Solving particle in a ring problem

Question

While solving the particle in a ring we get a general solution of the form:

$$\psi(x) = A\exp(imx) + B \exp(-imx)$$

Where $m=\left(\frac{2iE}{h}\right)^.5$. Imposing the boundary condition I get that $m$ should be an integer, but most of the books drop one of the terms in the general solution. Why is that? They write the solutionn as $\psi(x)=A\exp(i mx)$. I understand $m$ is integral, but this is obviously not the general solution. How are they getting it?

Answer 1

Since you haven't provided a reference, I can only guess why this is done. However, I am quite sure that it's just shorthand, as @baponkar points out in the comment above. $m$ can be either positive or negative, and so the linearly independent solutions can be compactly written as $e^{i m x}$ without loss of generality.

The solution that you have written is the solution to the time-independent Schrodinger Equation $$-\frac{\hbar^2}{2 M R^2}\frac{\partial^2\psi}{\partial\theta^2} = E\,\psi.$$

The solutions to this equation are $e^{im\theta}$ and $e^{-im\theta}$ where the boundary conditions force $m$ to be an integer. The more general state of definite energy can therefore be written as $$\psi_m(\theta) = A e^{i m \theta} + B e^{-i m \theta},$$ with an energy eigenvalue of $$E_m = \frac{\hbar^2 m^2}{2MR^2}.$$

As you can see, in general when $m \neq 0$, $\psi_m$ and $\psi_{-m}$ are not the same, but they give the same energy eigenvalue, and thus such pairs of states are degenerate. ($\psi_0$ is just a constant, and corresponds to the energy 0, and thus isn't degenerate.)

An arbitrary state of the system (not a state of definite energy) can be expressed as a linear combination of the states of definite energy $\psi_m$ in the following way:

$$\Psi(\theta,t) = \sum_{m=0}^\infty \left(A_m e^{im\theta} + B_m e^{-im\theta}\right) e^{-iE_mt/\hbar},$$

but since $E_m=E_{-m}$, you could just as well write it as $$\Psi(\theta,t) = \sum_{m=-\infty}^\infty C_m e^{im\theta}e^{-iE_mt/\hbar}.$$

You could interpret the last decomposition as decomposing the wavefunction in the basis $\{ e^{im\theta}, m\in\mathbb{Z}\},$ which is a perfectly valid basis that spans the space. (Any arbitrary periodic function can be decomposed in terms of complex exponentials.)

There is a slight advantage to using this basis, as I have pointed out in my answer to Wavefunction of a particle on a polar potential: the eigenfunctions $e^{im\theta}$ correspond not only to eigenfunctions of $\hat{H}$ but also of the ($z-$component of the) angular momentum $\hat{L}_z$. Thus, unlike $\psi_m$, the states $e^{im\theta}$ have a specific value of angular momentum, whose sign is given by the sign of $m$. (You could very crudely think of them as "right-rotating" or "left-rotating" solutions, depending on whether $m$ is positive or negative respectively.)

Answer 2

This is purely out of convenience.

The boundary conditions guarantee that $m$ must be an integer, and the energy $E=\frac{\hbar^2}{2ML^2} m^2$ (assuming $\hat H=\frac{1}{2M}\hat{P}^2$ in periodic boundary conditions in an interval of length $L$) is given by $m^2$, so both $m$ and $-m$. This then gives you two options:

• You can restrict your attention to $m\geq 0$, and keep two linearly-independent eigenstates, $e^{imx}$ and $e^{-imx}$. Except for $m=0$, where there is only one independent eigenstate.

• Alternatively, you can define your basis functions as $\psi_m(x)=e^{imx}$ for $m$ both positive, negative, and zero, each with energy $E_m=\frac{h^2}{2ML^2} m^2$, and keep in mind that the hamiltonian is degenerate and that the eigenspace for $E_m$ is $\mathrm{span}\{\psi_m,\psi_{-m}\}$.

  Note, in particular, that this fully correct, because it covers each linearly-independent eigenfunction exactly once.

There are probably some circumstances in which it makes sense to do the former, but in the vast majority of cases, it will be a significant hassle as compared to the simplicity of the second approach. As such, most people use the latter.