# Why does the triplet state $\dfrac{1}{\sqrt{2}}(\uparrow\downarrow+\downarrow\uparrow)$ have spin 1 and not 0?

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 $$\dfrac{1}{\sqrt{2}}(\uparrow\downarrow-\downarrow\uparrow)$$?

• Closely related question here. – knzhou Jan 29 '19 at 15:52
• – rob Jan 29 '19 at 19:19

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 $$s=1$$.
• 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? – PhysicsMan Jan 29 '19 at 16:28
If you apply the operator $$S^2$$ you get $$S(S+1)$$, so 0 or 2 respectively.
• When you apply $S^2$, you get the state back times 2, which is why the quantum number for its total spin is 1: $1(1+1)=2$. – G. Smith Jan 30 '19 at 21:21