Don't the spins in the state $\dfrac{1}{\sqrt{2}}(\uparrow\downarrow+\downarrow\uparrow)$ cancel each other so that the total spin is 0 just like for the singlet state $\dfrac{1}{\sqrt{2}}(\uparrow\downarrow-\downarrow\uparrow)$?
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3$\begingroup$ Closely related question here. $\endgroup$– knzhouCommented Jan 29, 2019 at 15:52
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1$\begingroup$ Another possible duplicate. $\endgroup$– rob ♦Commented Jan 29, 2019 at 19:19
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$\begingroup$ Try $S_-$ and $S_+$. $\endgroup$– my2ctsCommented Mar 17, 2021 at 17:36
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2 Answers
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This state, like $\frac{1}{\sqrt{2}}\left(\uparrow\downarrow -\downarrow\uparrow
\right)$, is an eigenstate of $L_z$ with $m_s=0$. However, if you act on
$$
\vert\psi\rangle =\frac{1}{\sqrt{2}}\left(\uparrow\downarrow +\downarrow\uparrow
\right)
$$
with $\hat L_+$ or $\hat L_-$ you will not get $0$. Rather, for instance,
$$
\hat L_+\vert\psi\rangle =\sqrt{2}\uparrow\uparrow
$$
with which is an eigenstate of $L_z$ with eigenvalue $1$. Since a state
with spin $0$ would be killed by $L_+$ and $L_-$, your state cannot have spin-$0$.
In fact since $\hat L_+\hat L_+\vert\psi\rangle=0$ you can deduce that $\hat L_+\vert\psi\rangle$ is proportional to the spin-$1$ state with $m_s=1$. Using $L_-\hat L_+\vert\psi\rangle$ will provide you with a state proportional to $\vert\psi\rangle$, which must have the same value of $s=1$.
answered Jan 29, 2019 at 16:02
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$\begingroup$ If I do a spin measurement I just get $\uparrow\downarrow$ 50% of the measurements for both the singlet and triplet.So what is the physical difference between the singlet $S_0= \dfrac{1}{\sqrt{2}}(\uparrow\downarrow-\downarrow\uparrow)$ and the triplet $T_0 = \dfrac{1}{\sqrt{2}}(\uparrow\downarrow+\downarrow\uparrow)$. Are they physically distinguishable? $\endgroup$ Commented Jan 29, 2019 at 16:28
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$\begingroup$ If you have an experiment that is sensitive to the relative phase - some sort of interference effects - you could pick up the difference. $\endgroup$ Commented Jan 29, 2019 at 16:32
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If you apply the operator $S^2$ you get $S(S+1)$, so 0 or 2 respectively.
