^{II} + 2 \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I}

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c.f. p.251:

\displaystyle{\sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I}

\displaystyle{= \sum_{n \in \mathbf{Z}^+} n \eta^{II} + 2 \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I}

\displaystyle{= \frac{-1}{12} (D - 2) + 2 \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I}

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Equation at Problem 14.5:

\displaystyle{\alpha' M_L^2}

\displaystyle{= \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}

\displaystyle{= \frac{1}{2} \left[ \frac{-1}{12} (D - 2) + 2 \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}

\displaystyle{= \frac{-1}{24} (D - 2) + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}

\displaystyle{= \frac{-1}{8} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}

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D = 10

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\displaystyle{\sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}

\displaystyle{= \sum_{r = - \frac{1}{2}, - \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A}

\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} (-r) \lambda_{r}^A \lambda_{-r}^A + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A}

\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ (-1) \lambda_{r}^A \lambda_{-r}^A + \lambda_{-r}^A \lambda_r^A \right]}

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\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ (-1) \lambda_{r}^A \lambda_{-r}^A + \lambda_{-r}^A \lambda_r^A \right]}

\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ \lambda_{-r}^A, \lambda_r^A \right]}

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Equation (14.29):

\displaystyle{\left\{ b_r^I, b_s^J \right\} = \delta_{r+s, 0} \delta^{IJ}}

\displaystyle{b_r^I b_s^J = - b_s^I b_r^J + \delta_{r+s, 0} \delta^{IJ}}

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\displaystyle{\sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}

\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ (-1) \lambda_{r}^A \lambda_{-r}^A + \lambda_{-r}^A \lambda_r^A \right]}

\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ (-1) \left( - \lambda_{-r}^A \lambda_r^A + \delta_{r-r, 0} \delta^{AA} \right) + \lambda_{-r}^A \lambda_r^A \right]}

\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ \lambda_{-r}^A \lambda_r^A + \lambda_{-r}^A \lambda_r^A - 1 \right]}

\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ 2 \lambda_{-r}^A \lambda_r^A - 1 \right]}

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\displaystyle{\sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}

\displaystyle{= - \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ b_{-r}^A b_r^A + \lambda_{-r}^A \lambda_r^A \right]}

\displaystyle{= - \frac{1}{2} \sum_{r = 1, 3, ...} r + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ b_{-r}^A b_r^A + \lambda_{-r}^A \lambda_r^A \right]}

\displaystyle{= \left[ - \frac{1}{24} + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left( b_{-r}^A b_r^A + \lambda_{-r}^A \lambda_r^A \right) \right]}

— This answer is my guess. —

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— Me@2018-08-06 10:23:48 PM

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2018.08.06 Monday (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

The fault of optimism

The illusion of peace

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The seemingly rational world is rendered by exceptional good parents.

The seemingly peaceful time is provided by exceptional expense of defense.

— Me@2011.08.19

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2018.07.18 Wednesday (c) All rights reserved by ACHK

神的旨意 2.2

這段改編自 2010 年 4 月 18 日的對話。

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合情合理,自圓其說的,就有機會真;無情無理,自相矛盾的,則必定為假。

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又例如,如果現在神明顯靈在你面前,你怎樣判斷,那真的是「神明」?

「祂」既可能其實是「邪靈」,亦可能只是你自己的幻覺而已。

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假設,現在神明顯靈,向你(甲)頒下旨意:

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神:你必須殺害無辜,將來死後,才可以上天堂;否則,我會罰你下地獄。

甲:我不想作惡。

神:那不是你想不想的問題。你必須奉行神的旨意。

甲:但是,為什麼要我殺害無辜呢?

神:我是神,而你只是人。神的心思,凡人不一定能夠理解。所以,作為凡人的你,應該心存謙卑,完全信任你的神,奉行神的旨意。

甲:不合理呀!

如果要殺害無辜,才可上天堂,而不殺害無辜,會被罰落地獄的話,那代表著,你天堂的居民,都是壞人;而你地獄的居民,反而是善人。

那樣,你的所謂「天堂」,根本是我的地獄。而你雖然自封為「神」,卻根本是我的「魔」。

— Me@2018-07-16 07:51:33 PM

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2018.07.18 Wednesday (c) All rights reserved by ACHK

Chain Rule of Differentiation

Consider the curve y = f(x).

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\displaystyle{\frac{d}{dx}} is an operator, meaning “the slope of the tangent of”. So the expression \displaystyle{\frac{dy}{dx}}, meaning \displaystyle{\frac{d}{dx} (y)}, is not a fraction.

In order words, it means the slope of the tangent of the curve y = f(x) at a point, such as point A in the graph.

d_2018_07_15__21_31_32_PM_

The symbol dx has no relation with the symbol \displaystyle{\frac{dy}{dx}}. It means \Delta x as shown in the graph. In other words,

dx = \Delta x

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The symbol dy also has no relation with the symbol \displaystyle{\frac{dy}{dx}}. It means the vertical distance between the current point A(x_0, y_0), where y_0 = f(x_0), and the point C on the tangent line y = mx + c, where m is the slope of the tangent line. In other words,

dy = m~dx

or

\displaystyle{dy = \left[ \left( \frac{d}{dx} \right) y \right] dx}

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The relationship of \Delta y and dy is that

\displaystyle{\Delta y = \frac{dy}{dx} \Delta x + \text{higher order terms}}

\displaystyle{\Delta y = \frac{dy}{dx} dx + \text{higher order terms}}

\Delta y = dy + \text{higher order terms}

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Similarly, for functions of 2 variables:

\displaystyle{\Delta f(x,y) = \frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y + \text{higher order terms}}

\displaystyle{df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy}

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For functions of 3 variables:

\displaystyle{df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz}

\displaystyle{\frac{df}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} + \frac{\partial f}{\partial z}\frac{dz}{dt}}

— Me@2018-07-15 09:30:29 PM

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2018.07.15 Sunday (c) All rights reserved by ACHK

Problem 14.4b2

Closed string degeneracies | A First Course in String Theory

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(b) State the values of \alpha' M^2 and give the separate degeneracies of bosons and fermions for the first five mass levels of the type IIA closed superstrings. Would the answer have the different for type IIB?

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Type IIB closed superstrings

Equation (14.85)

(NS+, NS+), (NS+, R-), (R-, NS+), (R-, R-)

— Me@2015.09.16 06:08 AM: Should be the same. But I am not sure whether I have missed something.

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f_{NS+}(x) = 8 + 128 \, x + 1152 \, x^{2} + 7680 \, x^{3} + 42112 \, x^{4} + ...

f_{R-}(x) = 8 + 128 x + 1152 x^{2} + 7680 x^{3} + 42112 x^{4} + ...

f_{NS-}(x) = \frac{1}{\sqrt{x}} + 36 \sqrt{x} + 402 x^{\frac{3}{2}} + 3064 x^{\frac{5}{2}} + ...

f_{R+}(x) = 8 + 128 x + 1152 x^{2} + 7680 x^{3} + 42112 x^{4} + ...

— Me@2018-07-14 09:41:10 PM

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2018.07.14 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

Mirror selves, 5.2

Anatta 3.3 | 無我 3.3

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You fight for existence, for being alive.

However, your existence is not “yours”.

The existence of you, is not your property.

The existence of you, is a property of the group you are in.

The existence of you, is a property of other people.

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To meaningfully say the statement “I exist”, you have to specify you exist with respect to whom.

To exist, you have to specify to exist in which people’s world.

— Me@2018-05-22 7:43 AM

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2018.07.02 Monday (c) All rights reserved by ACHK

神的旨意 2.1

這段改編自 2010 年 4 月 18 日的對話。

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(問:你閱讀過很多有關「瀕死經驗」的文章?)

可以這樣說。

如果你閱讀那些文章的話,要小心一點,因為那類文章良莠不齊——當中有些文章發人深省,有些則謊話連篇。

(問:那你怎樣分辨,「瀕死經驗」的文章之中,哪些是真,哪些為假?)

看看文中所說的,合不合理。

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例如,你怎樣知道,我說的話,是真還是假?

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合情合理,自圓其說的,就有機會真;無情無理,自相矛盾的,則必定為假。

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又例如,如果現在神明顯靈在你面前,你怎樣判斷,那真的是「神明」?

「祂」既可能其實是「邪靈」,亦可能只是你自己的幻覺而已。

— Me@2018-06-28 10:23:28 PM

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2018.06.30 Saturday (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

Quick Calculation 14.8.2

A First Course in String Theory

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What sector(s) can be combined with a left-moving NS- to form a consistent closed string sector?

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There are no mass levels in NS+, R+, or R- that can match those in NS-. So NS- can be paired only with NS-:

(NS-, NS-)

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f_{NS} (x)
= \frac{1}{\sqrt{x}} \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8
= \frac{1}{\sqrt{x}} g_{NS}(x)
= \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...

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g (\sqrt{x})
= \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8
= 1 + 8 \, \sqrt{x} + 36 \, x + 128 \, x^{\frac{3}{2}} + 402 \, x^{2} + 1152 \, x^{\frac{5}{2}} + 3064 \, x^{3} + ...

g (-\sqrt{x})
= \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8
= 1 -8 \, \sqrt{x} + 36 \, x -128 \, x^{\frac{3}{2}} + 402 \, x^{2} -1152 \, x^{\frac{5}{2}} + 3064 \, x^{3} + ...

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g (\sqrt{x}) + g (-\sqrt{x})
= 2(1 + 36 x + 402 x^{2} + 3064 x^{3} + ...)

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f_{NS-}(x)
= \frac{1}{2 \sqrt{x}} \left[ g (\sqrt{x}) + g (-\sqrt{x}) \right]
= \frac{1}{2 \sqrt{x}} \left[ \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8 + \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8 \right]
= \frac{1}{2 \sqrt{x}} \left[ 2(1 + 36 \, x + 402 \, x^{2} + 3064 \, x^{3} + ...) \right]
= \frac{1}{\sqrt{x}} + 36 \sqrt{x} + 402 x^{\frac{3}{2}} + 3064 x^{\frac{5}{2}} + ...

— Me@2018-06-26 07:36:41 PM

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2018.06.26 Tuesday (c) All rights reserved by 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

大學經濟

這段改編自 2010 年 4 月 18 日的對話。

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我猜想,當一個人改變存在型態時,會立刻或者將會,知道很多生時不知道的東西。但是,那些新知識,未必包括你想知道的東西。

比喻說,由中學升到大學,你將會學到,很多超過中學程度的知識。但是,如果你中學時,沒有讀過經濟科的話,單單是「升大學」本身,並不會令你,立刻獲得經濟科的知識。

大學生「由零開始學經濟學」,都同樣要花時間;分別是,通常而言,比中學生「由零開始學經濟學」,速度會高一點。

(問:不一定呀。中學生比較年青,腦袋理應高速一點。)

無錯。

方便起見,暫時用同一個人來比較,例如你。

「中學的你」可以因為腦袋較年青,學習新事物比「大學的你」較快。「大學的你」可能因為知識和經驗較多,學習新事物比「中學的你」較快。

視乎情況,因人而異,沒有一定的答案。

但是,至少你會同意一點:

如果你中學時,沒有讀過經濟科,在大學時要「由零開始學經濟學」的話,你會立刻看大學程度的經濟書,而不是由中學教科書開始學。

— Me@2018-06-05 11:54:51 AM

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2018.06.07 Thursday (c) All rights reserved by ACHK

Plato

trowawee 4 months ago

I’m a little frustrated at the tossed-off reference to Plato and Aristotle at the beginning – “The good life may have sufficed for Plato and Aristotle, but it is no longer enough.” – because I feel like that ignores the fact that both Plato and Aristotle, along with a lot of philosophers, actually had a lot to say about physical fitness. Plato was a champion wrestler, and both he and Aristotle viewed physical education as a fundamental component to living the good life. Xenophon quotes Socrates saying this:

“For in everything that men do the body is useful; and in all uses of the body it is of great importance to be in as high a state of physical efficiency as possible. Why, even in the process of thinking, in which the use of the body seems to be reduced to a minimum, it is matter of common knowledge that grave mistakes may often be traced to bad health.”

The whole article feels a little too mired in presentism, and ignorant of the history of self-improvement ideas.

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coldtea 4 months ago

>Plato was a champion wrestler

And the name Plato is a nickname — meaning “the broad/wide one” given to him for his broad shoulders because of that training and physical appearance. Real name: Aristocles.

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kernelbandwidth 4 months ago

It’s funny to consider that one of the canonically great philosophers in history is known essentially by the equivalent of his WWE wrestling name. It’s like if in the future there were classes taught on the philosophical ideas of “The Rock”.

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coldtea 4 months ago

Some other amusing related stuff: so, Plato, was called for for the ancient greek word for broad/wide.

Modern [English] words that stem from the same root: plateau, platitude, plat, plate — via French and Latin (plattus) from Greek (platis “flat, wide, broad”).

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danohu 4 months ago

Well, the first Pope was literally called The Rock (Peter). Jesus appointed him by saying “you are The Rock, and I’ll build my church on this rock”.

Exactly what he meant has led to centuries of debate between protestants and catholics.

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acangiano 4 months ago

“No man has the right to be an amateur in the matter of physical training. It is a shame for a man to grow old without seeing the beauty and strength of which his body is capable.”

― Socrates

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— Improving Ourselves to Death

— Hacker News

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2018.06.01 Friday ACHK

Problem 14.4b1.4

Closed string degeneracies | A First Course in String Theory

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What is the meaning of “With a = 1, ..., 8 and \bar b = \bar 1, ..., \bar 8, …”?

— Me@2015.09.14 12:11 PM

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p.315 “Explicitly, the eight states | R_a \rangle, a = 1, 2, ..., 8, with an even number of creation operators are … ”

p.316 “The eight states |R_{\bar{a}} \rangle, \bar a = \bar 1, \bar 2, ..., \bar 8, with an odd number of creation operators are … ”

— Me@2018-05-24 11:41:34 AM

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2018.05.24 Thursday (c) All rights reserved by ACHK