Timeline for Why is the second order perturbative correction to the ground state energy always down?
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| Apr 13, 2017 at 12:39 | history | edited | CommunityBot |
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| Mar 31, 2013 at 10:07 | comment | added | Qmechanic♦ | @ChickenGod: Nice question! Note that we have also used the Ansatz $\langle i^{(0)}|V|i^{(0)}\rangle=0$ to deduce that the eigenvalues of $H$ are more spread that the eigenvalues of $H^{(0)}$. Similarly, we could have used an "opposite" Ansatz $\langle i|V|i\rangle=0$ to deduce that the eigenvalues of $H^{(0)}$ are more spread that the eigenvalues of $H$. However, the latter "opposite" Ansatz is not natural, as it assume knowledge of the full system that we don't have. Thus the situation $H \leftrightarrow H^{(0)}$ is not symmetric, and no paradox arises. | |
| Mar 30, 2013 at 1:32 | comment | added | ChickenGod | For explanation IV), couldn't you invert the argument, pretending that you knew the eigenstates of the full Hamiltonian $H=H^{(0)}+V$ and treat $H^{(0)}=H-V$ perturbatively, and conclude that the eigenvalues of $H^{(0)}$ are more spread out that for $H$? | |
| Jul 16, 2012 at 23:45 | comment | added | Alfred Centauri | More or less related is that the (imaginary part of the) zeros of the Riemann zeta function display this repulsion phenomenon thus the conjecture that the zeros correspond to the eigenvalues of a Hermitian operator | |
| Jul 15, 2012 at 20:38 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
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| Jul 15, 2012 at 8:38 | vote | accept | Ten | ||
| Jul 15, 2012 at 8:36 | vote | accept | Ten | ||
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| Jul 14, 2012 at 16:41 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
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| Jul 14, 2012 at 16:40 | vote | accept | Ten | ||
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| Jul 14, 2012 at 16:07 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
added explanation; added 31 characters in body
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| Jul 14, 2012 at 15:45 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
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| Jul 14, 2012 at 15:28 | history | answered | Qmechanic♦ | CC BY-SA 3.0 |