When dealing with a system of two spin-$1/2$ particles, we use the singlet and triplet states as a common basis for the complete set of commuting observables $\{\hat S ^2,\hat S_z, \hat S_{z,1}, \hat S_{z,2}\}$, where $\hat {\mathbf S}=\hat {\mathbf S}_1+\hat {\mathbf S}_2$:

$$ \left.\begin{array}{ll} |1,1\rangle & =\uparrow \uparrow \\ |1,0\rangle & =\frac{1}{\sqrt{2}}(\uparrow \downarrow+\downarrow \uparrow) \\ |1,-1\rangle & =\downarrow \downarrow \\ \end{array}\right\} \quad s=1 \quad \text { (triplet) } \\ \left.|0,0\rangle=\frac{1}{\sqrt{2}}(\uparrow \downarrow-\downarrow \uparrow)\right\} \quad s=0 \quad \text { (singlet) } $$

It is said that the singlet state has lower energy than those states of the triplet, which I think makes sense since in the singlet state there are no unpaired particles, whereas in the triplet state the particles are unpaired. My question is: what is the energy associated to these states? And, are states of the triplet state all of the same energy?


1 Answer 1


It's not necessarily the case, It depends on your Hamiltonian. It's true for the case of Hyperfine interaction, where in addition to the Coulomb interaction, there exists another interaction between the electron and proton in the Hydrogen atom. The Hamiltonian describing this interaction, which is due to the magnetic moments of the two particles is, $$H_{\text{hf}}=A\mathbf{S}_1\cdot \mathbf{S}_2\ \ \ (A>0)$$ $H_\text{hf}$ splits the ground state into two levels: \begin{align*} E_+&= -\text{Ry}+\frac{\hbar^2A}{4}\\ E_- &= -\text{Ry}-\frac{3\hbar^2 A}{4} \end{align*}

and that corresponding states are triplets and singlet, respectively.

  • $\begingroup$ And, for example, if the two particles were two electrons behaving each one like a 1D quantum harmonic oscillator? Would the three states of the triplet state equally energetic? What would the contribution of the spin part be to the total energy of the system? $\endgroup$
    – Invenietis
    Commented May 22, 2021 at 10:43
  • 1
    $\begingroup$ In one or two dimensional, Hamiltonian doesn't have a spin-dependent part, so energy will be the same. $\endgroup$ Commented May 22, 2021 at 10:45

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