Undergrad Mean value of the nuclear tensor operator

[kvothe18]

Does anyone know how can you prove that the mean value of the tensor operator S₁₂ in all directions r is zero?

S₁₂ : http://prntscr.com/j3gn40

where s₁, s₂ are the spin operators of two nucleons.

 

[mfb]

It is linear in the individual spins. You can split one into an orthogonal and a parallel component (relative to the other). I would expect that it is easy to show based on that.

 

[samalkhaiat]

Does anyone know how can you prove that the mean value of the tensor operator S₁₂ in all directions r is zero?

S₁₂ : http://prntscr.com/j3gn40

where s₁, s₂ are the spin operators of two nucleons.

\langle \frac{3}{r^{2}} ( \vec{r} \cdot \vec{S}_{1}) ( \vec{r} \cdot \vec{S}_{2} ) \rangle = \frac{1}{4 \pi} \int d \Omega \frac{3}{r^{2}} \ ( \vec{r} \cdot \vec{S}_{1} ) ( \vec{r} \cdot \vec{S}_{2} ) = \frac{3}{4 \pi} \int d \Omega \ ( \hat{r} \cdot \vec{S}_{1}) ( \hat{r} \cdot \vec{S}_{2} ) . Now, the integral J (S_{1} , S_{2}) \equiv \int d \Omega \left( \hat{r} \cdot \vec{S}_{1} \right) \left( \hat{r} \cdot \vec{S}_{2} \right) \ , is scalar and linear in \vec{S}_{1} and \vec{S}_{2}. So it must be proportional to the scalar product \vec{S}_{1} \cdot \vec{S}_{2}, i.e., J (S_{1} , S_{2}) = C \ \vec{S}_{1} \cdot \vec{S}_{2} . To find the constant C, consider the special case \vec{S}_{1} = \vec{S}_{2} = \hat{z}: C \ \hat{z} \cdot \hat{z} = J (\hat{z} , \hat{z}) = \int d \Omega \left( \hat{r} \cdot \hat{z}\right)^{2} = \int d \Omega \cos^{2} \theta . This gives you C = \frac{4 \pi}{3} and, therefore, J(S_{1},S_{2}) = \frac{4 \pi}{3} \vec{S}_{1} \cdot \vec{S}_{2} = \frac{4 \pi}{3}\langle \vec{S}_{1} \cdot \vec{S}_{2} \rangle . \ \ \ \ (2) The last equality follows because \langle \vec{A} \cdot \vec{B} \rangle = \frac{1}{4 \pi} \int d \Omega \ \vec{A} \cdot \vec{B} = \vec{A} \cdot \vec{B}.

Substituting (2) in the first equation of this post, you find

\langle \frac{3}{r^{2}} ( \vec{r} \cdot \vec{S}_{1} ) ( \vec{r} \cdot \vec{S}_{2}) \rangle = \langle ( \vec{S}_{1} \cdot \vec{S}_{2} ) \rangle .