E = mc^2 from quantum mechanics? [Archive]

Bob Day

Jun1-04, 04:43 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\nCan E = mc^2 for a nonzero rest mass be derived\nstrictly from quantum mechanics? If so, how?\n\n-- Bob Day\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>Can E = mc^2 for a nonzero rest mass be derived
strictly from quantum mechanics? If so, how?

-- Bob Day

Mark

Jun4-04, 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>\nOn Tue, 01 Jun 2004 05:43:00 -0400, Bob Day wrote:\n\n>\n>\n> Can E = mc^2 for a nonzero rest mass be derived\n> strictly from quantum mechanics? If so, how?\n>\n> -- Bob Day\n\n\nNo. More like the other way around. The basic equations of quantum\nmechanics are derived by combining the laws of mechanics with some other\nideas, such as replacing momentum and position with operators.\n\nSo if you start with newtonian mechanics you use E = p^2/2m, replace the\nmomentum p and the energy E by operators, and end up with the Schrodinger\nequation (replacing E by an operator is a bit dodgy, but its an easy way\nto think about it).\n\nAlternatively, you can start with E^2 = p^2c^2 + m^2c^4 (this reduces to E\n= mc^2 for p=0, i.e. stationary particles), do the same tricks with\noperators, and you end up with the Klein-Gordon equation, a relativistic\nQM equation.\n\nWhat I am trying to say is that QM has to have some basic laws of\nmechanics built into it, rather than starting with QM and getting things\nlike E=mc^2 out.\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>On Tue, 01 Jun 2004 05:43:00 -0400, Bob Day wrote:

>
>
> Can E = mc^2 for a nonzero rest mass be derived
> strictly from quantum mechanics? If so, how?
>
> -- Bob Day

No. More like the other way around. The basic equations of quantum
mechanics are derived by combining the laws of mechanics with some other
ideas, such as replacing momentum and position with operators.

So if you start with newtonian mechanics you use E = p^2/2m, replace the
momentum p and the energy E by operators, and end up with the Schrodinger
equation (replacing E by an operator is a bit dodgy, but its an easy way
to think about it).

Alternatively, you can start with E^2 = p^{2c}^2 + m^{2c}^4 (this reduces to E= mc^2 for p=0, i.e. stationary particles), do the same tricks with
operators, and you end up with the Klein-Gordon equation, a relativistic
QM equation.

What I am trying to say is that QM has to have some basic laws of
mechanics built into it, rather than starting with QM and getting things
like E=mc^2 out.

davidoff404

Jun4-04, 02:48 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>Bob Day wrote:\n> Can E = mc^2 for a nonzero rest mass be derived\n> strictly from quantum mechanics? If so, how?\n>\n> -- Bob Day\n>\n\nStrictly speaking, yes, although it depends on how closely you want the\nderivation to remain in spirit to Einstein\'s original derivation in SR.\nFor example, Schrodinger\'s relativistic wave equation can be used when\nanalysing stationary states of hydrogen and solutions satisfying\nreasonable boundary conditions can be found for the energy values\n\nE = mc^2( 1 - a^2/2n^2 - a^4/2n^4(n/(l+1/2) - 3/4) + ... )\n\nwhere a, n, and l are, respectively, the fine structure constant, a\npositive-definite integer, and units of orbital angular momentum.\nHowever, this expression for the energy levels is considerably larger\nthan those observed by experiment -- as far as I know the discrepancy\nmakes itself obvious for the n=2 levels and above in Schrodingers\nequation. Of course, you could always use Sommerfeld\'s much more\naccurate estimate for the levels, or simply modify the above equation to\ntake account of the coupling between spin and orbital angular momentum\n(I\'m still talking about hydrogen here).\n\nIf one wants to look at the most accurate model one needs Dirac\'s\nrelativistic qm. Using Dirac\'s model one finds the hydrogen energy\nlevels are given by\n\nE = mc^2( 1 + a^2/( n - j - (1/2) + ( (j+1/2)^2 - a^2)^(1/2) )^2 )^(-1/2)\n\nwhere j are units of spin angular momentum. I think this result is due\nto Gordon, although I\'m not entirely sure about that. The energy levels\nto third order in a^2 agrees with experiment.\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>Bob Day wrote:
> Can E = mc^2 for a nonzero rest mass be derived
> strictly from quantum mechanics? If so, how?
>
> -- Bob Day
>

Strictly speaking, yes, although it depends on how closely you want the
derivation to remain in spirit to Einstein's original derivation in SR.
For example, Schrodinger's relativistic wave equation can be used when
analysing stationary states of hydrogen and solutions satisfying
reasonable boundary conditions can be found for the energy values

E = mc^2( 1 - a^2/2n^2 - a^4/2n^4(n/(l+1/2) - 3/4) + .[/itex].. )

where a, n, and l are, respectively, the fine structure constant, a
positive-definite integer, and units of orbital angular momentum.
However, this expression for the energy levels is considerably larger
than those observed by experiment -- as far as I know the discrepancy
makes itself obvious for the n=2 levels and above in Schrodingers
equation. Of course, you could always use Sommerfeld's much more
accurate estimate for the levels, or simply modify the above equation to
take account of the coupling between spin and orbital angular momentum
(I'm still talking about hydrogen here).

If one wants to look at the most accurate model one needs Dirac's
relativistic qm. Using Dirac's model one finds the hydrogen energy
levels are given by

[itex]E = mc^2( 1 + a^2/( n - j - (1/2) + ( (j+1/2)^2 - a^2)^(1/2) )^2 )^(-1/2)

where j are units of spin angular momentum. I think this result is due
to Gordon, although I'm not entirely sure about that. The energy levels
to third order in a^2 agrees with experiment.