Why is a sum remaining here?

Question

Substituting eqn (5.9) into the time-dependent equation gives $$ i\hbar \sum_n \dot c_n(t) |u_n\rangle e^{-iE_nt/\hbar} = \sum_n V(t) |u_n\rangle e^{-iE_nt/\hbar}c_n(t) \tag{5.10} $$ Now take the scalar product with $|u_p\rangle$ to find $$ i\hbar \dot c_p(t) = \sum_n V_{pn}(t) e^{-i\omega_{pn}t}c_n(t) \tag{5.11} $$ where the matrix element is $$ V_{pn}(t) = \left< u_p \middle | V \middle | u_n \right> \tag{5.12} $$

I am sure that there must be a simple reason for this that I am missing but I have tried to find one to no avail. I am trying to understand why applying the inner product in eqn 5.11 causes only the left hand side sum to vanish via the Kronecker delta. It seems to me that the sum on both sides should collapse to the cases in which $n=p$.

Answer 1

On the right hand side you don't have $\langle u_p|u_n\rangle $ products but the matrix elements of V, the terms $\langle u_p|V|u_n\rangle$. These are not equal to Kronecker delta, in general. So the sum does not reduce to the term containing $\langle u_p|V|u_p\rangle$ but should keep all the terms.

Answer 2

The difference is that the right side has an operator, $V$, acting on $|u_n\rangle$ while the left side has only scalars multiplying it. When an operator acts on this ket, it is no longer necessarily orthonormal to $\langle u_p|$.