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View Full Version : Degenerate vacua in QFT


pirillo
Dec14-04, 05:34 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>In the thread titled\n\n"The vacuum involving spontaneously broken Higgs fields"\n\nDr Baez says:\n\n"It goes back to the fact that in quantum field\ntheory, the canonical commutation relations have many inequivalent\nrepresentations. It turns out we can\'t think of all the different\nnoninvariant vacua as states in the same Hilbert space"\n\nI\'m wondering what does he mean by "the different vacua are not\nstates in the same hilbert space" ?\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">&nbsp;&nbsp;View this Usenet post in original ASCII form </a></div><P></jabberwocky>In the thread titled

"The vacuum involving spontaneously broken Higgs fields"

Dr Baez says:

"It goes back to the fact that in quantum field
theory, the canonical commutation relations have many inequivalent
representations. It turns out we can't think of all the different
noninvariant vacua as states in the same Hilbert space"

I'm wondering what does he mean by "the different vacua are not
states in the same hilbert space" ?