Timeline for Annihilation and creation operator - $\phi$ and $\pi$ for Klein-Gordon Field
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Jul 18, 2013 at 18:02 | history | suggested | Abhimanyu Pallavi Sudhir |
KGFs have KGE as their WE.
|
|
| Jul 18, 2013 at 17:59 | review | Suggested edits | |||
| S Jul 18, 2013 at 18:02 | |||||
| Mar 20, 2013 at 18:08 | vote | accept | J L | ||
| Mar 16, 2013 at 18:08 | comment | added | J L | @twistor59 I'm almost done with the question, could you check my answer (in particular the final remarks) Thanks, you are of great help, learning almost by myself QFT (working through Peskin) is being a kind of difficult experience, because I want to derive as many things as I'm able by myself-with the help of this site. | |
| Mar 16, 2013 at 17:32 | answer | added | J L | timeline score: 10 | |
| Mar 16, 2013 at 17:21 | history | edited | J L | CC BY-SA 3.0 |
deleted 685 characters in body
|
| Mar 16, 2013 at 17:00 | comment | added | twistor59 | Sorry I didn't see the prime! So your computation looks correct. w.r.to your final question: nothing else depends on x, so as far as the x integration is concerned, you treat everything else (inc. the p's) as constant, and you get the delta function. You could edit your question to its original form and post this computation as an answer. It's fine to post answers to your own questions. | |
| Mar 16, 2013 at 15:42 | comment | added | J L | @twistor59 thanks, but I don't see it clearly sorry. In the integral the exponential is $e^{i\vec{p}'\cdot\vec{x}}$ so it will be $\tilde{\phi}(\vec{p}')$. If I multiply it by $e^{i\vec{p}\cdot\vec{x}}$ I can't put it under the integral sign w.r.t $\vec{p}$. | |
| Mar 16, 2013 at 15:37 | comment | added | twistor59 | In the line just after EDIT 1: that's an expression for $\phi(p)$ i.e. $p$ is the Fourier transform variable. So in the next expression (with the two integral signs) you should use $p'$ when you expand $\phi(x)$ in terms of momentum. So your annihilation and creation operators will carry $p'$s | |
| Mar 16, 2013 at 15:31 | comment | added | J L | @twistor59 I edited the question | |
| Mar 16, 2013 at 15:30 | history | edited | J L | CC BY-SA 3.0 |
added 739 characters in body
|
| Mar 16, 2013 at 13:48 | comment | added | twistor59 | Yes (assuming you're using $x$ rather than $x'$ as in the current version of the question) | |
| Mar 16, 2013 at 13:42 | comment | added | J L | Ok so it goes as: multiply by $e^{i(p'x)}$ integrate with respect to $x$, and integrate with respect to $p$ with the help of $\delta(p-p')$? I'm going to edit the question to see if I did understood it right | |
| Mar 16, 2013 at 13:42 | comment | added | twistor59 | Yep in the line right after "the inverse Fourier transform" | |
| Mar 16, 2013 at 13:24 | comment | added | Michael | The argument of $\phi$ needs to be the integration variable, i.e. $x'$. | |
| Mar 16, 2013 at 13:15 | history | asked | J L | CC BY-SA 3.0 |