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An electron is orbiting two protons. With the Born-Oppenheimer approximation that the protons do not move, I'd write the Hamiltonian of the electron's movement as:

$$ \mathbf{H} = -\frac{\hbar^2}{2m}\nabla^2 + E_p$$


$$E_p = -\frac{e^2}{4\pi \epsilon_0}\left(\frac{1}{r_1}+\frac{1}{r_2}\right) $$

as the potential caused by the protons, with $r_1$ and $r_2$ denoting distances to the protons. Apparently it should be

$$E_p = -\frac{e^2}{4\pi \epsilon_0}\left(\frac{1}{r_1}+\frac{1}{r_2}-\frac{1}{r_0}\right), $$

where $r_0$ denotes the distance between the protons. How can that term be explained?

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The protons are interacting with each other, too, don't they? – Luboš Motl Apr 11 '12 at 8:32
@LubošMotl Why would their interaction with each other affect the electron? My suggestion for $E_p$ describes the potential felt by any test charge – Per Apr 11 '12 at 8:34
Lubos and Vijay are both giving the correct answer. The reason we care is this: think about $r_0$ as a variational parameter. Classically, the electron wants to be "close" to both protons, but the two protons repel one another. Having a quantum ground state energy that depends on $r_0$ lets you determine the ideal distance between the protons that minimizes the total energy. (i.e., just because we make the B.O. approximation that the protons don't move, that doesn't mean we should just put them wherever.) – wsc Apr 11 '12 at 14:13
Proton-proton interaction does not affect the electron, but it is involved into the total molecule energy. – Vladimir Kalitvianski Apr 11 '12 at 14:42
Thank you everyone, for helping me to understanding this. – Per Apr 12 '12 at 15:18
up vote 2 down vote accepted

The $\frac{e^2}{4\pi\epsilon_0}\frac{1}{r_0}$ term appears in the potential for the electron motion, as Luboš and Vijay point out, to keep the whole energy accounting in place so that the nuclear motion can be properly quantized. The key point is that this potential does not involve the electron coordinates, so that as far as the electronic wavefunction is concerned it acts like a constant and therefore does not affect the solution to the electronic eigenvalue equation.

If you do include that term, and write the potential energy of the molecule as $$E_p=-\frac{e^2}{4\pi\epsilon_0}\left(\frac{1}{r_1}+\frac{1}{r_2}-\frac{1}{r_0}\right)$$ then you can write the potential for the nuclear coordinates directly as $$E_n=\langle\psi(\mathbf{R}_1,\mathbf{R}_2)|E_p|\psi(\mathbf{R}_1,\mathbf{R}_2)\rangle$$ where the total wavefunction is split into electronic and nuclear parts as $$\langle \mathbf{r},\mathbf{R}_1,\mathbf{R}_2|\Psi\rangle=\langle\mathbf{r}|\psi(\mathbf{R}_1,\mathbf{R}_2)\rangle\langle\mathbf{R}_1,\mathbf{R}_2|\phi\rangle.$$ The Schrödinger equation for the nuclear coordinates is then $$\left(\frac{\mathbf{p}_1^2}{2M}+\frac{\mathbf{p}_2^2}{2M}+E_n\right)|\phi\rangle=E|\phi\rangle.$$

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Is'nt $$\frac{e^2}{4\pi\epsilon_0 \, r_0}$$ the (constant) electrostatic energy of repulsion between the two stationary protons at a separation $r_0$?


The full problem of $H_2^+$ molecule $\mathcal{H} \Psi = \mathcal{E} \Psi$ is given by the Hamiltonian $$\mathcal{H} = \frac{\mathbf{p}^2}{2m} + \frac{\mathbf{P}_1^2}{2M} + \frac{\mathbf{P}_2^2}{2M} - \frac{e^2}{4\pi\epsilon_0} \Big(\frac{1}{|\mathbf{r}_1|} + \frac{1}{|\mathbf{r}_2|} - \frac{1}{|\mathbf{R_1}-\mathbf{R_2}|} \Big)$$ where $\mathbf{r}$ is the position of the electron, $\mathbf{R}_i$ the nuclei location, and $\mathbf{r}_i = \mathbf{r} - \mathbf{R}_i$.

In the Born-Oppenheimer approximation for massive nuclei, we take $$\Psi(\mathbf{r}, \mathbf{R}_1, \mathbf{R}_2) = \phi(\mathbf{R}_1, \mathbf{R}_2) \, \psi(\mathbf{r}, \mathbf{R}_1, \mathbf{R}_2)$$ where it is assumed that the $\mathbf{R}_i$ are treated as constant parameters in $\psi$. The nuclear and electronic motions then decouple.

One can write the decoupled equations in two ways

(1) including the electrostatic energy of the protons in the electronic equation $$\Big[ \frac{\mathbf{p}^2}{2m} - \frac{e^2}{4\pi\epsilon_0} \Big(\frac{1}{|\mathbf{r}_1|} + \frac{1}{|\mathbf{r}_2|} - \frac{1}{|\mathbf{R_1}-\mathbf{R_2}|} \Big) \Big] \psi = E \psi$$ $$\Big[ \frac{\mathbf{P}_1^2}{2M} + \frac{\mathbf{P}_2^2}{2M} + E \Big] \phi = \mathcal{E} \phi$$

(2) including the electrostatic energy of the protons in the equations for the nuclei $$\Big[ \frac{\mathbf{p}^2}{2m} - \frac{e^2}{4\pi\epsilon_0} \Big(\frac{1}{|\mathbf{r}_1|} + \frac{1}{|\mathbf{r}_2|} \Big) \Big] \psi = E' \psi$$ $$\Big[ \frac{\mathbf{P}_1^2}{2M} + \frac{\mathbf{P}_2^2}{2M} + \frac{e^2}{4\pi\epsilon_0} \frac{1}{|\mathbf{R_1}-\mathbf{R_2}|} + E' \Big] \phi = \mathcal{E} \phi$$ where $E$ and $E'$ are the electronic energy contributions, and are functions of the $\mathbf{R}_i$.

In both the cases above, you solve the equation for $\psi$ assuming constant $\mathbf{R}_i$. The electronic energy $V_1 = E(\mathbf{R}_1, \mathbf{R}_2)$ then acts as an effective potential for the motion of the nuclei in case (1) above while in case (2) the nuclei experience an effective potential energy $V_2 = \frac{e^2}{4\pi\epsilon_0} \frac{1}{|\mathbf{R_1}-\mathbf{R_2}|} + E'(\mathbf{R}_1, \mathbf{R}_2)$.

It looks like your course material refers to case (1) above, and this is standard in many places, for e.g., Chap 7 of Atkins. I also found places where case (2) is considered.

In either case, the effective potential experienced by the nuclei (either $V_1$ or $V_2$) seems to have the same structure. So probably either case is fine though I would like to know if there is any subtle difference between them.

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Yes, but I don't understand why their repulsion would affect the potential. Each proton is already taken account for. – Per Apr 11 '12 at 8:32
You said "With the Born-Oppenheimer approximation that the protons do not move...". How are the protons take into account already? You also say "Apparently it should be..." Can you give a reference? – Vijay Murthy Apr 11 '12 at 8:49
Each proton gives rise to a potential proportional to $1/r_i$, and these potentials are additive. My reference is the answer to this question in Swedish course material; I'll translate it for you if you think it would help in finding what I'm misunderstanding. But both you and Lubos seem to agree that $1/r_0$ should indeed be there. – Per Apr 11 '12 at 8:52
@Per a translation would probably be a good thing to edit into the question, if the material implies that the $r_0$ term shouldn't be there. – David Z Apr 11 '12 at 15:14
@DavidZaslavsky: Sorry, I was unclear: the material agrees with everyone here that $r_0$ should be there, it's just that I don't (didn't) agree. – Per Apr 12 '12 at 15:12

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