# Working with indices, where Einstein summation applies

Problem:

Let $$H:={1\over 2m}\vec p\cdot \vec p -{1\over r}$$, where $r=(\vec r\cdot \vec r)^{1\over 2}, \,\,\,\vec r= (x_1,x_2,x_3)^T$ and $$\vec R:= {1\over m}\vec p\times \vec L -\hat r$$. I wish to show that $${\partial H\over \partial p_i}{\partial R_j\over \partial x_i}-{\partial H\over \partial x_i}{\partial R_j\over \partial p_i}=0$$

I have worked on it and got $${\partial H\over \partial p_i}={1\over m}p_i$$ $${\partial H\over \partial x_i}={x_i\over r}$$ $${\partial R_j\over \partial x_i}={1\over m} (p_np_n\delta_{ij}-p_ip_j)-{1\over r}\delta_{ij}+{1\over r^3}x_ix_j$$ $${\partial R_j\over \partial p_i}={1\over m} (\epsilon_{jik}L_k+x_jp_i-x_kp_k\delta_{ij})$$ So substituting these into $I:={\partial H\over \partial p_i}{\partial R_j\over \partial x_i}-{\partial H\over \partial x_i}{\partial R_j\over \partial p_i}$, I get $$I=({1\over m}p_i)({1\over m} (p_np_n\delta_{ij}-p_ip_j)-{1\over r}\delta_{ij}+{1\over r^3}x_ix_j)-({x_i\over r})({1\over m} (\epsilon_{jik}L_k+x_jp_i-x_kp_k\delta_{ij}))$$ $$={1\over m} ({1\over m} (p_np_np_j-p_ip_ip_j)-{1\over r }p_j+{1\over r^3} x_ip_ix_j-{1\over mr} (\epsilon_{jik}L_kx_i+x_jp_ix_i-x_kp_kx_j))$$ Out of the $7$ terms, the first $2$ and the last $2$ vanish. So we are left with $$I={1\over m}(-{1\over r }p_j+{1\over r^3} x_ip_ix_j-{1\over r} (x_ip_ix_j-x_ix_ip_j))$$

Unfortunately, I don't see why this should equal to $0$ and I can't find my mistake either. Any suggestions? Thanks.

• Note that you can use \left( and \right) to make the sizes of the parentheses come out right; they're rather ugly and hard to read when they're all small. – joriki Jul 22 '12 at 10:41

$\displaystyle {\partial H\over \partial x_i}={x_i\over r}$ is wrong: $\displaystyle{\partial r\over \partial x_i}={x_i\over r}$, but the term in $H$ is $-1/r$, so you get another $1/r^2$.
If you include the factor $1/r^2$, the remaining terms in your result cancel if you use $x_ix_i=r^2$.