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?
Best Answer
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 $$\frac{\hbar}{2}\left(\begin{matrix}1&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&-1 \end{matrix}\right)+\frac{\hbar}{2}\left(\begin{matrix}1&0&0&0\\0&-1&0&0\\0&0&1&0\\0&0&0&-1 \end{matrix}\right)$$