I'm currently using David J. Griffiths 'Introduction to Quantum Mechanics' to teach myself quantum mechanics and I had a quick question about the way he factors the Hamiltonian into the raising and lowering operators for the potential V(x)=(1/2)kx² On page 42 he writes the Hamiltonian as: (1/2)[p²+(mωx)²] and then he factors it into the raising and lowering operator as: a±=(1/√2ℏmω)(∓ip+(mωx)) My question is---does it make a physical difference how you factor the Hamiltonian? For example, I was always taught that a²+b² factors into (a+bi)(a-bi), but he factored it as (ai+b)(ai-b). Now I know that mathematically you get the same results, either way, once you distribute, but since we are talking about factoring the momentum operator and the (mωx) term, I just wanted to know if there's a particular reason why he did it that way---if it's okay to factor it my way. So to sum things up his way is: a±=(1/√2ℏmω)(∓ip+(mωx)) and I would have done: a±=(1/√2ℏmω)(p∓i(mωx)) is either way okay? Sorry for not using the format---I just thought it wasn't needed for my question.