In Sakurai, p. 452, discussing the permutation symmetry in a two fermions system, where a state ket is denoted by $\left| x_1,m_1;x_2,m_2 \right>$ it is said that the two particles permutation operator $P_{12}$ may be decomposed to a space permutation and a spin permutation -
$$ P_{12}=P_{12}^{space}P_{12}^{spin} $$

with the spin permuation being -

$$ p_{12}^{spin}= \frac{1}{2}(1+\frac{4}{\hbar^2}\boldsymbol{S}_1\cdot\boldsymbol{S}_2) $$

It is also menationed that -

$$ \boldsymbol{S}_1\cdot\boldsymbol{S}_2 = \begin{cases} \frac{\hbar^2}{4},\;triplet\\ \frac{-3\hbar^2}{4},\; singlet \end{cases} $$

what I don't understand here is how $\boldsymbol{S}_1\cdot\boldsymbol{S}_2$ is defined and how it's above values were calculated.

According to my understanding, assuming a spin in the $\hat{z}$ direction (Should we assume that?), the values should be $\frac{\hbar^2}{4}$ with the triplet state $\left| +-\right>+\left| -+\right>$ having the same values as the signlet state.


1 Answer 1


As explained here, the scalar product of two spins can be defined using the total spin of the system in the case where this value is known to us. The squared total spins is - $$ S_{total}^2=(S_1+S_2)\cdot(S_1+S_2) $$ So the scalar product of the spins is - $$ S_1\cdot S_2= \frac{1}{2}(S_{total}^2 - S_1^2 - S_2^2) $$

for spin $\frac{1}{2}$ particles the square spin is $S^2=\hbar^2s(s+1)=\frac{3\hbar^2}{4}$, so for the singlet state where $S_{total} = 0$ we get - $$ S_1\cdot S_2 = \frac{1}{2}(0-\frac{3\hbar^2}{4}-\frac{3\hbar^2}{4})=-\frac{3\hbar^2}{4} $$

and for the triplet state where $S_{total} = 1$ we get -

$$ S_1\cdot S_2= \frac{1}{2}(2\hbar^2-\frac{3\hbar^2}{4}-\frac{3\hbar^2}{4})=\frac{\hbar^2}{4} $$

  • $\begingroup$ Is there any physical interpretation why the result is different for the two states? $\endgroup$
    – Jhonny
    Mar 14, 2018 at 16:39
  • $\begingroup$ @Jhonny In the singlet state the total spin is 0 while in the triplet state it is 1. $\endgroup$
    – proton
    Mar 15, 2018 at 16:39

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