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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?

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  • $\begingroup$ Your $\chi$'s are Pauli spinors are they? $\endgroup$ – snulty Dec 1 '16 at 17:46
  • $\begingroup$ Yes, that' s right, $\endgroup$ – Peter123 Dec 1 '16 at 18:03
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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.

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

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

So,

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

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

$$\psi=\phi_{\text{orbital}}\cdot\chi_{\text{spin}}$$

or with kets,

$$\left|\psi\right\rangle=\left|\phi_{\text{orbital}}\right\rangle\otimes\left|\chi_{\text{spin}}\right\rangle$$

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}

and

\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.

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  • $\begingroup$ @Peter123 No problem, glad to help :) $\endgroup$ – snulty Dec 1 '16 at 22:00

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