3

For atomic orbitals:

E2 orb: $\binom{N-n_1}{n_2}$

E2 orb: $\binom{N-n_1-n_2}{n_3}$

E2 orb: $\binom{N-n_1-n_2-n_3}{n_4}$

...

En orb: $\binom{n_i}{n_i}$

now probability function is:

$P= N! \prod^{n}_{n=1}\frac{1}{n_{i}!}$

Why? In general?

[Update]

Every combination is greater than 1. So their product is greater than 1. How on earth can such multiplication lead to a probability function? Is the probability function scaled back to range $[0,1]$?

hhh
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  • An incomplete question ? – Sasha Jul 11 '11 at 22:55
  • @Sasha: you can find the definition of "combination" here. Orbitals are finite so are the states. $(a b)$ where $a$ is the amount of free electrons and $b$ is the free slots. For the next orbital, you have less electrons $N-n_{last}$ and less orbitals (one less) so they are connected. Now if you multiply them, you will see cancelling but why is it so and when is it so in general? How to quarantee such phenomenon in practice? – hhh Jul 11 '11 at 22:57
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    Maybe it would be a good idea to reproduce the relevant passages. I find your question highly unclear. – t.b. Jul 11 '11 at 23:13
  • Are you asking why $${N \choose n_1} \cdot {{N- n_1} \choose n_2} \cdot \cdots \cdot {{N- n_1 - n_2 \cdots -n_{k-1}} \choose n_{k}} = N! \prod_{i=1}^{k} \frac{1}{n_{i}!}$$ if $N = n_1 + \cdots + n_{k} \quad$? – t.b. Jul 11 '11 at 23:30
  • I'm voting to close. Also in view of your edit summary. – t.b. Jul 11 '11 at 23:44
  • yes. What is combination in latex? – hhh Jul 12 '11 at 00:02
  • To get $\displaystyle {N - n_1 - \cdots - n_{k-1} \choose n_k}$ write ${N - n_1 - \cdots - n_{k-1} \choose n_k}$. You can also right-click on the formula in my previous comment, choose "show source" and simply copy it (this works everywhere on this site, by the way). – t.b. Jul 12 '11 at 00:10
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    Or \binom{n}{k}. – Dylan Moreland Jul 12 '11 at 00:26
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    BOUNTY: People are overlooking the last assertion that the product is a probability function. I understand the term "probability function" in a such way that you give some input and then you get some output in some range usually scaled to $[0,1]$. I am totally lost with this. Even multinomial distribution does not lead to results in the range $[0,1]$. I will reward the bounty to a person who can sort out the last statement. What is the probability function? Why is it probability function? By which definition is it probability function? Please, vote this up so people see it. – hhh Aug 29 '11 at 20:35
  • Let me support @Theo's suggestion, made a few comments ago, to reproduce the passage which provoked your question. Offering a bounty will not change the fact that several good-willing experts tried to understand it--and failed. So: the source, please! – Did Aug 29 '11 at 20:53
  • @Didier Piau: very well, it is in different language. I have to do some translation. It will take some time. – hhh Aug 29 '11 at 20:54
  • Which language? – Did Aug 29 '11 at 20:58
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    @Didier Piau: please, read the page 10 on the lecture slides here, alert not in English but a lot of pictures. I am lost with it, why it would become a probability function -- the question here is pretty much explaining the page. The page deduce the probability function with multinomial ditribution (my understanding) but I cannot see how it can be right. It is apparently scaling the probability function to changing range, I am lost now what they mean here. Or I am just misunderstanding the term "todennakoisyssfunktio". Dictionary.... – hhh Aug 29 '11 at 21:12
  • hhh, @Didier: I notified Jyrki Lahtonen and Ilmari Karonen in the hope that they might be able to add some explanations and translations since I fail to see why the answer(s) are unsatisfactory. – t.b. Aug 29 '11 at 21:26
  • Hmmmm... The pictures were not of any help but it seems clear from the text that P enumerates the microstates corresponding to a given macrostate of a physical system. To get the probability of said macrostate one should normalize P by the total number of microstates, or something like that. All in all, the intended audience of your text does not seem to be a mathematical one. If rigor is mandatory, a more mathematically oriented introduction text could be preferable. – Did Aug 29 '11 at 21:38
  • @Theo, good idea. Google-translate is surprisingly good between Finnish and another F-language. – Did Aug 29 '11 at 21:41
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    @Theo, All: Very little to add. You seem to have gotten the point that the author seems to take certain liberties in using mathematical concepts. Note that in addition to this not being a probability function he freely differentiates w.r.t to a parameter ranging over natural numbers :-). Live with it. At times physicists do not give definitions, they give descriptions. Here the description is statistical. If $N$ is in the range of Avogadro's constant ($10^{23}$) it somehow works. In other words: the OP is experiencing a "culture shock". Try to get the idea, and don't get stuck in the language. – Jyrki Lahtonen Aug 30 '11 at 04:27

4 Answers4

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As André Nicolas shows, $$ \binom{N}{n_1}\binom{N-n_1}{n_2}\dots\binom{N-n_1-n_2-\dots-n_{k-1}}{n_k}=N!\prod_{i=1}^k\frac{1}{n_i!} $$ This is an integer; it is the number of ways to arrange $N$ distinct things into $k$ bins with $n_i$ things in bin $i$. If you sum this over all possibilities for the $\{n_i\}$, you get $k^N$ (the number of maps from $N$ things to $k$ bins). So you would get a probability distribution if you divide by $k^N$.

The multinomial theorem states $$ \left(\sum_{i=1}^kx_i\right)^N=\sum_{\sum_{i=1}^kn_i=N}N!\prod_{i=1}^k\frac{x_i^{n_i}}{n_i!} $$ Setting $x_i=1$ for $1\le i\le k$, we get that $$ k^N=\sum_{\sum_{i=1}^kn_i=N}N!\prod_{i=1}^k\frac{1}{n_i!} $$

robjohn
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  • you are right here if you divide by $k^{N}$ but the assertion is that you do not divide by $k^{N}$, it is the score of this problem. The statement has no such scaling. +1 because I think you are seeing the main problem here, hopefully other people can focus on this. – hhh Aug 29 '11 at 21:27
  • ...perhaps too much asked but I want to prove this rigorously. Very well, we have here symmetric polynomials. $(x_{1}+x_{2} +...+x_{k})^{2} = (X_{i}^{2} + 2(X_{i}X_{j}))$ where large $X$ is a generator thing (my own terminology, noticed in some very old Vaisala Number theory book or something like that, trying to find proper terminology...). And again multiplication, there must be some rules from number theory to kill this problem in a few second. – hhh Aug 29 '11 at 21:48
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    @hhh: It seems that an assertion is being made about the probabilistic distribution of the of states (counted as integers), from which a probability will be deduced (after dividing by the total). Since they are taking the derivative of the logarithm, any constant factor drops out. Perhaps the author is taking liberties with the term "probability function" or perhaps it loses something in the translation, but I think the final result is still valid. – robjohn Aug 29 '11 at 21:52
  • you are clever! Very misleading and dangerous statement, it is not the probability function at that point per se -- only the amount of occurrences. But I still want to rigorously understand your last part...calculating. – hhh Aug 29 '11 at 21:56
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    @hhh: You don't need to dig for Väisälä's number theory book (no need to leave out diacritical marks from letters, we have left the era of 7-bit ASCII here). You get the multinomial formula from the binomial formula using induction on the number of unknowns. Robjohn hits the nail on the head: the constant $1/N!$ disappears at the end of the day. – Jyrki Lahtonen Aug 30 '11 at 04:39
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As you correctly point out, the terminology on page 10 of the lecture notes you linked to is incorrect: $P$ is the number of microstates making up the given macrostate; it is not a probability, since it typically exceeds one.

However, dividing $P$ by the total number of microstates $n^N$ does give the probability of the given macrostate, under the assumption that all microstates are a priori equally likely. Since the constant $1/n^N$ is the same for all macrostates of the system, we may safely ignore it, as the author of the lecture notes does, when comparing the probabilities of different macrostates.

(If I were reviewing those notes, I'd also point out that the author seems rather excessively fond of the letter "n". Having $n$ and $N$ as system parameters is confusing enough, but when he then introduces the macrostate parameters $n_1$, $n_2$ and up to $n_n$...)

Ilmari Karonen
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  • Physicists are not as rigorous with their notation sometimes. I referred to some old notes, and at some places, this was called "unnormalized" probability. – kuch nahi Aug 30 '11 at 03:09
  • Yeah, the notation is awful. I have seen definitions of functions like $F(t)=\int_0^t f(t),dt$ in a chemistry book. The intention is clear (in the context at least), but my calculus students were reading that book :-). IOW don't hold your breath, if you expect the author to understand this bit of criticism. – Jyrki Lahtonen Aug 30 '11 at 04:41
  • Ilmari, @Jyrki: Thanks a lot for following up on my pings. I removed the off-topic comments to your answers. – t.b. Aug 30 '11 at 10:54
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Before the question is closed, I would like to give you a start.

Suppose $a+b+c+d=N$. Let us look at $$\binom{N}{a}\binom{N-a}{b}\binom{N-a-b}{c}\binom{N-a-b-c}{d}$$ (the last term is $1$, it is just there to make things look nice.)

Calculate, using the usual formula for $\binom{n}{k}$.

The first term is $$\frac{N!}{a!(N-a)!}.$$

The second term is

$$\frac{(N-a)!}{b!(N-a-b)!}.$$

The third term is

$$\frac{(N-a-b)!}{c!(N-a-b-c)!}.$$

Note that $N-a-b-c=d$.

Multiply, and observe the very nice cancellations! We get

$$\frac{N!}{a!b!c!d!}.$$

The "general" case solution is basically the same, except that all those subscripts tend to make things less obvious.

Added: By "the usual formula" for $\binom{n}{k}$ I mean

$$\binom{n}{k}=\frac{n!}{k!(n-k)!}.$$

Since the question has not yet been closed, I am adding a link to a Wikipedia entry which I think is quite well written, and which I hope you will give you all of the additional information you may need.

André Nicolas
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  • user6312: do I understand right that the probability function is just a multinomial distribution $P=\binom{N}{k_{1},k_{2},...,k_{m}}$? – hhh Jul 12 '11 at 09:56
  • There is no sense with it! It must be scaled to $[0,1]$ or you are using some different definition of probability function, anyway this is the assertion my lecture slides offer. – hhh Aug 29 '11 at 20:30
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    Naturally, as you observe, these are not probabilities, and any reference that calls them a probability function is just wrong. If you know the probabilities $p_1$, $p_2$, and so on up to $p_k$ (where $p_j$ is the probability of being in state $j$), then after $N$ trials, probability of $n_1$ in state $1$, $n_2$ in state $2$, and so on up to $n_k$ in state $k$ is $\frac{N!}{n_1!\cdots n_k!}p_1^{n_1}\cdots p_n^{n_k}$. That conceivably may be what you want, but I have a feeling it is not. (to be continued) – André Nicolas Aug 29 '11 at 21:43
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    (continued) More likely, use the fact that the sum of all these numbers, as $n_1$, $n_2$, and so on up to $n_k$ range over all choices with $n_1+\cdots+n_k=N$, is $k^N$. So if you divide each of the numbers we have been working with by $k^N$, you really do get probabilities. (This is the "equally likely" case of the multinomial distribution that I gave the formula for in the previous comment.) So divide all your numbers by $k^N$. – André Nicolas Aug 29 '11 at 21:46
  • @hhh: I forgot to start the above comments properly. Don't know how exactly the system works, and whether they are automatically sent to you. – André Nicolas Aug 29 '11 at 23:41
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The "probability function" in the notes is usually referred to as the number of microstates $\Omega$ (I was taught this way). By the postulate of equal a priori probability, the probability, of a particular macrostate $i$ is $$p_i = \frac{1}{\Omega}$$So the probability function is different from the actual probability, which is probably what caused the confusion.

Hence the $P$ in your notation is the number of ways $N$ particles can be classified into energy levels according to their energies (by the product rule, as your instructor and Andre above has done). Note that the analysis is classical as it assumes the particles filling the energy levels are distinct (come with labels).

You can find the same method in the Wikipedia article on Maxwell Boltzmann Distribution, except that it does not use confusing terminology of "Probability function".

$$\Omega = N!\prod \frac{g_i^{N_i}}{N_i!}$$ Where on your case all $g_i = 1$ as there are no degenerate levels involved. Your instructor derives the MB distribution by trying to maximize the above number (as a system naturally seeks the maximum number of microstates) given the constraints (no exchange of particles or energy of the system with the surroundings) using Lagrange Multipliers.

kuch nahi
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