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(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Homework Help: A quick question I had about the way the Hamiltonian is factored

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