# Single particle tunneling Hamiltonian

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?

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 $\Delta = a_\text{RL} = a_\text{LR}$.
• how can we say the first and last term do not describe an interaction, srry do not understand it. And how do you know $a_{RL}=a_{LR}$ ? Apr 10, 2016 at 13:39
• and how do we write if we have two particles in the double well ?. I think in that case there will be a 4D Hilbert space spanned by $\mid LL\rangle, \mid LR\rangle, \mid RL\rangle, \mid RR\rangle$ Apr 10, 2016 at 14:36
• @ss1729: The first and the last term just send a state to itself, that's not an interaction. And I wrote how I know $a_\text{RL} = a_\text{LR}$ - the Hamiltonian has to be self-adjoint. Apr 10, 2016 at 14:45