Timeline for complex version of Gaussian integral
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Jan 26, 2017 at 9:16 | comment | added | Philipp Wacker | Thanks for your help. Could you give a reference on analytical continuation of inner products from real Hilbert spaces to their complex expansion so I can read up on that? | |
| Jan 26, 2017 at 9:06 | vote | accept | Philipp Wacker | ||
| Jan 26, 2017 at 9:03 | comment | added | Philipp Wacker | OK, I see now what my mistake was. I mixed up the original real and the new artificial complex Hilbert space. I will try to figure out the rest. | |
| Jan 26, 2017 at 8:52 | comment | added | Philipp Wacker | yes. Although I'm not sure how to make sense of "$y$ is real". | |
| Jan 26, 2017 at 8:22 | comment | added | Carlo Beenakker | @FasEtNefas --- you only take $b$ complex, the $y$ is still real, right? so you are not integrating over a complex Hilbert space. | |
| Jan 26, 2017 at 6:41 | comment | added | Philipp Wacker | But still, if we start with a complex Hilbert space to begin with, then we can't just change the fact that our inner product needs to be positively definite. Right? Actually, I think that maybe there needs to be a slight modification to the original proof of DPZ, but I first need to clarify this (lest I make a fool of myself ;)). | |
| Jan 25, 2017 at 14:36 | comment | added | Carlo Beenakker | here $\langle\cdots,\cdots\rangle$ follows from analytical calculation of the real inner product, so indeed $\langle \alpha a,\beta b\rangle=\alpha\beta\langle a,b\rangle$ without any complex conjugation of the coefficients $\alpha,\beta$. Analytical continuation does not conserve the positivity, as you can immediately see because $\alpha^2<0$ for imaginary $\alpha$. | |
| Jan 25, 2017 at 14:14 | comment | added | Philipp Wacker | So on the premises stated in my previous comment, I see that your proposal works (as the calculation yields the "correct" result). My problem is that the $\langle \cdot,\cdot\rangle$ inner product is not positively definite anymore (but it needs to be as it is the inner product of a Hilbert space $H$). I suggest to slightly adapt this in the following way: Use the "normal" inner product, but write $\langle b_1 + ib_2, L\cdot (b_1 - ib_2)\rangle$, where $L$ is the complicated operator in the expression above. Then the calculation proceeds as I'd like it too. What do you think? | |
| Jan 25, 2017 at 13:50 | comment | added | Philipp Wacker | Now I'm confused: Is it $\langle c, b_1+ib_2\rangle = \langle c, b_1 \rangle + i \langle c, b_2\rangle$ or with the other sign? Because for the formula to yield the result I want in the end, I need this. Also, I need $\langle i b, ic\rangle = -\langle b, c\rangle$. So the key is to use a non-positive inner product? How can we justify this in this context? | |
| Jan 25, 2017 at 13:42 | comment | added | Philipp Wacker | Thanks, I will take a look at it and see whether it works out. | |
| Jan 25, 2017 at 13:36 | comment | added | Carlo Beenakker | added a clarification | |
| Jan 25, 2017 at 13:36 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
added 200 characters in body
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| Jan 25, 2017 at 13:32 | comment | added | Philipp Wacker | But how does this address my concern that the l.h.s. will have a nonzero imaginary part and the r.h.s will be purely real? | |
| Jan 25, 2017 at 13:23 | history | answered | Carlo Beenakker | CC BY-SA 3.0 |