Timeline for When constructing Hamiltonian matrix for many spins, what is the significance of the order of factors in the outer product?
Current License: CC BY-SA 3.0
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| Nov 21, 2017 at 22:04 | comment | added | ZeroTheHero | @DanielJordan the notation of the text is not so easy at first glance. | |
| Nov 21, 2017 at 22:03 | comment | added | Daniel Jordán | Thanks for your answer, it does explain the problem, I was just forgetting that the Hamiltonian written above only specifies interactions between first neighbors, so when the third particle was added the outer product was done from the right and did not neet to include any interaction between particles 1 and 3 (no periodic boundary conditions here). | |
| Nov 21, 2017 at 21:57 | vote | accept | Daniel Jordán | ||
| Nov 14, 2017 at 6:45 | history | edited | ZeroTheHero | CC BY-SA 3.0 |
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| Nov 14, 2017 at 3:54 | comment | added | Daniel Jordán | Yes, the $I_2$ being the 2x2 identity matrix I got, what and it makes sense to me to multiply it after the $H_2$ so as to expand the hilbert space. However, I don't understand why the tilde matrix has the identity first. I'm trying to wrap my head around this so that I can learn how to do any other construction later on. One thing I didn't mention is that the book uses the image of two spins side by side, where in the original Hamiltonian, the #1 operator acts on the "left" spin and the #2 operator on the "right" spin. Then the third spin is added to the right of the pair. | |
| Nov 14, 2017 at 2:13 | history | answered | ZeroTheHero | CC BY-SA 3.0 |