# Subspace and symmetry of 4 electrons

For 2 coupled electrons, the possible spin wavefunctions are:

\begin{align*} \chi_{1,1} = \upuparrows,\quad \chi_{1,-1}=\downdownarrows\ \end{align*} \begin{align*} \chi_{1,0}=\frac{1}{\sqrt2}(\uparrow\downarrow+\downarrow\uparrow)\quad(symmetry), \ and\quad\chi_{0}=\frac{1}{\sqrt2}(\uparrow\downarrow-\downarrow\uparrow)\quad(anti-symmetry). \end{align*} I'm now considering the spin wavefunction of 4 electrons, and trying to distinguish different cases yield the total spin equal to 0. Since

\begin{align*} \frac{1}{2}\otimes\frac{1}{2}\otimes\frac{1}{2}\otimes\frac{1}{2} = 2⊕1⊕1⊕1⊕0⊕0 \end{align*} I'm wondering how to spot the two '0' subspaces in the right-hand-side (direct sum part), I think like two electrons case, the total spin wavefunction, in this case, should be anti-symmetric, so should one of them correspond to $$\chi_{0}\otimes\chi_{0}\$$? (which is the superposition of 'superposition states'?) If so, what's the other '0' subspace? Also, the tensor product of which two spin wavefunctions would produce the '0' in other subspaces (like 1 and 2)?

Thanks!!

Frankly, If you only sat down and wrote down all 16 states involved, and confirmed the multiplets, you'd see them all together, including their symmetries, and dispel your magical misconception on symmetry and antisymmetry. (You'd have 6 states of total M=0, of which two combinations would amount to your singlets!)

In any case, distributing $$(1\oplus 0)\otimes (1\oplus 0)= (1\otimes 1 )\oplus (1\otimes 0 )\oplus(0\otimes 1 )\oplus(0\otimes 0 )$$ indicates your two singlets reside in the first and last summand, so in $$\chi_1 \otimes \chi_1$$ and $$\chi_0 \otimes \chi_0$$, both of them in the symmetric part of these Kronecker products.

To see the symmetry of the first, repeat what you did by composing two triplets (9 components): see that the quintet and the singlet in this product comprise the 6 symmetric components, while the triplet comprises the 3 antisymmetric ones.

The symmetry of the second one $$\chi_0 \otimes \chi_0$$ is evident. It is proportional to $$\uparrow\downarrow\uparrow\downarrow - \downarrow\uparrow\uparrow\downarrow +\downarrow\uparrow\downarrow \uparrow - \uparrow\downarrow \downarrow \uparrow .$$ Can you see how a spin raising (or lowering) operator annihilates it? (A singlet cannot have several m states.)

How it is orthogonal to the other singlet of the previous paragraph, $$\uparrow\downarrow\uparrow\downarrow + \downarrow\uparrow\uparrow\downarrow +\downarrow\uparrow\downarrow \uparrow + \uparrow\downarrow \downarrow \uparrow -2 \uparrow \uparrow\downarrow \downarrow -2 \downarrow \downarrow \uparrow\uparrow ,$$ also annihilated by a raising (or lowering) operator?

• Thanks!! Could you explain a bit more about why the first and summand should represent singlets?
– ZR-
Commented Sep 13, 2020 at 15:41
• The last summand is a singlet. In the first, $1\otimes 1=2\oplus 1\oplus 0$. Convince yourself its first and last terms are symmetric and the middle one antisymmetric. Is this what you want? Commented Sep 13, 2020 at 15:45
• Yes! Thank you so much!
– ZR-
Commented Sep 13, 2020 at 15:58
• I tried to write down all the possible states. Can I conclude the symmetry of total wavefunction by considering the symmetry of each part of the tensor product? For example, 0 is symmetric since the tensor product of two symmetric or anti-symmetric wavefunctions is symmetric? Also, $\chi_{1,1}\otimes\chi_{0,1}$ should be anti-symmetric. Is that right? Thanks!!
– ZR-
Commented Sep 13, 2020 at 20:26
• No. Write down the 6 states of M =0, so with two uparrows and two downarrows respectively. Two orthogonal linear combinations of them (which?) are annihilated by raising or lowering operators. They are isolated singlet subspaces. Do you want help finding them? You thus observe their mixed symmetries. Commented Sep 13, 2020 at 20:33