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 $$\begin{equation}\frac{p^2}{2m}=-W_H.\end{equation}$$ 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 $$\begin{equation}A\propto\frac{e^{-(\sqrt{2mW_H}/\hbar)R}}{R} \end{equation}$$ 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?


$$ \newcommand{\ket}[1]{\left| #1 \right\rangle} \newcommand{\bra}[1]{\left\langle #1 \right|} $$ Sooo. It's sometimes a little hard to connect with the way he does it here, because it's a long way from the way I (and I think most of us) think about QM. Not in interpretation or fact or anything, just that it's strange to think of a "negative energy particle hopping around." Let me explain what is meant by an energy shift, and hopefully you can understand how it connects. Let's say you start with a two-state system, where both states are disconnected. Call these states $\ket{1}$ and $\ket{2}$, so the Hamiltonian starts off looking like: $$ H = E_1 \ket{1}\bra{1} + E_2 \ket{2}\bra{2} $$ This is clearly two states, with two energies $E_1$ and $E_2$. In this example, let's even say that $E_1 = E_2 = E$. Now, suppose we add a coupling between these states with strength $a$, that is, the Hamiltonian will now allow for transitions between them: $$ H = E \ket{1}\bra{1} + E \ket{2}\bra{2} + a \ket{1} \bra{2} + a \ket{2} \bra{1} $$ (Assume $a$ is a real number so that $H$ is Hermitian.) Thing is, $\ket{1}$ is no longer an eigenstate of this system. In this case: $$ H \ket{1} = E \ket{1} + a \ket{2} $$ In matrix form, we have: $$ H = \begin{pmatrix} E & a \\ a & E \end{pmatrix} $$ You now have to diagonalize this. I'll spare the details, but the new eigenstates have energies of $E \pm a$. Therefore, the eigenstates of the system with coupling have different energies than the eigenstates without coupling. What Feynman's saying is this: when the electron can jump between atoms, it changes the energies of the system by having a coupling present. It changes the energy more if it can more easily "jump." Since the system will want to be in the lowest energy state, this means it will prefer to "let the electron jump" since this will let it put itself in the $E - a$ state and increase the value of $a$ (the coupling). However, this is balanced by the fact that protons repel each other, so you end up with a potential minimum, which is the binding length of $H_2^+$.

| cite | improve this answer | |
  • $\begingroup$ $$\begin{align}i\hbar\,\frac{d}{dt} {C_1}&=H_{11}C_1+H_{12}C_2,\\[1ex]i\hbar\,\frac{d}{dt} {C_2} &=H_{21}C_1+H_{22}C_2 \end{align}$$ 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? $\endgroup$ – user36790 Sep 24 '15 at 4:29
  • $\begingroup$ Yes. An example in atomic physics would be the AC Stark Shift. $\endgroup$ – zeldredge Sep 24 '15 at 4:30
  • $\begingroup$ 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? $\endgroup$ – user36790 Sep 24 '15 at 4:55
  • $\begingroup$ 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 $\lambda_1 + \lambda_2 = 2E.$ $\endgroup$ – CR Drost Sep 24 '15 at 5:19
  • $\begingroup$ @user36790 Right. The protons can lose energy by staying close together, but only up to a point, because if they get too close the Coulomb repulsion will overwhelm the shift. $\endgroup$ – zeldredge Sep 24 '15 at 13:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy