Single particle tunneling Hamiltonian

Question

In reference to Problem 9, Chapter 2 in Modern Quantum Mechanics by JJ Sakurai,

For a single particle tunneling in a 1D double well potential, with position eigenkets $\mid R\rangle$, $\mid L\rangle$. A general state can be written as: $$ \mid \alpha \rangle=\mid R\rangle\langle R\mid\alpha\rangle + \mid L\rangle\langle L\mid\alpha\rangle $$ The particle can tunnel through the barrier and it says that this tunneling effect can be characterized by the Hamiltonian: $$ H=\Delta\bigg(\mid L\rangle\langle R\mid+\mid R\rangle\langle L\mid\bigg) $$

With the given information the problem can be easily solved as $$ \Delta=+E\\ \mid E+\rangle=\frac{1}{\sqrt{2}}\big(\mid R\rangle+\mid L\rangle\big) $$ and $$ \Delta=-E\\ \mid E-\rangle=\frac{1}{\sqrt{2}}\big(\mid R\rangle-\mid L\rangle\big) $$

But how do we came to know the form of the Hamiltonian for tunneling is the one given above in the first place?

Answer 1

It's simply the most general kind of interaction Hamiltonian you can write down in this simplified two-level system. On the 2D Hilbert space spanned by $\lvert R\rangle,\lvert L \rangle$, the most general linear operator is written as $$ A = a_\text{RR}\lvert R\rangle\langle R\rvert + a_\text{RL}\lvert R\rangle\langle L \rvert + a_\text{LR}\lvert L\rangle\langle R\rvert + a_\text{LL}\lvert L\rangle\langle L\rvert$$ and the first and the last term do not describe an interaction, so we drop them. Self-adjointness of the Hamiltonian also forces $a_\text{RL} = a_\text{LR}^\ast$, and choosing them real, we are left with one free parameter, namely $\Delta = a_\text{RL} = a_\text{LR}$.