solution of hydrogen atom using Schrodinger's equation

[jumanicus]

In the conventional treatment of hydrogen atom using Schrodinger's
equation, we assume a spherically symmetric electric potential with the
source (proton) at the origin of the (spherical polar) coordinate
system. How can this be? Here we consider electron to be represeneted
by a wave function and proton as a particle located exactly at the
origin. Does this not contradict uncertainty principle?

[Chris H. Fleming]

jumanicus wrote:
> In the conventional treatment of hydrogen atom using Schrodinger's
> equation, we assume a spherically symmetric electric potential with the
> source (proton) at the origin of the (spherical polar) coordinate
> system. How can this be? Here we consider electron to be represeneted
> by a wave function and proton as a particle located exactly at the
> origin. Does this not contradict uncertainty principle?

Work it just like the classical version.
Put the center of mass in the origin.

[Jamie]

Yes! But it's okay, because the proton is so much more massive. From
the uncertainty relations,
dp dx ~ h
or,
m dv dx ~ h
and so an increase in mass, accompanied by a decrease in dx, is
allowed.

Of course, this introduces a small amount of additional error into the
calculation, but this is always going to be an approximate calculation
anyway.

[markwh04@yahoo.com]

jumanicus wrote:
> In the conventional treatment of hydrogen atom using Schrodinger's
> equation, we assume a spherically symmetric electric potential with the
> source (proton) at the origin of the (spherical polar) coordinate
> system. How can this be? Here we consider electron to be represeneted
> by a wave function and proton as a particle located exactly at the
> origin. Does this not contradict uncertainty principle?

(They're not spherically symmetric, except for the s states, where the
total angular momentum is 0; more on that later).

There's nothing per se that says that a total system cannot be hybrid
-- some degrees of freedom classical, some quantum. So, it's perfectly
possible to treat an object, like an atomic nucleus that is much more
heavy that the electrons around it, as a classical subsystem with
classical degrees of freedom, and the electrons as quantum. This
amounts to solving the equations of motion for the electron in a
classical or "external" field.

If you want to treat the nucleus as a quantum subsystem, then it will
be incorporated into the full quantum system and the result will be
that you will be solving a 2-body problem, instead of a 1-body problem.
The 2-body problem reduces to a combination of: inertial center of mass
motion + effective 1-body problem for coordinates relative to the
center of mass. Both the inertial motion and center of mass relative
motion are described by dynamical variables that are subject to the
Heisenberg principle.

In fact, you can go much further with this. The equations of motion for
the overall system are exactly the same as they are in the
corresponding classical Kepler problem, and the solution and solution
method carries over unchanged. In particular, the 6 variables
describing the center of mass motion are given by center of mass
position R = (X, Y, Z) and velocity V = (dX/dt, dY/dt, dZ/dt) such that
[X, dX/dt] = [Y, dY/dt] = [Z, dZ/dt] = i h-bar/M, where M is the
system's total mass; with all other commutators being 0. The center of
mass relative motion is described by coordinates (r = (x,y,z), v =
(dx/dt, dy/dt, dz/dt)) also subject to the Heisenberg principle,
[x,dx/dy] = i h-bar/m; etc., where m is the effective mass of the
center of mass relative orbital motion (m = m1 m2/M, where m1, m2 are
the masses of the 2 bodies). Out of these coordinates come the very
same constants of motion that arise in the Kepler problem:
L = m r x v; e = (v x L/m)/K - r/|r|
where ()x() denotes vector product (symmetrized, e.g. (AxB)_x = 1/2
(AyBz+BzAy) - 1/2 (AzBy+ByAz)). The constant K is, in MKS, Ze^2/(4 pi
epsilon_0) = Z alpha h-bar c, where Z is the charge of the nucleus and
alpha the fine structure constant.

The quantity e determines the shape of the orbit, and L its size; L is
also the angular momentum of the orbital system. Each of these take on
diagonal form for each orbital and the orbitals, in fact, are
eigenstates of these operators. I posted the eigenvalues of L and e for
the low lying orbitals a while back; the most interesting case is the
0s orbital, where L = e = 0 (which, if it were classical, would
represent a system where the electron had fallen into the atom's
nucleus).

These quantities have the following as the only non-zero commutators
[L_x, L_y] = i h-bar L_z; [L_x, e_y] = i h-bar e_z
[e_x, e_y] = i h-bar k L_z
The other relations derived by cyclic permutations in (x,y,z). The
constant k depends on the total energy of the system, as given by E =
1/2 mv^2 - K/r. It is positive for bound.

They are sobject to the relation
L.e = L_x e_x + L_y e_y + L_z e_z = 0
(This makes for 5 independent constants of motion. The 6th constant of
motion -- at least for the classical Kepler problem -- is the one that
is explicitly time dependent: it defines the time of closest passage in
the orbit).

You may recognize the commutators for L as describing those of a
typical system with a given angular momentum. A quantized system, if
finite dimensional, may only have the values m h-bar and l(l+1) h-bar^2
for L_z and |L|^2, with l = 0, 1/2, 1, 3/2, 2, ...; and m = -l, 1-l,
2-l, ..., l-2, l-1, l. The system above can be disentangled by adopting
the following:
J+ = (L + sqrt(k) e)/2; J- = (L - sqrt(k)
e)/2
Then each satisfies its own set of analogous relations
{J+x,J+y] = i h-bar J+z; [J-x,J-y] = i h-bar J-z;
etc.
with J+'s components now commuting with J-'s components. The constraint
L.e = 0 becomes |J+|^2 = |J-|^2. So, the only finite dimensional
representations are those where
|J+|^2 = |J-|^2 = (n/2)((n/2)+1) h-bar^2
with eigenvalues J+z = m+, J-z = m-, with both m+ and m- ranging
between -n and n and with n = 0, 1, 2, 3, ...

For n = 0, this leads to the state m+ = m- = 0 ... the 0s state. For n
= 1, you get for (m+,m-) the combinations (-h-bar/2,-h-bar/2),
(-h-bar/2, h-bar/2), (h-bar/2, -h-bar/2), (h-bar/2, h-bar/2). These are
the level 1 states, with m = m+ + m- ranging over the list (-1, 0, 0
and 1). These give you the 1s, 1px, 1py, 1pz states, though I forget in
what combination. For level 2 you get the 9 states associated with the
2nd energy level, etc.

[Boo]

jumanicus wrote:
> In the conventional treatment of hydrogen atom using Schrodinger's
> equation, we assume a spherically symmetric electric potential with the
> source (proton) at the origin of the (spherical polar) coordinate
> system. How can this be? Here we consider electron to be represeneted
> by a wave function and proton as a particle located exactly at the
> origin. Does this not contradict uncertainty principle?

The treatment in elementary QM texts obviously does contradict the uncertainty
principle to an extent, but it's really just an approximation : the hydrogen
nucleus is nearly 2000 time more massive than the electron so the uncertainty in
its location is proportionately smaller c.f. the electrons "orbit" and is just
ignored.

I've never seen an analytical treatment that treats the hydrogen atom as
anything other than a single electron in a coulomb potential, maybe someone
else here can point to a more realistic treatment ?

--
Boo

[paulaireilly]

> Here we consider electron to be represeneted
> by a wave function and proton as a particle located exactly at the
> origin. Does this not contradict uncertainty principle?

The proton's mass is about 1830 times the mass of the electron, so its
wavelength for a given momentum is about 1830 times smaller* than an
electrons. Thus we can construct a wave packet for the proton that is
so small compared to the electron's that it might as well be a point -
the error introduced by this approximation is neglible for hydrogen
(and indeed for all outer-shell electrons of atoms and molecules).
Inner shell electrons of massive atoms with high Z have small but
non-neglible corrections from this, I believe.

-pair

* In fact the extent of the nuclear wave packet (with edges defined in
some reasonable way, e.g, 99.99 % of the probability weight is inside
the boundary) is more than 1830 times smaller than the electron's wave
packet, it's about 100,000 times smaller. I believe this is due to the
stronger nuclear force. Note that this is *linear* extent. For
purposes of calculating the energy eigenvalues of the electron states
and so on, the volume ratio, about 10^15 (!) may be a better measure of
how neglible the nuclear size is.

[Cl.Massé]

"jumanicus" <vishwesh.muzumdar@rediffmail.com> a écrit dans le message de
news: 1147238944.041918.17830@g10g2000cwb.googlegroups.c om

> In the conventional treatment of hydrogen atom using Schrodinger's
> equation, we assume a spherically symmetric electric potential with the
> source (proton) at the origin of the (spherical polar) coordinate
> system. How can this be? Here we consider electron to be represeneted
> by a wave function and proton as a particle located exactly at the
> origin. Does this not contradict uncertainty principle?

The mass of the proton is 2000 times greater than the one of the electron,
the uncertainty on its position can then be neglected.

Anyway, the motion that is described is the one of a reduced mass around the
centre of mass.

--
~~~~ clmasse on free F-country
Liberty, Equality, Profitability.

[Arkadiusz Jadczyk]

On Thu, 11 May 2006 20:38:06 +0000 (UTC), "jumanicus"
<vishwesh.muzumdar@rediffmail.com> wrote:

>In the conventional treatment of hydrogen atom using Schrodinger's
>equation, we assume a spherically symmetric electric potential with the
>source (proton) at the origin of the (spherical polar) coordinate
>system. How can this be? Here we consider electron to be represeneted
>by a wave function and proton as a particle located exactly at the
>origin. Does this not contradict uncertainty principle?

Uncertainty principle, if you study its derivation, in details deals
with the products of dispersions of probability distributions that are
somewhat fuzzily related to the "measurement process". It does not
apply to the parameters used in our construction of Hamiltonians.

--

Arkadiusz Jadczyk
http://quantumfuture.net/quantum_future/jadpub.htm
--

[markwh04@yahoo.com]

markwh04@yahoo.com wrote:
> jumanicus wrote:
> (They're not spherically symmetric, except for the s states, where the
> total angular momentum is 0; more on that later).
>
> The quantity e determines the shape of the orbit, and L its size; L is
> also the angular momentum of the orbital system. Each of these take on
> diagonal form for each orbital and the orbitals, in fact, are
> eigenstates of these operators. I posted the eigenvalues of L and e for
> the low lying orbitals a while back; the most interesting case is the
> 0s orbital, where L = e = 0 (which, if it were classical, would
> represent a system where the electron had fallen into the atom's
> nucleus).

The article is in the thread, dated 2005 April 16:

http://groups.google.com/group/sci.physics.research/browse_thread/thread/e164664112229014/24f8d5a6e6c72dd9?q=Orbits+SO(4)&rnum=4#24f8d5a6e6c72dd9

The 2s orbital is an elliptical orbit with eccentricity e = 1/2; the 2p
orbitals have eccentricities of sqrt(3)/2 (for 3s, 3p and 3d, e is
respectively sqrt(2)/3, sqrt(6)/3 and sqrt(8)/3).

The commutators are:
[L^a,L^b] = i h-bar e^{ab}_c L^c
[L^a,e^b] = i h-bar e^{ab}_c e^c
[e^a,L^b] = i h-bar e^{ab}_c e^c
[e^a,e^b] = i h-bar e^{ab}_c (-2H/(K^2 m)) L^c
where
H = p^2/2m - K/r
is the Hamiltonian.

For |e| < 1, |e| = 1, |e| > 1 (respectively, elliptical, parabolic,
hyperbolic orbits), one has H < 0, H = 0, H > 0, thus giving you SO(4)
for elliptical orbits, Galilei for parabolic orbits and SL(2,C) ( =
Lorentz) for hyperbolic orbits. In all cases, there is the constraint
(L.e = e.L = 0).

The components e_z, L_z, and the squares e^2 and L^2 = h-bar^2 l(l+1)
are diagonal in the orbital states, as is |J+|^2 = |J-|^2 = h-bar^2
s(s+1).

For low-lying values of s, l, this results in the following:
s l e states & degeneracy
0 0 0 1s, 1
1/2 0 sqrt(3)/2 2p, 3
1/2 1 1/2 2s, 1
1 0 sqrt(8)/3 3d, 5
1 1 sqrt(6)/3 3p, 3
1 2 sqrt(2)/3 3s, 1
3/2 0 sqrt(15)/4 4f, 7
3/2 1 sqrt(13)/4 4d, 5
3/2 2 3/4 4p, 3
3/2 3 sqrt(3)/4 4s, 1
2 0 sqrt(24)/5 5g, 9
2 1 sqrt(22)/5 5f, 7
2 2 sqrt(18)/5 5d, 5
2 3 sqrt(12)/5 5p, 3
2 4 2/5 5s, 1

For the 2nd energy level (s = 1/2), one gets the decomposition into
(J+_z, J-_z) eigenstates:
2p+ = |++>
2p0 = (|+-> + |-+>)/sqrt(2)
2p- = |-->
and
2s = (|+-> - |-+>)/sqrt(2).

[Cl.Massé]

"jumanicus" <vishwesh.muzumdar@rediffmail.com> a écrit dans le message de
news: 1147238944.041918.17830@g10g2000cwb.googlegroups.c om

> In the conventional treatment of hydrogen atom using Schrodinger's
> equation, we assume a spherically symmetric electric potential with the
> source (proton) at the origin of the (spherical polar) coordinate
> system. How can this be? Here we consider electron to be represeneted
> by a wave function and proton as a particle located exactly at the
> origin. Does this not contradict uncertainty principle?

Other answer: as the probability of presence of the electron in the region
of the proton, above all for non s states, is very small, whether the field
source has a spatial extension or not doesn't change the potential.

--
~~~~ clmasse on free F-country
Liberty, Equality, Profitability.

[David M. Palmer]

In article <1147238944.041918.17830@g10g2000cwb.googlegroups.c om>,
jumanicus <vishwesh.muzumdar@rediffmail.com> wrote:

> In the conventional treatment of hydrogen atom using Schrodinger's
> equation, we assume a spherically symmetric electric potential with the
> source (proton) at the origin of the (spherical polar) coordinate
> system. How can this be? Here we consider electron to be represeneted
> by a wave function and proton as a particle located exactly at the
> origin. Does this not contradict uncertainty principle?

Conventionally, you use the 'reduced mass' method, where you separate
the problem by writing the equation in terms of a) the center of mass
and b) the separation between the electron and the proton.

You place the center of mass at the origin and solve for the wave
function of the separation. (Which is, to within 0.05% in the spatial
dimensions, the wave function of the electron.)

You can independently solve for the wave function of the center of
mass. Since energy is independent of center-of-mass location (for an
isolated H atom) this part of the solution drops out of the
calculations of most of the things you care about.

--
David M. Palmer dmpalmer@email.com (formerly @clark.net, @ematic.com)

[paulaireilly]

Also, even if the proton were only as massive as the electron, the
potential would still be of the same form but with a reduced mass. In
fact, positronium, made of an electron and a positron, has energy
levels analogous to hydrogen's, up to the effect of reduced mass and
magnetic energy level splits. It's surprisingly stable, with a
lifetime on the order of 10^-7 s, I believe - millions of 'orbits', so
it has a recognizable spectrum, hardly smeared at all.

[Bongi]

Can you please ouline for me the use of spherical polar coordinates when resolving the Schrodinger equation for hydrogen atom