Timeline for Why does a fermionic Hamiltonian always obey fermionic parity symmetry?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Mar 28 at 18:50 | comment | added | Alex Fischer | Why should we expect $H_n$ and $H_m$ to be uncorrelated for distant sites? If your state $|\psi\rangle$ is an entangled state, and $H_n$ and $H_m$ don't commute, then you would generically expect correlations when you measure the local Hamiltonian operator at distant sites. | |
| Jun 1, 2017 at 20:12 | history | bounty ended | tparker | ||
| Jun 1, 2017 at 8:59 | comment | added | Ruben Verresen | @tparker Thanks for pointing out that algebra mistake! I have reshaped the argument as to not rely on that equation. I do think the general message I was trying to make still holds. | |
| Jun 1, 2017 at 8:58 | history | edited | Ruben Verresen | CC BY-SA 3.0 |
correcting claim
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| Jun 1, 2017 at 8:49 | history | edited | Ruben Verresen | CC BY-SA 3.0 |
correcting claim
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| Jun 1, 2017 at 3:59 | comment | added | tparker | Also, there are states of indefinite fermion parity that have $\langle \psi | H_n | \psi \rangle$ for all $n$. For example, the state $$| \psi \rangle = \frac{1}{2} e^{i \phi} \left[ (\cos \theta + \sin \theta) (-| 00 \rangle + | 11 \rangle) + (-\cos \theta + \sin \theta) (| 0 1 \rangle + | 1 0 \rangle) \right]$$ for any angle $\theta$ for a two-fermion system. | |
| Jun 1, 2017 at 3:02 | comment | added | tparker | I can't reproduce your main equation. I get that for a two-fermion system, under the ordering convention $$| \psi \rangle = \sum_{i_1, i_2} c_{i_1, i_2} (a_1^\dagger)^{i_1} (a_2^\dagger)^{i_2} | 0 \rangle,$$ $\langle \psi | H_2 | \psi \rangle = 2\ \text{Re}(c_{01}^* c_{00} - c_{11}^* c_{10})$, but $\langle \psi | a_1 H_2 a_1^\dagger | \psi \rangle = 2\ \text{Re}(c_{10}^* c_{00})$. | |
| May 31, 2017 at 15:06 | history | answered | Ruben Verresen | CC BY-SA 3.0 |