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Triplet vs. Singlet diatomic molecules: Why the energy difference?

  1. Mar 17, 2009 #1
    I've been doing some research on singlet-triplet molecular states, and one things I can't perfectly settle is a rigorous demonstration of why the triplet state is a higher energy than the singlet.

    One way I can qualitatively understand why the singlet state has lower energy is this: If the atoms are far apart then it does not matter much whether the spin state is triplet or singlet. But when the atoms come close together to form a molecule then we can view the two valence electrons to have almost similar position co-ordinates. It is then necessary that Pauli exclusion be strictly obeyed, and for molecule formation it's would thus be much more favorable to have spins aligned oppositely. For a triplet state with spins pointing in the same direction it is impossible for the atoms to form a molecule due to Pauli exclusion principle.

    That, however, is an insufficient answer for me. It's too much "it makes sense if we want it too, but not a priori". Anyone have a good handle on this?
  2. jcsd
  3. Mar 17, 2009 #2


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    Well, it's not quite right either. Let's think about the simplest possible case, two hydrogen atoms come together to form a molecule.

    Either hydrogen atom is either in their doublet ground state 2S. When they come together they can combine in two possible ways, forming either a singlet [tex]^1\Sigma[/tex] state or a triplet [tex]^3\Sigma[/tex] state. Which one is lower in energy?

    Well, the resulting two-particle wavefunction can be written as:
    [tex]\Psi(r_1,r_2,s_1,s_2) = \varphi(r_1,r_2)\sigma(s_1,s_2)[/tex]
    Where r are the electronic coordinates, and phi and sigma are the spatial- and spin-coordinate parts of the wavefunction.

    Now, we're talking about electrons - fermions, here. So the overall wavefunction must be anti-symmetric with respect to the exchange of electrons, that is:
    [tex]\Psi(r_1,r_2,s_1,s_2) = -\Psi(r_2,r_1,s_2,s_1)[/tex]

    In the singlet state, where the spins are antiparallel, the spin-coordinate part of the wavefunction is asymmetrical, and hence, the spatial-coordinate part is symmetrical.
    In the triplet state, where the spins are parallel, then the spin-coordinate part of the wavefunction is symmetrical, and the spatial-coordinate part is asymmetrical.

    What does this mean in terms of energy? Well, since the triplet spatial wavefunction is anti-symmetrical, it must have a node where r1=r2. Since the singlet spatial wavefunction has no node, it is lower in energy.
  4. Nov 18, 2009 #3

    I am not sur but I think the triplet is more stable than the singulet. We have S0, with the more little energy, after T1 and after S1. We see for the same number of electronic state than the triplet 1 is less in energy that the singulet 1. In my course the reason is du to the spin correlation effect: the electrons have a electronic and magnetic part. If the spin are in the same direction, triplet state, they are a magnetic repulsion and the electrons are farther. This gives a less electronic repulsion, thus an lower energy state.

    I am not very convinced with this theory and I such more informations about spin correlation effect, but I don't find anything.....
  5. Nov 19, 2009 #4


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    The singlet state is usually lower in energy in diatomic molecules.

    Offhand the only diatomic I can think of that has a triplet ground state (at room temperature) is oxygen.
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