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I've studied that the spin space of an electron is a two-dimensions Hilbert space. A possible representation of this space can be constructed defining: $$\chi_+ = \begin{pmatrix}1 \\ 0 \end{pmatrix} \quad \chi_- = \begin{pmatrix}0 \\ 1 \end{pmatrix} $$
With this base the spin operators became: $$ \hat{S}_n = \frac{\hbar}{2} \sigma_n $$

where $\sigma_n$ is the nth Pauli matrix. If we take the system of two isolated electrons we have a four-dimension Hilbert space and a possible base for this space is: $$ \chi_+(1)\chi_+(2) \quad \chi_+(1)\chi_-(2) \quad \chi_-(1)\chi_+(2) \quad \chi_-(1)\chi_-(2) $$

(I'm ignoring indistinguishability and symmetrization principles because I think are not fundamentals in this question)

During lectures my professor didn't talked about possible representation of this space and what type of product is the one between the two spinors in the base we constructed.

I made some research and I've found out that this product is the tensor product between vector spaces which became the kronecker product in case of finite-dimension space. I applied this product to the base vectors and I've found this representation for the composite system: $$ \chi_+(1)\chi_+(2) = \begin{pmatrix}1\\0\\0\\0 \end{pmatrix} \quad \chi_+(1)\chi_-(2) = \begin{pmatrix}0\\1\\0\\0 \end{pmatrix} \quad \chi_-(1)\chi_+(2) = \begin{pmatrix}0\\0\\1\\0 \end{pmatrix} \quad \chi_-(1)\chi_-(2) = \begin{pmatrix}0\\0\\0\\1 \end{pmatrix} $$

which seemed to me pretty reasonable. Now I was trying to find a representation for the spin operators in this space but applying the kroneker product to the $\hat{S}_z$ operators in the two dimension space with itself: $$ \hat{S}_z \otimes \hat{S}_z = \frac{\hbar^2}{4} \begin{pmatrix}1 &0 &0 &0 \\ 0 &-1 &0 &0\\0& 0 &-1 &0\\ 0 &0 &0 &1\end{pmatrix} $$

But the correct operator should be: $$ \hat{S}_z = \hbar \begin{pmatrix}1 &0 &0 &0 \\ 0 &0 &0 &0\\0& 0 &0 &0\\ 0 &0 &0 &-1\end{pmatrix} $$

What am I doing wrong? How can I correctly build the representation of a composite system starting from the one of a single particle space?

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    $\begingroup$ Actually, two electron (fermion) system require additional statistical rule, so the Hilbert space is actually 3-d. It's tricky to get it into the 4-d rep. $\endgroup$ – RoderickLee Jul 19 '16 at 21:38
  • $\begingroup$ What statistical rule are you talking about? I know that for the symmetrization principle the $\chi_{+-}$ and $\chi_{-+}$ are in fact not acceptable and so you have to use singlet and triplet states. But however I count four possible state, one for s=0 and three for $s=1$. $\endgroup$ – skdys Jul 19 '16 at 21:44
  • $\begingroup$ check my answer below. $\endgroup$ – RoderickLee Jul 19 '16 at 21:46
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Spin operator of the total 2 electron system is tricky: the statistical requirement reduces the Hilbert space to a 3-d rather than 4-d version. Like a spin-1 system, the $S_z$, as represents in basis $|S_z=1,0,-1\rangle$ is (see http://quantummechanics.ucsd.edu/ph130a/130_notes/node247.html, for example) $$S_z=\hbar\left(\begin{matrix}1&0&0\\0&0&0\\0&0&-1 \end{matrix}\right)$$ And, considering $|S_z=1\rangle=|\uparrow\uparrow\rangle=(1,0,0,0)^T$, $|S_z=0\rangle=\sqrt{1/2}(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle)=\sqrt{1/2}(0,1,-1,0)^T$, $|S_z=-1\rangle=|\uparrow\uparrow\rangle=(0,0,0,-1)^T$, we can generalize this into 4-d: $$S_z=\hbar\left(\begin{matrix}1&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&-1 \end{matrix}\right)$$

A better explanation is: you would like to have "Total spin", which should be $S_z\otimes I_{2\times2}+I_{2\times2}\otimes S_z$, which equals to $$\frac{\hbar}{2}\left(\begin{matrix}1&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&-1 \end{matrix}\right)+\frac{\hbar}{2}\left(\begin{matrix}1&0&0&0\\0&-1&0&0\\0&0&1&0\\0&0&0&-1 \end{matrix}\right)$$

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  • $\begingroup$ I understand the reasoning, you started by the possible outcome of the measure and constructed the base by these vectors. But using only 3 dimensions how can I distinguish the singlet state from the triplet state with $m_s = 0$ ? Are not both represented by \begin{pmatrix} 0\\1\\0 \end{pmatrix} ? $\endgroup$ – skdys Jul 19 '16 at 21:55
  • $\begingroup$ @skdys firstly, there should be no triplet m_s=0 state; even if (when spin are localized), see my edited version $\endgroup$ – RoderickLee Jul 19 '16 at 21:56
  • $\begingroup$ Ok, but from where the $S_z\otimes I_{2\times2}+I_{2\times2}\otimes S_z$ came out? What rule I have to use to compose operators of the single system to obtain the one of the composite system? (If it is a question which require a long answer could you give me some references? I'm studying on Griffiths' book but this topic is not present) $\endgroup$ – skdys Jul 19 '16 at 22:03
  • $\begingroup$ @skdys it comes from the definition, where a total spin means sum over individual spin. I_2 means a unit matrix of the space other than this one. It's easily generalized to more particle. For textbook, Sakurai's modern quantum mechanics would be a good choice. Griffiths focuses on wave mechanics, which is not so straightforward in most cases. $\endgroup$ – RoderickLee Jul 19 '16 at 22:06
  • $\begingroup$ @skdys You are confusing the Cartesian product with the coproduct (Kronecker product= addition of angular momenta). The coproduct obeys the same Lie algebra, here SU(2), as the constituent Pauli matrices. See the question 147987. $\endgroup$ – Cosmas Zachos Jul 19 '16 at 22:33

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