How Do You Derive the Klein-Gordon Propagator from Commutation Relations?

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Homework Help Overview

The discussion revolves around the derivation of the Klein-Gordon (K-G) propagator, specifically in the context of quantum field theory (QFT). Participants are trying to clarify the original poster's question and the relevant material from Peskin's book.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are questioning the clarity and specificity of the original poster's inquiry regarding the K-G propagator. There are suggestions for the poster to elaborate on their specific difficulties and to reference particular sections of the text.

Discussion Status

The discussion is ongoing, with participants seeking more details from the original poster to facilitate a more productive dialogue. There is a recognition of the need for a clearer question to guide the discussion effectively.

Contextual Notes

There is mention of different editions of Peskin's book, which may lead to confusion regarding the content being referenced. Some participants express uncertainty about the original poster's familiarity with the subject matter, indicating a potential gap in understanding the K-G propagator and its derivation.

HadronPhysics
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Homework Statement
I am reading Peskin's book on Chapter 2. I have a question about deriving the K-G propagator
Relevant Equations
## \bra 0|[\phi(x), \phi(y)] |0 \ket = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2E_{p}} \left( e^{-ip(x-y)} - e^{ip(x-y)} \right) =\int \frac{d^3 p}{(2\pi)^3} \frac{1}{2E_{p}}e^{-ip(x-y)} + \int \frac{d^3 p}{(2\pi)^3} \frac{1}{-2E_{p}}e^{-ip(x-y)}##
$r$
 
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What is the question?
 
Standard derivation of the K-G propagator ? (o:))

Sorry I got no clue which book he means and got no clue what is that propagator. I suspect it has something to do with QFT of which I know very little...

All I know is that K-G probably stands for Klein - Gordon and I don't even know why I am replying to this post, I vent sleep well but I just can't sleep and I feel I got something to do.
 
Delta2 said:
Standard derivation of the K-G propagator ?
Should be a more specific question, about some detail or something.

HadronPhysics said:
Relevant Equations:: ## \bra 0|[\phi(x), \phi(y)] |0 \ket = \ldots ##
Also it should read ##$ \langle 0|[\phi(x), \phi(y)] |0 \rangle = ##
Delta2 said:
Sorry I got no clue which book he means and got no clue what is that propagator. I suspect it has something to do with QFT of which I know very little...
Yeah it is QFT
 
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HadronPhysics said:
Homework Statement:: I am reading Peskin's book on Chapter 2.
No, you are reading the 1995 book by Peskin and Schroeder. The distinction is important because there is a 2019 book authored solely by Peskin.

malawi_glenn said:
Should be a more specific question, about some detail or something.

Yes, what is the specific difficulty? Something on pages 29 - 31?
 
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George Jones said:
The distinction is important because there is a 2019 book authored solely by Peskin.
But chapter 2 in that book "concepts of elementary particle physics" does not deal with scalar quantum field theory :)
 
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HadronPhysics said:
Homework Statement:: I am reading Peskin's book on Chapter 2. I have a question about deriving the K-G propagator
Relevant Equations:: ## \bra 0|[\phi(x), \phi(y)] |0 \ket = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2E_{p}} \left( e^{-ip(x-y)} - e^{ip(x-y)} \right) =\int \frac{d^3 p}{(2\pi)^3} \frac{1}{2E_{p}}e^{-ip(x-y)} + \int \frac{d^3 p}{(2\pi)^3} \frac{1}{-2E_{p}}e^{-ip(x-y)}##

$r$
Welcome to PF!

Please elaborate what your specific question is and if it is an actual homework problem, please show your attempt at a solution first. Also, sometimes an answer can be found looking at relevant threads from the past which are pulled up at the bottom of this page. Thanks.
 
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HadronPhysics said:
Homework Statement:: I am reading Peskin's book on Chapter 2. I have a question about deriving the K-G propagator
Relevant Equations:: ## \bra 0|[\phi(x), \phi(y)] |0 \ket = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2E_{p}} \left( e^{-ip(x-y)} - e^{ip(x-y)} \right) =\int \frac{d^3 p}{(2\pi)^3} \frac{1}{2E_{p}}e^{-ip(x-y)} + \int \frac{d^3 p}{(2\pi)^3} \frac{1}{-2E_{p}}e^{-ip(x-y)}##

$r$
Thread has been closed as Substandard. The new OP has been asked to start a new thread with a better post, including details about what exactly is confusing them in this material. Thanks folks for trying.
 
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