Yes, the operation f(a) = (a) XOR (3a) is a doubling. 

f(a)= 2 a. 

Since I proved it for a single bit a_1 = 1, and an arbitrary bit b, a_b = 1, I think I unknowingly proved it for all numbers (!) since all numbers are superpositions of n bits. Since the relationship holds for all single bits in position b starting with a_n = a_b then it holds for all numbers: f(a_i + a_j + a_k ... ) = (2a_i + 2a_j + 2a_k) = 

2 (a_i+a_j + a+k ...)   

 

one thing needed to complete the proof is this: 

for the case:  a_n = 0 

f(a_n) = 0 = 2 a_n