John Baez
Apr28-04, 03:14 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\nIn article <030420041017457836%mtvaughn@neu.REMOVE.edu>,\nMik e Vaughn <mtvaughn@neu.REMOVE.edu> wrote approximately:\n\n>In article <c4dtbf\\$nl1\\$1@glue.ucr.edu>, baez@galaxy.ucr.edu (John Baez) wrote:\n\n>> There\'s also an independent completely arbitrary choice about whether\n>> time evolution goes like exp(-itH) or exp(itH). For some reason that\'s\n>> not so obvious - but is amusing to figure out - most physicists use\n>> exp(-itH).\n\n>How do you figure it out? It _is_ convenient in relativistic quantum\n>mechanics, having chosen exp(ip.x) as the representation for a plane\n>wave, but the exp(-iHt) may have come before that. (?)\n\nBefore relativistic quantum mechanics became popular, this sign\nwas built into Schroedinger\'s equation:\n\nd psi / dt = -i H psi\n\nor in other words:\n\ni d psi / dt = H psi\n\nSo, you can ask why Schroedinger picked the sign he did.\n\nI don\'t think I want to say my guess yet.\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>In article <030420041017457836%mtvaughn@neu.REMOVE.edu>,
Mike Vaughn <mtvaughn@neu.REMOVE.edu> wrote approximately:
>In article <c4dtbf$nl1$1@glue.ucr.edu>, baez@galaxy.ucr.edu (John Baez) wrote:
>> There's also an independent completely arbitrary choice about whether
>> time evolution goes like \exp(-itH) or \exp(itH). For some reason that's
>> not so obvious - but is amusing to figure out - most physicists use
>> \exp(-itH).
>How do you figure it out? It _is_ convenient in relativistic quantum
>mechanics, having chosen \exp(ip.x) as the representation for a plane
>wave, but the \exp(-iHt) may have come before that. (?)
Before relativistic quantum mechanics became popular, this sign
was built into Schroedinger's equation:
d \psi / dt = -i H \psi
or in other words:
i d \psi / dt = H \psi
So, you can ask why Schroedinger picked the sign he did.
I don't think I want to say my guess yet.
Mike Vaughn <mtvaughn@neu.REMOVE.edu> wrote approximately:
>In article <c4dtbf$nl1$1@glue.ucr.edu>, baez@galaxy.ucr.edu (John Baez) wrote:
>> There's also an independent completely arbitrary choice about whether
>> time evolution goes like \exp(-itH) or \exp(itH). For some reason that's
>> not so obvious - but is amusing to figure out - most physicists use
>> \exp(-itH).
>How do you figure it out? It _is_ convenient in relativistic quantum
>mechanics, having chosen \exp(ip.x) as the representation for a plane
>wave, but the \exp(-iHt) may have come before that. (?)
Before relativistic quantum mechanics became popular, this sign
was built into Schroedinger's equation:
d \psi / dt = -i H \psi
or in other words:
i d \psi / dt = H \psi
So, you can ask why Schroedinger picked the sign he did.
I don't think I want to say my guess yet.