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In quantum particle in a ring problem, the general solution for the wavefunction, with $k = R \sqrt{2 m E / \hbar^2}$, $R$ being the ring radius, $c_{+, -}$ being constants, $E$ the energy, and $m$ the particle mass, is

$$\psi(\theta) = c_{+} e^{i k \theta} + c_{-} e^{-i k \theta}$$

Periodic boundary conditions give the eigenvalues $k = n$, with $n=0, \pm 1, \pm 2, \ldots$.

At first glance the problem looks "underdetermined" - it seems that there are two unknowns ($c_{+}$ and $c_{-}$) but only one condition to find them, normalization of the wavefunction $\langle \psi \lvert \psi\rangle = 1$. But in my naive understanding, the problem is actually not underdetermined because one may set $c_{-} = 0$, as is seemingly done on Wikipedia in order to normalize the wavefunction.

Why may one set $c_{-} = 0$? Or, why do Wikipedia and textbooks (below) seemingly set $c_{-} = 0$? And how can we treat the general case with $c_{-} \ne 0$?

Even if we do not trust Wikipedia, most textbooks on introductory quantum look at the ring problem and do the same thing. For example:

  • "Introduction to Quantum Mechanics: in Chemistry, Materials Science, and Biology" By Sy M. Blinder
  • "Atkins' Physical Chemistry" By Peter Atkins, Julio de Paula
  • "Quantum Mechanics for Chemists" David O. Hayward
  • "Case studies in atomic physics 4" edited by E. McDaniel
  • "Quantum Chemistry" By John P. Lowe
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    $\begingroup$ Even though my comment seems to have disappeared, I still don't understand what you mean by "setting $c_- = 0$". There are two degenerate states, as e.g. the Wikipedia article clearly states, and one is with $c_+ = 0$, the other with $c_-=0$. As they are degenerate, for many cases it might suffice to look at only one of them, but what exactly is the issue here? Please quote a passage where this "setting $c_-=0$" occurs and explain what specifically is unclear about it. $\endgroup$ – ACuriousMind Apr 22 '16 at 15:31
  • $\begingroup$ @ACuriousMind By "setting" I mean "equating". Regarding "for many cases it might suffice to look at only one of them, but what exactly is the issue here?". Issue is cases where it might not suffice to look at only one of them. $\endgroup$ – Nigel1 Apr 23 '16 at 12:55
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This problem is actually the same problem as a free 1D particle, but with periodic boundary conditions. Here you have conserved $L_z$, while for a free 1D particle you have conserved $p_x$. In this case the solution is similarly

$$\psi(x)=A e^{ikx}+Be^{-ikx}.$$

You can set $A=0$ or $B=0$ if you want an eigenstate of momentum operator (which commutes with Hamiltonian). Or you can set $A=\pm B$ to get eigenstate of parity operator (which commutes with Hamiltonian, but doesn't with momentum operator).

It's what degeneracy is about: existence of pairs of conserved quantities with mutually non-commuting operators. Eigenstates of both operators from a pair must be eigenstates of Hamiltonian, but if they correspond to the same energy, they must either not change when acted on by the operators (thus allowing them to effectively commute for this state), or be degenerate.

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