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A quick question I had about the way the Hamiltonian is factored

  1. Jul 25, 2012 #1
    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:


    and then he factors it into the raising and lowering operator as:


    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:


    and I would have done:


    is either way okay?

    Sorry for not using the format---I just thought it wasn't needed for my question.
    Last edited: Jul 25, 2012
  2. jcsd
  3. Jul 25, 2012 #2


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    Let's use ##a_\pm## to stand for the standard choice of operators. Your alternative is to define

    $$b_+ = -i a_-, ~~~b_- = i a_+.$$

    We can work out expressions for the Hamiltonian, ground state, and general energy eigenstates in the new operators easily. The energy spectrum will be the same, but now ##b_+## annihilates the ground state and the eigenstates are formed by acting with ##b_-## on the ground state.

    Since the spectrum is the same, the two representations are equivalent. All we've done is changed the labels around. It's ok to use your representation, but in order to avoid confusion when communicating with other people, it's best to use the standard conventions.

    It turns out that the transformation ##b_+ = -i a_-## is a unitary transformation. More generally, if we were to write ##b_+ = U a_-##, then the Hamiltonian is invariant as long as ##U^* U =1##, which is obviously satisfied for ##U=-i##. More generally, any ##U=e^{i\theta}## leaves the Hamiltonian (and therefore the spectrum) invariant.

    This type of invariance under unitary transformations comes up often in quantum mechanics and is very important. It's a bit simple here, but is more useful when dealing with the 2d and 3d harmonic oscillators, which should be covered later in the text.
  4. Jul 26, 2012 #3
    Thanks for the input---I shall forge ahead using the standard convention.
    Last edited: Jul 26, 2012
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