Can a sum of products be split as a product of two sums? I have 
$$\sum_k^n P_k x_k$$   
Am I allowed to split it up into two sums so I have it like 
$$\sum_k^n P_k  \sum_k^nx_k$$
 A: $\qquad\qquad$ 
$$x_1\qquad\qquad\qquad\qquad x_2\qquad\qquad\qquad\qquad x_3\qquad\qquad\qquad\qquad x_4$$
Imagine that the vertical segments are your $P$'s. Then the product of sums is the area of the entire rectangle, whereas the sum of products is only the red area.
A: Assuming P depends on k and not n then the summation
$$
\sum_k P_k x_k
$$ 
Creates terms like
$$
P_1 x_1 + P_2 x_2 + \ldots + P_n x_n
$$
But, if you took $P_k$ out then you would have two independent sums:
$$
\sum_k P_k \sum_k x_k
$$
Producing two separate sums multiplying each other like
$$
(P_1 + P_2 + \ldots + P_n) (x_1 + x_2 + \ldots + x_n)
$$
And this obviously produces terms like $P_1 x_1, P_1 x_2,\ldots$ or as produced by the double sum written like:
$$
\sum_i \sum_j P_i x_j
$$
So the original summation is definitely not the same as this double sum.
A: http://tutorial.math.lamar.edu/Classes/CalcI/SummationNotation.aspx
^^^That offers a nice summary of summation notation and its properties. 
To answer your question specifically, you should not split this into two different summations. Rather, notice that P does not depend on k. You can treat P as a constant and simply move it outside the summation so that it multiples the result of your summation of $x_k$, which does depend on your variable k.  
A: If you just want to analyze the quantum, you can just use the Cauchy's inequality in n-dimensional Euclidean space, i.e.
$$
\left(\sum_{i=1}^n u_i v_i\right)^2 \leq\left(\sum_{i=1}^n u_i^2\right)\left(\sum_{i=1}^n v_i^2\right)
$$
