Timeline for Coordinate Bethe Ansatz
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 17 at 12:24 | comment | added | Jules Lamers | @Mathemechanique Glad to hear this was helpful. Re 2: that's right, with some thought you can actually see what the result is without evaluating any commutators. Thanks for accepting the answer, and feel free to it (the triangle pointing up on th left) as well. | |
| Sep 16 at 4:30 | vote | accept | Mathemechanique | ||
| Sep 16 at 4:30 | comment | added | Mathemechanique | Hi Jules, I got them all now. Thanks for your hints. For 2b, I didn't need to use the commutators to obtain the right answer. I never worked with spins and eigenvalue sums before, and when I realized that I should be viewing the operations in 2. in pairs, it made the whole problem much more obvious. Thank you for giving hints and not the answer outright. | |
| Sep 12 at 13:44 | comment | added | Jules Lamers | I don't think that'd be needed, this is a very standard computation. You can just lmk if my hints were enough. | |
| Sep 12 at 7:33 | comment | added | Mathemechanique | I will definitely post my solutions when I have had some time to go through everything. Thanks! | |
| Sep 11 at 18:48 | comment | added | Jules Lamers | It might help to replace the dummy index of one sum by $j$ and carefully use the commutation relations | |
| Sep 11 at 17:42 | comment | added | Mathemechanique | Hi Jules, thanks for the hints. I will try to get to them later today. As for 1, I thought the terms where I took the commutators with the second spin operator in each product were all zero due to i $\neq$ (i+1). This could be one of the points of my confusion. | |
| Sep 11 at 9:19 | history | edited | Jules Lamers | CC BY-SA 4.0 |
another small edit
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| Sep 11 at 6:36 | history | answered | Jules Lamers | CC BY-SA 4.0 |