# What is the (triplet) eigenstate for two electrons? [duplicate]

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Let's assume I have a harmonic oscillator which is one dimensional. What is my plan is to work the the two electron's spin states and my requirement is that they have to be in the triplet sates.

Lets also consider that wave function which is $$\Psi (r_1,r_2) \ \chi(1,2)$$

If the radial wave function is antisymmetric, the spin part has to be symmetric to have two electron triplet states under the given conditions.

There are three possible values of $$m_z$$ (1,0,-1) in the triplet states which are symmetric spin states. But if I asked to write down, how would be the spin state of the total function?

Is it going to be something like this:

$$\chi(1,2) = a |{\uparrow \uparrow} \rangle + b |{\downarrow \downarrow} \rangle+ c \frac{1}{\sqrt{2}}\Big(|{\uparrow \downarrow} \rangle + |{\uparrow \downarrow}\rangle\Big)$$

The reason I did this because I thought all the spin states could be possible.

## marked as duplicate by ZeroTheHero, Thomas Fritsch, Jon Custer, Aaron Stevens, John Rennie quantum-mechanics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 16 at 9:23

• The question in your title does not match the one in body of your question (which does not mention an operator, much less an eigenstate of one) – By Symmetry Aug 15 at 10:02
• What about now? – user193422 Aug 15 at 10:44
• Or can you suggest one from your understanding so I can better understand your question? – user193422 Aug 15 at 10:55
• I've introduced proper normalization on your states ─ this is the usual convention. – Emilio Pisanty Aug 15 at 13:08

Is it going to be something like this:

$$\chi(1,2) = a |{\uparrow \uparrow} \rangle + b |{\downarrow \downarrow} \rangle+ c \frac{1}{\sqrt{2}}\Big(|{\uparrow \downarrow} \rangle + |{\uparrow \downarrow}\rangle\Big)$$

Yep, that's correct. The triplet state manifold is (triply) degenerate, which means that it contains three equally-valid linearly-independent eigenstates. Any one of them can be chosen, or any suitable linear combination of them.

The state you've written is the most general form of the (pure) states in that eigenspace. In particular, this is a valid answer to

But if I asked to write down, how would be the spin state of the total function?

but the precise details depend on exactly what the question was phrased. Some state preparation procedures will produce a specific sub-choice of magnetic-quantum-number eigenvalue, which will set one of $$a$$, $$b$$ and $$c$$ to unity and the rest to zero; other preparation procedures will produce specific linear combinations of them.

In contrast, some preparation procedures (say, exciting from some ground state using collisional excitation with a beam of electrons of very precise energy) will be unable to generate enough coherence, and instead they will produce a mixed state of the three basis states in the triplet manifold.

So, the long answer is "it depends".

If the Hamiltonian is just one-body Hamiltonian in the form of harmonic oscillator without any additional term, I think you are right. In this case, all three triplet states are possible and are degenerate.

Let's say, the harmonic oscillator Hamiltonian gives you one-electron eigenstates $$\phi_i(r)$$, with $$i$$ labeling different states. When the radial part is antisymmetric, it is $$\Psi(r_1,r_2) = \phi_i(r_1)\phi_j(r_2) - \phi_j(r_1)\phi_i(r_2)$$ with $$\phi_i$$ different with $$\phi_j$$. Being fermions, the spin part $$\chi(1,2)$$ needs to be symmetric, and $$|{\uparrow\uparrow}\rangle$$, $$|{\downarrow\downarrow}\rangle$$ and $$|{\uparrow\downarrow}\rangle + |{\downarrow\uparrow}\rangle$$ are degenerate and free to choose. Together with the spin singlet state, you will have four two-electron states (assuming the two electrons stays at i and j). You can label these with the basis $$\left|S, S^z\right\rangle$$, where $$S$$ is the total spin and $$S^z$$ its $$z$$-component.

• Spin singlet: $$|0,0\rangle = [\phi_i(r_1)\phi_j(r_2) + \phi_j(r_1)\phi_i(r_2)] [\left|{\uparrow\downarrow}\right\rangle - \left|{\downarrow\uparrow}\right\rangle]$$

• Spin doublets: \begin{align} |1,1\rangle &= [\phi_i(r_1)\phi_j(r_2) - \phi_j(r_1)\phi_i(r_2)] \left|{\uparrow\uparrow}\right\rangle \\ |1,0\rangle &= [\phi_i(r_1)\phi_j(r_2) - \phi_j(r_1)\phi_i(r_2)] [\left|\uparrow\downarrow\right\rangle + \left|{\downarrow\uparrow}\right\rangle] \\ |1,-1\rangle &= [\phi_i(r_1)\phi_j(r_2) - \phi_j(r_1)\phi_i(r_2)] \left|{\downarrow\downarrow}\right\rangle \end{align}

Things may change when you involve other terms. Say the electron-electron Coulomb interaction term $$\frac{e^2}{|r_1-r_2|}$$. It will favor parallel spin states $$\left|1,\pm1\right\rangle$$ over antiparallel spin states. This effect is also known as direct exchange.