2006, 4

Posted on January 10, 2021 by rpflee

— Me@2021-01-10 06:11:40 PM

Ex 1.14 Lagrange equations for L’, 2

Posted on January 9, 2021 by rpflee

Structure and Interpretation of Classical Mechanics

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

~~~

$\displaystyle{x = F(t, x')}$

$\displaystyle{x' = G(t, x)}$

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

$\displaystyle{v' = \partial_0 G(t, x) + \partial_1 G(t,x) v}$

$\displaystyle{ \begin{aligned} \frac{\partial v'}{\partial v} &= \partial_1 G(t,x) = \frac{\partial x'}{\partial x} \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} &\partial_2 L \circ \Gamma[q] \\ &= \frac{\partial L'}{\partial v'} \frac{\partial x'}{\partial x} \\ \end{aligned}}$

The Lagrange equation:

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

$\displaystyle{ \begin{aligned} &\partial_1 L \circ \Gamma[q] \\ &= \partial_1 L' \circ \Gamma[q'] \\ &= \frac{\partial}{\partial x} L' (t, x', v') \\ &= \frac{\partial L'}{\partial x'} \frac{\partial x'}{\partial x} + \frac{\partial L'}{\partial v'} \frac{\partial v'}{\partial x} \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} D ( \partial_2 L \circ \Gamma[q]) - (\partial_1 L \circ \Gamma[q]) &= 0 \\ D \left( \frac{\partial L'}{\partial v'} \frac{\partial x'}{\partial x} \right) - \left( \frac{\partial L'}{\partial x'} \frac{\partial x'}{\partial x} + \frac{\partial L'}{\partial v'} \frac{\partial v'}{\partial x} \right) &= 0 \\ D \left( \frac{\partial L'}{\partial v'} \right) \frac{\partial x'}{\partial x} + \frac{\partial L'}{\partial v'} D \left( \frac{\partial x'}{\partial x} \right) - \left( \frac{\partial L'}{\partial x'} \frac{\partial x'}{\partial x} + \frac{\partial L'}{\partial v'} \frac{\partial v'}{\partial x} \right) &= 0 \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} &D \left( \frac{\partial x'}{\partial x} \right) \\ &= \frac{d}{dt} \left( \frac{\partial x'}{\partial x} \right) \\ &= \frac{\partial}{\partial t} \left( \frac{\partial x'}{\partial x} \right) \frac{\partial t}{\partial t} + \frac{\partial}{\partial x'} \left( \frac{\partial x'}{\partial x} \right) \frac{\partial x'}{\partial t} + \frac{\partial}{\partial v'} \left( \frac{\partial x'}{\partial x} \right) \frac{\partial v'}{\partial t} \\ &= \frac{\partial}{\partial t} \left( \frac{\partial x'}{\partial x} \right) + \frac{\partial}{\partial x} \left( \frac{\partial x'}{\partial x'} \right) \frac{\partial x'}{\partial t} + \frac{\partial}{\partial x} \left( \frac{\partial x'}{\partial v'} \right) \frac{\partial v'}{\partial t} \\ &= \frac{\partial}{\partial t} \left( \frac{\partial x'}{\partial x} \right) + \frac{\partial}{\partial x} \left( 1 \right) \frac{\partial x'}{\partial t} + \frac{\partial}{\partial x} \left( 0 \right) \frac{\partial v'}{\partial t} \\ &= \frac{\partial}{\partial x} \left( \frac{\partial x'}{\partial t} \right) + 0 + 0 \\ &= \frac{\partial v'}{\partial x} \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \left[ D \left( \frac{\partial L'}{\partial v'} \right) - \frac{\partial L'}{\partial x'} \right] \frac{\partial x'}{\partial x} + \frac{\partial L'}{\partial v'} \frac{\partial v'}{\partial x} - \frac{\partial L'}{\partial v'} \frac{\partial v'}{\partial x} &= 0 \\ \left[ D \left( \frac{\partial L'}{\partial v'} \right) - \frac{\partial L'}{\partial x'} \right] \frac{\partial x'}{\partial x} &= 0 \\ D \left( \frac{\partial L'}{\partial v'} \right) - \frac{\partial L'}{\partial x'} &= 0 \\ \end{aligned}}$

— Me@2021-01-06 08:10:46 PM

The square root of the probability, 4.3

Posted on January 6, 2021 by rpflee

Eigenstates 3.4.3

The indistinguishability of cases is where the quantum probability comes from.

— Me@2020-12-25 06:21:48 PM

In the double slit experiment, there are 4 cases:

1. only the left slit is open

2. only the right slit is open

3. both slits are open and a measuring device is installed somewhere in the experiment setup so that we can know which slit each photon passes through

4. both slits are open but no measuring device is installed; so for each photon, we have no which-way information

For simplicity, we rephrase the case-3 and case-4:

1. only the left slit is open

2. only the right slit is open

3. both slits are open, with which-way information

4. both slits are open, without which-way information

Case-3 can be regarded as a classical overlapping of case-1 and case-2, because if you check the result of case-3, you will find that it is just an overlapping of result case-1 and result case-2.

However, case-4 cannot be regarded as a classical overlapping of case-1 and case-2. Instead, case-4 is a quantum superposition. A quantum superposition canNOT be regarded as a classical overlapping of possibilities/probabilities/worlds/universes.

Experimentally, no classical overlapping can explain the interference pattern, especially the destruction interference part. An addition of two non-zero probability values can never result in a zero.

Logically, case-4 is a quantum superposition of go-left and go-right. Case-4 is neither AND nor OR of the case-1 and case-2.

We can discuss AND or OR only when there are really 2 distinguishable cases. Since there are not any kinds of measuring devices (for getting which-way information) installed anywhere in the case-4, go-left and go-right are actually indistinguishable cases. In other words, by defining case-4 as a no-measuring-device case, we have indirectly defined that go-left and go-right are actually indistinguishable cases, even in principle.

Note that saying “they are actually indistinguishable cases, even in principle” is equivalent to saying that “they are logically indistinguishable cases” or “they are logically the same case“. So discussing whether a photon has gone left or gone right is meaningless.

If 2 cases are actually indistinguishable even in principle, then in a sense, there is actually only 1 case, the case of “both slits are open but without measuring device installed anywhere” (case-4). Mathematically, this case is expressed as the quantum superposition of go-left and go-right.

Since it is only 1 case, it is meaningless to discuss AND or OR. It is neither “go-left AND go-right” nor “go-left OR go-right“, because the phrases “go-left” and “go-right” are themselves meaningless in this case.

— Me@2020-12-19 10:38 AM

— Me@2020-12-26 11:02 AM

It is a quantum superposition of go-left and go-right.

Quantum superposition is NOT an overlapping of worlds.

Quantum superposition is neither AND nor OR.

— Me@2020-12-26 09:07:22 AM

When the final states are distinguishable you add probabilities:

$\displaystyle{P_{dis} = P_1 + P_2 = \psi_1^*\psi_1 + \psi_2^*\psi_2}$

.

When the final state are indistinguishable,[^2] you add amplitudes:

$\displaystyle{\Psi_{1,2} = \psi_1 + \psi_2}$

and

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

.

[^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

.

$\displaystyle{ P_{ind} = P_1 + P_2 + \psi_2^*\psi_1 + \psi_2^*\psi_2 }$

$\displaystyle{ P_{\text{indistinguishable}} = P_{\text{distinguishable}} + \text{interference terms} }$

— Me@2020-12-26 09:07:46 AM

interference terms ~ indistinguishability effect

— Me@2020-12-26 01:22:36 PM

WHY NOT

Posted on January 5, 2021 by rpflee

You have only ONE life in this life.

WHY NOT do something extraordinary in order to help people to help others?

— Me@~2008

鐵達尼極限 2

Posted on January 5, 2021 by rpflee

尋覓 1.4

這段改編自 2010 年 10 月 14 日的對話。

對於沒有自己人生目標的人來說，

愛情有如患精神病，過不了「鐵達尼極限」（三日）。

— 黃子華

然後，又再深一層。

如果你要愛情事業成功，首先要令到自己，不需要愛情。

比喻說，如果你吃雪糕的原因是，你上了癮的話，吃雪糕就只能令你避免，暫時的不安，但不可以令你得到，真正的快樂。相反，如果你根本沒需要吃雪糕，而選擇吃雪糕，純粹是出於自由意志的話，你就可以真正享受得到，吃雪糕的樂趣。

那如何訓練到，自己不需要愛情呢？

剛才所講，

各人的心靈腦海之中，其實都有超過一個自己。只不過，眾多自我被困於，同一個頭顱之中。

如果你自己的眾多自我，也相處融洽時，你就不太需要外在愛情。相反，如果他們水火不容時，你就極有需要有愛情，用來逃避自己心中，那些眾多不友善的自我。

自己都不愛怎麼相愛怎麼可給愛人好處　– 林夕

人為什麼要有同伴？

比喻說，人自己一個時，太過孤獨，太過冰冷，所以，期望透過有同伴，去感受溫暖。

根據哲學家叔本華所講，人有如「寒冬裡的刺蝟」：一方面，牠們要走近對方，互相取暖；另一方面，走得太近，又會刺傷對方。

但是，如果有一個人，心中有一個太陽，自己會發熱發亮的話，他就不怕人情冷暖。亦即是話，如果你一個人時，生活已經十分精采的話，你就不需要愛情。

留意，「不需要」不代表「不應有」。一個人不需要愛情，而仍然選擇愛情的話，可以是因為技術上的問題，例如，剛巧有對象與他相愛，而他又想有自己的兒女。

Enjoy everything, need nothing, …

.

… especially for human relationships.

– Conversations with God

— Me@2021-01-03 04:18:55 PM

1954

Posted on January 2, 2021 by rpflee

— Photomyne colorization

— Me@2021-01-02 05:20:52 PM

Problem 2.5a

Posted on January 1, 2021 by rpflee

A First Course in String Theory

2.5 Constructing simple orbifolds

(a) Consider a circle $\displaystyle{S^1}$ , presented as the real line with the identification $\displaystyle{x \sim x + 2}$ . Choose $\displaystyle{-1 < x \le 1}$ as the fundamental domain. The circle is the space $\displaystyle{-1 < x \le 1}$ with points $\displaystyle{x = \pm 1}$ identified. The orbifold $\displaystyle{S^1/\mathbb{Z}_2}$ is defined by imposing the (so-called) $\displaystyle{\mathbb{Z}_2}$ identification $\displaystyle{x \sim -x}$ . Describe the action of this identification on the circle. Show that there are two points on the circle that are left fixed by the $\displaystyle{\mathbb{Z}_2}$ action. Find a fundamental domain for the two identifications. Describe the orbifold $\displaystyle{S^1/\mathbb{Z}_2}$ in simple terms.