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Simple question from Peskin Schroeder

  • Thread starter Neitrino
  • Start date
Gentlemen,
Could you help me please, Im sure it is not even worthy of your attention, but anyway...

In Peskin, Schroeder - from expresion [tex] <0|\phi(x)\phi(y)|0> [/tex] survives [tex] <0|a_p a_q^\dag|0> [/tex] so it creates one-particle state |q> at position y and another one-particle state | p> at postion x. But how do I intuitively see that causuality/propagation of particle between these positions is imbeded and considered in that expresion?

And another question [tex] <0|\phi(x)|p>=.......e^ipx[/tex] formula 2.42

it's said that it is a position-space representation of the state |p> just as in NR QM <x|p>, so it should be projection of single-particle |p> state onto the <x| baisis and what vectors/basis that state is projected on? where is that <x| vectors in 2.42

P.S. Im sure i missed something very simple in understanding of above and that's why posting such "silly" questions

Thanks in advance
 
(1) I think the expresion [tex] <0|\phi(x)\phi(y)|0> [/tex]
survives [tex] <0|a_p a_q^\dag|0> [/tex] means:
[tex] <0|a_p^\dag a_q^\dag|0> [/tex]=0 and [tex] <0|a_p a_q|0> [/tex]=0;
only [tex] <0|a_p a_q^\dag|0> [/tex] survives, of course p and q are arbitary,
not single p and single q. the final result will be an integral over all possible p or q.

(2) [tex] <0|\phi(x)=<x|[/tex], this is a simple calculation.
there is no special physical significance here, the author, I guess, just mentioned NR
case to make the formula be easily understood.
 
snooper007 said:
(2) [tex] <0|\phi(x)=<x|[/tex], this is a simple calculation.
Dear Snooper007 thks for ur reply..
but [tex] <0|\phi(x)[/tex] it is a complex conjugation of [tex] \phi(x)|0>[/tex] (as u mentioned in QM forum). So [tex] <0|\phi(x)=\int{\frac{d^3 p}{(2\pi)^3}\frac{1}{2E_p}e^{ipx}<p|[/tex]

but with [tex] <0|\phi(x)=<x|[/tex] Im confused <-How/why it's that?
 

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