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nymphidius
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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.
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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