Partial derivative potential energy of 'free' vibration

Question

I have this rather mathematical question about the calculation of the partial derivative of a potential energy function given by:

$$U(x_i)=\frac{1}{2}\sum_{i,j}\frac{\partial^2U(0)}{\partial x_i\partial x_j} x_ix_j.$$ Or if we use $b_{ij}$ for the Hessian : $$U(x_i)=\frac{1}{2}\sum_{i,j}b_{ij}x_ix_j.$$

I want to calculate the force: $k_i=\frac{-\partial U}{\partial x_i}$. This should be $-\sum_{j} b_{ji}x_j$.

My questions:

1. Why is the summation over the index $i$ gone?

2. Why are the indices of the hessian $b$ switched?

I hope someone can give me an answer. I put this in the physics section because it's a physics problem, but my question is actually purely mathematical.

Answer 1

1.) The differentiation operator acting will give rise to Kronecker-Deltas since

$\frac{\partial x_a}{\partial x_b}=\delta_{ab}$ This will kill one summation.

More specifially: $\frac{\partial U}{\partial x_a}=-1/2 \sum_{ij}b_{ij}(\delta_{ai}x_j+\delta_{aj}x_i)=-1/2( \sum_{j}b_{aj}x_j+\sum_{i}b_{ia}x_i)=-\sum_{j}b_{aj}x_j$. Rename j to be i and you're done.

2.) The Hessian matrix is a matrix of second-order partial derivatives; hence it is symmetric w.r.t. its indices, i.e. $b_{ij}=b_{ji}$.