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For the last part, I'm not sure what they mean by "explain how to form eigenstates of the total spin $\hat S^2$ and $S^z = S_1^z + S_2^z$. Are they simply referring to the spin singlet and tripplet states?

I know for tripplet states total spin is 1, the states are $|S^2,m_{s1} + m_{s2} \rangle$: |1,1>,|1,-1> and |1,0> and for the singlet states total spin is 0, the only state is |0,0>.

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You know that the total spin operator is given by

$$ {\bf S}^2 = ({\bf S_1} + {\bf S_2})^2 = {\bf S_1}^2 + {\bf S_2}^2 + 2 {\bf S_1}\cdot {\bf S_2}$$

and the most general state is given by

$$\left|s,m\right> = a\left|\uparrow\uparrow\right> + b\left|\downarrow\downarrow\right> + c\left|\uparrow\downarrow\right> + d\left|\downarrow\uparrow\right>$$

You need to show that you can choose coefficients $a,b,c,d$ such that

$${\bf S}^2\left|s,m\right> = \lambda \left|s,m\right>$$

This is turns out to be the singlet and triplet states you mention above. To answer the question it might be enough to take these states and show that they are indeed eigenstates of ${\bf S}^2$ and $S_z$ by explicitly calculating ${\bf S}^2\left|s,m\right>$ and $S_z\left|s,m\right>$. Or you can calculate ${\bf S}^2\left|s,m\right>$ on the most general state and show that the only way it can be an eigenstate is if its the singlet or the triplet state.

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  • $\begingroup$ Also, you can derive these states based on addition of two spin angular momenta of spin-half particles by working out the clebsch-gordon coeffients. $\endgroup$
    – user44840
    Jun 19, 2014 at 13:34
  • $\begingroup$ Yes, and I think thats probably the best way to do it. $\endgroup$
    – Winther
    Jun 19, 2014 at 19:30

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