Timeline for Two point function of a free scalar field in Euclidean space-time
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| Apr 13, 2017 at 12:40 | history | edited | CommunityBot |
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| May 4, 2016 at 16:24 | comment | added | asv | @user1504: Say $\phi(z)$ contains $\log z$ as a summand which is a multi-valued function. But $\partial \phi$ is well defined. | |
| May 4, 2016 at 16:24 | comment | added | asv | @user1504: Regarding (non-)existence of operators $\phi(x)$. In the CFT case as in di Francesca et al., my impression is that the real field $\phi(z,\bar z)$ (in their notation) satisfying $\Delta \phi(z,\bar z)=0$ does exist on the cylinder. However when this harmonic function is presented as a sum of holomorphic and anti-holomorphic functions $\phi(z,\bar z)=\phi(z)+\bar\phi(\bar z)$ globally on cylinder the summands are not well defined due to non-simply connectedness of cylinder. | |
| May 4, 2016 at 16:05 | comment | added | asv | @user1504: I am not sure what you mean by contact terms for non-time-ordered two-point function. In Minkowski case there are no contact terms. In Euclidean case $D(x)$ is well defined only in a half-space, and this was my main mistake from the vary beginning as I thought $D(x)$ is a well defined generalized function on the whole space $\mathbb{R}^D$ (see my answer below). Perhaps if you write the formula you mean explicitly including the contact terms, it will be more clear to me. | |
| May 4, 2016 at 15:50 | comment | added | user1504 | 1) I encourage you not to say $\phi$ exists when it does not. Wightman was proving a no-go theorem, not advocating the abandoment of Hilbert space. 2) Not sure if this came across clearly, but I was pointing to an error in your third equation. $\Delta D(x,y) \neq 0$ because of these contact terms. Keeping the contact term provides another path to your answer below. | |
| May 4, 2016 at 13:06 | vote | accept | asv | ||
| May 4, 2016 at 13:02 | vote | accept | asv | ||
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| May 4, 2016 at 12:28 | answer | added | asv | timeline score: 3 | |
| May 3, 2016 at 5:17 | comment | added | asv | @user1504: 1) As it follows from the comments below the operator $\phi$ (or rather $\partial \phi$) does exist at least sometimes in 2d. 2) For the usual correlation function (defined say via path integral) of course there is a contact term, you are right. But my question was about operator formalism in Euclidean case. | |
| May 2, 2016 at 21:19 | comment | added | user1504 | 1) There is no Euclidean operator $\phi(x)$ to impose $\Delta \phi(x) = 0$ on. In the Euclidean setting, you don't have operators. You have stochastic classical fields. 2) It is not true that $(\Delta_x \phi)(x) \phi(y) = \Delta_x(\phi(x) \phi(y)$. There is a non-trivial correction, called a 'contact term' coming from $x =y$. | |
| Apr 29, 2016 at 17:33 | answer | added | Igor Khavkine | timeline score: 7 | |
| Apr 29, 2016 at 17:00 | history | edited | asv | CC BY-SA 3.0 |
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| Apr 29, 2016 at 16:10 | history | edited | asv | CC BY-SA 3.0 |
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| Apr 29, 2016 at 15:04 | history | edited | asv | CC BY-SA 3.0 |
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| Apr 29, 2016 at 13:22 | history | edited | asv | CC BY-SA 3.0 |
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| Apr 29, 2016 at 13:09 | comment | added | asv | @AbdelmalekAbdesselam: I do not understand your claim. In the post I gave an argument that $\Delta D=0$ and I do not see how one should get $\delta$-function. Is anything wrong with my argument? Moreover in the analogous case of Minkowski metric one has $\Box D=0$ (see Remark). In this case one would get the $\delta$ function if one had the time ordered product in the definition of $D$ (which I do not assume). | |
| Apr 29, 2016 at 12:44 | comment | added | Abdelmalek Abdesselam | The function $D(x)$ should not satisfy $\Delta D=0$ but rather $\Delta D=\delta$. It's a Green's function of the Laplacian. So the correct conclusion is multiple of $|x|^{2-D}$ instead of multiple of the constant function equal to 1. | |
| Apr 29, 2016 at 12:17 | history | edited | asv | CC BY-SA 3.0 |
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| Apr 29, 2016 at 12:09 | history | asked | asv | CC BY-SA 3.0 |