Timeline for Physical meaning of the Yang-Baxter equation
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 30, 2020 at 10:57 | comment | added | Léo S. | Perfect then I got it. Thanks a lot for taking the time to answer all my questions | |
| Apr 30, 2020 at 10:54 | comment | added | user245141 | yes. as you said - as the particles are indistinguishable, and at the end we implicitly take the full symmetric form of the wave function, it doesn't matter if we solve for different orderings of the positions of $x_i$ or the momenta $k_i$, both are equivalent. So I agree with your description in the last two comments. If you want to see an example where the Yang-Baxter equation is not satisfied trivially as in the Lieb-Liniger model you can examine the Bethe Ansatz solution of the Kondo model | |
| Apr 30, 2020 at 10:41 | comment | added | Léo S. | Second, particles are bosons so the wave function is independent under the permutations of the particles ($x_i$'s), so we may assume a specific ordering, and look what happens at the limit, when two particles meet. this given a relation between between $A_{12}$ and $A_{21}$, and since the same relation holds for any pair of particles, we can deduce a relation between $A_{123}=A_{321}$, but since there are two different ways to do it, they must be equal is we hope to have a global solution. | |
| Apr 30, 2020 at 10:35 | comment | added | Léo S. | Thanks. I've been comparing your answer with [tqm.courses.phy.cam.ac.uk/docs/lectures/LiebLinigerModel/…. First, I probably misunderstood $\theta$ (or your model is different) because it seems to me that given an ordering of the $x_i$'s only one term is non-zero (hence the feeling that each separate term is a solution). This is different in equation (13), where the sum is over all the permutations of the phases, and not the positions ($x_i$'s). Permuting the phases make sense from a static point of view, using the fact that particles are indistinguishable. | |
| Apr 29, 2020 at 18:37 | history | edited | user245141 | CC BY-SA 4.0 |
added 158 characters in body
|
| Apr 29, 2020 at 18:11 | comment | added | user245141 | I've edited my answer accordingly | |
| Apr 29, 2020 at 15:34 | history | edited | user245141 | CC BY-SA 4.0 |
added 2162 characters in body
|
| Apr 29, 2020 at 12:08 | comment | added | Léo S. | Thanks. There is still something I don't get though, but explaining myself was to long for a comment so I edited my original question. | |
| S Apr 29, 2020 at 11:38 | history | suggested | Léo S. | CC BY-SA 4.0 |
fixed minor mistake
|
| Apr 29, 2020 at 11:33 | review | Suggested edits | |||
| S Apr 29, 2020 at 11:38 | |||||
| Apr 29, 2020 at 10:07 | comment | added | user245141 | Hm. I'm not sure if this intuition works for me. It seems that you are thinking of a dynamical process where the particles scatter into a new ordering. The point is that we are looking at a static solution, where all different ordering of the particles coexist. The solution is a superposition of all different possible ordering of the particles. As each of this orderings must have a well-defined amplitude, different ways to relate them to each other must agree. It's not that "we may as well consider them equal" but they must be equal otherwise we don't have a valid solution. | |
| Apr 28, 2020 at 18:19 | vote | accept | Léo S. | ||
| Apr 28, 2020 at 18:18 | comment | added | Léo S. | Thanks a lot, this is a very clear explanation. It tried to go further and find some intuition for this consistency (I know it's dangerous to always look for physical intuition in quantum mechanics, but I try anyway). Here is what I came up with: "we only know the initial and final states of the system, that is 123 and 321. As a general rule in quantum mechanics, the system should be the sum of the two paths going from 123 to 321. In particular, there is no way to distinguish between the two: the system is both at the same time, and so we may as well consider there are equal". Is this correct? | |
| Apr 28, 2020 at 15:59 | history | answered | user245141 | CC BY-SA 4.0 |