3
$\begingroup$

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

$\endgroup$
3
  • 3
    $\begingroup$ Closely related question here. $\endgroup$
    – knzhou
    Commented Jan 29, 2019 at 15:52
  • 1
    $\begingroup$ Another possible duplicate. $\endgroup$
    – rob
    Commented Jan 29, 2019 at 19:19
  • $\begingroup$ Try $S_-$ and $S_+$. $\endgroup$
    – my2cts
    Commented Mar 17, 2021 at 17:36

2 Answers 2

6
$\begingroup$

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 $s=1$.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$
    – PhysicsMan
    Commented Jan 29, 2019 at 16:28
  • $\begingroup$ 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
1
$\begingroup$

If you apply the operator $S^2$ you get $S(S+1)$, so 0 or 2 respectively.

$\endgroup$
2
  • $\begingroup$ When you apply $S^2$, you get the state back times 2, which is why the quantum number for its total spin is 1: $1(1+1)=2$. $\endgroup$
    – G. Smith
    Commented Jan 30, 2019 at 21:21
  • $\begingroup$ @G. Smith Sure, I corrected my answer. $\endgroup$
    – my2cts
    Commented Jan 30, 2019 at 22:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.