This has gone through the mail, so simple text formatting is used, but it should read fairly clearly anyway.
> Some months ago I went to the big trouble of
> converting the SWE from Cartesians to sphericals. I thought that was it but now, as I
> begin to assemble the final wavefunction for hydrogen, I find I need the volume element dV
> for integrals. Understanding how this comes out as r^2.sin(theta).dr.d(theta).d(phi) is
> not altogether easy! My books have the standard diagram and talk about increasing r, theta
> and phi by dr, d(theta) and d(phi) but it`s really not easy to see how to reach the
> expression for dV from the diagram.
> Looking on the net I see that there is another way via `the Jacobian`
> (matrices/determinants) but this looks a little complicated (but I will go there if I have
> to!).
> If I have:
> x = r.sin(theta).cos(phi)
> y = r.sin(theta).sin(phi)
> z = r.cos(theta)
> is there a way of finding dx, dy and dz from these expressions and then simply
> multiplying them together to get dV?? It`s got to be better than working it out from the
> diagram. Perhaps if you show me how to get `dx` I should be okay to get the others. I have
> bad memories of this stuff - I remember feeling I was swimming in
> variables and never being able to find the right way through.
The Jacobian is the "right" way to do it, a mathematician would say.
It works for ANY coordinate change and it is a simple and automatic
formula.
Basically, changing from a set of variables A to a set of variables B you
change your integration in the natural generalization of what you have for
simple 1-dimensional integrals:
integral f(A) dA = integral f(A(B)) |dA/dB| dB
where now |dA/dB| is the absolute value of the determinant of the Jacobian
matrix of derivatives dA_i / dB_j
(for 1-dimensional integral that would be the absolute value of the
derivative dA/dB.
The physicist way to look at it with the small increments has always left
me skeptical... And I'm a physicist!
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> It is common to write the assoc. Laguerre polys as L superscript 2l+1, subscript n+l. Now
> I know what l and n are individually but what is the significance of the (2l+1) and the
> (n+l)? One of these may be something to do with the `order` of the poly. What is meant by
> `order` here? And is the other to do with the `degree` of the polys. Again, what exactly
> is `degree`?
> Secondly, when we are looking at the normalization which bit is the so-called `weight
> function`? Is it the exp-(p/2), is it the p^l, is it the p^2 factor in the volume
> element, or is it some combination of these?
> These are pretty trivial questions I know but I need to get them straight before I get any
> deeper into the normalization.
Briefly, once you have separated the radial from the angular motion, for a
general potential U(r) you have a radial equation whose negative-energy
solutions you label with a quantum number n_r (_ indicating subscript).
This number n_r=0,1,2,3,.. has the meaning of the "number of nodes" of the
wavefunct for the one-dimensional radial equation.
By nodes one means points where the function passes through 0.
This happens also for the hydrogenic potential U(r)=-Z/r.
The radial eigenfuncts in this special case are written in terms of a special
function, the hypergeometrc function.
The "important" quantum numbers are therefore n_r , l and m_l
(plus, if you want, m_s for spin).
Then one can make combinations of those fundamental q.n.
For example, as the energy of the bound states of this special potential is
given by
E(n_r,l) = Z^2 e^2 / (2 a_0) * 1/ (n_r + l + 1)^2
it is convenient to define n = n_r + l + 1 , so that the expression for the
energy of the levels look neater.
However the q.n. with a clear and fundamental meaning are those listed above.
Now, for traditional reasons one notices that for the physical values of
the quantum numbers, fixed by the boundary conditions, the hypergeometrc
function appearing in the radial wavef. is actually related to some
polynomial which were already known for previous reasons. These Laguerre
polynomials are often defined with two indexes, like
L_a^b (x)
(^ indicating superscript).
The degree of a polynomial, as defined in high school calculus is the maximum
power of the variable (x here) appearing there.
The degree of L_a^b (x) happens to be a-b (to convince yourself check the
definition of L_a^b in books of special functions).
As a result, having a=l+n and b=2l+1 (from the relation between hypergeometric
and Laguerre), we have in the radial function a polynomial of degree
a-b= n-l-1 = n_r
Indeed, you might remember that the number of zeros of a polynomial of degree
n_r (i.e. the roots of the equat. P(x)=0) is at most n_r.
Actually, one can show that the L_a^b (x) have exactly n_r roots, corresponding
to the nodes of the radial wavefunction as I explained above.
Things to remember:
1) The "principal" quantum number n fixes the energy of the so-called shell.
2) The number of radial nodes n_r = n-l-1 changes within a "shell".
3) Given n, higher-l functions have less radial nodes (but, of course, more
angular ones!)
Some reference about the hypergeometric function:
http://www.efunda.com/math/hypergeometric/hypergeometric.cfm
And more details in standard books on special functions such as
M. Abramowitz, I. A. Stegun, eds., Handbook of Mathematical
Functions with Formulas, Graphs, and Mathematical Tables, National
Bureau of Standards Applied Mathematics Series, vol. 55, US
Government Printing Office, Washington, DC (1964).
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> I have come across spherical Bessel functions
> and have been interested to find out what they are and their relation to
> ordinary Bessel functions. First question is about ordinary
> Bessel functions and the ODE from which they derive.
> There seem to be two definitions of this equation, one with a constant
> and one without. Thus:
> x^2 y`` + x y` + (x^2 - n^2) y = 0
> and
> x^2 y`` + x y` + (k^2 x^2 - n^2) y = 0
>
> Now I guess I should not be concerned that one has the constant k^2
> and the other doesn`t but wouldn`t this make a difference in terms of
> the solutions?
Once I wasted 3 days'work to discover that there was different definitions of
the Bessel functions in 2 different textbooks of quantum mechanics (I believe
it was the Schiff and the Messiah).
Sadly, there's no universal definition, you have to decide yourself which one
you find more convenient, and always check the conventions.
> Secondly, the spherical Bessel functions. From what I`ve found out so far these arise when we take
> the SWE in spherical coordinates and then let the potential equal zero so that we are dealing with a
> free particle. My question is, is the resulting `spherical Bessel equation` so called because it is
> a simple Bessel equation but has origins in
> spherical coordinates?
Yes, this is the origin of the name. Different wavefunctions come up if you
do the same exercise in cylindrical coordinates (or equivalently in a plane).
I think these are called cylindrical (or maybe circular) Bessel functions.
> Finally, when we want to solve the SWE for hydrogen we are forced to convert from Cartesian coords
> to spherical coords on account of the potential term (which involves square roots). When looking at
Or, more simply, since the potential is spherically symmetric.
You would find spherical coords convenient for any spherically symmetric
potential, there is nothing special about 1/r for this choice.
> a free particle we end up with the spherical Bessel equation, but if the potential is zero could we
> not just put zero in the original SWE and solve in Cartesians? In other words if we have zero
> potential can we solve the SWE in Cartesians without needing to go to sphericals?
Of course, this is what one usually does.
However, there are reasons for solving the V=0 Schroedinger problem (a very
special case spherically symmetric potential) ALSO in spherical
coordinates, other than pure masochism.
There are actually 2 reasons that come to my mind for doing that.
1) Any potential which is only a function of r = sqrt(x^2+y^2+z^2)
and is CONSTANT everywhere except in a finite set of discontinuity points has
eigenwaves which, in each of the "shells" where the potential is constant, are
spherical Bessel functions times spherical harmonics. Think for example of a
potential spherical well with V(r) = V0 for r< r0 and V(r)=0 for r >=r0.
Actually, this sort of potential is very useful, because you can always
approximate any (almost any, actually) spherical potential with a stepwise
potential of this kind: thus it could be a neat trick to find numerical
approximate solutions of SWE for arbitrary potentials...
2) This approach is extremely useful in the theory of scattering of
Schroedinger waves (representing e.g. electrons or neutrons) against
spherically symmetric potentials (representing e.g. nuclei). It is a very
important subject in quantum mechanics, not a very easy one, though.
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> Thanks, that`s very useful. So we pick our coordinates to suit what might
> be coming next, eg. if we are interested in scattering then we pick
> spherical coordinates, because scattering is suited best to spherical
> coords. One thing I`m not sure about, though, is whether the set of plane
> waves are the same as the spherical Bessel functions. In other words,
> exp ikr is seems very different to spherical Bessel functions. But as we
> are actually solving the same equation (Cartesian and spherical) we must
> get the same answer, which must mean that the two solutions/sets of
> solutions for the two cases are in fact the same thing, although they look
> very different.
> Is this correct?
They ARE different ;-)
The reason is degeneracy.
States with the same energy are degenerate.
For example, the three 2p orbitals (m=-1,0,1) of the hydrogen atom are degenerate.
When you have degenerate states you can always combine them to get other
eigenstates with the same energy. For example the 2p/sub x/ orbital you find
drawn in many books is actually the combination [2p(m=-1) + 2p(m=1)]/sqrt(2)
It is just a matter of convenience to choose one basis or another in a
degenerate linear space.
In the case of a free particle, you know that the energy is
hbar^2 |k|^2 /(2*mass)
Clearly it does not depend on the direction of k, only on its modulus.
Therefore all the states with the same |k| have the same energy.
So, one can combine them at pleasure if this is convenient.
The Bessel version of the free-particle states are indeed linear combinations
of the degenerate states for each |k|. Note that only degenerate states are
mixed in, not states with different values of |k|.
The coefficients of these linear combinations are given in most books of
quantum mechanics (and also of optics, as the same formalism applies for
scattering of electromagnetic waves).
There is also an inverse formula, expressing the spherical Bessel function
times Y_lm as a linear combination of plane waves.
All this is nothing more than a basis change in a linear space (of functions),
of the type explained in Sakurai's first chapter.
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> I have been revisiting angular momentum. In particular I have been
> looking at the treatment given in `Molecular Quantum Mechanics` (2nd Ed.,
> P58) by P.W. Atkins. Atkins has what seems to be a different approach - he
> starts by looking first at a particle confined to a ring, then a particle
> on a sphere, then a particle in a central potential, then, as a final step,
> he combines all three solutions and gets the hydrogen wavefunctions. For
> the first bit, the particle confined to a ring, he uses cylindrical
> coordinates. He takes the cylindrical 3D Laplacian and sets z=0 which gives
> him a sort of 2D Laplacian.
> This is:
> (1/r)partial d/dr(r partial d/dr) + (1/r^2)partial d^2/dphi^2
> [then he sets `r` as a constant, incidentally]
> However, I thought that taking the spherical Laplacian and letting
> theta = pi/2 would give me the same 2D
> Laplacian as the one above, but evidently it doesn`t.
> With sphericals we end up with:
> (1/r^2) partial d/dr[(1/r^2) partial d/dr] + (1/r^2)partial d^2/dphi^2.
The above is incorrect, the correct expression (for theta=pi/2) is
(1/r^2) partial d/dr[(r^2) partial d/dr] + (1/r^2)partial d^2/dphi^2
Let's forget from now on the d^2/dphi^2 term, which is equal in both expressions.
If you evaluate the derivative of the product, you get
(1/r^2) partial d/dr[(r^2) partial d/dr = partial d^2/dr^2 + (2/r) partial d/dr
Instead, if you do the same with the expression derived from cylindrical
coordinates you get
(1/r)partial d/dr(r partial d/dr) = partial d^2/dr^2 + (1/r) partial d/dr
You see the two are indeed different, by a factor 2 in the second term.
> So it seems that the two are different with respect to the `r` part. My
> question is do you know why we get the wrong result if we take the
> spherical Laplacian and let theta=pi/2. Initially I thought it might be
> because of the `r` is differently defined in the two cases i.e. the
> spherical `r` has a z component whereas the cylindrical `r` is only in x
> and y, but if we are letting z=0 then the `r` becomes the same in both
> cases.
> Do you know where I`m going wrong?
You are right then, the 2 expressions for the Laplacian differ!
The reason behind it is that when you put theta=pi/2 in the spherical laplacian
you are not actually FIXING z to 0.
The expression for r = sqrt(x^2+y^2+z^2) in spherical coordinates is different
from the expression r = sqrt(x^2+y^2) in cylindrical coordinates.
This implies that partial d/dr in the 2 cases has a different meaning.
In practice, in spherical coordinates the derivative partial d/dr is also
varying z, not only x and y: this accounts for the extra (1/r) partial d/dr
term.
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> I am revisiting the radial equation from the point of view solving it
> using the `reduced method`, ie. using the early substitution `rR` to get a
> simpler equation to solve, as opposed to the `other way` which solves for R
> rather than rR. After I have completed the task I will try and see which of
> the two methods is the easiest/most instructive.
> But this question concerns what happens at an early stage in the
> process. After making the rR substitution we get the `reduced equation`
> thus:
> -hbar^2/2m . d^2u/dr^2 + (hbar^2/2.m.r^2) . l(l+1) - Ze^2 . u /4.pi.epsilon(0).r = E.u
>
> This is the equation where terms 2 and 3 are called (in combination) the
> `effective potential` and the l(l+1) term is called the `centrifugal
> potential`. This question is about the nasty case of l=0. In this case the
> centrifugal potential term is zero so we have only attractive forces on the
> electron. And as r goes to zero the attraction increases. French and Taylor
> p522 says that there is a finite probability density of finding the
> electron in the nucleus.
> How do we get out of this one? I cannot find a satisfactory answer in any
> of my books. [I`ll quote Atkins below]. What is the explanation of the fact that
> the electron should be right in the nucleus - if the attraction increases
> with 1/r then it can`t be anywhere else! Is this an example of how we can use the
> Heisenberg UP to get us out of the problem?
>
> This is what Atkins says: `when l=0 the electron has no angular momentum
> (no motion) to fling it away from the nucleus; there is no repulsive part of
> the effective potential energy. When l=0 there is a non-zero probability of
> finding the electron at the nucleus. This is an echo of the shape of the
> orbit in the Bohr theory in which the electron swings in a straight line
> through he nucleus, and which he dismissed as unacceptable` And he leaves
> it there, so no really good explanation of what`s going on.
> Hope this question makes sense,
Well, the probability for the el. to be exactly at r=0 is, strictly
speaking, 0.
You have a nonzero probability DENSITY |R(0)|^2.
However yes, for the states with nonzero R(0), the integration of
r^2 |R(0)|^2 over any small but finite sphere gives a probability which is
proportional to the volume of this sphere, and not to some higher power of
the volume (as it happens for l>0 states).
So in this sense the electron has a nonzero probability to stay at the
nucleus.
You can convince yourself that this is quite OK, and that the electron is
not "falling into the nucleus" by evaluating the kinetic and potential
energies of some NORMALIZED exponential state R(r) ~ exp(-r/b).
You will see that for small b (very concentrated states close to the nucleus)
the potential energy gets more and more negative (of course!), but at the same
time the kinetic term is growing larger and larger. In the opposite limit of
large b (spread out state, the potential energy is close to 0 and the kinetic
energy is small as well.
The value of b that optimizes (makes smallest) the total (kinetic+potential)
energy is indeed Bohr's radius.
[NOTE: this calculation is an example of variational estimate of the ground
state wavefunction, a special one indeed, as you happen to find the exact
one among the family of wavefunctions you try - usually the best one is
only approximate instead...]
This balance between kinetic energy (positive and increasing with
localization) and potential energy (negative and getting more negative as
localization increases) is indeed a consequence of Heisenberg UP.
You cannot localize an electron too much without it having a very large
kinetic energy.
However, the kinetic energy grows as the inverse size of the wavefunc,
i.e. as b^-2. Therefore, an attractive potential which grows attractive
faster close to r=0 (like -1/r^3, say) would win and suck "down" the
quantum particle and concentrate it to r=0 (with infinitely large negative
total energy).
> Hi Nick,
> I have been through the radial equation in exquisite detail, but I`ve got almost to the end and realised that there`s a piece of the
> jigsaw missing. The fully normalised wavefunction I get is missing a component and I`m having trouble tracing where it`s gone. I have
> solved the radial equation using both of the common methods, the `R` and the `u=rR` methods (and used the same substitutions in both).
> Just concentrating
> on the `R` method, my wavefunction comes out as:
> R(p) = SQRT[(n-l-1)! / 2n.[(n+l)!]^3] exp(-p/2) p^l L, where L is the assoc. poly.
> But the corresponding correct version is:
> R(r) = minus SQRT[(2Z/na)^3 (n-l-1)! / 2n.[(n+l)!]^3] exp(-p/2) p^l L
> so the difference is simply the (2Z/na) where `a` is the Bohr radius, and the minus. But where does this extra bit come from? The
> definitions I used at the start of solving the radial equation must have something to do with it. They were:
> alpha = [-2.mu.E]^1/2
> p = 2.alpha.r / hbar
> lambda = mu.Z.e^2 / 4.pi.epsilon_0 alpha.hbar
> [these seem to be common substitutions in my books]
> Now I guess the root of this problem is converting R(p) to R(r), but how do I do this? It seems like it`s a simple case of just
> inserting in (2Z/na), and that`s that, but it`s not clear to me how this works.
> Then I will have finally finished the radial equation! (apart from dealing with the minus sign!)
The minus sign is a trivial phase factor. You remember you can always
multiply any wavefunc. by exp(i phi), where phi is an arbitrary real
number, and get another equivalent wavefunc. Here you take phi=pi.
For the
SQRT[(2Z/na)^3]
factor the question is slightly more serious. This has to do with
normalization. Take a solution of an eigenvalue equation like that for
R(r): if you multiply it by any factor, you still get a solution, with the
same energy.
To decide which is the right factor (up to a phase!) you have to normalize the
wavefunction. i.e. you have to impose that the integral of the wavefunction
over all space is =1. This is like saying that the probability to find the
particle somewhere has to be unity.
For the radial function you have an additional r^2 factor from the polar
varibles,
so the condition here is integral(from 0 to +infinity) dr r^2 |R(r)|^2 = 1
This fixes the "right" factor. Note that dimensionally psi has to be a
length^(-3/2), so that its square is a probability density...
> Do you know why in hydrogen the S_1/2 level is sandwiched between the P_1/2 and
> P_3/2 levels
> whereas for lithium and the higher alkalis the P levels are at distinctly
> higher energies than the S level.
> Why doesn't hydrogen follow the same pattern as the alkalis?
This level ordering is well known, for example you find it at page 50
of my book.
The reason of the splitting is completely different for H and for the
alkalis:
For H nonrelativistic quantum mechanics (c=infinity) puts all levels with
the same n and different L at the same energy.
The detailed calculation of the correction to leading order in 1/c (i.e. in
fact order alpha^2, where alpha is the fine-structure constant) makes the
energy of the levels depend on the total angular momentum J, besides n. No
dependence on l should occur, thus S_1/2 and P_1/2 should be exactly
degenerate. They are eventually split by extra tiny effects of quantum
electromagnetism, sketched qualitatively at
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/lamb.html
In the alkalis the reason for the S-P splitting is completely different: it
is a big effect which has nothing to do with relativity (it would be there
even if c was infinity). In fact this splitting is at the basis of the
construction of the periodic table, and is one of the most important
concepts in atomic physics.
You may find it sketched at pag. 39 of my notes, in the discussion of
Fig. 1.24
Thanx to David!
Comments and debugging are welcome!
| created: 30 Oct 2001 | last modified: 3 Jun 2018 by Nicola Manini |