Timeline for Differentiating Propagator, Green's function, Correlation function, etc
Current License: CC BY-SA 3.0
18 events
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| Nov 1, 2019 at 16:48 | comment | added | Mikhail Skopenkov | Although incorrect in general, the formula $G_R(x,x')=Θ(x_0−x'_0)K(x,x')$ is correct e.g. for wave equation, see this answer for a detailed proof and further references. | |
| Oct 30, 2019 at 12:11 | comment | added | Mikhail Skopenkov | The formula $G_R(x,x′)=Θ(x_0−x′_0)K(x,x′)$ is indeed incorrect is general. By the obvious reason that when the operator L is multiplied by 2, a kernel remains a kernel (whatever boundary conditions are) but a Green function cannot remain a Green function. | |
| Oct 28, 2019 at 16:58 | comment | added | jak | physics.stackexchange.com/questions/510673/… | |
| Oct 28, 2019 at 16:58 | comment | added | jak | this answer is incorrect as far as I see. In particular, in Wightman function is not equal to the delta distribution in the equal time limit. This is a hallmark of QFT as discussed in this answer physics.stackexchange.com/questions/133701/… But only if the Wightman function is equal to the delta distribution in the equal time limit, the product of the Wightman function times the Heaviside function fulfills the defining condition of a Green function. This is spelled out more explicitly here | |
| Apr 27, 2019 at 13:25 | comment | added | knzhou | @tparker Replying for posterity/for others: a very simple example is a harmonic oscillator: $e^{i \omega t}$ is a kernel, but $e^{i \omega t} \theta(t)$ is a Green's function. The jump caused by $\theta(t)$ really is the whole fundamental difference. | |
| Nov 10, 2018 at 3:21 | comment | added | tparker | Could you give an example of a distinct Green's function and kernel for a single linear differential operator? | |
| Nov 3, 2017 at 0:12 | comment | added | user143410 | You've stated that the Wightman function is a kernel and kernels are defined either by Dirichlet boundary conditions $\lim_{x→x′}K(x,x′)=δ(x−x′)$ or Neumann boundary conditions $\lim_{x→x′}∂K(x,x′)=δ(x−x′)$. So what boundary conditions define the Wightman function? If the Wightman function satisfies the Dirichlet boundary condition, then this would suggest it is a parametrix--i.e., differs from a fundamental solution by smooth function which evidently goes to zero in the coincidence limit. Is this true? What does this smooth function look like for the Wightman function? | |
| Jun 29, 2017 at 0:03 | comment | added | Jahan Claes | @josh "If one uses the kernel with Neumann boundary conditions on a time-slice boundary"... Shouldn't this be Dirichlet boundary? | |
| Jul 1, 2012 at 16:21 | history | bounty ended | Nikolaj-K | ||
| Mar 2, 2012 at 15:21 | vote | accept | Nikolaj-K | ||
| Feb 15, 2012 at 19:39 | vote | accept | Nikolaj-K | ||
| Feb 15, 2012 at 19:40 | |||||
| Feb 12, 2012 at 17:50 | comment | added | josh | To see how quantities like $W(x,x^\prime)=\langle0|\varphi(x)\varphi(x^\prime)|0\rangle$ satisfy the right differential equations and boundary conditions, read about Schwinger-Dyson Equations in QFT. And don't forget that when you canonically quantize a Klein-Gordon field, the canonical momentum $\pi = \partial_t\varphi$ and so $[\varphi(x,t),\partial_t\varphi(x^\prime,t)] = i\hbar\delta(x-x^\prime)$. This will matter in getting the right boundary conditions on the time-slice boundary. | |
| Feb 12, 2012 at 17:46 | comment | added | josh | In how many dimensions you take the limit (i.e. just time or time and space) is sort of a matter of terminology, due to the fact that the $\delta$ function is zero everywhere except one point. For the limit of the heat kernel, for example, all I'm getting at is that if the two time coordinates approach one another and the spatial points are not equal, the result vanishes. But if they are equal and then the time coordinates are made to approach, you get a quantity that behaves like a $d$-dimensional $\delta$ function. | |
| Feb 11, 2012 at 23:46 | comment | added | Nikolaj-K | Very nice answer. I wonder why when you introduce the Kernel, the $lim$ is taken to be w.r.t. the same arguments $x$ and $x'$ as the delta function, but later, you only use times. Also, in statistical mechanics, is the correlation function (which depend on the correlation length and which specify how macroscopic the effect are) a Green(s) function? I don't see any differential equations there. That's generally the problem I think, that I read the name Greens function, where there are no Differential equations and deltas around. Lastly, what about the functions characterizing susceptibilities? | |
| Feb 11, 2012 at 21:28 | history | edited | josh | CC BY-SA 3.0 |
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| Feb 10, 2012 at 17:49 | history | edited | josh | CC BY-SA 3.0 |
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| Feb 10, 2012 at 17:34 | history | edited | josh | CC BY-SA 3.0 |
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| Feb 10, 2012 at 17:22 | history | answered | josh | CC BY-SA 3.0 |