sage
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Continuing discussion of quantum mechanics
Let's continue in a new post.
There seems to be quite some confusions (maybe on my part as well). Therefore, I would like to set some basic concepts straight before thinking further. This may not address all your problems, but I think it is a good idea to start basic.
I appologize ahead if you are already very familiar with all of this,
1) The definition of free particle is that the Hamiltonian of the
system is free. Any solution of such a Lagrangian is called free
particle. Let's take this as a definition for now (hold your
opinion until you have worked through the rest of the post).
2) Interference is different from interaction. Only a coherent state
could interfer. It could only interfer with itself. We can write a
state as
|psi>=|1>+|2>
then, the probability distribution of such a state is (let's take
x-representation. it does not have to be position. )
<psi|x><x|psi>=|<x|1>|^2 + |<x|2>|^2+2Re(<1|x><x|2>)
the third term is the interference effect. It obviously does not
come from any interaction. |psi> could be a free particle. The
interference comes from the fact that we write any state as a
superposition of two coherent state |1> and |2>.
Superposition is indeed the magic of a lot of the quantum
effect. Coherent sum is the source, and the only
source, for the quantum interference.
3) Two particle state could be understood in similar ways. However, in
this case, it is important to understand the concept of identical
particles. Two particles are identical if there is no interaction
could tell them apart. This is the same in both classical and
quantum mechanics.
Being identical, there is nothing to label particle. therefore, the
notion of this particle and that particle breaks down. Two particle
form a single coherent state. the number of particles in that state
is 2. However, give you any piece of that state, you could not say
this is particle one and that is particle two.
Therefore, the concept of this particles wave-function overlap with that
particle's wave-function of produce interference is not correct. if
they are distinct state, they will not interfer. the state interfer
only when they are in a coherent sum. And, coherent sum only comes
from identical particles.
Suppose again there are two states ('here' and 'there', if you
want) which we write as |1> and |2>. A two particle state is (I
will take two bosons as an example. I am also a bit sloppy on notations)
|psi>=|1>|2>+|2>|1>
Similar to the previous case, the probability distribution is (in
terms of wave-functions)
<psi|x1,x2><x1,x2|psi>= |<x1|1>|^2 |<x2|2>|^2 + |<x1|2>|^2
|<x2|1>|^2 + <2|x1><x1|1> <1|x2><x2|2>+c.c
the first two terms are independent probability distributions. the
thrid term is interference. if one insist that first ket is
particle one (not a physical label, just a convention), then, we
see that it is the interference between two first kets and between
two second kets. Therefore, even if we label them by convention, it
is not the interference between two particles. It is the
interference between a particle and itself due to the requirement
of a coherent sum.
When there is no overlap between |1> and |2>, this result reduces
to two independent particles. In this case, one can choose to treat
them either as identical, or as distinct (labelled by
coordinate). There is no physical difference.
-----------------------------------------------------------------------
Now some misc. comments
1) superposition is a coherent sum. It is very different from a simple
addition of all components. For example, a plane-wave, could be
written as a set of eigenstates of angular momentum (spherical
bessel function in the radial directs times spherical harmonics for
angular variable.) However, it is a plane-wave, very different from
any of the angular momentum eigenstates.
another example, I can write a finite size wave-packet as a sum of
plane-waves. However, this is still a finite size wave-packet. very
different from any plane-waves.
2) Long distance entanglement, or Bell inequality, has been
experimentally established. there is no reason to suspect things
will be any different over even larger distances.
3) A side comment on causality. even we could propagate signal outside
the lightcone, it does not necessarily break causality. If I can
send a signal instantaneously, as long as nobody can send it back
before I send it, there is no breaking of causality.
| 发表时间：2005-07-12, 08:46:14 |
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HPC
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Re: Continuing discussion of quantum mechanics
one issue not clear to me:
when we consider two particle, I think we only make some approximation such as in-going states or out-going states. because the interaction is always of existence.
Faith, Fashion and Fancy.
Welcome to 我的域名：http://hongbaozhang.blog.edu.cn
| 发表时间：2005-07-12, 09:34:58 |
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元江
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Re: Continuing discussion of quantum mechanics
Thanks sage for your posting. Let's leave this topic for now, I will come back to discuss with net friends here more about this topic.
道可道非常道
名可名非常名
| 发表时间：2005-07-12, 13:25:47 |
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元江
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Re: Continuing discussion of quantum mechanics
2) Interference is different from interaction. Only a coherent state
could interfer. It could only interfer with itself. We can write a
state as
|psi>=|1>+|2>
------------------------------------------------
A question, truly a question.
If I have two identicle particles, I know one is "here" marked by |1>, another particle is "there" and I mark it by |2>, both |1> and |2> are solutions of the Shcrodinger equation. Since Shcrodinger equation is linear, so the addition of two solutions is also a solution of Shcrodinger equation, i.e. |psi> , is also a solution of
Shcrodinger equation. Can I construct |psi> as you wrote here as a solution of two particle system? If |1> and |2> are both localized (amplitude covers a finite space), then, when |1> and |2> are far away from each other, their amplitudes do not overlap, the coherence between two particles (two wave sources) is zero. When the overlap of two amplitudes are not zero, these two particles interfere with each other.
道可道非常道
名可名非常名
| 发表时间：2005-07-12, 16:02:11 |
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sage
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Re: Continuing discussion of quantum mechanics
>If I have two identicle particles, I know one is "here" marked by |1>, another particle is "there" >and I mark it by |2>, both |1> and |2> are solutions of the Shcrodinger equation. Since >Shcrodinger equation is linear, so the addition of two solutions is also a solution of Shcrodinger >equation, i.e. |psi> , is also a solution of
>Shcrodinger equation. Can I construct |psi> as you wrote here as a solution of two particle >system? If |1> and |2> are both localized (amplitude covers a finite space), then, when |1> and >|2> are far away from each other, their amplitudes do not overlap, the coherence between two >particles (two wave sources) is zero. When the overlap of two amplitudes are not zero, these >two particles interfere with each other.
you cannot. very roughly speaking, |1> or |2> is only one-particle state. their coherent sum is still one particle state.
the two particle state should be a direct product |1>|2>+permutation.
very naively, you can convince yourself this way,
two particle hamiltonian is
p_1^2/2m +p_2^2/2m
wavefuntion of |1>+|2> is <x|1>+<x|2>, it is not a solution of the schrodinger equation.
on the other hand, <x_1|1><x_2|2> is a solution, as is obvious from separation of variables.
| 发表时间：2005-07-12, 22:09:52 |
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星空浩淼
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Re: Continuing discussion of quantum mechanics
建议大家回复时，把引用的打上引号，免得误会。本人用中文，就不用了。
2) Interference is different from interaction. Only a coherent state
could interfer. It could only interfer with itself.
不过，不同方式的相干叠加，决定粒子的去向。例如在一种相干叠加方式下，粒子在某处出现的概率极大，而改变一下相干叠加方式，会使得粒子在这个地方出现的概率为零。例如AB效应，其实就是通过改变粒子叠加方式来体现势存在时产生的物理效应，它的目的就是表明势是有物理效应的，不只是数学方便的工具。
另外，如果波函数处于相干叠加态，跟不处于相干叠加态时，对Lagrangian中的相互作用项有不同的贡献，此时相干叠加的存在就等效于存在一种作用。
Therefore, the concept of this particles wave-function overlap with that
particle's wave-function of produce interference is not correct. if
they are distinct state, they will not interfer. the state interfer
only when they are in a coherent sum. And, coherent sum only comes
from identical particles.
全同粒子的对称性，已经不是一种数学概念，而是能带来非平凡的物理效应：交换势。至于算不算相干叠加态，我感觉这只是语言说法上的问题，例如可以把它看成是由全同粒子对称性规律支配下的相干叠加，全同粒子对称性给叠加方式附加了额外的条件。
Now some misc. comments
1) superposition is a coherent sum. It is very different from a simple
addition of all components. For example, a plane-wave, could be
written as a set of eigenstates of angular momentum (spherical
bessel function in the radial directs times spherical harmonics for
angular variable.) However, it is a plane-wave, very different from
any of the angular momentum eigenstates.
another example, I can write a finite size wave-packet as a sum of
plane-waves. However, this is still a finite size wave-packet. very
different from any plane-waves.
我严重不同意这种看法。平面波展开成柱面波时，它就对应无穷多个柱面波的相干叠加；当它展开成无穷多个球面波的叠加时，它就看作无穷多个球面波的相干叠加，至于平面波不是柱面波或球面波，这是当然的，但跟上面这个不矛盾。好比你不能因为房子不是砖头，就否定房子是砖头构成的。
2) Long distance entanglement, or Bell inequality, has been
experimentally established. there is no reason to suspect things
will be any different over even larger distances.
这个没有看明白。
3) A side comment on causality. even we could propagate signal outside
the lightcone, it does not necessarily break causality. If I can
send a signal instantaneously, as long as nobody can send it back
before I send it, there is no breaking of causality.
这个我不同意。如果你send a signal instantaneously （outside
the lightcone），另一个人收到信号，那么跟你有相对速度的另外一个观察者看来, 收到信号的人在你发送信号之前就收到了信号，即时序是颠倒的！这也是“因果律破坏”的一个含义。
唯有与时间赛跑，方可维持一息尚存
| 发表时间：2005-07-13, 05:12:02 |
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星空浩淼
发表文章数: 1743
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Re: Continuing discussion of quantum mechanics
A question, truly a question.
If I have two identicle particles, I know one is "here" marked by |1>, another particle is "there" and I mark it by |2>, both |1> and |2> are solutions of the Shcrodinger equation. Since Shcrodinger equation is linear, so the addition of two solutions is also a solution of Shcrodinger equation, i.e. |psi> , is also a solution of
Shcrodinger equation. Can I construct |psi> as you wrote here as a solution of two particle system? If |1> and |2> are both localized (amplitude covers a finite space), then, when |1> and |2> are far away from each other, their amplitudes do not overlap, the coherence between two particles (two wave sources) is zero. When the overlap of two amplitudes are not zero, these two particles interfere with each other.
元江兄这个问题，在说法上有问题，但内容上是教材上的标准说法。全同粒子不能编号，所以元江兄的问题应该这样说：
“有两个粒子，它们的波包重叠时，在重叠区域，全同对称性决定的叠加方式产生了交换势；如果它们的波包不重叠，就不会存在交换势。”
唯有与时间赛跑，方可维持一息尚存
| 发表时间：2005-07-13, 05:18:57 |
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星空浩淼
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Re: Continuing discussion of quantum mechanics
you cannot. very roughly speaking, |1> or |2> is only one-particle state. their coherent sum is still one particle state.
the two particle state should be a direct product |1>|2>+permutation.
very naively, you can convince yourself this way,
two particle hamiltonian is
p_1^2/2m +p_2^2/2m
wavefuntion of |1>+|2> is <x|1>+<x|2>, it is not a solution of the schrodinger equation.
on the other hand, <x_1|1><x_2|2> is a solution, as is obvious from separation of variables.
SAGE兄上面说法还是有漏洞多多：
1）如果<x_1|1><x_2|2> is a solution，那么<x|1>+<x|2>必定也是一个解。事实上，<x_1|1><x_2|2>中的<x|1>或者<x|2>退化成一个常数时，就变成<x|2>或者<x|1>。
2）不考虑全同粒子对称性时，<x|1>+<x|2>是通常的解，只有考虑到全同粒子对称性时，|1>|2>+permutation是解。在不考虑全同粒子对称性时后者可以退化成前者。
3）|1> or |2> is one-particle state. their coherent sum 称为双粒子态而不是单粒子态，你只能说称为“单态”或“一个量子纯态”，而不是“单粒子态”。
唯有与时间赛跑，方可维持一息尚存
| 发表时间：2005-07-13, 05:33:20 |
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sage
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Re: Continuing discussion of quantum mechanics
SAGE兄上面说法还是有漏洞多多：
1）如果<x_1|1><x_2|2> is a solution，那么<x|1>+<x|2>必定也是一个解。事实上，<x_1|1><x_2|2> 中的<x|1>或者<x|2>退化成一个常数时，就变成<x|2>或者<x|1>。
2）不考虑全同粒子对称性时，<x|1>+<x|2>是通常的解，只有考虑到全同粒子对称性时，|1>|2>+permutation是解。在不考虑全同粒子对称性时后者可以退化成前者。
3）|1> or |2> is one-particle state. their coherent sum 称为双粒子态而不是单粒子态，你只能说称为“单态”或“一个量子纯态”，而不是“单粒子态”。
I disagree.
1) the wave-function of a two particle state must have two sets of independent coordinates. in other word, it must be a product hilbert space. The hamilonian should be just
\partial_x1^2/2m + \partial_x2^2/2m
it is impossible to even think <x|1>+<x|2> to be a solution.
2) |1>+|2> is not the same as a product space when one of the factor become constant.
a product space is |1>|1'>+|2'>|2>
when one of the wave-function become constant, it is still
|1>|c>+|c'>|2>
there is no interference between |1> and |2> in this state. which is very different from |1>+|2>.
3) we could construct a particle number operator, (a sensible should be just) acting on |1>+|2> will give you 1.
| 发表时间：2005-07-13, 08:11:37 |
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sage
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Re: Continuing discussion of quantum mechanics
我严重不同意这种看法。平面波展开成柱面波时，它就对应无穷多个柱面波的相干叠加；当它展开成无穷多个球面波的叠加时，它就看作无穷多个球面波的相干叠加，至于平面波不是柱面波或球面波，这是当然的，但跟上面这个不矛盾。好比你不能因为房子不是砖头，就否定房子是砖头构成的。
please read carefully before you reply. I did not say anything different from what you said.
the only thing wrong for the brick-house analogy is that coherent sum is not a simple addition.
这个我不同意。如果你send a signal instantaneously （outside
the lightcone），另一个人收到信号，那么跟你有相对速度的另外一个观察者看来, 收到信号的人在你发送信号之前就收到了信号，即时序是颠倒的！这也是“因果律破坏”的一个含义。
time is just a local label, it has no meaning. unless you can show this is a close time curve, there is no violation of causality.
| 发表时间：2005-07-13, 08:23:25 |
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星空浩淼
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Re: Continuing discussion of quantum mechanics
我上面发完最后一个帖子，出门时意识到自己犯了一个错误：即第二条，“在不考虑全同粒子对称性时后者可以退化成前者”错了。就算要退化，也是退化成<x_1|1><x_2|2>之类的形式，而不是退化成<x|1>+<x|2>。
不过，如果<x_1|1><x_2|2> is a solution，那么单独的<x|1>或<x|2>的确也是一个解，因此<x|1>+<x|2>必定也是一个解，正因为正如你所说，the wave-function of a two particle state must have two sets of independent coordinates. 所以<x|1>对x_2求偏导为零，同理<x|2>对x_1求偏导为零,所以单独的<x|1>或<x|2>的确也是一个解。
事实上，常数可以看作波函数的平凡解。
唯有与时间赛跑，方可维持一息尚存
| 发表时间：2005-07-13, 08:29:19 |
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星空浩淼
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Re: Continuing discussion of quantum mechanics
time is just a local label, it has no meaning. unless you can show this is a close time curve, there is no violation of causality.
为了避免因果悖论，信号是不能超光速的，这是共识，至于它如何会导致a close time curve, 我不清楚.利用 是否产生close time curve来分析因果律问题，属于相对论专家级水平，对于非相对论专业的，只有设计悖论来判断。
如果sage兄发射一颗子弹打死拉登，那子弹是超光速的，那么另一个观察者可能看到sage兄还没开枪时，拉登已经中弹身亡。这时候这个观察者赶紧拿一块板子挡在拉登面前，sage兄发出的子弹终于来到，但是被板子挡住了。这样就有了一个悖论：拉登到底该不该死？
唯有与时间赛跑，方可维持一息尚存
| 发表时间：2005-07-13, 08:46:08 |
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sage
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Re: Continuing discussion of quantum mechanics
不过，如果<x_1|1><x_2|2> is a solution，那么单独的<x|1>或<x|2>的确也是一个解，因此<x|1>+<x|2>必定也是一个解，正因为正如你所说，the wave-function of a two particle state must have two sets of independent coordinates. 所以<x|1>对x_2求偏导为零，同理<x|2>对x_1求偏导为零,所以单独的<x|1>或< x|2>的确也是一个解。
事实上，常数可以看作波函数的平凡解。
again, wave-function reduce to a constant is not what |1>+|2> mean
|1>+|2> means that |1> and |2> are in a coherent sum.
what is a solution in the constant wave-function case is
<x1|1>+<x2|2>
and there is no interference bewteen them.
It is , therefore, very different from |1>+|2>
| 发表时间：2005-07-13, 08:49:05 |
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星空浩淼
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Re: Continuing discussion of quantum mechanics
呵呵，我的意思就是<x1|1>+<x2|2>，一直写成了|1>+|2>，
唯有与时间赛跑，方可维持一息尚存
| 发表时间：2005-07-13, 08:59:00 |
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元江
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Re: Continuing discussion of quantum mechanics
Make the notation clear
Solution 1:
\pai_1(x,y)=f(x-k1)exp(-ik1 y)
Solution 2:
\psi_2(x,y)=f(x-k2)exp(-ik2 y)
Both of them are solutions of S-equation
(-\partial^2_x + (i\partial_y -x )^2)\psi(x,y)=\epsilon \psi(x,y)
This equation is reduced from a 3D S-equation for a charged particle in external magnetic field. This equation generates something like Landau Energy Level .
if \psi_1 and \psi_2 are degenerate (\epsilon are the same),
\Psi = \psi_1 + \psi_2
is also a solution of the above equation.
Now, I view \pai_1(x,y) as a wave source centralized at position k1, it corresponds to a particle. I view \pai_2(x,y) as a wave source centralized at position k2, it corresponds to another particle. I want to see what the interference spatial pattern will loke like ? I calculate |\Psi|^2.
May be this treatment of the solutions does not fit into the current well-known formalism of many-particle system, it does not use number operator space and does not use Hilbert product space. It does not matter, formalism comes after physics. My question: is this a valid physics question to look at?
I am very appreciated for the discussion on this forum, sage and 星空 are both sharp on those topics. I am sorry I did not put my thought in this clear way from the begining since I thought this is so obvious. ONly after this discussion, I realize that the wave-source interference idea does deviates from the traditional formalism of many particle system.
道可道非常道
名可名非常名
| 发表时间：2005-07-13, 10:40:22 |
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sage
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Re: Continuing discussion of quantum mechanics
Make the notation clear
Solution 1:
\pai_1(x,y)=f(x-k1)exp(-ik1 y)
Solution 2:
\psi_2(x,y)=f(x-k2)exp(-ik2 y)
Both of them are solutions of S-equation
(-\partial^2_x + (i\partial_y -x )^2)\psi(x,y)=\epsilon \psi(x,y)
==================================================================
confused about your notation, is k1, k2 location or momentum?
| 发表时间：2005-07-13, 11:43:31 |
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元江
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Re: Continuing discussion of quantum mechanics
There is a natural length scale in the S-equation, called \xi, in fact, it is in similar to the one when we learn hamonic ociallator solution. After I devide every coordinates by this \xi, the coordinate space is dimensionless, the k1 (or k2) is
also dimentionless (y is devided by \xi, the wave vector K1 is multiplied by \xi, so I mark it as k1), from the phase factor exp(-ik1 y), you can say it is a wave-vector,
from f(x-k1), you can say it is a position which is the center of the amplitude, f.
Therefore, we can either say k1 is the center of a wave-source or k1 is a wave-vector.
道可道非常道
名可名非常名
| 发表时间：2005-07-13, 11:51:53 |
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sage
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内力值: 535/535
Re: Continuing discussion of quantum mechanics
the fact that there are both x and y in \psi_1 means that, strictly speaking, it is a two particle state. more specifically, you nedd two (not one) measurement of position to completely collapse the wave-function. Therefore, you cannot call it a particle.
| 发表时间：2005-07-13, 13:07:15 |
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元江
发表文章数: 228
武功等级: 逍遥拳
(第五重)
内力值: 200/200
Re: Continuing discussion of quantum mechanics
What it should be called may not be that important, right?
星空学友有何评论吗？
道可道非常道
名可名非常名
| 发表时间：2005-07-13, 13:51:29 |
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sage
发表文章数: 1125
武功等级: 天山六阳掌
(第六重)
内力值: 535/535
Re: Continuing discussion of quantum mechanics
>What it should be called may not be that important, right?
=====================================================
yes and no, depending on what conclusion you want to draw from it.
for this system, I am not sure what you are trying to show. if you want to talk about interference between two wave packets, it is important to remember it is not one-particle interfere with another.
| 发表时间：2005-07-13, 15:05:25 |
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sage
发表文章数: 1125
武功等级: 天山六阳掌
(第六重)
内力值: 535/535
Re: Continuing discussion of quantum mechanics
为了避免因果悖论，信号是不能超光速的，这是共识，至于它如何会导致a close time curve, 我不清楚.利用 是否产生close time curve来分析因果律问题，属于相对论专家级水平，对于非相对论专业的，只有设计悖论来判断。
如果sage兄发射一颗子弹打死拉登，那子弹是超光速的，那么另一个观察者可能看到sage兄还没开枪时，拉登已经中弹身亡。这时候这个观察者赶紧拿一块板子挡在拉登面前，sage兄发出的子弹终于来到，但是被板子挡住了。这样就有了一个悖论：拉登到底该不该死？
this is not the correct picture. you did not interpret this example correctly.
however, it has nothing to do with quantum mechanics. let's discuss it in another post.
| 发表时间：2005-07-13, 15:07:27 |
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星空浩淼
发表文章数: 1743
武功等级: 九阳神功
(第五重)
内力值: 617/617
Re: Continuing discussion of quantum mechanics
回元江兄：我要过阵子再来，这次进来就是为了回帖说明一下。
一般说来，我的意见跟sage兄不会有多少分歧，最多是从其他角度做一些补充，以便于进一步理解。
唯有与时间赛跑，方可维持一息尚存
| 发表时间：2005-07-13, 23:54:20 |
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