# Solving particle in a ring problem

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?

• Can you show your calculation for $m$? Nov 11, 2020 at 8:02
• As m is either positive or negative integer so we can write single exponential term. Nov 11, 2020 at 8:03
• and How you getting $x$, It's a two-dimensional problem. Try working in polar coordinate. en.wikipedia.org/wiki/Particle_in_a_ring Nov 11, 2020 at 8:04
• Apr 22, 2021 at 13:00

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.)

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.