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Consider the two cases outlined in an introductory text of QM:

Case 1: Two particles (of spin 2 and spin 1) are in a box and the total spin is 3 and its $z$ component is $0$, then a measurement of $S^{(1)}_z$ could return the values $\hbar$ (with probability $\frac{1}{5}$), or $0$ (with probability $\frac{3}{5}$), or $- \hbar$ (with probability $\frac{1}{5}$). This is given as

$$ |3 0 \rangle = \frac{1}{\sqrt{5}}|2 1 \rangle|1 -1\rangle + \sqrt{\frac{3}{5}}|2 0 \rangle |1 0 \rangle + \frac{1}{\sqrt{5}}|2 -1 \rangle|1 1 \rangle.$$

Case 2: If you put particles of spin $\frac{3}{2}$ and spin $1$ in the box, and you know that the first has $m_1= \frac{1}{2}$ and the second has $m_2 = 0$ (so $m$ is necessarily $\frac{1}{2}$) and you measure the total spin $s$ you could get $\frac{5}{2}$ (with probability $\frac{3}{5}$), or $\frac{3}{2}$ (with probability $\frac{1}{15}$), or $\frac{1}{2}$ (with probability $\frac{1}{3}$). Again, the sum of the probabilities is $1$. This is given as:

$$ |\frac{3}{2} \frac{1}{2} \rangle |1 0 \rangle = \sqrt{\frac{3}{5}}|\frac{5}{2} \frac{1}{2} \rangle + \sqrt{\frac{1}{15}}|\frac{3}{2} \frac{1}{2} \rangle - \sqrt{\frac{1}{3}}|\frac{1}{2} \frac{1}{2} \rangle$$

Question: Why in the first case is there a probability associated with individual components and in the second probabilities associated with total spin? Does this have something to do with the way the operators of spin of the individual particles commute with the operator of the total spin and total $z$ component?

Thanks.

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  • $\begingroup$ Is that possible related with entanglement state ? $\endgroup$ – Jack Dec 7 '16 at 8:35
  • $\begingroup$ @Jack Only marginally so. $\endgroup$ – G. Bergeron Dec 11 '16 at 11:43
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There are two bases in which you could express the state of the total system. The first is defined by the total-spin operator $S^2$, where $S = S_{(1)} + S_{(2)}$. The second basis is defined by the spin for each particle, given by the operators $S^2_{(1)}$ and $S^2_{(2)}$.

In case 1, you wrote the state $|30\rangle$ that has a well defined total spin: that is, an eigenstate of $S^2$. The equation you wrote expresses this state in a different basis, the basis of eigenstates of $S_{(1)}^2$ and $S_{(2)}^2$.

In case 2, you wrote the state $|\frac{3}{2} \frac{1}{2}\rangle |10\rangle$, which is an eigenstate of $S_{(1)}^2$ and $S_{(2)}^2$, and you expanded it in the eigenbasis of $S^2$.

Both sets of bases are completely valid, and any given state of the system can be expressed in either basis. However, since the operator $S^2$ does not commute with $S_{(1)}^2$ or $S_{(2)}^2$, any eigenstate of $S^2$ must be expanded as a superposition in the other basis, and vice versa. This is why both case 1 and case 2 involve probability amplitudes.

EDIT: As G. Bergeron points out, the coefficients in the conversion between these two bases (in your equations, these are the $1/\sqrt{5}$ type factors) are called Clebsch–Gordan coefficients.

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  • $\begingroup$ Would be nice to talk briefly about Clebsch-Gordan decomposition. $\endgroup$ – G. Bergeron Dec 11 '16 at 4:27
  • $\begingroup$ Good point. I made an edit to note that the Clebsch-Gordan coefficients describe quantitatively how to change between the two bases. $\endgroup$ – Harry Levine Dec 11 '16 at 4:38

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