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The Hamiltonian is composed of the following terms
{$$ \tag{1} {\cal H} = \underbrace{K_1 + V_{1A} + V_{1B}}_{{\cal H}_1} + \underbrace{K_2 + V_{2A} + V_{2B}}_{{\cal H}_2} + V_{12}$$}
where {$K = -\hbar^2\nabla^2/2m$}, {$V_{i\alpha} = -e^2/4\pi\epsilon_0|\mathbf r_i-\mathbf r_\alpha|$} with {$i=1,2$} electrons and {$\alpha=A,B$} protons, and {$V_{1,2}=e^2/4\pi\epsilon_0|\mathbf r_i-\mathbf r_2|$}.
In the following we omit the vector symbols for the coordinates as a shorthand. In first approximation we disregard {$V_{1,2}$}. The problem becomes separable, since {${\cal H}_i$} operate only on the {$i$}-th electron, i.e. the state of the two electrons is the product of their independent states {$|1\,2\rangle = |1\rangle\|2\rangle$}, each, say, the 1s ground state of hydrogen
{$$\tag{2} {\cal H} |1\rangle\|2\rangle = ({\cal H}_1 + {\cal H}_2)|1\rangle\|2\rangle = E |1\rangle\|2\rangle$$}
Molecular Orbital (MO) solution
The Hamiltonian for each of the two independent electron {${\cal H}_I$}, is that if a H{$_2^+$} ion, and we have already obtained the solution. A rough but simple approximation is to neglect the overlap {$S=\langle A|B\rangle$}
{$$ |i\rangle = \frac {|A\rangle\pm|B\rangle}{\sqrt 2} $$}
where {$|A,B\rangle$} is the hydrogen atom solution on proton A or B respectively and the plus and minus sign refer to bonding and antibonding states. Their eigenenergies are {$E_{\pm}\epsilon_0 + V_c \mp t$} where {$\epsilon_0$} is the hydrogen ground state, {$V_c=\langle A|V_B|B\rangle$} and {$t=-\langle A|V_B|B\rangle$}. In this approximation the eigenvalue of the total energy is simply {$E=2E_\pm$}. Correct treatment yields a qualitatively similar result
{$$E= 2\frac {\epsilon_0 + V_c - St }{1+S^2}$$}
Valence Bond solution
The MO ground state solution is the product four terms: {$|AA\rangle + |AB\rangle+ |BA\rangle+ |BB\rangle$}. Born-Oppenheer approximation is implied, i.e. the solution is the same, formally, irrespective of the distance {$R_{AB}=R_A-R_B$}. This implies that when protons are brought apart the states remain the same, which is unphysical: one would expect them to separate in an atom plus an ion, H and H{$^+$}. The way around this is to choose an ansatz that forces a solution of the type {$|AB\rangle+ |BA\rangle$}. Each term in this solution, upon separation of A and B, lets the single electron localize either on A or on B. This ansatz is called Valence Bond and the energy eigenvalue is the same.
Electron repulsion
The repulsion was ignored above, despite the fact that it has the magnitude of the included attractive potentials. A handwaving argument is that repulsion keeps electrons apart, therefore their interaction energy is smaller. If we include {$V_{12}$} the problem is not analytically solvable, so the first order correction is to include just the average value of the repulsive potential over the independent electron solutions, {$\langle 12|V_{12}|12\rangle$}, which is a two electron term that does not factorize. In this case using the MO or the VB states gives different answers. Moreover, although the operators do not depend on spin, one must be careful and formally include the spin state.
The MO ground state is spatially symmetric under exchange of electrons, whereby for instance {$|AB\rangle$} becomes {$|BA\rangle$} (remember, {$|AB\rangle$} means that the first electron is in the H 1{$s$} state at A an the second at B). Therefore the spin state must be antisymmetric, the so-called singlet
{$$ |S\rangle = \frac {|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle}{\sqrt 2} $$}
There is a single value of the correction,
{$$ \langle 12|V_{12}|12\rangle = \frac {e^2}{4\pi\epsilon_0}\int d\mathbf r_1 \int d\mathbf r_1\frac {|\psi(\mathbf r_1)|^2|\psi(\mathbf r_2)|^2}{|\mathbf r_i-\mathbf r_2|}$$}
Viceversa the VB state comes in two forms that do not have the same energy. They are the spatially symmetric {$|AB\rangle+ |BA\rangle$}, again a singlet, and the spatially antisymmetric {$|AB\rangle - |BA\rangle$}, a spin triplet
{$$ |T\rangle = \begin{cases} & |\uparrow\uparrow\rangle \\ & \frac {|\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle}{\sqrt 2} \\ & |\downarrow\downarrow\rangle \end{cases} $$}
Sorgenti:
For H2+ see http://www.pci.tu-bs.de/aggericke/PC4e/Kap_II/H2-Ion.htm
For H2 and rotational see MIT course5-61 undergrad Chem Phys 2007
For Roto-vibrations see also Wikipedia, hyperphysics and MIT spectroscopy
- Notes in Stato Solido/ RDR lecture notes / Molecole 1
- ione molecolare H2+, schizzo
- molecola H2
- livelli rotazionali
- livelli vibrazionali
- spettroscopia come illustrazione
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