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Ground level energies (Particle in a box vs Harmonic Osc.)

  1. Jan 4, 2013 #1

    I have a tiny question that has confused me.

    Currently I'm reading about potential wells, harmonic oscillators, the free particle in quantum physics.
    If I just take the particle in a box as an example you have a region where the potential is zero, and you have some walls/boundaries where the potential is infinite, so the particle cannot escape.
    The energy levels of the particle in a box is given by:

    En = pn2 / 2m,
    for n = 1, 2, 3...

    So far so good.

    But when I get to the harmonic oscillator, the energy levels is given by:
    En = (n + ½)h-bar*ω,
    for n = 0, 1, 2...

    And then my book just writes: "Note that the ground level of energy in the harmonic oscillator is n = 0, not n = 1..."

    So my question is, why is that ? Any particular reason, or just something I have to accept ? :)

    Thanks in advance.

  2. jcsd
  3. Jan 4, 2013 #2


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    The quantum number is a parameter that has to be viewed in light of the way that the wavefunctions have been written down, as well as the physics of the problem. For the particle in a box, we typically take

    [tex]\psi_n(x) = A \sin (n \pi x/L).[/tex]

    With this choice, [itex]n=0[/itex] corresponds to [itex]\psi_0=0[/itex]. The corresponding probability density also vanishes, so this physically corresponds to having no particle at all in the box. So we discard it on physical grounds and take [itex]n=1,\ldots[/itex].

    However, it would not be wrong to take

    [tex]\psi'_n(x) = A \sin ((n-12) \pi x/L)[/tex]

    as our solutions. Then we would determine that [itex]n=12[/itex] is an unphysical solution, while [itex]n=11[/itex] is equivalent to [itex]n=13[/itex], etc., so we should take [itex]n=13,\ldots[/itex] to have a complete set of solutions. This choice is more cumbersome and confusing compared to the textbook one, so it's not used in practice.

    Similarly, for the harmonic oscillator, using the index set [itex]n=0,\ldots[/itex] is the most convenient choice when you actually write down expressions for the wavefunctions. In particular, if you study the formalism of raising and lowering operators ([itex]a^\dagger,a[/itex]), it is a very physically natural one to use, since [itex]n[/itex] then counts the number of raising operators that are acting on the ground state.

    In other cases, the choice of parameter is even more concretely suggested by the physics of the problem. For example, angular momentum states are chosen so that the quantum number [itex]l=0[/itex] for the state with zero angular momentum. As in the particle in a box case, we could choose a different parametrization, but the physical interpretation would be more confusing.
  4. Jan 5, 2013 #3
    In case of harmonic oscillators,it is convenient to use creation and annihilation operator(you can see it elsewhere )
    defining |p>=a|H'>(a is annihilation operator and |H'> is an eigenket)
    <p|p>=<H'|a*a|H'>,now a*a (a* is creation operator) can be written as H'-(1/2)h-ω
    since<p|p> and<H'|H'> is positive,so it implies
    E≥ (1/2)h-ω
  5. Jan 5, 2013 #4
    I could be wrong, but I assume that if you are asking the question you are asking, you are not skilled in Dirac notation and Hermetian conjugates yet, so the previous explanations may not sing so loud. So I will give you my equally freshman understanding of the situation, which is that the n for the particle in the box solution COULD be zero, but that would lead to a trivial solution where nothing would happen, so we leave that option out. We do not have that same problem with the harmonic oscillator because the added 1/2 in the term gives a non-trivial solution even when n=0.
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