<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 wrote:\n> Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<cba9bj\\$u0r\\$1@lfa222122.richmond.edu>...\ n>\n>>David wrote:\n>\n>>>Can anyone suggest what can be said about (say) 0+1 dimensional\n>>>quantum mechanics in the limit where kinetic terms are suppressed?\n>\n>>This is just the infinite mass limit of a system. The system becomes\n>>very slow and behaves essentially classically, if started in a coherent\n>>state.\n>\n>\n> I think I have become confused, because carrying out this process\n> would naively seem to result in a non-propagating theory in which the\n> wavefunction converges to \\delta(x_0) for all time, for a particle\n> which has initial conditions x=x_0 at t=t_0.\n\nOf course, if you have an object with truly infinite mass, one needs an\ninfinite amount of energy to move it. The classical behavior appears in\nthe case where the mass is just very large. Infinite mass (i.e., no kinetic\nterm) is unphysical.\n\n> But, this isn\'t what I\n> would like to understand as a classical limit, in which I would prefer\n> kinetic terms to be still present. Have I misunderstood?\n>\n> Despite my confusion, then, am I correct in supposing that the\n> amplitude for a particle to propagate from x_a to x_b, in a quantum\n> theory without kinetic terms, is just \\delta(x_a - x_b), appropriately\n> normalized by the volume of space?\n\nThe explicit solution of the Schroedinger equation\ni hbar psidot(x) = V(x) psi(x)\nwith psi = psi_0 at t=0 is\npsi(x,t)= exp(-i t V(x)/hbar) psi_0(x),\nwhich is boring.\n\n\nArnold Neumaier\n\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>David wrote:
> Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<cba9bj$u0r$1@lfa222122.richmond.edu>...
>
>>David wrote:
>
>>>Can anyone suggest what can be said about (say) 0+1 dimensional
>>>quantum mechanics in the limit where kinetic terms are suppressed?
>
>>This is just the infinite mass limit of a system. The system becomes
>>very slow and behaves essentially classically, if started in a coherent
>>state.
>
>
> I think I have become confused, because carrying out this process
> would naively seem to result in a non-propagating theory in which the
> wavefunction converges to \delta(x_0) for all time, for a particle
> which has initial conditions x=x_0 at t=t_0.
Of course, if you have an object with truly infinite mass, one needs an
infinite amount of energy to move it. The classical behavior appears in
the case where the mass is just very large. Infinite mass (i.e., no kinetic
term) is unphysical.
> But, this isn't what I
> would like to understand as a classical limit, in which I would prefer
> kinetic terms to be still present. Have I misunderstood?
>
> Despite my confusion, then, am I correct in supposing that the
> amplitude for a particle to propagate from x_a to x_b, in a quantum
> theory without kinetic terms, is just \delta(x_a - x_b), appropriately
> normalized by the volume of space?
The explicit solution of the Schroedinger equation
i \hbar psidot(x) = V(x) \psi(x)
with \psi = \psi_0 at t=0 is
\psi(x,t)= \exp(-i t V(x)/\hbar) \psi_0(x),
which is boring.
Arnold Neumaier