# What did Feynman mean by 'energy shift' here?

I was reading Feynman's Lectures III's Chapter 10: Other Two-State Systems. There he discussed about hydrogen molecular ion having two base states:

base states http://www.feynmanlectures.caltech.edu/img/FLP_III/f10-01/f10-01_tc_big.svgz

The amplitude of the molecule to go from $|1\rangle$ to $|2\rangle$ is $-A$.

I, then, came to this stanza

In our theory of the $\text{H}_2^+$ ion we have discovered an explanation for the mechanism by which an electron shared by two protons provides, in effect, an attractive force between the two protons which can be present even when the protons are at large distances. The attractive force comes from the reduced energy of the system due to the possibility of the electron jumping from one proton to the other. In such a jump the system changes from the configuration (hydrogen atom, proton) to the configuration (proton, hydrogen atom), or switches back. We can write the process symbolically as $$(H,p)⇌(p,H)$$. The energy shift due to this process is proportional to the amplitude $A$ that an electron whose energy is $−W_H$ (its binding energy in the hydrogen atom) can get from one proton to the other. For large distances $R$ between the two protons, the electrostatic potential energy of the electron is nearly zero over most of the space it must go when it makes its jump. In this space, then, the electron moves nearly like a free particle in empty space—but with a negative energy! [...] the amplitude for a particle of definite energy to get from one place to another a distance r away is proportional to $$\begin{equation*} \frac{e^{(i/\hbar)pr}}{r}, \end{equation*}$$, where $p$ is the momentum corresponding to the definite energy. In the present case (using the non-relativistic formula), $p$ is given by $$$$\frac{p^2}{2m}=-W_H.$$$$ This means that $p$ is an imaginary number, $$\begin{equation*} p=i\sqrt{2mW_H} \end{equation*}$$ (the other sign for the radical gives nonsense here). We should expect, then, that the amplitude $A$ for the $\text{H}_2^+$ ion will vary as $$$$A\propto\frac{e^{-(\sqrt{2mW_H}/\hbar)R}}{R}$$$$ for large separations $R$ between the two protons. The energy shift due to the electron binding is proportional to $A$, so there is a force pulling the two protons together which is proportional—for large $R$—to the derivative with respect to $R$.

I couldn't conceive what he was meaning by energy shift in the present context. Can anyone explain me what Feynman meant by 'energy shift' here?

• \begin{align}i\hbar\,\frac{d}{dt} {C_1}&=H_{11}C_1+H_{12}C_2,\\[1ex]i\hbar\,\frac{d}{dt} {C_2} &=H_{21}C_1+H_{22}C_2 \end{align} where $H_{21}=H_{12}= -A$. So, the the previous energy level $E_0$ has been split to $E_0 +A$ & $E_0-A$ & the difference between these two is $2A$. Is this the energy shift? – user36790 Sep 24 '15 at 4:29
• So, an attractive force acts between the two protons which bring them close together & is proportional to the derivative of $A$ until the repulsive electrostatic force becomes prominent, is it so? – user36790 Sep 24 '15 at 4:55
• Re: "I'll spare the details", the determinant is a product of eigenvalues and the trace is the sum, so this 2x2 matrix is really easy, as you can just read off $\lambda_1\lambda_2 = E^2 - a^2$ while $\lambda_1 + \lambda_2 = 2E.$ – CR Drost Sep 24 '15 at 5:19