和 Ben 一起去买水

今天我去买水,Ben也要求同行,我们两个就高高兴兴的去了。到了卖水的机器面前,Ben 负责投币,我负责搬运。Ben总是很淘气,有很多鬼主意:机器上最大的按钮给你五加仑水,需要 $1.75。如果你投入两块钱,机器就给你一个一夸特的硬币 $0.25。我自己已经习惯了这些设定,每次都是插入两美元纸币,按下按钮,然后拿回我的硬币。Ben插入纸币的时候就问我,如果放入3块钱,那会找回5个一夸特的硬币吗?我一想这倒是一个把纸币换成硬币的好办法,只不过自己每次都来去匆匆,即使有想尝试的念头,也只是一闪而过,从来没有认真执行:万一把机器搞坏了怎么办?还是先忙正事吧。不过Ben既然有此念头,我想了一下就说那你试试吧。结果是机器根本不再接受多余的钱:如果你已经插入了两美元,无论是投入更多的硬币、或者插入更多的纸币,都会被机器自动退回。这个设计还是很聪明啊。

Yetao的新球拍

Yetao最近喜欢上打乒乓球,开始用正胶,可能是觉得击打比较舒服,后来看我和本的反胶球拍击球威力很大,就也想换个反胶。我手头只有一块 Hadraw SK 贴着快两年旧的 Tenergy 05-FX,拍子有点重,而且胶皮也老了,就给她换了一个 Joola X-Plode Sensitive,也是以击打为主。粘的时候旧板子除胶可真是费了一番功夫!

拍子粘好了: Hadraw SK + 2 * X-Plode Sensitive = 182 克

我的拍子:Timo Boll ZLF + 2 * Evolution MX-S (软胶面,硬海绵)= 192 克

Ben的拍子:Timo Boll Allround + 2 * Sapphira = 161 克

看来Ben的拍子还是最轻呀。

找教练训练了5天乒乓球接发球

最近痛感自己的乒乓球接发球太差,正好利用休假的时间安排了五天乒乓球训练,每天两小时,说说自己训练下来的感受。

  • 乒乓球接发球很难。这五天训练下来,花掉800块钱。接发球的水平远远没有达到自己想要的水平。
  • 专业强度的训练,对于业余爱好者太难。五天训练下来。我的身体都吃快吃不消了。
  • 接发球正手位比反手位难,尤其是短球。因为横版正手搓球大多不太好,但是上步用反手,对于步伐的要求又太高了。
  • 接发球不能大力出奇迹,台内短球基本上全靠手腕手指发力,否则极易接球弧线太长出界。

总的来讲,自己的下旋球接的还可以,正反手都可以搓一搓,并且能够根据顺旋转和逆旋转来调整版型。但是上旋球就难很多:反手还可以手腕发力摩擦一板,正手打的一塌糊涂,只会推送。特别的,教练说我击打的感觉很差,这说起来都有一些黑色幽默:打乒乓球这么久,结果到最后反而只会摩擦,不会打了。

error function

昨天做 《Introduction to Quantum Mechanics》上的一道题目,即归一化下面的波函数:

    \[\Psi(x, t) = A e^{-a(\dfrac{mx^2}{h} + it)}\]

搞了一会儿,才发现自己不会计算 e^{-x^2} 的不定积分:熟悉的积分计算规则全部都用不上!上网查了一会,才发现这就是大名鼎鼎的 Error Function:

    \[erf(x) = \dfrac{2}{\sqrt \pi} \int_0^{x} e^{-t^2}dt\]

它的图像是:

这是一个不能用初等解析函数表示的函数,在统计学和量子力学中有着广泛的应用。对于我上面的问题,根据归一化的基本要求我们有:

    \[\int_{-\infty}^{+\infty} |\Psi|^2 dx = A^2 \int_{-\infty}^{+\infty} e^{\dfrac{-2amx^2}{h}} dx = 1\]

如果我们假定:

    \[u = \sqrt{\dfrac{2am}{h}} \cdot x\]

那么:

    \[dx = \dfrac{1}{\sqrt{\dfrac{2am}{h}}} \cdot du\]

按照erf 的形式展开有:

    \[\begin{split} 1 &= \int_{-\infty}^{+\infty} |\Psi|^2 dx = A^2 \int_{-\infty}^{+\infty} e^{\dfrac{-2amx^2}{h}} dx \\ &= A^2 \int_{-\infty}^{+\infty} e^{-(\sqrt{\dfrac{2am}{h}} \cdot x)^2} dx \\ &= A^2 \cdot \dfrac{1}{\sqrt{\dfrac{2am}{h}}} \int_{-\infty}^{+\infty} e^{-u^2} du \\ &= \dfrac{\sqrt \pi}{2} A^2 \dfrac{1}{\sqrt{\dfrac{2am}{h}}} \cdot \dfrac{2}{\sqrt \pi} erf(u) \Biggr |_{-\infty}^{+\infty} \\ &= \dfrac{\sqrt \pi}{2} A^2 \dfrac{1}{\sqrt{\dfrac{2am}{h}}} \cdot 2 \end{split}\]

这样就计算出

    \[A = \sqrt \sqrt{\dfrac{2am}{h \pi}}\]

动量对于时间的导数等于势能对于距离的导数

自己最近在努力学习量子力学,看到Ehrenfest’s theorem: the expected values in Quantum Mechanics obeys the classical laws. 现在证明动量对于时间的导数等于势能对于距离的导数,即:

(1)   \begin{equation*} \dfrac{d \langle p \rangle }{dt} = \langle -\dfrac{\partial V}{\partial x} \rangle \end{equation*}

First by definition, we have:

(2)   \begin{equation*} \dfrac{d \langle p \rangle }{dt} = -ih \int_{-\infty}^{\infty} \dfrac{\partial}{\partial t}(\Psi^* \dfrac{\partial \Psi}{\partial x})dx \end{equation*}

(3)   \begin{equation*} \langle \dfrac{\partial V}{\partial x} \rangle = \int_{-\infty}^{\infty} \dfrac{\partial V}{\partial x} |\Psi|^2 dx \end{equation*}

How do we proof the equality? First we notice that the norm of the wave function can be expressed the following:

(4)   \begin{equation*} |\Psi|^2 = \Psi^* \cdot \Psi \end{equation*}

so (3) can be expanded as:

    \[\label{potential} \langle \dfrac{\partial V}{\partial x} \rangle = \int_{-\infty}^{\infty} \dfrac{\partial V}{\partial x} \Psi^* \Psi dx \end{equation}\]

With the following integration by parts technique:

(5)   \begin{equation*} \int_{a}^{b} {u \cdot \dfrac{dv}{dx}} dx = - \int_{a}^{b} {v \cdot \dfrac{du}{dx}} dx + uv \Biggr|_{a}^{b} \end{equation*}

and also notice that at -\infty and +\infty, the wave function \Psi must be zero, so (3) can be further reduced to:

(6)   \begin{equation*} \begin{split} & \langle \dfrac{\partial V}{\partial x} \rangle \\ &= \int_{-\infty}^{\infty} \dfrac{\partial V}{\partial x} \Psi^* \Psi dx \\ &= - \int_{-\infty}^{\infty} V \dfrac{\partial (\Psi^* \Psi) }{\partial x} dx + V \Psi \Psi^* \Biggr|_{- \infty}^{\infty} \\ &= - \int_{-\infty}^{\infty} V (\Psi \dfrac{\partial \Psi^*}{\partial x} + \Psi^* \dfrac{\partial \Psi}{\partial x})dx \end{split} \end{equation*}

Where should we go from here for the proof? Notice that in (2) we still have the derivative to time, so based on the basic wave function:

    \[i h \dfrac {\partial \Psi}{\partial t} = - \dfrac{h^2}{2m} \dfrac{{\partial}^2 \Psi}{{\partial} x^2}+ V \Psi\]

we can covert the derivative of \Psi and \Psi^* to time to their derivative to space as the following:

(7)   \begin{equation*} \dfrac{\partial \Psi}{\partial t} = \dfrac{i h}{2m} \dfrac{\partial^2 \Psi}{\partial x^2} - \dfrac{i}{h} V \Psi \end{equation*}

(8)   \begin{equation*} \dfrac{\partial \Psi^*}{\partial t} = - \dfrac{i h}{2m} \dfrac{\partial^2 \Psi^*}{\partial x^2} + \dfrac{i}{h} V \Psi^* \end{equation*}

With these intermediate results, we can further expand the 2 as the following:

(9)   \begin{equation*} \begin{split} & \dfrac{d \langle p \rangle }{dt} \\ =& -ih \int_{-\infty}^{\infty} \dfrac{\partial}{\partial t}(\Psi^* \dfrac{\partial \Psi}{\partial x})dx \\ =& -ih \int_{-\infty}^{\infty} [\dfrac{\partial \Psi^*}{\partial t} \dfrac{\partial \Psi}{\partial x} + \Psi^* \dfrac{\partial}{\partial t} (\dfrac{\partial \Psi}{\partial x}) ] dx \\ =& -ih \int_{-\infty}^{\infty} [\dfrac{\partial \Psi^*}{\partial t} \dfrac{\partial \Psi}{\partial x} + \Psi^* \dfrac{\partial}{\partial x} (\dfrac{\partial \Psi}{\partial t}) ] dx \end{split} \end{equation*}

Applying integration by parts for the second term inside the integration above, and notice that the constant term is 0, we can have:

    \[\int_{-\infty}^{\infty} \Psi^* \dfrac{\partial}{\partial x} (\dfrac{\partial \Psi}{\partial t}) dx = - \int_{-\infty}^{\infty} \dfrac{\partial \Psi^*}{\partial x} \dfrac{\partial \Psi}{\partial t} dx\]

Replace this result to the above equation, we will continue to have:

    \[\dfrac{d \langle p \rangle }{dt} = -ih \int_{-\infty}^{\infty} [\dfrac{\partial \Psi^*}{\partial t} \dfrac{\partial \Psi}{\partial x} - \dfrac{\partial \Psi^*}{\partial x} \dfrac{\partial \Psi}{\partial t}] dx\]

Now, replace the time derivative of wave function with space derivative, and rearrange terms, we can have:

(10)   \begin{equation*} \begin{split} & \dfrac{d \langle p \rangle }{dt} \\ =& -ih \int_{-\infty}^{\infty} [\dfrac{\partial \Psi^*}{\partial t} \dfrac{\partial \Psi}{\partial x} - \dfrac{\partial \Psi^*}{\partial x} \dfrac{\partial \Psi}{\partial t}] dx \\ &= \int_{-\infty}^{\infty} V \Psi^* \dfrac{\Partial \Psi}{\partial x} dx - \dfrac{h^2}{2m} \int_{-\infty}^{\infty} \dfrac{\partial^2 \Psi^*}{\partial x^2} \dfrac{\Partial \Psi}{\partial x} \\ &+  \int_{-\infty}^{\infty} V \Psi \dfrac{\Partial \Psi^*}{\partial x} dx - \dfrac{h^2}{2m} \int_{-\infty}^{\infty} \dfrac{\partial^2 \Psi}{\partial x^2} \dfrac{\Partial \Psi^*}{\partial x} \\ &= \langle -\dfrac{\partial V}{\partial x} \rangle - \dfrac{h^2}{2m} \int_{-\infty}^{\infty} \dfrac{\partial^2 \Psi^*}{\partial x^2} \dfrac{\partial \Psi}{\partial x} - \dfrac{h^2}{2m} \int_{-\infty}^{\infty} \dfrac{\partial^2 \Psi}{\partial x^2} \dfrac{\partial \Psi^*}{\partial x} \end{split} \end{equation*}

The last thing we need to notice is that by applying integration by parts and ignore zero constant terms, we can have:

    \[\int_{-\infty}^{\infty} \dfrac{\partial^2 \Psi^*}{\partial x^2} \dfrac{\partial \Psi}{\partial x} = - \int_{-\infty}^{\infty} \dfrac{\partial^2 \Psi}{\partial x^2} \dfrac{\partial \Psi^*}{\partial x}\]

so the last two terms in above equation cancel each other and we prove the desired theorem:

(11)   \begin{equation*} \dfrac{d \langle p \rangle }{dt} = \langle -\dfrac{\partial V}{\partial x} \rangle \end{equation*}

 

隔壁奶奶家的桃子

隔壁奶奶的前院有一棵桃树,今年的桃结的又大又红。前几天我和太太散步路过,不由得垂诞欲滴。可是这毕竟不是自己的东西,总是不好意思直接伸手。太太看出来我的意思,笑眯眯的跟我说,她和邻家奶奶关系很好,帮我说说就行。结果当天晚上桌上就有了一大包奶奶送过来的桃子,吃得我很开心。才过了几天,今天早上奶奶又亲自登门,邀请我拿个大袋子过去再摘一些桃子,我很不好意思,又很高兴。摘桃子的时候,奶奶说起来她看到很多路过的年轻人,也不和主人打招呼动手就摘,让她不高兴。我就说那样摘桃的人也心里恐怕有些惴惴,哪里像我这样一边大大方方的摘,主人还在一遍帮忙,吃到肚子里也算是是心安理得,我们两个人就都笑起来了,真好。

教Ben打乒乓球

我很喜欢打乒乓球,觉得这项爱好有很多优点,自然也想把这项运动传给孩子们。Yetao还好,每天都能打一会儿,但是Ben不愿意和我打,我好说歹说,他就是不愿意。今天Ben邀请我一块儿玩儿 Dodge Ball,他主动开出条件可以陪我打五分钟乒乓球。打乒乓球的时候,我就故意把很多球打飞,结果Ben赢了很多球,这下子来了兴趣。打完五分钟,主动还要再打十分钟,后来干脆把闹钟停下来。打了一会儿,我说今天已经够多啦,休息一下吧。结果Ben还说今天晚些时候能不能再打一次。看来Ben好胜心很强,只要能一直赢球,他就有兴趣。这倒是一个把他引上乒乓球的好办法。想来以前我和他打的时候,总是对他教训太严,这也不对,那也不对,缺少鼓励,所以他就渐渐没了兴趣。看来真的是要因材施教啊。

今天我教他练了一会儿正手攻球和反手攻球,Yetao也跑来参与。大家玩的很开心。希望以后每天都能这样。

How to Disable Google Doodle on Google Chrome New Tab Page

By default Google Chrome will show Google Doodle on its New Tab Page. It is OK for me if Google wants to push some of its ideology to users by using Google Doodle, but recently I found Google goes too far by starting using animation-based Doodle, which is very annoying: Every time I open a new tab, I can not stop paying some attentions to the animated doodle, which is a constant distraction. To add insults to hurts, Google removed the old chrome flag which can disables Doodle on NTP. I really can not understand why Google does this other than being arrogant to its users.

After suffering from the animated Doodle for while, I decided to search for a solution. And eventually I found the –google-doodle-url flag on this page, which you can use if you start Google Chrome from a terminal and tell it to only use a static Google Doodle. For example, on my Macbook, I can run the following command from a terminal:

/Applications/Google\ Chrome.app/Contents/MacOS/Google\ Chrome \
--google-doodle-url="https://www.gstatic.com/chrome/ntp/doodle_test/ddljson_android0.json"

and no longer see the animated Doodle. Vola!

Hopefully Google’s won’t take this away again, otherwise we have to switch to Firefox completely.