Timeline for Can we cut/sew path integrals along surfaces which aren't everywhere spacelike?
Current License: CC BY-SA 4.0
9 events
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| Mar 25, 2022 at 2:26 | comment | added | pb1729 | Yes, $U_{\Sigma\to2}$ acts trivially on $\Sigma_2^L$. Overall we can write it as $I\otimes A(J)$, where $A$ is some operator depending on $J$. The identity operator $I$ corresponds to the action on $\Sigma_2^L$. Then write $U_{1\to\Sigma}$ with tensor indices to match up with those of $I$ and $A(J)$: $B_i^{jk}$. The overall product with an initial state $\phi_1^i$ is then: $\delta_j^lA(J)_k^mB_i^{jk}\phi_i=A(J)_k^mB_i^{lk}\phi_i$. | |
| Mar 23, 2022 at 10:10 | comment | added | nodumbquestions | Great, we agree that $J$ doesn't affect $U_{1\to\Sigma}$. Do you also agree that $U_{\Sigma\to 2}$ acts trivially on $\Sigma^2_L$, regardless of the choice of $J$? I'm asking because I find your answer very vague as it stands. | |
| Mar 23, 2022 at 0:10 | comment | added | pb1729 | Nope, $J$ doesn't effect $U_{1\to\Sigma}$. As you say, there's not any way it could affect it. Just the product of $U_{\Sigma\to2}U_{1\to\Sigma}$ is affected. (Along with $U_{\Sigma\to2}$ itself.) | |
| Mar 22, 2022 at 9:44 | comment | added | nodumbquestions | I appreciate the edit, but I don't see how it addresses my concern. Can you clarify: are you claiming that changing $J$ affects $U_{1\to\Sigma}$? If so, how is this possible given that $J$ isn't contained in the domain of the first path integral? | |
| Mar 22, 2022 at 4:33 | comment | added | pb1729 | The intuition with the integral is that we're integrating the product of two functions, and if changing $J$ changes one of the functions, then it will change the result of the overall integral. I've edited my answer to provide a bit more explanation. | |
| Mar 22, 2022 at 4:32 | history | edited | pb1729 | CC BY-SA 4.0 |
Add more clarification and explanation, also add a diagram
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| Mar 21, 2022 at 17:29 | comment | added | nodumbquestions | In other words, I don't see why the equation you wrote describes "how the influence of $J$ on $\Sigma$ interacts with the influence of $\Sigma$ on $\Sigma^2_L$." As far as I can see, there is no influence of $J$ on $\Sigma$, since none of the source lies in the domain of the path integral defining $U_{1\to\Sigma}$. | |
| Mar 21, 2022 at 17:26 | comment | added | nodumbquestions | I don't think this answers my question. You've simply rewritten $U_{1\to 2}=U_{\Sigma\to 2}U_{1\to\Sigma}$ in terms of matrix elements. But the issue remains: $U_{1\to\Sigma}$ doesn't depend on $J$, and $U_{\Sigma\to 2}$ acts trivially on $\Sigma_2^L$; hence the final state on $\Sigma_2^L$ doesn't depend on $J$. | |
| Mar 20, 2022 at 22:45 | history | answered | pb1729 | CC BY-SA 4.0 |