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Consider a composite of two spin $1/2$ system. The question was to obtain the eigenvalues and to determine the number of eigenstates. Usually one tended to approach by taking the matrix product in the form of $$ \begin{pmatrix} 1 \otimes S_{2z} &0\\ 0& -1 \otimes S_{2z} \end{pmatrix} = \begin{pmatrix} 1&0&0&0\\ 0&-1&0&0\\ 0&0&-1&0\\ 0&0&0&1\\ \end{pmatrix} $$ and solve with respect to the eigenvalue and eigenstates. (Related answer: Total spin of two spin-$1/2$ particles )

However, if one take into account of the total spin, such as $$\vec S^2 - S_z^2$$, where $\vec S=\vec S_1 +\vec S_2$ and $S_z= S_{1z}+S_{2z}$ shown above, though the eigenvalue remained the same, the total number of states were $6$ instead of $4$.

This seemed to suggest that the extension from a single spin $1/2$ system to a composite of two spin $1/2$ system needed further justification, and could not be treated as a simple tensor product, but rather to work with a basis of $|j_1,m_1\rangle \otimes |j_2,m_2\rangle$ or $|J,M\rangle$ (essentially the Clebsch–Gordan coefficients)

How to write $S_{1z}\otimes S_{2z}$ properly? Why in the second case the simple tensor product treatment could not work?

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    $\begingroup$ How were there 6 states? the Hilbert states dimension is 4, and should remain 4. It means that probably the 6 states you got are not linearly independent $\endgroup$
    – user275556
    Commented Dec 9, 2021 at 10:09
  • $\begingroup$ it just means that you have 6 numbers in your matrix, not that the dimension of the matrix is 6 (if I understand you correctly). I will try to write an answer maybe $\endgroup$
    – user275556
    Commented Dec 9, 2021 at 10:13
  • $\begingroup$ The space of two spin-1/2 particles is 4-dimensional. It's still super unclear why you think the number of states is 6. @yyy's answer is perfectly correct for the total spin of 2 spin-1/2 particles. What you are calling the total $j$ ($\hat{J}^2$) is what yyy is calling $s$ ($\hat{S}^2$). $\endgroup$
    – march
    Commented Dec 9, 2021 at 16:47
  • $\begingroup$ Footnotes: The question was WRONG. There was a mistake in the solution manual, where there was a mixed up the quantum numbers from the basis $|j_1,j_2,J,M\rangle$ and the quantum number from the basis $|j_1,m_1\rangle\otimes|j_2,m_2\rangle$ and thereby obtained "more" states. The actual eigenstates were between two $4$ by $4$ matrix representation. $\endgroup$ Commented Dec 9, 2021 at 20:53

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The dimension of the Hilbert space in the stated setup is four, and thus all matrix representations of the operators should be of dimension four. This is independent of basis representation.

To wit: let's start with the spin-addition basis $|J, M\rangle$ where we have the $4$ states $|J=1, M=1\rangle$, $|J=1, M=0\rangle$, $|J=1, M=-1\rangle$ and $|J=0, M=0\rangle$, and we associate them with the basis column vectors in that order (that is $|J=1, M=1\rangle = (1, 0, 0, 0)^T$ etc.). Then $$\vec{S}^2 = \hbar^2\begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} $$ and $$S_z^2 = \hbar^2\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} $$

Now we can work in the separated basis with the four basis states $|m_1=\hbar/2, m_2=\hbar/2\rangle$, $|m_1=\hbar/2, m_2=-\hbar/2\rangle$, $|m_1=-\hbar/2, m_2=\hbar/2\rangle$, $|m_1=\hbar/2, m_2=-\hbar/2\rangle$ and we associate them again with the basis column vectors in that order. The matrix representation of $S_z^2$ is still diagonal $$S_z^2 = \hbar^2\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$ but we get non-diagonal terms for the representation of $\vec{S}^2$. However calculating $\langle m_1, m_2 | \vec{S}^2 | m_1', m_2' \rangle$ is quite straight-forward, using CG. We have two states that can only have non-zero matrix elements with themselves (as they are also states in the momentum-added basis): $$ \langle m_1=\uparrow, m_2=\uparrow | \vec{S}^2 | m_1', m_2' \rangle = 2\hbar^2\delta_{m_1', \uparrow}\delta_{m_2',\uparrow}$$ $$ \langle m_1=\downarrow, m_2=\downarrow | \vec{S}^2 | m_1', m_2' \rangle = 2\hbar^2\delta_{m_1', \downarrow}\delta_{m_2',\downarrow}$$ and have a non-diagonal matrix elements when we look at the subspace associated with total $S_z = 0$: $$ \langle m_1=\uparrow, m_2=\downarrow | \vec{S}^2 | m_1=\uparrow, m_2=\downarrow \rangle = \langle m_1=\downarrow, m_2=\uparrow | \vec{S}^2 | m_1=\downarrow, m_2=\uparrow \rangle = \hbar^2$$ $$ \langle m_1=\hbar/2, m_2=\downarrow | \vec{S}^2 | m_1=\downarrow, m_2=\uparrow \rangle = \hbar^2$$ so we end up getting $$\vec{S}^2 = \hbar^2\begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 2 \end{pmatrix} $$

You can now convince yourself that the two representations are related to one another by basis transformations, which is given exactly by the CG coefficients (which is not surprising - this is how we got the matrices in the first place).

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  • $\begingroup$ I'm probably missing something very fundamental. Can you explicitly write these 6 states please? $\endgroup$
    – user275556
    Commented Dec 9, 2021 at 10:51
  • $\begingroup$ There's a total spin quantum number $j$ involved, in addition to the $m_1,m_2$ (en.wikipedia.org/wiki/… ) $\endgroup$ Commented Dec 9, 2021 at 11:05
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    $\begingroup$ in the separated basis the total $j$ is not a good quantum number. Again - can you explicitly write the 6 states please? I think it would greatly help me understand what do you mean and where is my mistake $\endgroup$
    – user275556
    Commented Dec 9, 2021 at 11:10
  • $\begingroup$ The solution manual was wrong, it mixed up the quantum numbers from the basis $|j_1,j_2,J,M\rangle$ and the quantum number from the basis $|j_1,m_1\rangle\otimes|j_2,m_2\rangle$ and thereby obtained "more" states. Your solution were correct. $\endgroup$ Commented Dec 9, 2021 at 20:50

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