-1
$\begingroup$

How do we get from the expression

$$\frac{1}{2} (3(\vec{r} \cdot \vec{r}')^2-r'^2 r^2)$$

to

$$\sum_{i,j=1}^3 r_ir_j \frac{1}{2} (3r'_ir'_j-r'^2\delta_{ij})$$

If someone could go step by step and explain I will be very thankful.

$\endgroup$
1
  • 1
    $\begingroup$ Hi Darkenin. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$ – Qmechanic May 4 '20 at 10:03
2
$\begingroup$

You need to write the scalar products $(\vec{r}\cdot \vec{r'})$ and $r^2\equiv (\vec{r}\cdot \vec{r})$ explicitly. Knowing that: $$\vec{a}\cdot \vec{b} = \sum_{i=1}^3 a_i b_i$$ you can rewrite your quantity as: $${1 \over 2} \left(3\left(\sum_{i=1}^3 r_i r'_i\right)^2 - r'^2 \sum_{i=1}^3 r_i r_i\right)=$$ Now you can write the sum squared as the product of two identical sums with different dummy indices $i,j$: $$={1 \over 2} \left(3 \sum_{i=1}^3 r_i r'_i \sum_{j=1}^3 r_j r'_j - r'^2 \sum_{i,j=1}^3 r_i r_i\right)=$$ and join the two sums as one sum over two indices. Plus, you can rewrite the term in the second sum using the Kronecker $\delta$: $$={1 \over 2} \left(3 \sum_{i,j=1}^3 r_i r'_i r_j r'_j - r'^2 \sum_{i,j=1}^3 r_i r_j \delta_{ij}\right)=$$ Finally you can take both the sum and the common factor $r_ir_j$ out of the parentheses: $$=\sum_{i,j=1}^3r_i r_j{1 \over 2} \left( 3 r'_i r'_j - r'^2 \delta_{ij}\right)$$

$\endgroup$

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