I'm interested in a proof that the singlet and triplet states of Helium are eigenstates of $\hat S^2$ and $\hat S_z$.

What I know:

In Helium, we have two electrons. Let one electron be in an excited state and the other one in ground state. Now, there are two possible configurations: both electrons have a parallel spin or they have an antiparallel spin.

For the case of parallel spins, we get a total spin of $s = 1$. Thus, the magnetic quantum number can be $m_s = -1, m_s = 0\,\,\text{or}\,\,m_s = 1$. This corresponds to the triplet states. These are given by

$|1,1\rangle = \chi^+ (1) \chi^+ (2)$

$|1,0\rangle = \frac{1}{\sqrt{2}}(\chi^+ (1) \chi^- (2) + \chi^- (1) \chi^+ (2))$

$|1,-1\rangle = \chi^- (1) \chi^- (2)$

For the case of antiparallel spins, we get a total spin of $s = 0$. Thus, the magnetic quantum number is $m_s = 0$. This corresponds to the singlet state. It is given by

$|0,0\rangle = \frac{1}{\sqrt{2}}(\xi^+ (1) \xi^- (2) - \xi^- (1) \xi^+ (2))$.

So far, all seems clear to me, but I can't find a proof that the singlet and triplet states of Helium are eigenstates of $\hat S^2$ and $\hat S_z$.

Could anyone of you help me, please?

  • $\begingroup$ Your $\chi$'s are Pauli spinors are they? $\endgroup$
    – snulty
    Dec 1, 2016 at 17:46
  • $\begingroup$ Yes, that' s right, $\endgroup$
    – Peter123
    Dec 1, 2016 at 18:03

1 Answer 1


So just thinking about Spin and not the helium atom for the moment,

$$S=S_1+S_2, \quad S^2=S_1^2+2S_1\cdot S_2+S_2^2, \quad S_z=S_{1z}+S_{2z}$$

We can also use that $S_+=S_x+iS_y$ and $S_-=S_x-iS_y$ that

$$S_1\cdot S_2=S_{1z}S_{2z}+\frac{1}{2}\left(S_{1+}S_{2-}+S_{1-}S_{2+}\right)$$

Anyway what will be the old spin basis will be common eigenvectors of the commuting observables $$\{S_1^2, S_2^2,S_{1z},S_{2z}\}$$

These could be notated e.g. $\left|\uparrow\uparrow\right\rangle=\left|\uparrow\right\rangle\otimes\left|\uparrow\right\rangle=\chi^+(1)\chi^+(2)$. In the new spin basis we take the vectors to be common eigenvectors of the commuting observables

$$\{S_1^2, S_2^2,S^2,S_{z}\}$$

We just change basis and as you have it in the new basis the eigenstates are

\begin{align} \left|1,1\right\rangle&=\left|\uparrow\uparrow\right\rangle\\ \left|1,0\right\rangle&=\frac{1}{\sqrt{2}}\left(\left|\uparrow\downarrow\right\rangle+\left|\downarrow\uparrow\right\rangle\right)\\ \left|1,-1\right\rangle&=\left|\downarrow\downarrow\right\rangle\\ \left|0,0\right\rangle&=\frac{1}{\sqrt{2}}\left(\left|\uparrow\downarrow\right\rangle-\left|\downarrow\uparrow\right\rangle\right)\\ \end{align}

You can check using the definitions of $S_z$ and $S^2$ that theses are eigenvectors, e.g.

$$S_z\left|1,1\right\rangle= (S_{1z}+S_{2z})\left|\uparrow\uparrow\right\rangle=S_{1z}\left|\uparrow\uparrow\right\rangle+S_{2z}\left|\uparrow\uparrow\right\rangle$$

$$S_1\left|\uparrow\uparrow\right\rangle=\left(S_1\left|\uparrow\right\rangle\right)\otimes \left|\uparrow\right\rangle=\frac{\hbar}{2}\left|\uparrow\uparrow\right\rangle$$ $$S_2\left|\uparrow\uparrow\right\rangle=\left|\uparrow\right\rangle\otimes \left(S_2\left|\uparrow\right\rangle\right)=\frac{\hbar}{2}\left|\uparrow\uparrow\right\rangle$$


$$S_z\left|1,1\right\rangle= \hbar\left|\uparrow\uparrow\right\rangle=\hbar \left|1,1\right\rangle$$

Another example is $S^2$, \begin{align} S^2 \left|1,1\right\rangle&=\left(S_1^2+S_2^2 +2S_{1z}S_{2z}+S_{1+}S_{2-}+S_{1-}S_{2+}\right)\left|\uparrow,\uparrow\right\rangle\\ &=\left(\frac{3}{4}\hbar^2+\frac{3}{4}\hbar^2 +2\frac{\hbar}{2}\frac{\hbar}{2}+0+0\right)\left|\uparrow,\uparrow\right\rangle\\ &=\left(\frac{8}{4}\hbar^2\right)\left|\uparrow,\uparrow\right\rangle\\ &=2\hbar^2\left|1,1\right\rangle\\ &=(s)(s+1)\hbar^2\left|1,1\right\rangle\quad \text{where } s=1. \end{align}

You get the second line similar to the last example and using how the single spin operators act on a single spin, in short:

\begin{align} S^2\left|\uparrow\right\rangle&=\frac{3}{4}\hbar^2\left|\uparrow\right\rangle\\ S_z\left|\uparrow\right\rangle&=\frac{\hbar}{2}\left|\uparrow\right\rangle\\ S_+\left|\uparrow\right\rangle&=0\\ S_-\left|\downarrow\right\rangle&=0 \end{align}

Anyway, back to the helium atom, depending on the levels of approximation (order in perturbation theory etc) you might try write say the ground state as


or with kets,


Written as such the spin operator on acts on the $\chi$ part of the wave function, so if this is a spin singlet or triplet then it is and eigenvalue of $S^2$ and $S_z$, e.g. if $\left|\chi\right\rangle=\left|s,m\right\rangle$ then

\begin{align} S^2\left|\psi\right\rangle&=S^2\left(\left|\phi_{\text{orbital}}\right\rangle\otimes\left|s,m\right\rangle\right)\\ &=\left|\phi_{\text{orbital}}\right\rangle\otimes\left(S^2\left|s,m\right\rangle\right)\\ &=s(s+1)\hbar^2\left|\phi_{\text{orbital}}\right\rangle\otimes\left|s,m\right\rangle\\ \end{align}


\begin{align} S_z\left|\psi\right\rangle&=S_z\left(\left|\phi_{\text{orbital}}\right\rangle\otimes\left|s,m\right\rangle\right)\\ &=\left|\phi_{\text{orbital}}\right\rangle\otimes\left(S_z\left|s,m\right\rangle\right)\\ &=m\hbar\left|\phi_{\text{orbital}}\right\rangle\otimes\left|s,m\right\rangle\\ \end{align}

I don't know all of the details of the exact calculations of Helium ground and excited states, or the energies however.

  • $\begingroup$ @Peter123 No problem, glad to help :) $\endgroup$
    – snulty
    Dec 1, 2016 at 22:00

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