View Full Version : Schroedinger equation in "wrong" coordinates
David Macmanus
Apr22-04, 02:49 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Solving the Schroedinger equation for the hydrogen atom is relatively\nsimple after the early step of translating the equation from\nCartesiancoordinates to spherical coordinates which then allows a\nstraightforward separation of variables to give three simple (ish)\nequations. But what I can`t seem to be able to find out is whether it is\nactually possible to solve the equation using Cartesian\ncoordinates i.e. the x, y and z rather than the r, theta, phi. Now, I\nappreciate that it might be madness to attempt such a feat when it is\nrelatively simple to go over to spherical coordinates, but I`d really\nlike to know if it is *theoretically possible* to solve it in Cartesian\ncoordinates. Can anyone say definitively whether it be done (if you were\na maths genius, for example)?\nThanks,\nDavid.\n\n\n--\nPosted via Mailgate.ORG Server - http://www.Mailgate.ORG\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Solving the Schroedinger equation for the hydrogen atom is relatively
simple after the early step of translating the equation from
Cartesiancoordinates to spherical coordinates which then allows a
straightforward separation of variables to give three simple (ish)
equations. But what I can`t seem to be able to find out is whether it is
actually possible to solve the equation using Cartesian
coordinates i.e. the x, y and z rather than the r, \theta, \phi. Now, I
appreciate that it might be madness to attempt such a feat when it is
relatively simple to go over to spherical coordinates, but I`d really
like to know if it is *theoretically possible* to solve it in Cartesian
coordinates. Can anyone say definitively whether it be done (if you were
a maths genius, for example)?
Thanks,
David.
--
Posted via Mailgate.ORG Server - http://www.Mailgate.ORG
Dirk Van de moortel
Apr23-04, 03:13 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n"David Macmanus" <macmanus@tripos.com> wrote in message news:33f0323d9493528e98f1d9ff3164b1d3_35661@mygate .mailgate.org...\n> Solving the Schroedinger equation for the hydrogen atom is relatively\n> simple after the early step of translating the equation from\n> Cartesiancoordinates to spherical coordinates which then allows a\n> straightforward separation of variables to give three simple (ish)\n> equations. But what I can`t seem to be able to find out is whether it is\n> actually possible to solve the equation using Cartesian\n> coordinates i.e. the x, y and z rather than the r, theta, phi. Now, I\n> appreciate that it might be madness to attempt such a feat when it is\n> relatively simple to go over to spherical coordinates, but I`d really\n> like to know if it is *theoretically possible* to solve it in Cartesian\n> coordinates. Can anyone say definitively whether it be done (if you were\n> a maths genius, for example)?\n> Thanks,\n> David.\n\nSee http://www.missioncollege.org/depts/physics/P4poe/P4D/Hydrogen.htm\nand probably a lot more on\nhttp://www.google.com/search?q=solving+schroedinger+equation+cartesian+c oordinates\n\nEnjoy,\nDirk Vdm\n[identically but separately replied in unmoderated group sci.math]\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>"David Macmanus" <macmanus@tripos.com> wrote in message news:33f0323d9493528e98f1d9ff3164b1d3_35661@mygate .mailgate.org...
> Solving the Schroedinger equation for the hydrogen atom is relatively
> simple after the early step of translating the equation from
> Cartesiancoordinates to spherical coordinates which then allows a
> straightforward separation of variables to give three simple (ish)
> equations. But what I can`t seem to be able to find out is whether it is
> actually possible to solve the equation using Cartesian
> coordinates i.e. the x, y and z rather than the r, \theta, \phi. Now, I
> appreciate that it might be madness to attempt such a feat when it is
> relatively simple to go over to spherical coordinates, but I`d really
> like to know if it is *theoretically possible* to solve it in Cartesian
> coordinates. Can anyone say definitively whether it be done (if you were
> a maths genius, for example)?
> Thanks,
> David.
See http://www.missioncollege.org/depts/physics/P4poe/P4D/Hydrogen.htm
and probably a lot more on
http://www.google.com/search?q=solving+schroedinger+equation+cartesian+c oordinates
Enjoy,
Dirk Vdm
[identically but separately replied in unmoderated group sci.math]
Hendrik van Hees
Apr23-04, 03:14 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>David Macmanus wrote:\n\n> Solving the Schroedinger equation for the hydrogen atom is relatively\n> simple after the early step of translating the equation from\n> Cartesiancoordinates to spherical coordinates which then allows a\n> straightforward separation of variables to give three simple (ish)\n> equations. But what I can`t seem to be able to find out is whether it\n> is actually possible to solve the equation using Cartesian\n> coordinates i.e. the x, y and z rather than the r, theta, phi. Now, I\n> appreciate that it might be madness to attempt such a feat when it is\n> relatively simple to go over to spherical coordinates, but I`d really\n> like to know if it is *theoretically possible* to solve it in\n> Cartesian coordinates. Can anyone say definitively whether it be done\n> (if you were a maths genius, for example)?\n\nYou can solve the hydrogen problem without any introduction of\ncoordinates from the algebra of operators alone. In fact that was the\nfirst way how this was done, after Heisenberg (during a stay on\nHelgoland where he tried to get rid of a bad attac of hay fever) had\ninvented modern quantum mechanics in 1925 in its form as matrix\nmechanics. He could not figure out the hydrogen, so he concentrated on\nthe harmonic oscillator with anharmonic perturbations.\n\nIt was in fact Pauli, who solved the hydrogen problem within matrix\nmechanics in a completely algebraic way. The trick is to use the\ndynamical symmetry of the (non-relativistic) Kepler problem, which\ngives rise to a special additional conserved vector, the socalled\nRunge-Lenz Vector.\n\n--\nHendrik van Hees Cyclotron Institute\nPhone: +1 979/845-1411 Texas A&M University\nFax: +1 979/845-1899 Cyclotron Institute, MS-3366\nhttp://theory.gsi.de/~vanhees/ College Station, TX 77843-3366\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>David Macmanus wrote:
> Solving the Schroedinger equation for the hydrogen atom is relatively
> simple after the early step of translating the equation from
> Cartesiancoordinates to spherical coordinates which then allows a
> straightforward separation of variables to give three simple (ish)
> equations. But what I can`t seem to be able to find out is whether it
> is actually possible to solve the equation using Cartesian
> coordinates i.e. the x, y and z rather than the r, \theta, \phi. Now, I
> appreciate that it might be madness to attempt such a feat when it is
> relatively simple to go over to spherical coordinates, but I`d really
> like to know if it is *theoretically possible* to solve it in
> Cartesian coordinates. Can anyone say definitively whether it be done
> (if you were a maths genius, for example)?
You can solve the hydrogen problem without any introduction of
coordinates from the algebra of operators alone. In fact that was the
first way how this was done, after Heisenberg (during a stay on
Helgoland where he tried to get rid of a bad attac of hay fever) had
invented modern quantum mechanics in 1925 in its form as matrix
mechanics. He could not figure out the hydrogen, so he concentrated on
the harmonic oscillator with anharmonic perturbations.
It was in fact Pauli, who solved the hydrogen problem within matrix
mechanics in a completely algebraic way. The trick is to use the
dynamical symmetry of the (non-relativistic) Kepler problem, which
gives rise to a special additional conserved vector, the socalled
Runge-Lenz Vector.
--
Hendrik van Hees Cyclotron Institute
Phone: +1 979/845-1411 Texas A&M University
Fax: +1 979/845-1899 Cyclotron Institute, MS-3366
http://theory.gsi.de/~vanhees/ College Station, TX 77843-3366
Arnold Neumaier
Apr24-04, 11:17 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>David Macmanus wrote:\n> Solving the Schroedinger equation for the hydrogen atom is relatively\n> simple after the early step of translating the equation from\n> Cartesiancoordinates to spherical coordinates which then allows a\n> straightforward separation of variables to give three simple (ish)\n> equations. But what I can`t seem to be able to find out is whether it is\n> actually possible to solve the equation using Cartesian\n> coordinates i.e. the x, y and z rather than the r, theta, phi. Now, I\n> appreciate that it might be madness to attempt such a feat when it is\n> relatively simple to go over to spherical coordinates, but I`d really\n> like to know if it is *theoretically possible* to solve it in Cartesian\n> coordinates. Can anyone say definitively whether it be done (if you were\n> a maths genius, for example)?\n\nTake the solution in spherical coordiantes and undo the substitution and you\nget the solution in Cartesian coordinates. So it is possible but foolish.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>David Macmanus wrote:
> Solving the Schroedinger equation for the hydrogen atom is relatively
> simple after the early step of translating the equation from
> Cartesiancoordinates to spherical coordinates which then allows a
> straightforward separation of variables to give three simple (ish)
> equations. But what I can`t seem to be able to find out is whether it is
> actually possible to solve the equation using Cartesian
> coordinates i.e. the x, y and z rather than the r, \theta, \phi. Now, I
> appreciate that it might be madness to attempt such a feat when it is
> relatively simple to go over to spherical coordinates, but I`d really
> like to know if it is *theoretically possible* to solve it in Cartesian
> coordinates. Can anyone say definitively whether it be done (if you were
> a maths genius, for example)?
Take the solution in spherical coordiantes and undo the substitution and you
get the solution in Cartesian coordinates. So it is possible but foolish.
Arnold Neumaier
Arnie King
Apr27-04, 03:52 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\nMy understanding is that Schroedinger\'s equation with the 1/r\npotential is not separable in cartesian coordinates.\n\nFor it to be separable, the potential would have to be a sum of three\nfunctions each depending on only one of x, y, z. (This is the case\nwith the isotropic harmonic potential, V(r)=-kr^2, but -1/r cannot be\nso written.)\n\nThe hydrogen equation is however separable in parabolic cylinder\ncoordinates. I think Merzbacher\'s book discusses it. The separability\nin more than one system has something to do with the "extra" constant\nof motion (the Runge vector mentioned in a previous post), and with\nthe famous degeneracy in hydrogen that allows states of different l\n(angular momentum quantum number) to have the same energy.\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>My understanding is that Schroedinger's equation with the 1/r
potential is not separable in cartesian coordinates.
For it to be separable, the potential would have to be a sum of three
functions each depending on only one of x, y, z. (This is the case
with the isotropic harmonic potential, V(r)=-kr^2, but -1/r cannot be
so written.)
The hydrogen equation is however separable in parabolic cylinder
coordinates. I think Merzbacher's book discusses it. The separability
in more than one system has something to do with the "extra" constant
of motion (the Runge vector mentioned in a previous post), and with
the famous degeneracy in hydrogen that allows states of different l
(angular momentum quantum number) to have the same energy.
Danny Ross Lunsford
Apr27-04, 01:41 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Hendrik van Hees wrote:\n\n> It was in fact Pauli, who solved the hydrogen problem within matrix\n> mechanics in a completely algebraic way. The trick is to use the\n> dynamical symmetry of the (non-relativistic) Kepler problem, which\n> gives rise to a special additional conserved vector, the socalled\n> Runge-Lenz Vector.\n\nYes, one of the most beautiful papers ever written:\n\nZeitschrift fur Physik 36 (1926) p.336.\n\nThere is a translation into English in the book\n\n"Sources of Quantum Mechnics", B.L. van der Waerden, editor (Dover 1968).\n\nThe Lenz vector may be visualized as a dimensionless vector along the\nmajor axis of the orbit with direction away from the (fixed) force\ncenter, and length equal to eccentricity of the orbit.\n\nThis paper had an effect on physicists equal to that of Einstein\'s\npapers on gravity. Pauli first sent the calculation to Heisenberg, who\nreplied "I need not write how overjoyed I was with the theory of the\nhydrogen atom..." The general acceptance of quantum mechanics begins\nwith this paper.\n\nOne cannot overstate the calculational mastery of Pauli.\n\n-drl\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Hendrik van Hees wrote:
> It was in fact Pauli, who solved the hydrogen problem within matrix
> mechanics in a completely algebraic way. The trick is to use the
> dynamical symmetry of the (non-relativistic) Kepler problem, which
> gives rise to a special additional conserved vector, the socalled
> Runge-Lenz Vector.
Yes, one of the most beautiful papers ever written:
Zeitschrift fur Physik 36 (1926) p.336.
There is a translation into English in the book
"Sources of Quantum Mechnics", B.L. van der Waerden, editor (Dover 1968).
The Lenz vector may be visualized as a dimensionless vector along the
major axis of the orbit with direction away from the (fixed) force
center, and length equal to eccentricity of the orbit.
This paper had an effect on physicists equal to that of Einstein's
papers on gravity. Pauli first sent the calculation to Heisenberg, who
replied "I need not write how overjoyed I was with the theory of the
hydrogen atom..." The general acceptance of quantum mechanics begins
with this paper.
One cannot overstate the calculational mastery of Pauli.
-drl
Gentil Correa
Apr27-04, 01:44 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>"David Macmanus" <macmanus@tripos.com> wrote in message news:<33f0323d9493528e98f1d9ff3164b1d3_35661@mygat e.mailgate.org>...\n> Solving the Schroedinger equation for the hydrogen atom is relatively\n> simple after the early step of translating the equation from\n> Cartesiancoordinates to spherical coordinates which then allows a\n> straightforward separation of variables to give three simple (ish)\n> equations. But what I can`t seem to be able to find out is whether it is\n> actually possible to solve the equation using Cartesian\n> coordinates i.e. the x, y and z rather than the r, theta, phi. Now, I\n> appreciate that it might be madness to attempt such a feat when it is\n> relatively simple to go over to spherical coordinates, but I`d really\n> like to know if it is *theoretically possible* to solve it in Cartesian\n> coordinates. Can anyone say definitively whether it be done (if you were\n> a maths genius, for example)?\n> Thanks,\n> David.\n\nHi!\n\nChanging coordinates is a "change of language", so there must be\nan unitary operator which performs the trick. If you care to find this\noperator, then, starting from the solution in spherical coordinates,\nyou end up with the cartesian solution. Now, if what you want is to arrive\nat the cartesian solution without using other coordinates even at intermediate\nsteps, this seems to me rather uninteresting. It is like trying to solve\na differential equation forbidding oneself to change variables. However,\nit must be possible to do it ("if I were a maths genius").\n\nBest wishes,\nG.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>"David Macmanus" <macmanus@tripos.com> wrote in message news:<33f0323d9493528e98f1d9ff3164b1d3_35661@mygate.mail gate.org>...
> Solving the Schroedinger equation for the hydrogen atom is relatively
> simple after the early step of translating the equation from
> Cartesiancoordinates to spherical coordinates which then allows a
> straightforward separation of variables to give three simple (ish)
> equations. But what I can`t seem to be able to find out is whether it is
> actually possible to solve the equation using Cartesian
> coordinates i.e. the x, y and z rather than the r, \theta, \phi. Now, I
> appreciate that it might be madness to attempt such a feat when it is
> relatively simple to go over to spherical coordinates, but I`d really
> like to know if it is *theoretically possible* to solve it in Cartesian
> coordinates. Can anyone say definitively whether it be done (if you were
> a maths genius, for example)?
> Thanks,
> David.
Hi!
Changing coordinates is a "change of language", so there must be
an unitary operator which performs the trick. If you care to find this
operator, then, starting from the solution in spherical coordinates,
you end up with the cartesian solution. Now, if what you want is to arrive
at the cartesian solution without using other coordinates even at intermediate
steps, this seems to me rather uninteresting. It is like trying to solve
a differential equation forbidding oneself to change variables. However,
it must be possible to do it ("if I were a maths genius").
Best wishes,
G.
John T Lowry
Apr27-04, 01:48 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>A reference for the matrix method of solution is H.S. Green\'s Matrix\nMechanics, pp. 88-90.\n\nI suppose David meant analytic solution, not just solution, since "most"\ndifferential equations, under reasonable boundary or initial conditions,\nhave solutions. My (incomplete) understanding is that changing variables\n(simpler case going from two variables x, y to one v=y/x) may make an\nintractable (analytically!) d.e. solvable.\n\nJohn\n--\nJohn T Lowry, PhD\nFlight Physics\n5217 Old Spicewood Springs Rd, #312\nAustin, Texas 78731\n(512) 231-9391\njlowry100@earthlink.net\n"Hendrik van Hees" <hees@comp.tamu.edu> wrote in message\nnews:c69emj\\$9gf58\\$2@ID-71437.news.uni-berlin.de...\n> David Macmanus wrote:\n>\n> > Solving the Schroedinger equation for the hydrogen atom is relatively\n> > simple after the early step of translating the equation from\n> > Cartesiancoordinates to spherical coordinates which then allows a\n> > straightforward separation of variables to give three simple (ish)\n> > equations. But what I can`t seem to be able to find out is whether it\n> > is actually possible to solve the equation using Cartesian\n> > coordinates i.e. the x, y and z rather than the r, theta, phi. Now, I\n> > appreciate that it might be madness to attempt such a feat when it is\n> > relatively simple to go over to spherical coordinates, but I`d really\n> > like to know if it is *theoretically possible* to solve it in\n> > Cartesian coordinates. Can anyone say definitively whether it be done\n> > (if you were a maths genius, for example)?\n>\n> You can solve the hydrogen problem without any introduction of\n> coordinates from the algebra of operators alone. In fact that was the\n> first way how this was done, after Heisenberg (during a stay on\n> Helgoland where he tried to get rid of a bad attac of hay fever) had\n> invented modern quantum mechanics in 1925 in its form as matrix\n> mechanics. He could not figure out the hydrogen, so he concentrated on\n> the harmonic oscillator with anharmonic perturbations.\n>\n> It was in fact Pauli, who solved the hydrogen problem within matrix\n> mechanics in a completely algebraic way. The trick is to use the\n> dynamical symmetry of the (non-relativistic) Kepler problem, which\n> gives rise to a special additional conserved vector, the socalled\n> Runge-Lenz Vector.\n>\n> --\n> Hendrik van Hees Cyclotron Institute\n> Phone: +1 979/845-1411 Texas A&M University\n> Fax: +1 979/845-1899 Cyclotron Institute, MS-3366\n> http://theory.gsi.de/~vanhees/ College Station, TX 77843-3366\n>\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>A reference for the matrix method of solution is H.S. Green's Matrix
Mechanics, pp. 88-90.
I suppose David meant analytic solution, not just solution, since "most"
differential equations, under reasonable boundary or initial conditions,
have solutions. My (incomplete) understanding is that changing variables
(simpler case going from two variables x, y to one v=y/x) may make an
intractable (analytically!) d.e. solvable.
John
--
John T Lowry, PhD
Flight Physics
5217 Old Spicewood Springs Rd, #312
Austin, Texas 78731
(512) 231-9391
jlowry100@earthlink.net
"Hendrik van Hees" <hees@comp.tamu.edu> wrote in message
news:c69emj$9gf58$2@ID-71437.news.uni-berlin.de...
> David Macmanus wrote:
>
> > Solving the Schroedinger equation for the hydrogen atom is relatively
> > simple after the early step of translating the equation from
> > Cartesiancoordinates to spherical coordinates which then allows a
> > straightforward separation of variables to give three simple (ish)
> > equations. But what I can`t seem to be able to find out is whether it
> > is actually possible to solve the equation using Cartesian
> > coordinates i.e. the x, y and z rather than the r, \theta, \phi. Now, I
> > appreciate that it might be madness to attempt such a feat when it is
> > relatively simple to go over to spherical coordinates, but I`d really
> > like to know if it is *theoretically possible* to solve it in
> > Cartesian coordinates. Can anyone say definitively whether it be done
> > (if you were a maths genius, for example)?
>
> You can solve the hydrogen problem without any introduction of
> coordinates from the algebra of operators alone. In fact that was the
> first way how this was done, after Heisenberg (during a stay on
> Helgoland where he tried to get rid of a bad attac of hay fever) had
> invented modern quantum mechanics in 1925 in its form as matrix
> mechanics. He could not figure out the hydrogen, so he concentrated on
> the harmonic oscillator with anharmonic perturbations.
>
> It was in fact Pauli, who solved the hydrogen problem within matrix
> mechanics in a completely algebraic way. The trick is to use the
> dynamical symmetry of the (non-relativistic) Kepler problem, which
> gives rise to a special additional conserved vector, the socalled
> Runge-Lenz Vector.
>
> --
> Hendrik van Hees Cyclotron Institute
> Phone: +1 979/845-1411 Texas A&M University
> Fax: +1 979/845-1899 Cyclotron Institute, MS-3366
> http://theory.gsi.de/~vanhees/ College Station, TX 77843-3366
>
Roland Franzius
Apr27-04, 02:06 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n\n> David Macmanus wrote:\n>\n>>Solving the Schroedinger equation for the hydrogen atom is relatively\n>>simple after the early step of translating the equation from\n>>Cartesiancoordinates to spherical coordinates which then allows a\n>>straightforward separation of variables to give three simple (ish)\n>>equations. But what I can`t seem to be able to find out is whether it is\n>>actually possible to solve the equation using Cartesian\n>>coordinates i.e. the x, y and z rather than the r, theta, phi. Now, I\n>>appreciate that it might be madness to attempt such a feat when it is\n>>relatively simple to go over to spherical coordinates, but I`d really\n>>like to know if it is *theoretically possible* to solve it in Cartesian\n>>coordinates. Can anyone say definitively whether it be done (if you were\n>>a maths genius, for example)?\n>\n>\n> Take the solution in spherical coordiantes and undo the substitution and you\n> get the solution in Cartesian coordinates. So it is possible but foolish.\n\nNot at all imho.The introduction of the cartesian representation of the\nrotation group in reduced homogeneous polynomials of (x,y,z)/r makes\nthings much clearer and avoids 100 pages study of associate Legendre\nfunctions. Since the underlying symmetry group of H is an energy-scaled\nversion of O(4) one certainly should avoid obscuring special coordinates\nlike (r, theta, phi). The reason is that no procedure in solving the\nclassical Schrödinger H-problem has any relevant extension to more\nrealistic situations afais.\n\n\n\n--\n\nRoland Franzius\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> David Macmanus wrote:
>
>>Solving the Schroedinger equation for the hydrogen atom is relatively
>>simple after the early step of translating the equation from
>>Cartesiancoordinates to spherical coordinates which then allows a
>>straightforward separation of variables to give three simple (ish)
>>equations. But what I can`t seem to be able to find out is whether it is
>>actually possible to solve the equation using Cartesian
>>coordinates i.e. the x, y and z rather than the r, \theta, \phi. Now, I
>>appreciate that it might be madness to attempt such a feat when it is
>>relatively simple to go over to spherical coordinates, but I`d really
>>like to know if it is *theoretically possible* to solve it in Cartesian
>>coordinates. Can anyone say definitively whether it be done (if you were
>>a maths genius, for example)?
>
>
> Take the solution in spherical coordiantes and undo the substitution and you
> get the solution in Cartesian coordinates. So it is possible but foolish.
Not at all imho.The introduction of the cartesian representation of the
rotation group in reduced homogeneous polynomials of (x,y,z)/r makes
things much clearer and avoids 100 pages study of associate Legendre
functions. Since the underlying symmetry group of H is an energy-scaled
version of O(4) one certainly should avoid obscuring special coordinates
like (r, \theta, \phi). The reason is that no procedure in solving the
classical Schrödinger H-problem has any relevant extension to more
realistic situations afais.
--
Roland Franzius
Arnold Neumaier
Apr28-04, 02:27 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Roland Franzius wrote:\n> Arnold Neumaier wrote:\n>\n>> David Macmanus wrote:\n>>\n>>> Solving the Schroedinger equation for the hydrogen atom is relatively\n>>> simple after the early step of translating the equation from\n>>> Cartesiancoordinates to spherical coordinates which then allows a\n>>> straightforward separation of variables to give three simple (ish)\n>>> equations. But what I can`t seem to be able to find out is whether it is\n>>> actually possible to solve the equation using Cartesian\n>>> coordinates i.e. the x, y and z rather than the r, theta, phi. Now, I\n>>> appreciate that it might be madness to attempt such a feat when it is\n>>> relatively simple to go over to spherical coordinates, but I`d really\n>>> like to know if it is *theoretically possible* to solve it in Cartesian\n>>> coordinates. Can anyone say definitively whether it be done (if you were\n>>> a maths genius, for example)?\n>>\n>>\n>>\n>> Take the solution in spherical coordiantes and undo the substitution\n>> and you\n>> get the solution in Cartesian coordinates. So it is possible but foolish.\n>\n>\n> Not at all imho.The introduction of the cartesian representation of the\n> rotation group in reduced homogeneous polynomials of (x,y,z)/r makes\n> things much clearer and avoids 100 pages study of associate Legendre\n> functions. Since the underlying symmetry group of H is an energy-scaled\n> version of O(4) one certainly should avoid obscuring special coordinates\n> like (r, theta, phi). The reason is that no procedure in solving the\n> classical Schrödinger H-problem has any relevant extension to more\n> realistic situations afais.\n\nRealistic cases such as chemical reactions of small molecules\nare almost always done in suitable angle coordinates.\nOnly for H itself is the O(4) symmetry relevant.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Roland Franzius wrote:
> Arnold Neumaier wrote:
>
>> David Macmanus wrote:
>>
>>> Solving the Schroedinger equation for the hydrogen atom is relatively
>>> simple after the early step of translating the equation from
>>> Cartesiancoordinates to spherical coordinates which then allows a
>>> straightforward separation of variables to give three simple (ish)
>>> equations. But what I can`t seem to be able to find out is whether it is
>>> actually possible to solve the equation using Cartesian
>>> coordinates i.e. the x, y and z rather than the r, \theta, \phi. Now, I
>>> appreciate that it might be madness to attempt such a feat when it is
>>> relatively simple to go over to spherical coordinates, but I`d really
>>> like to know if it is *theoretically possible* to solve it in Cartesian
>>> coordinates. Can anyone say definitively whether it be done (if you were
>>> a maths genius, for example)?
>>
>>
>>
>> Take the solution in spherical coordiantes and undo the substitution
>> and you
>> get the solution in Cartesian coordinates. So it is possible but foolish.
>
>
> Not at all imho.The introduction of the cartesian representation of the
> rotation group in reduced homogeneous polynomials of (x,y,z)/r makes
> things much clearer and avoids 100 pages study of associate Legendre
> functions. Since the underlying symmetry group of H is an energy-scaled
> version of O(4) one certainly should avoid obscuring special coordinates
> like (r, \theta, \phi). The reason is that no procedure in solving the
> classical Schrödinger H-problem has any relevant extension to more
> realistic situations afais.
Realistic cases such as chemical reactions of small molecules
are almost always done in suitable angle coordinates.
Only for H itself is the O(4) symmetry relevant.
Arnold Neumaier
M Khomo
Apr29-04, 05:37 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\nGentil Correa wrote:\n> "David Macmanus" <macmanus@tripos.com> wrote in message news:<33f0323d9493528e98f1d9ff3164b1d3_35661@mygat e.mailgate.org>...\n>\n>>Solving the Schroedinger equation for the hydrogen atom is relatively\n>>simple after the early step of translating the equation from\n>>Cartesiancoordinates to spherical coordinates which then allows a\n>>straightforward separation of variables to give three simple (ish)\n>>equations. But what I can`t seem to be able to find out is whether it is\n>>actually possible to solve the equation using Cartesian\n>>coordinates i.e. the x, y and z rather than the r, theta, phi. Now, I\n>>appreciate that it might be madness to attempt such a feat when it is\n>>relatively simple to go over to spherical coordinates, but I`d really\n>>like to know if it is *theoretically possible* to solve it in Cartesian\n>>coordinates. Can anyone say definitively whether it be done (if you were\n>>a maths genius, for example)?\n>>Thanks,\n>>David.\n>\n>\n> Hi!\n>\n> Changing coordinates is a "change of language", so there must be\n> an unitary operator which performs the trick. If you care to find this\n> operator, then, starting from the solution in spherical coordinates,\n> you end up with the cartesian solution. Now, if what you want is to arrive\n> at the cartesian solution without using other coordinates even at intermediate\n> steps, this seems to me rather uninteresting. It is like trying to solve\n> a differential equation forbidding oneself to change variables. However,\n> it must be possible to do it ("if I were a maths genius").\n>\n> Best wishes,\n> G.\n>\nI think you\'re closest to my understanding of the \'difficulty\'\nsatisfying the question directly. Whenever we \'solve\' an equation, we\'re\nusing Arithmetic. When we do so with differential equations, we do so\nalong what they call Exact sequences, wherein the diferrential (or\nboundary) operator takes us from one integrable (flat) space to the\nnext. That is a geometrical feature of the solution to problem, which we\ncannot really wish away without also losing the ability to \'solve\'.\n\nIf one wants a \'change of coordinates\' to \'Cartesian\' just for the heck\nof it, I think there is an isomorphic transformation associated with\nRiemman, but I think Riemman attributed to Gauss: The stereographic\nprojection. So the spherical problem can be turned into a \'Cartesian\none\' on the stereographic projection plane (say the equatorial plane).\n\nProblem is, the projection pole maps to infinity, so the minimal\ncovering for it is two projections, say one rooted at the north pole\nexcluding the north pole, and another rooted in the south pole,\nexcluding the south pole, and with yet another change of coordinates to\nsmoothly switch projections at their selected \'overlap\' at say the\nequatorial plane. So given that atlas covering of the sphere, yes, there\nis a \'Cartesian\' solution, but then you have to respect the Spherical\ngeometry and thereby only deal with a Complex field. I do not know if\nthis still meets the OP\'s requirements; but at last reading, the\nprojection from spherical vector to \'Cartesian\' used the transformation:\n\n(theta,rho,radius)-> (2^-2(x-iy),-z,-2^-2(x+iy))\n\nAll that Hermitian Vector space stuff...\n\nMK\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Gentil Correa wrote:
> "David Macmanus" <macmanus@tripos.com> wrote in message news:<33f0323d9493528e98f1d9ff3164b1d3_35661@mygate.mail gate.org>...
>
>>Solving the Schroedinger equation for the hydrogen atom is relatively
>>simple after the early step of translating the equation from
>>Cartesiancoordinates to spherical coordinates which then allows a
>>straightforward separation of variables to give three simple (ish)
>>equations. But what I can`t seem to be able to find out is whether it is
>>actually possible to solve the equation using Cartesian
>>coordinates i.e. the x, y and z rather than the r, \theta, \phi. Now, I
>>appreciate that it might be madness to attempt such a feat when it is
>>relatively simple to go over to spherical coordinates, but I`d really
>>like to know if it is *theoretically possible* to solve it in Cartesian
>>coordinates. Can anyone say definitively whether it be done (if you were
>>a maths genius, for example)?
>>Thanks,
>>David.
>
>
> Hi!
>
> Changing coordinates is a "change of language", so there must be
> an unitary operator which performs the trick. If you care to find this
> operator, then, starting from the solution in spherical coordinates,
> you end up with the cartesian solution. Now, if what you want is to arrive
> at the cartesian solution without using other coordinates even at intermediate
> steps, this seems to me rather uninteresting. It is like trying to solve
> a differential equation forbidding oneself to change variables. However,
> it must be possible to do it ("if I were a maths genius").
>
> Best wishes,
> G.
>
I think you're closest to my understanding of the 'difficulty'
satisfying the question directly. Whenever we 'solve' an equation, we're
using Arithmetic. When we do so with differential equations, we do so
along what they call Exact sequences, wherein the diferrential (or
boundary) operator takes us from one integrable (flat) space to the
next. That is a geometrical feature of the solution to problem, which we
cannot really wish away without also losing the ability to 'solve'.
If one wants a 'change of coordinates' to 'Cartesian' just for the heck
of it, I think there is an isomorphic transformation associated with
Riemman, but I think Riemman attributed to Gauss: The stereographic
projection. So the spherical problem can be turned into a 'Cartesian
one' on the stereographic projection plane (say the equatorial plane).
Problem is, the projection pole maps to infinity, so the minimal
covering for it is two projections, say one rooted at the north pole
excluding the north pole, and another rooted in the south pole,
excluding the south pole, and with yet another change of coordinates to
smoothly switch projections at their selected 'overlap' at say the
equatorial plane. So given that atlas covering of the sphere, yes, there
is a 'Cartesian' solution, but then you have to respect the Spherical
geometry and thereby only deal with a Complex field. I do not know if
this still meets the OP's requirements; but at last reading, the
projection from spherical vector to 'Cartesian' used the transformation:
(\theta,\rho,radius)-> (2^-2(x-iy),-z,-2^-2(x+iy))
All that Hermitian Vector space stuff...
MK
David Macmanus
Apr30-04, 02:03 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>"Arnie King" <PanSynthesis@netscape.net> wrote in message\nnews:41400681.0404261009.566cc801@posting .google.com\n\n> My understanding is that Schroedinger\'s equation with the 1/r\n> potential is not separable in cartesian coordinates.\n>\n> For it to be separable, the potential would have to be a sum of three\n> functions each depending on only one of x, y, z. (This is the case\n> with the isotropic harmonic potential, V(r)=-kr^2, but -1/r cannot be\n> so written.)\n>\n> The hydrogen equation is however separable in parabolic cylinder\n> coordinates. I think Merzbacher\'s book discusses it. The separability\n> in more than one system has something to do with the "extra" constant\n> of motion (the Runge vector mentioned in a previous post), and with\n> the famous degeneracy in hydrogen that allows states of different l\n> (angular momentum quantum number) to have the same energy.\n\nThanks. That\'s actually pretty helpful. So really, solving it using\nparabolic cylinder coordinates\' is a similar sort of procedure as going\nover to sphericals -but because 1/r cannot be written as a sum of three,\nit\'s not possible to solve it in cartesians. I guess that this\neffectively answers my question - it\'s not doable in cartesians!!\nUnless someone knows different.............?????\nBig thanks to all respondents.\nDavid.\n\n\n\n\n--\nPosted via Mailgate.ORG Server - http://www.Mailgate.ORG\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Arnie King" <PanSynthesis@netscape.net> wrote in message
news:41400681.0404261009.566cc801@posting.google.c om
> My understanding is that Schroedinger's equation with the 1/r
> potential is not separable in cartesian coordinates.
>
> For it to be separable, the potential would have to be a sum of three
> functions each depending on only one of x, y, z. (This is the case
> with the isotropic harmonic potential, V(r)=-kr^2, but -1/r cannot be
> so written.)
>
> The hydrogen equation is however separable in parabolic cylinder
> coordinates. I think Merzbacher's book discusses it. The separability
> in more than one system has something to do with the "extra" constant
> of motion (the Runge vector mentioned in a previous post), and with
> the famous degeneracy in hydrogen that allows states of different l
> (angular momentum quantum number) to have the same energy.
Thanks. That's actually pretty helpful. So really, solving it using
parabolic cylinder coordinates' is a similar sort of procedure as going
over to sphericals -but because 1/r cannot be written as a sum of three,
it's not possible to solve it in cartesians. I guess that this
effectively answers my question - it's not doable in cartesians!!
Unless someone knows different.............?????
Big thanks to all respondents.
David.
--
Posted via Mailgate.ORG Server - http://www.Mailgate.ORG
Arnie King
May6-04, 07:52 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\n\n"David Macmanus" <macmanus@tripos.com> wrote in message news:<f3de83bbffd120ce3d0e3906df83e2e8.35661@mygat e.mailgate.org>...\n> "Arnie King" <PanSynthesis@netscape.net> wrote in message\n> news:41400681.0404261009.566cc801@posting.google.c om\n>\nmomentum quantum number) to have the same energy.\n>\n> Thanks. That\'s actually pretty helpful. So really, solving it using\n> parabolic cylinder coordinates\' is a similar sort of procedure as going\n> over to sphericals -but because 1/r cannot be written as a sum of three,\n> it\'s not possible to solve it in cartesians. I guess that this\n> effectively answers my question - it\'s not doable in cartesians!!\n> Unless someone knows different.............?????\n> Big thanks to all respondents.\n> David.\n\nI researched this a little and found my previous remarks are not quite\nright.\n\nThe equation is separable in parabolic coordinates, not parabolic\ncylinder coordinates. The usual names for the parabolic coordinates\nare phi, eta, and xi, where phi is the same azimuth coordinate used in\nspherical coords. A surface of constant xi is a concave-up paraboloid\nof revolution, the z-axis being the symmetry axis. As xi gets large\nthe paraboloid asymtotes to the xy plane. Similarly, a surface of\nconstant eta is a concave-down paraboloid of revolution about the\nz-axis. See Arfken or Morse & Feshbach.\n\nIn these coordinates, one has 1/r = 1/(eta + xi). So it isn\'t exactly\nas I said (that 1/r is a sum of funtions that each depend on only one\ncoordinate). But, in the attempt to separate variables, one multiplies\nthrough by (eta + xi), at which point one does have the sum I spoke\nof, and the equation is found to separate because of it.\n\nArfken has a note that the Schroedinger equation for hydrogen is also\nseparable in oblate spheroidal coordinate (or maybe it was prolate).\n\nIf you aren\'t so familiar with separation of variable you might just\ntry some examples to see how it works (or doesn\'t). Say, by working it\nout for the isotropic harmonic oscillator (V=-(x^2+y^2+z^2)) in\ncartesian, and comparing with what happens when you try V=-1/r in\ncartesian.\n\nSchiff discusses the algebraic approach for hydrogen. Working out the\nhydrogen Schroedinger equation in parabolic coordinates doesn\'t lead\nto any new insight (but it is useful in scattering problems). Instead,\nin the algebraic approach, one gets insight into the higher symmetry.\nOne uses the "extra" constant of motion, the Runge-Lenz vector M, as\nanother generator of symmetry operations (just like L is the generator\nof rotations). It turns out that the three components of M and the\nthree components of L together generate a six parameter symmetry group\nfor hydrogen. It is not a spatial symmetry, but a more abstract\ndynamical symmetry. For bound states, this six-parameter group is\nisomorphic to the rotation group in four dimensions. But for\npositive-energy states, it is isomorphic to the Lorentz group!\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>"David Macmanus" <macmanus@tripos.com> wrote in message news:<f3de83bbffd120ce3d0e3906df83e2e8.35661@mygate.mail gate.org>...
> "Arnie King" <PanSynthesis@netscape.net> wrote in message
> news:41400681.0404261009.566cc801@posting.google.c om
>
momentum quantum number) to have the same energy.
>
> Thanks. That's actually pretty helpful. So really, solving it using
> parabolic cylinder coordinates' is a similar sort of procedure as going
> over to sphericals -but because 1/r cannot be written as a sum of three,
> it's not possible to solve it in cartesians. I guess that this
> effectively answers my question - it's not doable in cartesians!!
> Unless someone knows different.............?????
> Big thanks to all respondents.
> David.
I researched this a little and found my previous remarks are not quite
right.
The equation is separable in parabolic coordinates, not parabolic
cylinder coordinates. The usual names for the parabolic coordinates
are \phi, \eta, and \xi, where \phi is the same azimuth coordinate used in
spherical coords. A surface of constant \xi is a concave-up paraboloid
of revolution, the z-axis being the symmetry axis. As \xi gets large
the paraboloid asymtotes to the xy plane. Similarly, a surface of
constant \eta is a concave-down paraboloid of revolution about the
z-axis. See Arfken or Morse & Feshbach.
In these coordinates, one has 1/r = 1/(\eta + \xi). So it isn't exactly
as I said (that 1/r is a sum of funtions that each depend on only one
coordinate). But, in the attempt to separate variables, one multiplies
through by (\eta + \xi), at which point one does have the sum I spoke
of, and the equation is found to separate because of it.
Arfken has a note that the Schroedinger equation for hydrogen is also
separable in oblate spheroidal coordinate (or maybe it was prolate).
If you aren't so familiar with separation of variable you might just
try some examples to see how it works (or doesn't). Say, by working it
out for the isotropic harmonic oscillator (V=-(x^2+y^2+z^2)) in
cartesian, and comparing with what happens when you try V=-1/r in
cartesian.
Schiff discusses the algebraic approach for hydrogen. Working out the
hydrogen Schroedinger equation in parabolic coordinates doesn't lead
to any new insight (but it is useful in scattering problems). Instead,
in the algebraic approach, one gets insight into the higher symmetry.
One uses the "extra" constant of motion, the Runge-Lenz vector M, as
another generator of symmetry operations (just like L is the generator
of rotations). It turns out that the three components of M and the
three components of L together generate a six parameter symmetry group
for hydrogen. It is not a spatial symmetry, but a more abstract
dynamical symmetry. For bound states, this six-parameter group is
isomorphic to the rotation group in four dimensions. But for
positive-energy states, it is isomorphic to the Lorentz group!
Doug Goncz
May12-04, 05:22 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Alistair I. M. Rae takes the first year student in quantum mechanics through\nthe one dimensional, Cartesian, and polar Shroedinger wave equations nicely in\nhis Quantum Mechanics, which I have and recommend.\n\nI\'d like to write a wave equation for any one of the strings on my guitar....\n\nI am familiar with the two-point boundary condition and some of the\ndifferential equations needed to describe the string\'s motion after a\ndisturbance (being plucked, picked, or strummed).\n\n\nYours,\n\nDoug Goncz ( ftp://users.aol.com/DGoncz/ )\n\nRead about my physics project at NVCC:\nhttp://groups.google.com/groups?q=dgoncz&scoring=d plus\n"bicycle", "fluorescent", "inverter", "flywheel", "ultracapacitor", etc.\nin the search box\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Alistair I. M. Rae takes the first year student in quantum mechanics through
the one dimensional, Cartesian, and polar Shroedinger wave equations nicely in
his Quantum Mechanics, which I have and recommend.
I'd like to write a wave equation for any one of the strings on my guitar....
I am familiar with the two-point boundary condition and some of the
differential equations needed to describe the string's motion after a
disturbance (being plucked, picked, or strummed).
Yours,
Doug Goncz ( ftp://users.aol.com/DGoncz/ )
Read about my physics project at NVCC:
http://groups.google.com/groups?q=dgoncz&scoring=d plus
"bicycle", "fluorescent", "inverter", "flywheel", "ultracapacitor", etc.
in the search box
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