Theodore George Erler IV
Aug30-04, 04:57 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\nTo see that\n|p> = Exp[ (-1/4)p^2 + p a^\\dag - (1/2) a^\\dag a^\\dag ] |0>\nis the right expression for the momentum eigenstate, you need to show two\nthings:\n\n1) p_op|p>=p|p>\n2) <p|p\'> = delta(p-p\')\n\nThe first one is easy to show: just write p_op in terms of a, a^\\dag and\nact it on the above expression. There is nothing tricky about this, in\nthis context "a" just works like a a derivative "d/da^dag ".\n\nNumber 2) is not quite as easy. To prove it you need the following formula\nfor contracting squeezed states:\n\n<0|exp( ta + (u/2) a^2)exp(va^dag + (w/2) (a^\\dag)^2)|0>\n= (1-uw)^(-1/2)exp( (2tv + wt^2 + uv^2)/2(1-uw) )\n\nThe easiest way to prove this formula is to insert a resolution of the\nidentity in terms of an overcomplete set of coherent states (see for\nexample Shankar\'s textbook). However, this formula doesn\'t give us the\nanswer yet, since for our case u=w=1, and the expression is singular. This\nof course is not a surprise, since the answer needs to be a delta\nfunction. To prove that it is actually a delta function, we need to\nregulate our expression for |p> and show that in the limit w->1 the\ncontraction formula converges, in the sense of distributions, to the delta\nfunction. This is how you determine the normalization, perticularly the\nexp(-(1/4)p^2) factor.\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>To see that
|p> = \Exp[ (-1/4)p^2 + p a^\dag - (1/2) a^\dag a^\dag ] |0>
is the right expression for the momentum eigenstate, you need to show two
things:
1) p_{op}|p>=p|p>2) <p|p'> = \delta(p-p')
The first one is easy to show: just write p_{op} in terms of a, a^\dag and
act it on the above expression. There is nothing tricky about this, in
this context "a" just works like a a derivative "d/da^dag ".
Number 2) is not quite as easy. To prove it you need the following formula
for contracting squeezed states:
<0|\exp( ta + (u/2) a^2)\exp(va^dag + (w/2) (a^\dag)^2)|0>= (1-uw)^(-1/2)\exp([/itex] (2tv [itex]+ wt^2 + uv^2)/2(1-uw) )
The easiest way to prove this formula is to insert a resolution of the
identity in terms of an overcomplete set of coherent states (see for
example Shankar's textbook). However, this formula doesn't give us the
answer yet, since for our case u=w=1, and the expression is singular. This
of course is not a surprise, since the answer needs to be a \delta
function. To prove that it is actually a \delta function, we need to
regulate our expression for |p> and show that in the limit w->1 the
contraction formula converges, in the sense of distributions, to the \delta
function. This is how you determine the normalization, perticularly the
\exp(-(1/4)p^2) factor.
|p> = \Exp[ (-1/4)p^2 + p a^\dag - (1/2) a^\dag a^\dag ] |0>
is the right expression for the momentum eigenstate, you need to show two
things:
1) p_{op}|p>=p|p>2) <p|p'> = \delta(p-p')
The first one is easy to show: just write p_{op} in terms of a, a^\dag and
act it on the above expression. There is nothing tricky about this, in
this context "a" just works like a a derivative "d/da^dag ".
Number 2) is not quite as easy. To prove it you need the following formula
for contracting squeezed states:
<0|\exp( ta + (u/2) a^2)\exp(va^dag + (w/2) (a^\dag)^2)|0>= (1-uw)^(-1/2)\exp([/itex] (2tv [itex]+ wt^2 + uv^2)/2(1-uw) )
The easiest way to prove this formula is to insert a resolution of the
identity in terms of an overcomplete set of coherent states (see for
example Shankar's textbook). However, this formula doesn't give us the
answer yet, since for our case u=w=1, and the expression is singular. This
of course is not a surprise, since the answer needs to be a \delta
function. To prove that it is actually a \delta function, we need to
regulate our expression for |p> and show that in the limit w->1 the
contraction formula converges, in the sense of distributions, to the \delta
function. This is how you determine the normalization, perticularly the
\exp(-(1/4)p^2) factor.