# Simple question from Peskin Schroeder

1. Sep 6, 2005

### Neitrino

Gentlemen,
Could you help me please, Im sure it is not even worthy of your attention, but anyway...

In Peskin, Schroeder - from expresion $$<0|\phi(x)\phi(y)|0>$$ survives $$<0|a_p a_q^\dag|0>$$ 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 $$<0|\phi(x)|p>=.......e^ipx$$ 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

2. Sep 8, 2005

### snooper007

(1) I think the expresion $$<0|\phi(x)\phi(y)|0>$$
survives $$<0|a_p a_q^\dag|0>$$ means:
$$<0|a_p^\dag a_q^\dag|0>$$=0 and $$<0|a_p a_q|0>$$=0;
only $$<0|a_p a_q^\dag|0>$$ 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) $$<0|\phi(x)=<x|$$, 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.

3. Sep 8, 2005

### Neitrino

Dear Snooper007 thks for ur reply..
but $$<0|\phi(x)$$ it is a complex conjugation of $$\phi(x)|0>$$ (as u mentioned in QM forum). So $$<0|\phi(x)=\int{\frac{d^3 p}{(2\pi)^3}\frac{1}{2E_p}e^{ipx}<p|$$

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