Interpretations of quantum mechanics and 组合数学 (Fall 2017)/Problem Set 1: Difference between pages

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{{original research|date=February 2013}}
*每道题目的解答都要有<font color="red" size=4>完整的解题过程</font>。中英文不限。
{{more sources|date=May 2014}}
{{Quantum mechanics}}
In [[quantum mechanics]], the mathematical formalism is very difficult to interpret physically. However, there are many ideas about the interpretations and meanings of quantum mechanics. There are no facts to prove any interpretation over the others, but there are some that are more accepted than others.


==Background material==
== Problem 1 ==
''Main article: [[Quantum mechanics]]''
#有<math>k</math>种不同的明信片，每种明信片有无限多张，寄给<math>n</math>个人，每人一张，有多少种方法？
#有<math>k</math>种不同的明信片，每种明信片有无限多张，寄给<math>n</math>个人，每人一张，每个人必须收到不同种类的明信片，有多少种方法？
#有<math>k</math>种不同的明信片，每种明信片有无限多张，寄给<math>n</math>个人，每人收到<math>r</math>张不同的明信片（但不同的人可以收到相同的明信片），有多少种方法？
#只有一种明信片，共有<math>m</math>张，寄给<math>n</math>个人，全部寄完，每个人可以收多张明信片或者不收明信片，有多少种方法？
#有<math>k</math>种不同的明信片，其中第<math>i</math>种明信片有<math>m_i</math>张，寄给<math>n</math>个人，全部寄完，每个人可以收多张明信片或者不收明信片，有多少种方法？


The main ideas of quantum mechanics are the postulates of Schrödinger and Heisenberg. The [[Schrödinger equation]] is a [[partial differential equation]] that describes the [[wavefunction]] of an object.<ref name=hofstra>{{cite web|title=Quantum Mechanics: The Uncertainity Principal, 1925 - 1927|url=http://aip.org/history/heisenberg/p08.htm|publisher=American Institute of Physics|accessdate=3 May 2014|year=1998-2014}}</ref> The equation can be given by
== Problem 2 ==
Find the number of ways to select <math>2n</math> balls from <math>n</math> identical blue balls, <math>n</math> identical red balls and <math>n</math> identical green balls.
* Give a combinatorial proof for the problem.
* Give an algebraic proof for the problem.


<math>i\hbar \frac{\partial \Psi}{\partial t}=\frac{-h}{2m}\nabla^2 \Psi + V(x)\Psi (x)</math>
== Problem 3 ==
*一个长度为<math>n</math>的“山峦”是如下由<math>n</math>个"/"和<math>n</math>个"\"组成的，从坐标<math>(0,0)</math>到<math>(0,2n)</math>的折线，但任何时候都不允许低于<math>x</math>轴。例如下图：


The basic meaning of this equation is that a particle, such as an [[electron]], is not ''just'' a point-like particle, but also a type of wave. The philosophical implications will be explored shortly. Another fundamental of quantum mechanics is the [[Heisenberg uncertainty principle]].<ref name=hofstra /> This bizarre theory is the idea that the position and the [[momentum]] of an object cannot both be known. The greater the certainty of the ''position'' of an object, the less the certainty of the ''momentum'' of the object. The mathematical formulation of this is given by
    /\
  /  \/\/\    /\/\
  /        \/\/    \/\/\
  ----------------------
:长度为<math>n</math>的“山峦”有多少？


<math>\Delta x\Delta p>\frac{\hbar}{2}</math>
*一个长度为<math>n</math>的“地貌”是由<math>n</math>个"/"和<math>n</math>个"\"组成的，从坐标<math>(0,0)</math>到<math>(0,2n)</math>的折线，允许低于<math>x</math>轴。长度为<math>n</math>的“地貌”有多少？


This can further be generalized by stating that
== Problem 4==
李雷和韩梅梅竞选学生会主席，韩梅梅获得选票 <math>p</math> 张，李雷获得选票 <math>q</math> 张，<math>p>q</math>。我们将总共的 <math>p+q</math> 张选票一张一张的点数，有多少种选票的排序方式使得在整个点票过程中，韩梅梅的票数一直高于李雷的票数？等价地，假设选票均匀分布的随机排列，以多大概率在整个点票过程中，韩梅梅的票数一直高于李雷的票数。


<math>\Delta X_1\Delta X_2>\frac{[X_1, X_2]}{2i}</math>
==Problem 5==
A <math>2\times n</math> rectangle is to be paved with <math>1\times 2</math> identical blocks and <math>2\times 2</math> identical blocks. Let <math>f(n)</math> denote the number of ways that can be done. Find a recurrence relation for <math>f(n)</math>, solve the recurrence relation.


Where <math>[X_1, X_2]</math> is the operator of <math>X_1</math> and <math>X_2</math>. This law also gives rise to an uncertainty between [[energy]] and time, which can be expressed in the same way as the relation between momentum and position.
== Problem 6 ==
* 令<math>s_n</math>表示长度为<math>n</math>，没有2个连续的1的二进制串的数量，即
*:<math>s_n=|\{x\in\{0,1\}^n\mid \forall 1\le i\le n-1, x_ix_{i+1}\neq 11\}|</math>。
:求 <math>s_n</math>。


==Probability waves==
*令<math>t_n</math>表示长度为<math>n</math>，没有3个连续的1的二进制串的数量，即
Another important fact of quantum mechanics is that the electron behaves in a very weird way. At first, no one really knew what the wavefunction meant physically. Max Born, a theoretical physicist, explained that the wavefunction is a ''probability wave.'' In other words, wherever the wave is denser, that is where the particle is most likely found, but it won't necessarily be found there. The way to find the probability
*:<math>t_n=|\{x\in\{0,1\}^n\mid \forall 1\le i\le n-2, x_ix_{i+1}x_{i+2}\neq 111\}|</math>。
(<math>P_{[a,b]}</math>) of the position of the particle in the region <math>a<x<b</math> is given by
*#给出计算<math>t_n</math>的递归式，并给出足够的初始值。
*#计算<math>t_n</math>的生成函数<math>T(x)=\sum_{n\ge 0}t_n x^n</math>，给出生成函数<math>T(x)</math>的闭合形式。


<math>\int_a^b \! |\Psi (x, t)|^2 dx=P_{[a,b]}</math>
注意：只需解生成函数的闭合形式，无需展开。


For example, if <math>P_{[a,b]}</math> is equal to .5, then there is a 50% chance of finding the particle within that region. This shows us that the location of a particle probabilistic; one can never say that the particle will ''definitely'' be found at a certain point in space, but rather, one can only give the probability of finding the particle within that region.
== Problem 7 ==
 
Let <math>a_n</math> be a sequence of numbers satisfying the recurrence relation:
==Copenhagen interpretation==
:<math>p a_n+q a_{n-1}+r a_{n-2}=0</math>
The most well-accepted interpretation of quantum mechanics is the idea called '''Copenhagen interpretation.''' This interpretation builds upon the probability-wave notion, but brings in a radical new idea called the '''superposition principle'''. The best way to explain this principle is by showing it mathematically. If the functions <math>\Psi_1, \Psi_2, \Psi_3, ... \Psi_n</math> are solutions of the Schrödinger equation, then the superposition of those wavefunctions is also a solution. i.e.,
with initial condition <math>a_0=s</math> and <math>a_1=t</math>, where <math>p,q,r,s,t</math> are constants such that <math>{p}+q+r=0</math>, <math>p\neq 0</math> and <math>s\neq t</math>. Solve the recurrence relation.
 
<math>\Theta = c_1\Psi_1 + c_2\Psi_2 + c_3\Psi_3 + ... c_n\Psi_n</math>  
 
Where <math>\theta</math> is the superposition of the various wavefunctions. This idea implies that a particle occupies every possible wavefunction it can. This implies that a particle occupies more than one position ''at the same time.'' In other words, a particle exists in at least two different positions simultaneously. When an observer comes and actually measures the position of the particle, something called the '''wavefunction collapse''' occurs. So when someone observes the particle, the following happens:
 
<math>\Theta</math>→<math>\Psi_n</math>
 
In simple terms: when there is no observation or observer, then a particle occupies many positions simultaneously; when an observation takes place, the wavefunction collapses and the particle exists only in one position.
 
==Many-worlds interpretation==
The '''many-worlds interpretation''' is by far the most fantastic interpretation of quantum mechanics. This interpretation says that rather than the wavefunction collapsing, each possibility actually occurs, but in separate universes. This means that the universes branch off for each possibility.<ref name=HSW>{{cite web|title=Do parallel universes really exist?|url=http://science.howstuffworks.com/science-vs-myth/everyday-myths/parallel-universe2.htm|publisher=HowStuffWorks website|accessdate=3 May 2014|author=Josh Clark|year=1998-2014}}</ref>  
 
==Quantum determinism==
The interpretation presented by [[Albert Einstein]] himself, states that the outcome of some random event is predetermined. So, rather than a particle existing as a probability wave, this interpretation says that the particle only exists only in one position, but we just perceive it to be a probability. This idea is much less popular, but nonetheless mentionable.
 
==Which one is right?==
So, of the three main interpretations of quantum mechanics, which one is correct? Physicists seem to think that the Copenhagen interpretation is the most likely, but no one is for sure.
 
== References ==
{{Reflist}}
 
[[Category:Quantum mechanics]]

Revision as of 13:14, 17 September 2017

  • 每道题目的解答都要有完整的解题过程。中英文不限。

Problem 1

  1. 有[math]\displaystyle{ k }[/math]种不同的明信片，每种明信片有无限多张，寄给[math]\displaystyle{ n }[/math]个人，每人一张，有多少种方法？
  2. 有[math]\displaystyle{ k }[/math]种不同的明信片，每种明信片有无限多张，寄给[math]\displaystyle{ n }[/math]个人，每人一张，每个人必须收到不同种类的明信片，有多少种方法？
  3. 有[math]\displaystyle{ k }[/math]种不同的明信片，每种明信片有无限多张，寄给[math]\displaystyle{ n }[/math]个人，每人收到[math]\displaystyle{ r }[/math]张不同的明信片（但不同的人可以收到相同的明信片），有多少种方法？
  4. 只有一种明信片，共有[math]\displaystyle{ m }[/math]张，寄给[math]\displaystyle{ n }[/math]个人，全部寄完，每个人可以收多张明信片或者不收明信片，有多少种方法？
  5. 有[math]\displaystyle{ k }[/math]种不同的明信片，其中第[math]\displaystyle{ i }[/math]种明信片有[math]\displaystyle{ m_i }[/math]张，寄给[math]\displaystyle{ n }[/math]个人，全部寄完，每个人可以收多张明信片或者不收明信片，有多少种方法？

Problem 2

Find the number of ways to select [math]\displaystyle{ 2n }[/math] balls from [math]\displaystyle{ n }[/math] identical blue balls, [math]\displaystyle{ n }[/math] identical red balls and [math]\displaystyle{ n }[/math] identical green balls.

  • Give a combinatorial proof for the problem.
  • Give an algebraic proof for the problem.

Problem 3

  • 一个长度为[math]\displaystyle{ n }[/math]的“山峦”是如下由[math]\displaystyle{ n }[/math]个"/"和[math]\displaystyle{ n }[/math]个"\"组成的，从坐标[math]\displaystyle{ (0,0) }[/math]到[math]\displaystyle{ (0,2n) }[/math]的折线，但任何时候都不允许低于[math]\displaystyle{ x }[/math]轴。例如下图：
   /\
  /  \/\/\    /\/\
 /        \/\/    \/\/\
 ----------------------
长度为[math]\displaystyle{ n }[/math]的“山峦”有多少？
  • 一个长度为[math]\displaystyle{ n }[/math]的“地貌”是由[math]\displaystyle{ n }[/math]个"/"和[math]\displaystyle{ n }[/math]个"\"组成的，从坐标[math]\displaystyle{ (0,0) }[/math]到[math]\displaystyle{ (0,2n) }[/math]的折线，允许低于[math]\displaystyle{ x }[/math]轴。长度为[math]\displaystyle{ n }[/math]的“地貌”有多少？

Problem 4

李雷和韩梅梅竞选学生会主席，韩梅梅获得选票 [math]\displaystyle{ p }[/math] 张，李雷获得选票 [math]\displaystyle{ q }[/math] 张，[math]\displaystyle{ p\gt q }[/math]。我们将总共的 [math]\displaystyle{ p+q }[/math] 张选票一张一张的点数，有多少种选票的排序方式使得在整个点票过程中，韩梅梅的票数一直高于李雷的票数？等价地，假设选票均匀分布的随机排列，以多大概率在整个点票过程中，韩梅梅的票数一直高于李雷的票数。

Problem 5

A [math]\displaystyle{ 2\times n }[/math] rectangle is to be paved with [math]\displaystyle{ 1\times 2 }[/math] identical blocks and [math]\displaystyle{ 2\times 2 }[/math] identical blocks. Let [math]\displaystyle{ f(n) }[/math] denote the number of ways that can be done. Find a recurrence relation for [math]\displaystyle{ f(n) }[/math], solve the recurrence relation.

Problem 6

  • 令[math]\displaystyle{ s_n }[/math]表示长度为[math]\displaystyle{ n }[/math]，没有2个连续的1的二进制串的数量，即
    [math]\displaystyle{ s_n=|\{x\in\{0,1\}^n\mid \forall 1\le i\le n-1, x_ix_{i+1}\neq 11\}| }[/math]。
求 [math]\displaystyle{ s_n }[/math]。
  • 令[math]\displaystyle{ t_n }[/math]表示长度为[math]\displaystyle{ n }[/math]，没有3个连续的1的二进制串的数量，即
    [math]\displaystyle{ t_n=|\{x\in\{0,1\}^n\mid \forall 1\le i\le n-2, x_ix_{i+1}x_{i+2}\neq 111\}| }[/math]。
    1. 给出计算[math]\displaystyle{ t_n }[/math]的递归式，并给出足够的初始值。
    2. 计算[math]\displaystyle{ t_n }[/math]的生成函数[math]\displaystyle{ T(x)=\sum_{n\ge 0}t_n x^n }[/math]，给出生成函数[math]\displaystyle{ T(x) }[/math]的闭合形式。

注意：只需解生成函数的闭合形式，无需展开。

Problem 7

Let [math]\displaystyle{ a_n }[/math] be a sequence of numbers satisfying the recurrence relation:

[math]\displaystyle{ p a_n+q a_{n-1}+r a_{n-2}=0 }[/math]

with initial condition [math]\displaystyle{ a_0=s }[/math] and [math]\displaystyle{ a_1=t }[/math], where [math]\displaystyle{ p,q,r,s,t }[/math] are constants such that [math]\displaystyle{ {p}+q+r=0 }[/math], [math]\displaystyle{ p\neq 0 }[/math] and [math]\displaystyle{ s\neq t }[/math]. Solve the recurrence relation.

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