Energy level splitting in two-state systems I was learning basic quantum mechanics using Feynman Lectures on Physics. In Chapter 8 Feynman has described ammonia inversion due to tunnelling. Feynman has first used the base states as the two possible orientations of ammonia molecule in space. Due to tunnelling effects he has also explained how the opposite configurations of the molecule affect each other. Then he has constructed another set of base states with definite energies $E-A$ and $E+A$ using the superpositions of the base states. It is these base states that I am confused with. Even though I understood the mathematics what do these two states represent physically? How did the two base states interfere to give two other states with different energy levels? Is there an intuitive explanation or a better way to think about it?
These are the chapters I'm referring to: Chapter 8, Chapter 9
I realized that I have a problem in understanding energy splitting in two state systems in general. So I changed the title.
 A: Physically these states represent two possible orientations of ammonia in space: the N atom could be on the two sides of the triangle formed by the hydrogen atoms. Later Feynman discusses in a similar fashion the states in a benzene ring, and many quantum mechanics texts also treat lifting of degeneracy in a two-well potential and the binding in a hydrogen molecule...
This plethora of examples is actually a preview of a more general physical property: the symmetry. Specifically, we deal here with descrete symmetries, which are expressed via point groups. In molecular physics they are used extensively to characterize the energy states and spectra of molecules without actually doing energy calculations! And this applies to molecules by far more complex than the ammonia.
Taken beyond the point groups, the symmetry analysis also forms the basis of crystallography (one has to incorporate the translational symmetry into the picture), while the continuous symmetries lead us to conservations laws (energy, momenta, and angular momenta.)
The real aces in symmetry analysis are usually people working in QFT. I recall Weinberg starting his book with an innocent remark that all particles are just the irreducible representations of the certain symmetry group... but I leave it to the experts to complete my answer on this aspect.
Update
The states with energies $E\pm A$ are actually the symmetric and antisymmetric states, which, following the Feynman's treatment, are superpositions of the original two states with energy $E$:
$$
|\psi_{\pm}\rangle = \frac{1}{\sqrt{2}}\left(|\psi_1\rangle \pm |\psi_2\rangle\right)
$$
There is an important caveat here: the symmetry of these states is guessed correctly, but otherwise the solution is wrong! There are serious mathematical difficulties in formulating the solution in terms of states $\psi_{1,2}$ that are not true eigenstates and the tunneling element. In some contexts this approach is referred to as transfer matrix hamiltonian.
The symmetry of the states is however guessed correctly! Which is why a more advanced but more general treatment in terms of the group symmetry is necessary.
A: There are two important bases here which you shouldn't confuse with eachother. Firstly you have the energy eigenbasis. The vectors of this basis can be found by diagonalizing your Hamiltonian which is the same as solving the time independent Schrödinger equation. The other basis I will call the basis of interest. In this basis $|1\rangle$ could be a state localized below the ammonia molecule and $|2\rangle$ could be above the ammonia molecule.
To make this clear I will consider a simpler example: the double square well (DSW). I will focus on the first two energy states similar to the treatment of the ammonia molecule. In the DSW the potential looks like this

In the case that $V_0=\infty$ this problem reduces to two separate infinite square well potentials and the solutions look like this

Here $a=3,L=3$. The solutions look a bit crooked because they are solved numerically. For the $V_0=\infty$ case both states have the same energy. I will call the left state $|s\rangle$ for being symmetric and the right state $|a\rangle$ for being antisymmetric. These states are eigenstates so if you evolve each of states in time they stay constant.
With a bit of fantasy you can translate this example to the ammonia molecule: negative $x$ corresponds to the nitrogen being below the ammonia molecule and positive $x$ to being above the molecule. Right now these states represent superpositions of being above and below the molecule at the same time. By taking the linear combinations $|L\rangle=\frac{1}{\sqrt 2}(|s\rangle-|a\rangle),|R\rangle=\frac{1}{\sqrt 2}(|s\rangle+|a\rangle)$ we can create a state that is localized on one side ($L$ for left, $R$ for right). These states look like this:

Because $|a\rangle$ and $|s\rangle$ have the same energy our new states $|L\rangle,|R\rangle$ are also eigenstates of the Hamiltonian. So if the particle starts in $|L\rangle$ it will stay in $|L\rangle$. Now for a real molecule $V_0$ is generally not infinite. If we let $V_0$ become finite we get some overlap between the energy states and this changes their shape. It also changes the energy so $|a\rangle,|s\rangle$ are no longer degenerate. The new states look like this:

Here $V_0=4$. Now $|s\rangle$ has a slightly lower energy $|a\rangle$ has slightly higher energy. Now $|L\rangle,|R\rangle$ are no longer eigenstates of the Hamiltonian. If you start in $|L\rangle$ the state will switch to $|R\rangle$ after a while. The particle will tunnel from left to right.
You can write down an approximate Hamiltonian for these states. With $V_0=\infty$ you get
$$H=E_0|a\rangle\langle a|+E_0|s\rangle\langle s|$$
After setting $V_0=4$ you get
$$H=(E_0-\delta)|a\rangle\langle a|+(E_0+\delta)|s\rangle\langle s|$$
where I should note that $|a\rangle,|s\rangle$ are now different vectors. The fact that the enery of both states changes by the same amount $\delta$ is an approximation which holds if $\delta$ is small. If you write this Hamiltonian in the $|L\rangle,|R\rangle$ basis you get off-diagonal elements:
$$H=E_0|L\rangle\langle L|+E_0|R\rangle\langle R|+\delta|L\rangle\langle R|+\delta|R\rangle\langle L|$$
You can check this by plugging in the definition of $|L\rangle,|R\rangle$. So when you see a matrix like
$$H=\pmatrix{E_0&\delta\\\delta&E_0}$$
you should be aware in which basis it is written. To connect this to my first paragraph $|a\rangle,|s\rangle$ form the energy eigenbasis and $|L\rangle,|R\rangle$ form the 'basis of interest'.
