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Why in the particle in a box model do the values of n begin at 1 but in the harmonic oscillator they begin at 0? I understand what the wave-functions and their corresponding probabilities look like and that the PIB has 0 nodes for n=1 which means the number of nodes is n-1, so for n = 0 it would have -1 nodes which is physically unreasonable. But I do not understand why the PIB cannot have n=0 whereas the Harmonic oscillator can.

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  • $\begingroup$ PIB cannot have n=0 precisely for the reason that you said yourself just a sentence ago: that the corresponding $\psi$ would not be physically meaningful. If that is not enough of a justification for you, then what is? $\endgroup$ – Ivan Neretin Dec 4 '19 at 14:48
  • $\begingroup$ This still doesn't answer the question of why PIB cannot have an n=0 whereas the HO can. All you've stated is that we can change the rules by redefining things. But redefining things doesn't change the facts. PIB can never have n = 0 because this would mean the energy is zero, which means the particle is not moving so momentum is 0 and we can know the position, but this violates the uncertainty principle. But why can HO have an n=0?? $\endgroup$ – Harley McFarlen Dec 4 '19 at 22:46
  • $\begingroup$ Ivan Neretin, an explanation to WHY something is not physically meaningful. Thanks $\endgroup$ – Harley McFarlen Dec 4 '19 at 22:47
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    $\begingroup$ $n=0$ is not a fact; it is a convention, and as such, can be changed, like Buck Thorn said. $n$ is not a physical value at all. Also, look at the HO state with $n=0$: its energy is not $0$. $\endgroup$ – Ivan Neretin Dec 5 '19 at 5:28
  • $\begingroup$ Do you realize that there is a zero point energy associated with the harmonic oscillator? That is, when n=0 the energy is not zero? $\endgroup$ – Buck Thorn Dec 5 '19 at 17:56
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Quantum numbers pop up when you solve Schrödingers wave equation for systems whose solutions are discrete due to the presence of a restraining potential. There is some degree of arbitrariness to what the range of quantum numbers might be, because it depends on how you define the quantum number.

You could just as well define the quantum number and associated energy levels for a 1D pib as $$\begin{align} E_{n'}&=k(n'+1)^2 \\&=k n'^2 +2kn' + k\end{align} $$ where $$k=\frac{\pi^2 \hbar^2}{2mL^2}$$ Here $n'$ defines a "new" quantum number related to the original one as $n'=n-1$ and therefore ranges from 0 to $\infty$.

For the harmonic oscillator, you could similarly rewrite the energy as

$$E_n = \hbar \omega n'$$

where $n'$ ranges in integer increments from $\frac12$ to $\infty$.

Physically speaking, the non-zero ground-state energy of both the pib and harmonic oscillators is usually attributed to the Heisenberg Uncertainty Principle, because a particle with no energy would have no velocity and therefore would be completely unrestrained in position. There are of course plane waves, but this would be a strange plane wave, with no amplitude.

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