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Why are singlet states representative of the $s$-orbitals, and similarly why are triplet states representative of $p$-orbitals? I learned about these states as resulting from having a total spin of a total amount-but I don't see why these orbitals would have spin limits.

In particular, in class we were shown that for a system of two electrons it either must exist with spin 1, or spin zero, the former being a triplet, the latter being a singlet. My professor said that these correspond to p and s orbitals, although I don't know why there would be total spin limitations on these orbitals.

How do we prove, then, that P-orbitals should exist in triplets?

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For single particles, S orbitals have angular momentum quantum number $\ell = 0$, and P orbitals have $\ell = 1$. The magnetic quantum number $m$ runs from $-\ell$ to $+\ell$ in integer steps, so S orbitals can only have one value of $m=0$ (hence a singlet state), and P orbitals can have $m=-1,0,1$ (hence a triplet of states). Is this what you are asking?

Edit

What I think your instructor meant is along the following lines. For a two-electron state, the combined wavefunction of the two electron system needs to be antisymmetric under exchange of electrons. The total wavefunction is the tensor product of the spin and spatial parts, so if one is symmetric, the other needs to be antisymmetric.

Note that the singlet $|\downarrow \uparrow\rangle - |\uparrow \downarrow\rangle$, with $s=0$, is antisymmetric under particle exchange, so the wavefunction of the electrons needs to be symmetric. So we can arrange $\Psi(x_1, x_2)$ for the spatial part of the wavefunction to be symmetric. The theory of addition of angular momentum tells us that states with even angular momentum are symmetric under exchange, so in this case $\ell = \ell_1 + \ell_2 = 0, 2, \dots$ is allowed. For the S state ($\ell = 0$), we can just write $\Psi(x_1, x_2) = \psi_s(x_1) \psi_s(x_2)$ where $\psi_s$ is a single-particle wave-function with $\ell_{1,2} = 0$.

For the triplet states, with $s=1$, each spin state is symmetric under particle exchange, so we need to arrange a spatial wavefunction that is antisymmetric. Allowable spatial wavefunctions have $\ell = 1, 3,\dots$. The simplest is $\ell=1$ which is a symmetrized linear combination of $\psi_s$ and $\psi_p$. The exact form of the wavefunction depends on $m_\ell$.

Note though that we are doing all of this from the point of view of the spins. So that while the $S$ state has both particles individually in an $S$ state, the two-particle $P$ state has individual particles in a superposition of $S$ and $P$ states.

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  • $\begingroup$ Possibly... In class we were shown that for a system of two electrons it either must exist with spin 1, or spin zero, the former being a triplet, the latter being a singlet. My professor said that these correspond to p and s orbitals, although I don't know why there would be total spin limitations on these orbitals. $\endgroup$ – user24082 Dec 5 '13 at 20:45
  • $\begingroup$ @Anonymous I think I'm starting to understand your concern. There are triplets, and there are triplets. P orbitals are triplets in that there are 3 degenerate levels, and S orbitals are singlets in the same sense. However, independent of atoms and orbitals, two spin-1/2 particles together can be viewed either as $|1/2,m_1\rangle \otimes |1/2,m_2\rangle$ (the tensor product of two identical Hilbert spaces) or as $|0,0\rangle \oplus |1,m\rangle$ (the direct sum of a singlet Hilbert space and a triplet one). This latter idea doesn't really connect with atomic orbitals, however. $\endgroup$ – user10851 Dec 5 '13 at 21:01
  • $\begingroup$ How do we prove, then, that P-orbitals should exist in triplets? $\endgroup$ – user24082 Dec 5 '13 at 21:33
  • $\begingroup$ I've amended my answer to prove that (spin) triplets should exist in P-orbitals. Is this ok? :) $\endgroup$ – lionelbrits Dec 5 '13 at 23:44

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