1. okey im working with bra-teks now, or dirac notions as its also known.
Im going to start to state some facts ive learned (to see if they are right)
<A| = |A>* (* detonates the complex conjugate)
<A| |B> = <A|B> = <A|B>*
<A|B> = (|A> , |B>)

(<A| C) |B> = <A| (C |B>) = <A|C|B>
where C is somekinda operator. this is how far ive come now. my question is,, exacly what is bra and teks? i understand <A| is bra and |A> is tek but exacly what is it? and what is a operator in the last part?

2.

3. Hi zelos

Do you know what a vector space and its dual is? A ket is a vector from the vector space in question (in QM we are dealing with a complex hilbert space) and a bra is a member of the dual space of vector space (the space of continuous linear functionals). It is similar (ok the same) as covariant and contravariant vectors. The correspondance comes from the Riesz representation theorem.

The A in your equations (note that <sup>*</sup> means taking duals in some of the lines and Conjugate transpose in others - which amounts to the same thing if your vector space is finite dimensional, but QM is not!) is another linear function A : V -> V where V is your vector space. All observables in QM are such operators (with some side clauses but thats not important now, all observables are operators but not all operators are observables - what is missing is self adjointness and correct transformation between different observers).

Making sense?

4. so bra is just a vector space and ket is its complex conjugate? ok i can understand that, but id like some more explinations on operators

5. A ket is a vector from some vector space while a bra is a covector from the dual space of that vector space.

What more do you want to know about linear operators, they are just linear maps from V to V. So they take a vector and give you another vector (for example take differentiation as your operator on the vector space of smooth functions f:R->R)

6. so it simply changes a vector to another vector? ok thx

7. The dual space is the space of all continuous linear functionals on that vector space - this is where bra's live.

8. seems like mathematics simply keep adding things just to cause confusion and keep unwanted people away.

9. ??? What has been added here ???

I dont see how the idea of a dual space causes confusion.

10. well, me neither, just need to grasp it and when im done its no biggie

11. Havent you done linear algebra?

There is quite a bit of maths you need to do before you start QFT, what is your actual mathematics background?

12. im a quick learner
ive studied vector algebra, know it well.
Studing tensors, bra-ket(obviusly) and more. My biggest challange, if any, is to remember what all symbols and stuff means in what way with what. except from that its not hard

13. Originally Posted by Zelos
im a quick learner
ive studied vector algebra, know it well.
Studing tensors, bra-ket(obviusly) and more. My biggest challange, if any, is to remember what all symbols and stuff means in what way with what. except from that its not hard
How did you do vector algebra without handling the dual space?

14. got no idea since i dont know what dual space is in swedish :S and therefor cant say if i have really done it

15. Originally Posted by Zelos
so
bra is just a vector space and ket is its complex conjugate? ok i can
understand that, but id like some more explinations on
operators
No, a bra, written <a| is an element of the space V* dual to the space V whose elements are kets, written |a>. As river_rat explained, V* is the space of linear functionals on V. (a functional is, very loosely speaking, a function that is also a vector, or do I mean v.v?). It is not right to think of kets as complex conjugates of bras.

Anyway, be careful when talking about complex conjugates of vectors. You should really think of complex conjugates of vector components. Best (in my view) to divorce the two concepts, that is conjugates of vector components and dual vectors. (It is unfortunate, to say the least, that the conjugate of the complex number z is denoted z* and the dual to the space V is denoted V*). Again, I think it is best to separate the two concepts (in fact, when using bra-ket notation, it is usual to use the over-bar to denote the complex conjugate).

(Hmm. I think there is a sense in which the dual space restores symmetry under the requirement of linearity over the complex field. I'll see if I can feret it out of my addled brain).

Anyway. Let <a| be in V* and |b> be in V, then <a|b> is defined as a scalar. If our vector space is a metric space (i.e. we know what we mean by "angle" and "length") then this scalar is said to be an inner product.

Like, suppose |v> and |w> are elements in V. Then we know that |v>·|w> is scalar, the scalar product if you like. There will be for each V, some dual space V* such that, if an inner product on V is defined then for all <v| in V* and |w> in V, <v|w> = |v>·|w>. But bear in mind that even if V is not a metric space, it will still have a dual.

See why I hate the Dirac notation? (Why not just use f<sub>v</sub>(w) = v·w?)

Anyway, as far as operators, just to get you motivated. There are two trivial operators on V:

the identity operator I: V→ V: for all v in V, Iv = v

the zero operator 0: V → V: for all v in V, 0v = 0.

In general one says that a linear operator A acting on elements v, w of V is a transformation such that, for scalars a and b, A(a|v> + b|w>) = aA|v> + bA|w>

16. No, a bra, written <a| is an element of the space V* dual to the space V whose elements are kets, written |a>. As river_rat explained, V* is the space of linear functionals on V. (a functional is, very loosely speaking, a function that is also a vector, or do I mean v.v?). It is not right to think of kets as complex conjugates of bras.
Its okay to think of them as complex conjugates as long as your vector space is finite dimensional - I think it boils down to defining the functional on a given basis for the vector space and proceeding from there.

Anyway. Let <a| be in V* and |b> be in V, then <a|b> is defined as a scalar. If our vector space is a metric space (i.e. we know what we mean by "angle" and "length") then this scalar is said to be an inner product.
I think you can drop the metric from here - you just need an innerproduct for lengths and angles to make sense (do angles make sense in just a metric space?)

Like, suppose |v> and |w> are elements in V. Then we know that |v>·|w> is scalar, the scalar product if you like. There will be for each V, some dual space V* such that, if an inner product on V is defined then for all <v| in V* and |w> in V, <v|w> = |v>·|w>. But bear in mind that even if V is not a metric space, it will still have a dual.
Yep, and describing the duals of general infinite dimensional vector spaces can be a pain - that is why most analysis either happens in a banach space or a hilbert space etc.

17. Originally Posted by river_rat

Its okay to think of them as complex conjugates as long as your vector space is finite dimensional - I think it boils down to defining the functional on a given basis for the vector space and proceeding from there.
"Boils down" is right! I agree in general though. The only point I wanted to make clear was that they are basically unrelated concepts. And of course, the choice of basis for the dual space is somewhat arbitrary, an ugly thing to do in general.

I think you can drop the metric from here - you just need an inner product for lengths and angles to make sense (do angles make sense in just a metric space?)
Oops, I mis-spoke. No, that's a lie, I mis-thought. "Length" and "angle" make perfect sense in an inner product space, even when there is no metric. The radic of the inner poduct of a vector with itself is its "length". Likewise the linear dependence condition needs no metric. Just as well, for...

most analysis either happens in a banach space or a hilbert space etc.
Which are of course function spaces.

Anyways, Zelos asked about linear operators. Let's see if I can trawl the memory banks to come up with the most impressive of these, the Hermitian operator, which is important in the most scary place I've ever been, the quantum world.

18. My work lives in the realm of topological algebras (doing my masters on topologising semigroups) - which can be as scary as the world of QFT i tell you!

I have a question here on bijective "rules" (as you dont have functions here) from proper classes to sets - are these things possible or do they serve as contradictions?

19. Originally Posted by river_rat
My work lives in the realm of topological algebras (doing my masters on topologising semigroups)
Are we talking Lie/Clifford algebras here? I know a bit about these guys, but less than I would like
I have a question here on bijective "rules" (as you dont have functions here) from proper classes to sets - are these things possible or do they serve as contradictions?
You're asking me?? You flatter me.

20. Umm, just checked. Topological algebras have nothing to do with Lie or Clifford. Ignore me.

21. Originally Posted by Guitarist
Are we talking Lie/Clifford algebras here? I know a bit about these guys, but less than I would like
Closer to lie groups- expect the operations are not smooth (so no manifold structure, no vector fields so no "lie algebra's") just continuous. Topological group theory is the basis for abstract harmonic analysis which extends fourier things to general spaces.

Topological semigroup theory has applications in number theory and ramsey theory - the algebra on beta N is quite powerful and has some interesting consequences depending on your set theoretic starting points.

You're asking me?? You flatter me.
The ideas are not complex - i just need to see if i havent missed something trivial that blows up my contradiction. But that is work for another night. Have you handled ordinal numbers and transfinite induction before?

22. Ha! Well so far I have proved to be allergic to number theory, but you are welcome to try and change this.

But look, I really like vector spaces, so let me ramble on a bit, starting with the inner product. It's rather elementary at first, I apologize if anyone finds it patronizing. Just for the moment I'm going to drop the bra-ket notation, for which off topic-ness I apologize. Thus v, w,... will be vectors a, b,... will be scalar, complex in general (I'll explain why in a bit)

If it helps anyone, it is OK for now to think of vectors as arrows (the posh name is directed line segment) but in general this is not a good idea.

Right, we know what it means to multiply two numbers, but, because vectors have the dual attributes of magnitude and "direction" there are two different forms of mutiplication. The vector (or tensor) product is somewhat akin to scalar mutiplication, so lets not bother with that here.

Let's think about "direction". In order not to be absolutist about it, by "direction" I mean relative "direction" which is just another way of saying the angle they make with each other, usually measured in radians i.e. whole fractions of 2π. Consider v and w. We want to know the "projection" of v in the direction of w. Like, how long a shadow does v cast on w.

It is obvious that this depends on their relative "lengths" and the angle between them. Let's assume for the moment they are of equal magnitude, so all we care about is the angle, call it φ. Then it is obvious that as φ approaches π/2 i.e. 90 degrees, the shadow grows shorter, and when φ = π/2 it vanishes.

We may write this down. Let's call our shadow some function f of φ such that at π/2, f(φ) = 0. The cosine is our man here. So (vw)cosφ = 0. All we need to do now is define that equation as v·w = 0, it being understood that this implies v and w are perpendicular in some sense. The posh term is "orthogonal". v·w is called the inner (or scalar or dot) product of v and w. Note that the RHS of the inner product is scalar.

Before proceeding, let's ask what happens if we use the same trick and try to find v·v. Well here v "covers" v, and so if v·v = a, which is a number, the inner product of a vector with itself gives us some information about "length". By an accepted abuse of notation, v·v is often written v², and so √(v·v) is taken as the "length" of v. This defined as ||v|| = √(v·v).

To resume. Note first I have made no mention of coordinate systems, or dimensions. These are of course related, and are entirely arbitrary and man-made constructs as ordinarily used. Arbitrariness is in general considered rather unhygenic, and we don't need to be dirty here.

The condition that v·w = 0, it's called the vanishing of the inner product, implies there is no way we can talk about v in terms of w and vice versa. This condition is thus called linear independence. Although in a 3-space it may be intuitively obvious there are no more that 3 linearly independent vectors, it's impossible to prove (I think), and in any case we can have no intuition about an n-space.

Nothing for it then but to define the dimension of our vector space as the number of linearly independent vectors we can find. This is nice. It implies any other vector x will be linearly dependent on this set, and we can know almost all we need to know about x by asking it (this set, that is). It also implies something rather more subtle. Say we have n linearly independent vectors. As soon as I introduce the (n + 1)th vector, I am free to demand that any one of the n + 1 vectors be dependent on the remaining n vectors. Weird, eh? The linearly independent, orthogonal set of vectors (there may be an infinity of them) about which this is always true are called a basis for the space.

So we have rolled the notions of dimension and coordinate into one, as we suspected should be possible, without any recourse to arbitrariness! Aren't we clever?

23. mmm, sorted it out myself (it suddenly dawned on me again what a cardinal number was if you assume choice - duh!) though it does seem like i am using a rocket powered sledgehammer to kill a fly!

24. ... two different forms of mutiplication
there is 3 vector multilications
Scalar
Vector
Matrix

25. Originally Posted by Zelos
there is 3 vector multilications
Scalar
Vector
Matrix
Well, personally I wouldn't include matrix multiplication as a seperate class. Vectors are matrices, after all, and the difference between the inner product and the vector product is merely a matter of how matix multiplication is set up, i.e. whether you transpose one of the matrices or not. I was hoping to say more on that.

Hey, let's do it now.
I talked at excrutiating length about the inner product, and I want to get to the notion of a dual vector. First thing we need to recognize is that the inner product is not symmetric: v·w not equal w·v. This can be seen if we pursue our Micky Mouse notion of shadows. If v and w are unequal in "length", then unless v·w = 0 (orthogonality), then the projection of v on w is not the same as the projection of w on v.

But, in a complex space we may have v·w = (v·w)*, where the star denotes complex conjugation. The reasoning here is simply that we
require ||v|| = √v·v to be real and positive definite, so we may take it as a defintion.

By the way, I promised you this. All definitions and theorems in vector space theory assume a complex underlying field. This is done entirely for reasons of economy, as is is possible, in fact essential, to regard the reals as a subset of the complexes: simply set the coefficient on the imaginary unit to zero. In this case, the conjugate becomes redundant, but not nonsensical.

With all this behind us, let u, v, w, x be vectors in a space V, a, b scalar.

Suppose v =a(w) + b(x), then by linearity I may have
u·v = a(u·w) + b(u·x). But now suppose that
u = a(w) + b(x), then because u·v = (v·u)*, I find that
v·u = (u·v)* = {a(u·w) + b(u·x)}*, and I can transfer the conjugate to
the coefficients thus
v·u = a*(u·w) + b*(u·x). In other words, u·v is linear wrt a and b, but v·u is linear in a* and b*. This asymmetry is not pretty, so, in order to restore it we define another space V* called the dual to V, whose elements are the linear functionals f<sub>v</sub> such that f<sub>v</sub>(w) is defined equal to v·w.

We now may have, by virtue of the functionality of the elements of V*,

(af<sub>v</sub> + bf<sub>w</sub>)x = af<sub>v</sub>(x) + bf<sub>w</sub>(x) and
f<sub>v</sub>(aw + bx) = af<sub>v</sub>(w) + bf<sub>v</sub>(x).

Linearity is restored all round.
In bra-ket notation, f<sub>v</sub>(w) ≡ <v|w>

OK, this is a bit rough and ready, but I hope gives a taste of what a
dual vector is. Oh, but take care. You may find these fellows called
co-vectors or even 1-forms. Take especial care of the latter: some
people use as a synonym for dual vector, others to mean the
dual/covector field, but these seem mostly to be physicists.
Tsk! You have been warned.

26. okey, now see this
i have the hamiltonian operator (i´ll be using atomic units for simplicity)
wich is -1/2D - Z/r
D is laplacian function thing.
Lets call that operator H (guess why)

And i have wavefunction Y (closes letter) Y containt constants but has the variable r in the e<sup>-r</sup>
Y=Z<sup>1,5</sup>/sqrt(pi) * e<sup>-r</sup>

<Y|H|Y> = 1/2 Z<sup>2</sup>

Now id like this explained quite detailed since all my atempts to do this failed. espacialy the laplacian part shall be speceficly detailed

THank you

27. Originally Posted by Zelos
okey, now see this
i have the hamiltonian operator (i´ll be using atomic units for simplicity)
wich is -1/2D - Z/r
D is laplacian function thing.
Lets call that operator H (guess why)

And i have wavefunction Y (closes letter) Y containt constants but has the variable r in the e<sup>-r</sup>
Y=Z<sup>1,5</sup>/sqrt(pi) * e<sup>-r</sup>

<Y|H|Y> = 1/2 Z<sup>2</sup>

Now id like this explained quite detailed since all my atempts to do this failed. espacialy the laplacian part shall be speceficly detailed

THank you
Well, I simply do not understand your question. But as I am not a physicist, maybe that's allowed.

But - I do know what the Laplacian operator is. Before I even try, tell me. Does this "∂" render as the partial for you? How about ∂²?

Let me know, because I don't think I can do it in words alone. Especially as I haven't found a way to render the del operator. Anybody know?

28. my question is that how do u get <Y|H|Y> to become ½Z<sup>2</sup>.
id like to see it step by step since ive tried and it becomes all wrong. its spherical coordinates you work with here

i used atomic units to simpliify it so its only mathematical stuff left. Could you show how that simplification is done? especially around laplacian stuff

29. Okay, for notational convincing let ∂<sub>x</sub> be the partial derivative of some quantity w.r.t x (with the obvious generalisation to higher order derivatives).

In atomic units our Hamiltonian is

H = -1/2 Nabla^2 - Z/r and Phi = Z<sup>1.5</sup> / Sqrt(pi) Exp(-Z r)

First notice the error in your phi, you forgot the Z in the exponent. In spherical coordinates we have that Nabla^2 = 2/r ∂<sub>r</sub> + ∂<sub>rr</sub> and that a general integral over the entire space of a rotationally invariant function f(r) is given by 4 pi (Integral from 0 to infinity of ) </sub>r<sup>2</sup>f(r)dr

Doing all the necessary derivatives and the easy integration by parts gives you the solution you are looking for.

30. you forgot the Z in the exponent.
ops

In spherical coordinates we have that Nabla^2 = 2/r ∂r + ∂rr and that a general integral over the entire space of a rotationally invariant function f(r) is given by 4 pi (Integral from 0 to infinity of ) r2f(r)dr

Doing all the necessary derivatives and the easy integration by parts gives you the solution you are looking for.
okey you lsot me here. that laplacian symbol (its right, right? since someone told me it was it)
have i found to be 1/r<sup>2</sup> ∂(r<sup>2</sup> ∂f/∂r)/∂r
is this right? (F is the phi equation) (when radius is the only variable)
just for sake of easyness, lets call that L

31. I think there is another error as well Zelos which i forgot to mention

<Phi| H |Phi> = -1/2 Z^2 (which corresponds to the correct sign for the ground energy of a hydrogen atom).

What i gave as the laplacian and what you gave are the same thing (check it).

Post your steps and lets work through what you have done - you will learn more that way then if i just give you the answer.

32. yes i know, but ok i´ll give my work after school today so you can see it.

33. okey im finaly done with school. here comes my work. to simplify writing laplacian symbol = L
integral symbol S

oh eyah i found a error in my wave equation.
Y=Z<sup>1,5</sup> * e<sup>-zr </sup> (Y instant of phi)
that sqr(pi) was only for none-atomic units
<Y|H|Y> = <Y| (H|Y>)

(H|Y>) = -½ L - Z/r | Z<sup>1,5</sup> * e<sup>-zr </sup> >
L (Z<sup>1,5</sup> * e<sup>-zr </sup>) = 1/r<sup>2</sup> ∂(r<sup>2</sup> ∂(Z<sup>1,5</sup> * e<sup>-zr </sup>)/∂r)/∂r
=Z<sup>1,5</sup>/r<sup>2</sup> ∂(r<sup>2</sup> ∂(e<sup>-zr </sup>)/∂r)/∂r = Z<sup>1,5</sup>/r<sup>2</sup> (2r ∂(e<sup>-zr </sup>)/∂r + r<sup>2</sup> ∂<sup>2</sup>(e<sup>-zr </sup>)/∂r<sup>2</sup>

∂(e<sup>-zr </sup>)/∂r = -z e<sup>-zr </sup>
∂<sup>2</sup>(e<sup>-zr </sup>)/∂r<sup>2</sup> = z<sup>2</sup> e<sup>-zr </sup>

Z<sup>1,5</sup>/r<sup>2</sup> (-2rz e<sup>-zr </sup>+ r<sup>2</sup> z<sup>2</sup> e<sup>-zr </sup> = Z<sup>1,5</sup>(z<sup>2</sup> e<sup>-zr </sup> - 2z e<sup>-zr </sup> / r)=Z<sup>1,5</sup>e<sup>-zr </sup>(z<sup>2</sup> - 2z/ r)=

-½ L - Z/r | Z<sup>1,5</sup> * e<sup>-zr </sup> > =
-Z<sup>1,5</sup>e<sup>-zr ½</sup>(z<sup>2</sup> - 2z/ r) - Z<sup>1,5</sup> * e<sup>-zr </sup> > = -Z<sup>1,5</sup>e<sup>-zr </sup>(z<sup>2</sup>/2 - z/ r + z/ r)=-Z<sup>1,5</sup>e<sup>-zr </sup> z<sup>2</sup>/2)=-½Z<sup>3,5</sup>e<sup>-zr </sup>

<Y|H|Y> = S<sub>0</sub><sup>infinite</sup> (Y (H|Y>)) dr = [-½Z<sup>3,5</sup>e<sup>-zr </sup>*Z<sup>1,5</sup> * e<sup>-zr </sup>]<sub>0</sub><sup>infinite</sup>=[-½Z<sup>5</sup>e<sup>-2zr </sup>]<sub>0</sub><sup>infinite</sup>=-½Z<sup>5</sup>[e<sup>-2zr </sup>]<sub>0</sub><sup>infinite</sup> = -(e<sup>-infinite</sup> -e<sup>-0</sup> )/2Z * -½Z<sup>5</sup>=-½Z<sup>5</sup>/2Z = Z<sup>4</sup>/4

this is how far ive come, its the square of the answer im looking for, why is it square and where did i make the error?

34. Okay for starters

<Y| H |Y> is not equal to the integral from 0 to infinity of (Y (H|Y>)) dr in this example. You need to do a 3 dimensional volume integral in spherical coordinates.

35. that is all that is and in the part im dealing with it is from 0 to infinite. I found on later pages (4 to be exact) they used intergral from 0-infinite. so thats how they have done. so except that where does it get squared?the integral from 0-infinite is there since, well the electron has to be somewhere :S there is a chance its at a infinite distance from the nuclues, its just a very small chance

36. The 0 to infinity bounds are correct, but the actual integral you are computing is wrong.

What is the inner product in this space?

37. <Y|H|Y>, Y do u see higher up and H aswell

38. Um, what i am asking is how would you calculate the inner product.

<U|V> denotes the inner product but how do you calculate it - you have not worked that out yet.

39. i dont udnerstand your question

40. Okay - how would you calculate

<exp(-Z r)|exp(-Z r)> where r denotes the radius in spherical coordinates?

41. Thanks goodness I'm not a physicist, if that's how you spend your time!

But, seriously, linking text, even of the simplest form like "it therefore follows" or "we know from blah-blah" would be helpful to outsiders.

And what is Z here, by the way?

I have read that in this this sort of situation, the Lebesgue integral applies. Is this right? I know waht this guy is, but would someone care to explain why? Is it a physics reason, or have I missed something?

Gosh, I feel out of my depth here.

42. Z is the charged the electrons in question feel, since this is helium Z lies somewhere between 1 and 2

river rat i wodul do like in my papers
<exp(-Z r)|exp(-Z r)> = S<sub>0</sub><sup>infinite</sup> exp(-Z r)*exp(-Z r) dr = S<sub>0</sub><sup>infinite</sup> exp(-2 Z r) dr
they did like this in ym paper for a thing about right after this specefic one so i can assume they did the same here since it gives the best result. also william gave me the first idea you could use integral here

43. Hi guitarist - the lebesgue measure comes in as it is the standard inner product on function spaces (and that is where QM lives).

Zelos - you have to use an integral, so you are correct there. But you have to use the correct one.

<Y| H |Y> = the integral over all space of Y* H(Y) dV

Where dV is the volume element. Now what is an integral over all space in spherical coordinates? Well fwhat you need is the correct volume element, which happens to be dV = r<sup>2</sup>sin(phi) dr d(phi) d(theta)

44. yes i know that, but i skipped the CC (complex conjucate) since its only real numbers we deal with here. and the angles did i skip since they arent involved here. the intergrale of them are nothing they are constant

45. Well you need those constants (and the sqrt of pi in your original wave equation - it is a normalisation factor) and even ignoring the angles your integration is incorrect as it is not in spherical coordinates.

46. how can they be wrong when the papers im working from done include the angles since the entire electron cloud is the same in all angles? They dont use it then whats wrong? why is it squared?

47. Zelos - look at your integral, look at the integral in spherical coordinates - compare and see what you missed. Your answer is very, very wrong.

48. okey then, i dont find what i do wrong, so i´ll write exacly like in my paper and you can do the stuff and tell me where it goes wrong

<Z<sup>1,5</sup>e<sup>-Zr</sup>|-½L - Z/r|Z<sup>1,5</sup>e<sup>-Zr</sup>>
L is laplacian symbol Ive tried, not succeded dotns ee where the errror is, could you try now and see what answer you get and show me how you did? for someone as good in mathematics this should be a piece of cake and dont take to much time

49. that is not wrong - but when you worked out the inner product you did not do it properly.

50. please show me then what i did wrong and how it should be instant of just saying it is wrong. Show also in the equation how to do it right since its easier to understand then

51. Okay - firstly your integral has to be (Integral from 0 to infinity) of r<sup>2</sup>exp(-2 Z r) dr (ignoring all the constants).

52. okej, why? where does the r² come from?

53. okay - take the volume element in cartesian coordinates dV = dx dy dz and convert to spherical coordinates. What do you get? Now integrate a function in spherical coordinates f(r, theta, phi) = exp(-2 Z r).

If you go through the steps you will see where the r<sup>2</sup> comes from.

54. i have done this atleast a dussin times i dont see where the r² comes from. so please show me where it goes wrong instant of saying its wrong. try building a car without any knowledge how it shall be and the only help you have is a dude saying "its wrong" if its wrong, you wont come very far

55. ok, you were right on using all the intergrals, now where does the r² comes from?

56. In spherical coordinates the volume element is

dV = r^2 sin(phi) dr d(theta) d(phi)

The r^2 comes from that.

57. ok, thank you that i didnt know. now i might finaly be able to solve it. THank you.
Could you please maybe explain why it is like that and how/where it comes from?
dV = r^2 sin(phi) dr d(theta) d(phi)
couldnt this be written as dV = 2*pi r^2 sin(phi) dr d(phi) ?

58. also how do i intergral this formula?
SS e<sup>-2Z(R<sub>1</sub>-R<sub>2</sub>)</sup>/|R<sub>1</sub>-R<sub>2</sub>| dR<sub>1</sub> dR<sub>2</sub>

im not sure about the bold stuff but isnt it a vector?
(Once again S detonates integral symbol)

59. Originally Posted by Zelos
ok, thank you that i didnt know. now i might finaly be able to solve it. THank you.
Could you please maybe explain why it is like that and how/where it comes from?
dV = r^2 sin(phi) dr d(theta) d(phi)
couldnt this be written as dV = 2*pi r^2 sin(phi) dr d(phi) ?
No - you can write it as 4*pi r^2 dr (use fubini's thm). It comes form the standard change of variable formula from cartesian co-ords to spherical co-ords.

60. good, how about integral of vectors? like in that formula i showed you

61. mmm, never seen that type of integral.

Ill see if i can work it out.

62. id appriciate it

63. where did you see that integral?

64. its the last part of the hamiltonian of helium wich represent the mutal repuslation of both electrons

65. did they give an answer?

66. yes, lets see, its first (Z<sup>3</sup>)<sup>2</sup> times the integral, and the answer they gives is 5/8Z
(maybe in the first formula a pi should be in, not sure in atomic untis

67. ah - had an ahah moment.

To understand the integral write it out in its cartesian coordinates (it is six dimensional) - though how to actually integrate it i have not worked out yet.

68. To understand the integral write it out in its cartesian coordinates (it is six dimensional)
explain and perhaps show?

69. Okay the answer you want is

(I havent worked that out yet though)

The notation is a bit of a pain to type here zelos - let R<sub>1</sub> = (x<sub>1</sub>, y<sub>1</sub>, z<sub>1</sub>) and R<sub>2</sub> = (x<sub>2</sub>, y<sub>2</sub>, z<sub>2</sub>)

Then S exp(-2Z (R1-R2)) / |R1-R2| dR1 dR2 =

S exp(-2Z (Sqrt(x<sub>1</sub>^2 + y<sub>1</sub>^2 + z<sub>1</sub>^2) + Sqrt(x<sub>2</sub>^2 + y<sub>2</sub>^2 + z<sub>2</sub>^2)) ) / Sqrt((x<sub>1</sub> - x<sub>2</sub>)^2 + (y<sub>1</sub> - y<sub>2</sub>)^2 + (z<sub>1</sub> - z<sub>2</sub>)^2) dx<sub>1</sub>dy<sub>1</sub>dz<sub>1</sub>dx<sub>2</sub>dy<sub>2</sub>dz<sub>2</sub>

70. yes, e/a in atmoic units(wich ive previusly used) = 1
now how do u deal with that?

71. http://mathworld.wolfram.com/VectorIntegral.html
this might help some, found it on the net

72. Hi zelos

Remembered how to do that integral (2nd year physics strikes again) though it is a real mess and it is easier to just "take it on faith".

Ill sketch the steps. (assume r<sub>1</sub> > r<sub>2</sub>. In the integral you have to split into two cases to handle this)

1/|r<sub>1</sub> - r<sub>2</sub>| = (r<sub>1</sub>^2 +r<sub>2</sub>^2 - 2 r<sub>1</sub> r<sub>2</sub> Cos(theta))<sup>-0.5</sup> = Sum P<sub>n</sub>(Cos(theta) (r<sub>2</sub> / r<sub>1</sub>)<sup>n</sup>

Where theta is the angle between the two different vectors and P<sub>n</sub>(x) is the legendre polynomials. You can then expand these as spherical harmonic functions and then some aweful integration and the like gets you the result.

73. ok, and how did you get this result?

74. Which result?

The expansion into legendre polynomials comes from the generating function for the legendre polynomials. To expand into spherical harmonics use the additive theorem for spherical harmonics.

75. now lets take it one level easier

76. as i see this is getting pretty nasty, hows it going river?

77. (r1^2 +r2^2 - 2 r1 r2 Cos(theta))-0.5
how do u calculate the angle ebtween the vectors?
and also explaine asier how u got this

78. Hey Zelos

Been moving so have not had a lot of time to post here. The expression you asked for comes from taking the definition of the norm in terms of the inner product and expanding.

To explain the legendre polynomial and harmonic function stuff is a mission - rather read the wikipedia and mathworld articles and if you have any questions shout. You should see most of this in any good electrodynamics book btw as multipole expansions use a similar trick.

79. ive been thinking and that equation of your is acctualy Law of cosines
wich ive also been thinking on, but how do u get the angle between the 2 points?

80. Yep it is the law of cosines - though i prefer to just use the inner product and the fact that the cosine of the angle between two unit vectors is defined to be their inner product. Points dont have angles between them, we are working in a vector space.

To get the angle between the two unit vectors in spherical coordinates just sub into the expression

(v<sub>1</sub>, v<sub>2</sub>) = x<sub>1</sub>x<sub>2</sub> + y<sub>1</sub>y<sub>2</sub> + z<sub>1</sub>z<sub>2</sub>

the spherical coordinate transformations (with r = 1) to get that

Cos(alpha) = Sin(phi<sub>1</sub>)Sin(phi<sub>2</sub>)Cos(Theta<sub>1</sub> - Theta<sub>2</sub>) + Cos(phi<sub>1</sub>)Cos(phi<sub>1</sub>)

Where alpha is the angle between the two vectors in question. This does not simplify the integral though - that is why you use the legendre polynomial trick.

81. riverrat, i worked out the same thing in school acctualy hurray i did right.

btw i think you made a misstage, you seem to have accidently put sin on cos and cos on sin, the one we have in common is cos(a-b) and ium rpetty sure im right in this case

Cos(alpha) = Cos(phi<sub>1</sub>)Cos(phi<sub>2</sub>)Cos(Theta<sub>1</sub> - Theta<sub>2</sub>) + Sin(phi<sub>1</sub>)Sin(phi<sub>2</sub>)

is the one i get

well how do we proceed from here now?

82. I double checked and my expression is right - how did you derive yours. Oh and phi is the angle between the z-axis and the vector in question.

Your expression does not work if phi<sub>1</sub> is 0 and theta<sub>1</sub> is pi/2 (i.e. the angle between the vector and the z-axis) for example :

Cos(phi<sub>2</sub>)Sin(Theta<sub>2</sub>) is not equal to Cos(phi<sub>2</sub>)

Have you read up on the legendre polynomials and harmonic functions?

83. using normal dot production on vectors with the spherical coordinates

Spherical coordinates: r,a,b

r<sub>1</sub>r<sub>2</sub>(Cos(a<sub>1</sub>)*Cos(a<sub>2</sub>)*Cos(b<sub>1</sub>)*Cos(b<sub>2</sub>) + Sin(a<sub>1</sub>)*Sin(a<sub>2</sub>)*Cos(b<sub>1</sub>)*Cos(b<sub>2</sub>) + Sin(b<sub>1</sub>)*Sin(b<sub>2</sub>))
then with some fixing i get
r<sub>1</sub>r<sub>2</sub>(Cos(b<sub>1</sub>)*Cos(b<sub>2</sub>)*Cos(a<sub>1</sub> - a<sub>2</sub>) + Sin(b<sub>1</sub>)*Sin(b<sub>2</sub>))

84. You are not using standard spherical coordinates

What are you transformation equations?

85. x= r cos(a)cos(b)
y= r sin(a)cos(b)
<= r sin(b)

http://mathworld.wolfram.com/SphericalCoordinates.html
just like they ahve

86. Okay - where did you get those equations?

The standard ones are

x = r Cos(theta) Sin(phi)

y = r Sin(theta) Sin(phi)

z = r Cos(phi)

your "b" is not a standard angle but is rather the angle from the x-y plane to the vector and not measured from the z-axis.

87. http://mathworld.wolfram.com/SphericalCoordinates.html
here
it says like i do, but if both angles shoudlnt r=x then logicly? and not z

88. it does? Read it again.

89. okey my bad:S where the hell did i get this form then? anyhow, why is it so that when both angles = 0 that z=r and not x=r that seems more logical and traditional
seems most logical to expand polar coordinates instant of changing

lets procced after that anyhow

90. Originally Posted by Zelos
okey my bad:S where the hell did i get this form then? anyhow, why is it so that when both angles = 0 that z=r and not x=r that seems more logical and traditional
seems most logical to expand polar coordinates instant of changing
It is expanding polar co-ordinates though. The r and theta are the same as before. phi is the new angle needed for R^3. Setting phi=0 to point in the direction of the z-axis means phi ranges from 0 to pi. If you wanted to set phi=0 to be the direction of the x-y plane, then phi would range from -pi/2 to pi/2, this might not be considered as 'nice' ('nice' is arbitrary of course). Either coordinate system will work fine, though there's no real advantage of using a different one than everyone else unless you enjoy confusion.

91. Why doesnt it make sense? phi is measured from the z-axis so if phi = 0 then you are on that axis.

92. you usualy measure from the x-axis as im famial with it

93. theta is measured from the x-axis. Anyway, what do you want to look at?

94. okey lets drop this, lets continue with the problem i´ll accept its from Z-axis and work with it. anyhow, we have that now what?

95. Have you read up on legendre polynomials and harmonic functions yet?

96. yes i have
Sum P<sub>n</sub>(Cos(theta) (r<sub>2</sub> / r<sub>1</sub>)<sup>n</sup>

just dont get how u got this thou

97. To explain generating functions here is a bit of a pain due to the notation problems. Take it as given that

(1 - 2xk + x^2)^-0.5 = Sum P<sub>n</sub>(x) k<sup>n</sup>

98. river rat, if its hard, use Word and send it to me, u know my mail

99.

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