Why is the ground state of a particle in a box called $n=1$?
For a quantum harmonic oscillator, the ground state in most sources is referred to as $n=0,$ and this state has zero nodes.
For a particle in a box, the ground state in most sources is called $n = 1.$ It has zero nodes.
So why is the ground state called $n=1$ for the particle in a box and $n=0$ for the harmonic oscillator?
quantum-mechanics terminology conventions
$\begingroup$ what sense could one draw from the designation of n=0 to the ground state . If we say that the particle is having a de broglie wavelength of say x units then the standing waves created (talking in classical terms) should have some integral multiple of x/2 as the length of the box and obviously if understood this way what can N=0 possibly mean. I am not properly trained in quantum physics so please forgive me if what I talked seems rubbish $\endgroup$ – Aditya Garg Mar 25 at 19:31
$\begingroup$ It's purely a matter of convention/choice. One can always define n'=n-1 to shift the beginning quantum number. $\endgroup$ – Bill N Mar 25 at 20:17
$\begingroup$ One is in FORTRAN and one is in C. $\endgroup$ – Jon Custer Mar 26 at 0:50
$\begingroup$ @JonCuster Fortran hasn't been FORTRAN for about 30 years; the name was officially changed in 1990. $\endgroup$ – Kyle Kanos Mar 26 at 22:02
$\begingroup$ @KyleKanos - surprisingly, that was an iOS autocorrect thing. Perhaps my phone is just in tune with my heavy Fortran use in the 80s and was just going all retro on me... $\endgroup$ – Jon Custer Mar 26 at 22:18
Discrete eigenvalues can always be labeled by the integers, so we can always pick up $E_n$ with $n$ starting from $0$ or $1$, it is just a matter of convenction. I suppose this is where your question originates. For the harmonic oscillator the energies are given by $\hbar \omega (m +1/2)$ with $m=0,1,\ldots$ so it is natural to pick $n=m$ starting from $0$.
For the particle in the box, note that the problem is equivalent to a "free" particle with certain boundary conditions (wavefunction zero at the border). In this sense it is no surprise that "momentum" is conserved (people speak of quasi-momentum in this case). In this picture the Hamiltonian is simply
$$H = \frac{p^2}{2m}.$$
If you set the box as in $[0,L]$ you can write the eigenfunctions as
$$\psi_k(x) \propto \sin(kx)$$
where $k$ is this quasi-momentum to be determined. In this way one of the boundary conditions is automatically satisfied. To satisfy the other one we must impose $\sin(Lk)=0$ which in principle leads to $k= m\pi/L$, with $m\in \mathbb{Z}$. However, $k$ and $-k$ single out the same quantum state (a quantum state is an equivalence class of vectors which differ by a phase). Moreover $k=0$ is not allowed because the zero wave-function is not normalizable. Hence we are led to
$$k= \frac{n\pi}{L}, \ \ n = 1,2,\ldots$$
In a sense, physically, this tells us that the state with zero momentum (a state that does not move left or right) is not allowed. This is clearly a quantum mechanical effect (a fixed particle is perfectly allowed classically). You can think of its origin in the Heisenberg's uncertainty principle. Since $H\propto p^2$ we can obviously label the energies with the same label that we use for the momentum (another good notation would be simply to label them with the quasi-momentum itself as in $E_k$ and remember what is the spectrum of $p$, $\sigma(p)$.
I suppose you knew all of this but I don't think you can say much more.
Let me briefly comment on the number of nodes of the wavefunctions.
For a Hamiltonian of the form
$$ H= \frac{p^2}{2m} + V(x) = - \frac{\hbar^2}{2m} \frac{\partial^2}{\partial^2 x} +V(x) $$
a theorem by Courant and Hilbert (generally called nodal theorem) states that the ground state wave function has no nodes (can be chosen to be everywhere non-negative) and the $n$-th level has precisely $n-1$ nodes. Moreover if the potential goes to infinity as $|x| \to \infty$, then the eigenvalues form a discrete unbounded sequence. I imagine variations of this theorem exist for $\mathbb{R}^d$. So the behavior you observed for the number of nodes is in fact quite general.
edited Apr 2 at 2:36
lcvlcv
$\begingroup$ I realize I didn't comment on the nodes of the groundstate wavefunction, which is a consequence of Perron-Frobenius theorem but don't think is related to your question. $\endgroup$ – lcv Mar 25 at 19:52
$\begingroup$ "For the harmonic oscillator the energies are given by $\hbar \omega (m +1/2)$ with $m=0,1,\ldots$ so it is natural to pick $n=m$ starting from $0$" ─ it would be equally natural to have $\hbar \omega (m -1/2)$ with $m=1,2,\ldots$. That's not an argument that holds any water. $\endgroup$ – Emilio Pisanty Mar 26 at 10:32
$\begingroup$ @emiliopisanty I guess it is because there is a simple expression for the number operator $N=a^\dagger a$ whose eigenvalues are the integers including zero. If you find a simple expression for the operator whose eigenvalues are the positive integers I imagine that notation could become popular (no pun intended). You propose to label the eigenvalues with the number of nodes. It's a nice convention. We would mostly start from then. Even for the hydrogen though atom whose radial part has no zeros in the ground state. $\endgroup$ – lcv Mar 26 at 19:22
$\begingroup$ There is an equally nice expression in terms of $N' = aa^\dagger$, which could well have been chosen as the "number" operator. $\endgroup$ – Emilio Pisanty Mar 26 at 19:40
$\begingroup$ But no, I'm not "proposing" anything. The conventions are already there and they are at this stage essentially set in stone, for good or ill. $\endgroup$ – Emilio Pisanty Mar 26 at 19:40
As with all questions of the form "why does this convention...", the answer is ultimately "because conventions are arbitrary".
Still, you are correct in noticing that the harmonic oscillator does buck an overall trend (which has instances in the hydrogen atom and the particle-in-a-box problem, among a few others) where eigenstates are numbered as $1,2,3,\ldots$, with integers starting at unity. So what is it about the harmonic oscillator that makes it more convenient to change that trend? (After all, if it wasn't more convenient, why change the trend?)
The reason is simple ─ the $n$th wavefunction of the harmonic oscillator has the form $$ \psi_n(x) = H_n(x) e^{-x^2/2} $$ (modulo constants), where $H_n(x)$ is a polynomial of degree $n$, starting with degree zero (a constant) for the ground state and then moving up with the eigenstate number. The Morse potential, which has a similar numbering scheme, also has this type of polynomial dependence.
This isn't a hard-and-fast rule, though, and it's likely that there are other systems with other types of exceptions. Conventions are tricky things, and they are social objects - it doesn't make too much sense to obsess over why the details settled to where they are.
Emilio PisantyEmilio Pisanty
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