Tag Info

New answers tagged

1

I don't know if that is the case, but you may have been misguided by the formalism. For convenience, the language of second quantization is often utilized in non-relativistic many body physics. So you may have stumbled upon creation and annihilation operators in that context. Nevertheless, even if written in second quantization, the Hamiltonians of many ...


0

Note that this is a two-site problem, so the many-body Hilbert space is finite (and actually just four). Therefore one can simply express all the operators in terms of $4\times 4$ matrices. Let us order the many-body basis states as $|s_1\rangle=a_{1\uparrow}^\dagger a_{2\downarrow}^\dagger|\Omega\rangle$, $|s_2\rangle=a_{2\uparrow}^\dagger ...



Top 50 recent answers are included