Solution of hydrogen atom : legendre polynomials

218
0
I was messing around with the [tex]\theta[/tex] equation of hydrogen atom. OK, the equation is a Legendre differential equation, which has solutions of Legendre polynomials. I haven't studied them before, so I decided to take closed look and began working on the most simple type of Legendre DE. And the story is as follows:

The book says:
[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = \delta_{mn} \frac{2}{2n+1}[/tex]

So I sat and tried derieving it. First, I gather an inventory that might be useful:

[tex](1-x^2)P_n''(x) - 2xP_n'(x) + n(n+1) = 0[/tex]
[tex][(1-x^2)P_n'(x)]' = -n(n+1)P_n(x)[/tex]

[tex]P_n(-x) = (-1)^n P_n(x)[/tex]
[tex]P_n'(-x) = -(-1)^n P_n'(x)[/tex]

And started the job. Now...
[tex][(1-x^2)P_n'(x)]' = -n(n+1)P_n(x)[/tex]
multiplying both sides with [tex]P_m(x)[/tex] and integrating from -1 to 1
[tex]\int_{-1}^{1} [(1-x^2)P_n'(x)]' P_m(x) dx = \int_{-1}^{1} -n(n+1)P_n(x) P_m(x) dx[/tex]
applying a partial to the lefthand side
[tex](1-x^2)P_n'(x) P_m(x) |_{-1}^{1} - \int_{-1}^{1} (1-x^2)P_n'(x) P_m'(x) dx = \int_{-1}^{1} -n(n+1)P_n(x) P_m(x) dx[/tex]
switching n and m's, ve get another equation like it
[tex](1-x^2)P_m'(x) P_n(x) |_{-1}^{1} - \int_{-1}^{1} (1-x^2)P_m'(x) P_n'(x) dx = \int_{-1}^{1} -m(m+1)P_m(x) P_n(x) dx[/tex]

substituing these

[tex][-n(n+1) + m(m+1)]\int_{-1}^{1} P_n(x) P_m(x) dx = [(1-x^2)(P_n'(x) P_m(x) - P_m'(x) P_n(x))] |_{-1}^{1}[/tex]

now, from the inventory
[tex]P_n(-x)P_m'(-x) = -P_n(x) P_m'(x)[/tex]
the righthand side becomes
[tex]\lim_{x \to 1} 2[(1-x^2)(P_n'(x) P_m(x) - P_m'(x) P_n(x))[/tex]
and in it's final shape
[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = \lim_{x \to 1} 2(1-x^2) \frac {(P_n'(x) P_m(x) - P_m'(x) P_n(x))}{-n(n+1) + m(m+1)}[/tex]

For [tex]n \neq m[/tex], there's no problem, it's straightly 0. But for [tex]n = m[/tex]...
And everything starts going wrong. By stating
[tex](1-x^2)P_n'(x) = \int -n(n+1)P_n(x) dx[/tex]
i'm making a little change

[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = \lim_{x \to 1} 2 \frac {P_m(x) (\int -n(n+1)P_n(x) dx) - P_n(x) (\int -m(m+1)P_m(x) dx))}{-n(n+1) + m(m+1)}[/tex]

Now I plug [tex]P_n(x) = P_m(x)[/tex], [tex]x=1[/tex] and [tex]P_n(1) = 1[/tex] and state

[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = \lim_{x \to 1} 2 \frac {P_m(x) (\int -n(n+1)P_n(x) dx) - P_n(x) (\int -m(m+1)P_m(x) dx))}{-n(n+1) + m(m+1)}[/tex]

[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = 2 (\int P_n(x) dx)_{x=1}[/tex]
which is the nonsense of the day.

Obviously something is wrong. But where's the error?
Thanks in advance.

P.S.: I made an awful mistake https://www.physicsforums.com/showthread.php?p=1092694" in math forum, where I was welcomed with the words "utter and complete nonsense" with no further actual explanation. Well, back to home!
 
Last edited by a moderator:
1,667
0
gulsen said:
I was messing around with the [tex]\theta[/tex] equation of hydrogen atom. OK, the equation is a Legendre differential equation, which has solutions of Legendre polynomials. I haven't studied them before, so I decided to take closed look and began working on the most simple type of Legendre DE. And the story is as follows:

The book says:
[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = \delta_{mn} \frac{2}{2n+1}[/tex]

So I sat and tried derieving it. First, I gather an inventory that might be useful:

[tex](1-x^2)P_n''(x) - 2xP_n'(x) + n(n+1) = 0[/tex]
[tex][(1-x^2)P_n'(x)]' = -n(n+1)P_n(x)[/tex]

[tex]P_n(-x) = (-1)^n P_n(x)[/tex]
[tex]P_n'(-x) = -(-1)^n P_n'(x)[/tex]

And started the job. Now...
[tex][(1-x^2)P_n'(x)]' = -n(n+1)P_n(x)[/tex]
multiplying both sides with [tex]P_m(x)[/tex] and integrating from -1 to 1
[tex]\int_{-1}^{1} [(1-x^2)P_n'(x)]' P_m(x) dx = \int_{-1}^{1} -n(n+1)P_n(x) P_m(x) dx[/tex]
applying a partial to the lefthand side
[tex](1-x^2)P_n'(x) P_m(x) |_{-1}^{1} - \int_{-1}^{1} (1-x^2)P_n'(x) P_m'(x) dx = \int_{-1}^{1} -n(n+1)P_n(x) P_m(x) dx[/tex]
switching n and m's, ve get another equation like it
[tex](1-x^2)P_m'(x) P_n(x) |_{-1}^{1} - \int_{-1}^{1} (1-x^2)P_m'(x) P_n'(x) dx = \int_{-1}^{1} -m(m+1)P_m(x) P_n(x) dx[/tex]
At this moment you obtain the kronecker delta if you notice that the boundary terms are zero. Now, for the normalization, I guess you can do that by solving the equation of motion and doing some partial integrations.
 
Last edited:
218
0
Equation of motion? Schrödinger equation? Well, this's simplified (without the reciprocal [tex]1+x^2[/tex] term) version of the actual equation, so I can't.

As a matter of fact, there has been a sloppiness in my post (which Hurkly pointed out at maths forum, but well), where I thought to be obvious. Here's the tedious correction: i'm thinking the limit of [tex]lim_{n \to m}[/tex] and [tex]x \to 1[/tex]. Hopefully, this will let me say [tex]\frac{-n(n+1) + m(m+1)}{-n(n+1) + m(m+1)} = 1[/tex].
 
Last edited:
1,667
0
gulsen said:
Equation of motion? Schrödinger equation? Well, this's simplified (without the reciprocal [tex]1+x^2[/tex] term) version of the actual equation, so I can't.

As a matter of fact, there has been a sloppiness in my post (which Hurkly pointed out at maths forum, but well), where I thought to be obvious. Here's the tedious correction: i'm thinking the limit of [tex]lim_{n \to m}[/tex] and [tex]x \to 1[/tex]. This will let me say [tex]\frac{-n(n+1) + m(m+1)}{-n(n+1) + m(m+1)} = 1[/tex].
(1) The differential equation for the legendre polynomials obviously (which is a constraining equation on P_n(x), but might be regarded as a second order equation of motion when you interpret x as a time coordinate, or as an eigenfunction problem for the corresponding second order differential operator.).
(2) up to the point where I commented your message, everything was fine. You should start from there and redo the rest.
 
218
0
Actually it's likely to be fine further, since I get a good answer for n=m.
I'm not sure where the error could be...I tried working it out for it several times, but I get to the same place. Obviously I have a bad habit, some kind of mistake I repeat. Eh, that's why I've posted it in the first place ^-^'

(note: for that argument -being more tedious, because Hurkly pointed out i'm quite sloppy- I define [tex]n = m + \delta[/tex] and consider the limit of [tex]\delta \to 0[/tex]. So [tex]-(n)(n+1) + m(m+1) = -(m+\delta)(m+\delta+1) + m(m+1) = -\delta((2m+1) + \delta)[/tex]. And my limit is [tex]\lim_{\delta \to 0} \frac{\delta((2m+1) + \delta)}{ \delta((2m+1) + \delta)} = 1[/tex])
 

matt grime

Science Advisor
Homework Helper
9,392
3
Why don't you just take the cases m=n and m=/=n separately rather than introducing unnecessary limits? A limit process which, incidentally, is not at all guaranteed to give the correct answer, and is not justifiable. (n and m are supposed to be integers anyway, so it doesn't make sense to take limits like that.)

Incidentally, if something is nonsense then why not call it that? It wasn't an insult, though it should be noticed that you gave a lengthy and incorrect rebuttal of someone elses critique first:


If you think it this way, you cannot compute a derivative as well, since you're dividing with 0 when computing it as well. Note that I'm dividing 0 with 0, which is indefinite, and from than line on, all I did was to get rid of this indefinition.
If that is how you're going to post don't be so touchy about the replies you get (or misleading, as you are being in this thread) by implying you were being picked on unfairly. What you posted could easily be taken in the same vein in which you took criticism of your argument.
 
Last edited:
141
0
Once again :

Look at the generating function for Legendre polynomials and derive the normalization condition from that. This is exactly the same advice I gave you on the math forum, but probably you were in too much of a rage to be able to read at that point.
 
141
0
Generating function:

[tex]
\left(1-2tx+t^2\right)^{-1/2} = \sum_{n=0}^{\infty}P_n\left(x\right)t^n
[/tex]

  1. Square both sides, and integrate from [itex]x=-1[/itex] to [itex]x=1[/itex].
  2. Use the orthogonality relation you have derived previously to kill cross terms of the type [itex]P_n\left(x\right)P_m\left(x\right)[/itex] so that you are left on the RHS with a single summation over [itex]n[/itex], of something like [itex]\int_{-1}^{1}dx \left[P_n(x)\right]^2 t^{2n}[/itex]
  3. Perform the integration on the LHS so that you're left with some function of [itex]t[/itex].
  4. Expand the LHS in a power series of [itex]t[/itex] and compare the coefficients on the LHS and the RHS.
Do the above and you will get your normalization factor.
 

Hurkyl

Staff Emeritus
Science Advisor
Gold Member
14,829
14
[tex]lim_{n \to m}[/tex]
As the others pointed out, for this to make any sense, you are going to have to generalize the notion of "Legendre polynomial", since the standard definition only defines them for (nonegative?) integers.

There's another problem with this: notice that (where I'm treating m and n as real variables) (I'm assuming that we are staying away from n = -1/2)

[tex]\delta_{mn} \frac{2}{2n+1}[/tex]

is discontinuous at [itex]m = n[/itex]. In fact, it's identically zero on [itex]m \neq n[/itex] giving us:

[tex]\lim_{m \rightarrow n} \delta_{mn} \frac{2}{2n+1} = 0[/tex]

So whatever generalization of the Legendre polynomials you decide to use, they will either not be continuous functions of m and n, or they will not satisfy the relation

[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = \delta_{mn} \frac{2}{2n+1}[/tex]
 
141
0
Hurkyl said:
As the others pointed out, for this to make any sense, you are going to have to generalize the notion of "Legendre polynomial", since the standard definition only defines them for (nonegative?) integers.
That's true, I wrote something like this in the other thread. If he's going to take an integer and add on some continous part then start continously tending this to zero, then he dosen't have Legendre ``polynomials'' any more, but in fact an infinite series which should diverge at [itex]x=\pm 1[/itex] (I think - I'm not entirely sure here). If this is the case then the integral of [itex][P_{\nu}(x)]^2[/itex] on the range [itex][-1,1][/itex] will just diverge. So it's pointless to talk about this integral for some non-integer Legendre function. I don't have Abramovitz and Stegun handy so I can't check the correctness of this but I think I'm right, from what I remember from Legendre polynomials - the only reason why the Polynomials don't diverge at [itex]x=\pm 1[/itex] is because they're polynomials, i.e. the series it terminated at some at some point, so at [itex]x=\pm 1[/itex] they're always finite.

--edit:

http://mathworld.wolfram.com/LegendreDifferentialEquation.html

It looks like the first solution (the Legendre function) is finite at at [tex]x=\pm 1[/itex], while the second solution diverges (Legendre function of the second kind).

http://www.math.sfu.ca/~cbm/aands/page_333.htm

It appears here that the relation for negative argument doesn't hold when the degree of the Legendre function is not an integer.
 
Last edited:

Meir Achuz

Science Advisor
Homework Helper
Gold Member
2,147
53
gulsen said:
[tex][(1-x^2)P_n'(x)]' = -n(n+1)P_n(x)[/tex]
To prove orthogonality, multiply lthis equation by P_m(x) and integrate from -1 to 1. An integration by parts then proves orthogonality.
The factor 2/(2n+1)can be found by integrating the (generating function)^2.
This derivation is given in many textbooks.
 
218
0
Thanks Hurkyl, I understood my error.
So, the problem is my incomplete knowledge about Legendre polynomials. Well, I leaped directly proving the orthonormality condition. I just -implicity- assumed non-integer incides are possible.

Meir, thanks but I've already found various proofs of this in some textbooks, which usually invokes Rodrigues formula, something I didn't know. I just tried to work a "bare" proof.

And let me say something personal, matt.

3x0 = 5x0
Diving this expression by zero gives 3=5 ??
What would you say when someone tried to lecture you on calculus & computing a limit rather than saying Legendre polynomials' indices are not continous, therefore limit does not exists.

And anoher guy, with a jerky attitude replies "utter complete nonse", with saying nothing further. Was it I in the first place who personalized the matter, or the guy who made a flame-only post. Still, everone seem to be happy with him, and I am to blame for implying this's unfair.
But I'm not implying that this's unfair, I'm saying this's something worse. If you're still saying I'm misleading, then so be it, I mislead you.

And I look at the guy who in the first place, rather than saying my assumption on Legendre polynomials' continuty is wrong, he rather said "no, error is you're dividing with zero". Though it turned out that the problem is not my calculus but ignorance on Legendre polynomials --and, yes, I still haven't solved the DE, so I still don't know why ns are integers, but I will.

And finally "Look, I'm not a freshman, and I know my calculus that much." -- clearly you don't.
Give me a reason not to say "go **** yourself" to that guy. If you'd like to blame me for my anger, I'm sorry, but I'm not the type who can bear that much arrogance. Critization is helluva far from "utter and complete" insults.
 
Last edited:
141
0
Let me just put the entire post which was made here:

jpr0 said:
The problem is that you're working with Legendre polynomials, which implies here that your n and m are integers. Now if you want to "tend n to m", without simply setting n=m, then you will be working with non-integer Legendre functions where the above recursion relations simply do not work (and they probably even diverge at x=1, x=-1 anyway).

The difference between a derivative and doing what you're doing is that the small parameter (h) in the definition of "derivative" ((f(x+h)-f(x))/h) can be continously taken to zero, when the leading term in f(x+h)-f(x) goes linearly to zero as a function of h. How can you continously let one integer tend to another? You can't.

You asked in your original post what was wrong with what you were doing. I pointed out your error. In another post I also suggested a method of proving the orthogonality relation based on the generating function, which is what you will see in most maths texts - sorry if this wasn't good enough for you.

And finally "Look, I'm not a freshman, and I know my calculus that much." -- clearly you don't.
The last line was made after you had told us all to go and **** ourselves.
Here is the post you made before mine:

gulsen said:
Look, I'm not a freshman, and I know my calculus that much. Stop lecturing me on that, it goes very funny.
Derivative:
[tex]\lim_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}[/tex]
If you directly set [tex]\Delta x = 0[/tex] which is the thing you ultimately do, you get a 0/0 which is not a solution, this is where you divide by at term that goes to 0 when finding a derivative.
Since you're too pedantic, you'll probably tell me division by limit of a term that goes to zero is different from division by 0, but that name was coined by jpr0 in the first place for my limit, and none of you give a **** to it.

Tell me how this is limit

[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = \lim_{x \to 1} 2 \frac {P_m(x) (\int -n(n+1)P_n(x) dx) - P_n(x) (\int -m(m+1)P_m(x) dx))}{-n(n+1) + m(m+1)}[/tex]

which is said to be "wrong because i'm dividing by 0" is different than

[tex]\lim_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}[/tex]
I appreciate why most of great physicist hate mathematicians. They seem to know you well. I have asked a simple question, yet what you say is everything far from solution. Most questions I raise at physics department goes smooth, while almost all questions I raise right here goes haywire.

Now go **** yourselves guys. I'll this question at QP forum. Even if I can't get a correctly reply, I know that chances are very low that I'll meet such arrogance and idiocy. Have fun, and lock the topic.
I really don't understand what your problem is. You posted your initial analysis, and then asked us to find the error. I pointed out to you that the error was division by zero. You then claimed that the division by zero was in fact not an error, because this is exactly what you do when you take a derivative - you divide by zero. Therefore, if a derivative is valid, so too was your arithmetic. It was then pointed out, yet again, that diving by zero is just plain wrong. At this point I suggested that you take the generating function and work out your normalization condition from that. You then again claimed that when you take a derivative, you're ultimately dividing some expression by zero, while claiming to know your stuff in calculus 'cos you're not a freshman, then demanded to know the difference between a derivative and your expression in terms of "division by zero", and that we should all go an **** ourselves. I then told you that when you're taking a derivative you're continuosly tending some parameter to zero, when the leading term in the expression for the derivative becomes independent of this parameter, whereas in your case, you're working with integers, and cannot do this, and that if you wanted to do such a limiting procedure you would have to work with non-integer Legendre functions when your analysis probably stops working at line 1. I then told you that you don't know your division by zero, er, i mean, calculus as well as you think you do.

So to address your first point:

"What would you say when someone tried to lecture you on calculus & computing a limit rather than saying Legendre polynomials' indices are not continous, therefore limit does not exists."

Why would I tell you that "taking the limit of n to m doesn't exist here because because n and m are integers" when you hadn't even taken any limits? You had simply divided by f(n) - f(m) and then set n=m. This is division by zero. Even afterwards, when you started talking about computing derivatives, you stated that when you take derivative you divide "zero by zero" (which, if this was the case, all derivatives would take the same value - whatever you compute 0/0 to be). Now, sorry if people tell you that you're understanding of a derivative is wrong when you say things like this - but it's true.

For your second point:

'And I look at the guy who in the first place, rather than saying my assumption on Legendre polynomials' continuty is wrong, he rather said "no, error is you're dividing with zero". Though it turned out that the problem is not my calculus but ignorance on Legendre polynomials --and, yes, I still haven't solved the DE, so I still don't know why ns are integers, but I will.'

So what you're telling me is that I should have somehow known that when you divided your equation by f(n) - f(m) and set n=m, that in fact you weren't just setting n=m, you were instead taking a limit? Ok, fine, let's assume for a second that I should have known this. Why then, in your next post didn't you say "yes, but you see, what I'm doing here is no different to taking a derivative, because I'm tending n to m, which essentially gives me a parameter that is tending to zero, much like the case of a derivative" rather than "yes, but you see, i'm diving by zero, just as you divide by zero when you take a derivative"?! Why? So no, the problem wasn't your knowledge of Legendre polynomials, nor was it even your knowledge of taking a derivative, but in fact your understanding of arithmetic division by zero.

Now your final point:

"Give me a reason not to say "go **** yourself" to that guy"

Because that guy told you what was wrong with your analysis, how to approach the problem in a different way, and even wrote down the steps you had to take in this different analysis to get the answer you wanted, despite your constant abuse and your inability to take criticism of arithmetic division by zero.

The problem wasn't that we were "arrogant" in the forum - rather, you were.
 
Last edited:

reilly

Science Advisor
1,070
0
There must be thousands of books dealing with Legendre polynomials, and othogonal polynomials in general -- going back at least to Maxwell's time. once stuck, it is advisable to go to the books. I won't comment on the flawed proof, but I will note that there's a substantial difference between legendre polynomials and legendre functions, in which the "n" of the polynomials can become in Legendre Functions "z",an arbitrary complex number. Two Legendre functions with differing "z"s are not orthogonal unless the "z" are integers.

Note that 'complex angular momentum" is sometimes used in particle physics, and involves Legendre Functions of complex order -- these were used by Arnold Sommerfeld around 1900, in his thesis, to solve some very difficult problems in radiation, which were relevant to the development of the new phenomena of radio. There's long and rich history for legendre functions and polynomials.

I think, as an ex-professor, that you should be commended for your work to get the appropriate orthogonality relations. However, when stuck go to the experts, the literature, the net. If I'm not mistaken, your error probably has to do with not getting the right values for everything at + or - one. Taking limits when you are dealing with integers is tricky -- in fact with Legendre functions -- the Legendre Function is not well defined on the real axis from -1 to +1, both of which are branch points, and the standard branch cut goes from -1 to +1. So the limits, x->1 and "z"->n. where n is an integer are problematic

I use the Handbook of Mathematical Functions by Abramowitz and Stegen for info,(1000+ pages) on special functions. (I think some of this book is available on the Web). It has everything you want to know, and much more, on almost every function you can image -- legendre, ellipitic, bessel, hypergeometric and confluent hypergeometric, and on and on. It typically gives great details on differential equations. generating functions, recursion relations, integrals and the like. everything you could ever want to know about Legendre Polynomials will be found in Chapter 22, and about Legendre Fuctions in Chap 8.

If people diss you for trying, then write them off -- you don't want to deal with them. And, of course,in this forum you'll encounter all kinds -- some know what they are doing, others do not. It's usually fairly easy to distinguish between them

Almost any text on mathematical physics. electromagnetism, or quantum mechanics will have plenty of material on legendre polynomials, laguerre polynomials, etc.

Also note that there are operator techniques for finding the orthogonality relations of L. polynomials in relation to Angular Momentum Theory -- see Edmonds book on that topic.

Regards.
Reilly Atkinson
 
141
0
Reilly you should look at his initial post, and see who was dissing who.

https://www.physicsforums.com/showthread.php?p=1092694

We had told him pretty much everything you said in this post here, and in addition, exactly how to derive the relation he wanted. He had a problem with us because, after asking us to point out his error, we pointed out his error.
 

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top