# Quantum Dot

1. Homework Statement
A “quantum dot” is a semiconductor device that may be modeled as an electron with an effective mass m that is confined in an infinite spherical well. Suppose that the spherical well has radius a and the electron has effective mass $m = f m_e$, where f is some real number and $m_e$ is the mass of the electron.

What is the energy difference between the ground state and the first excited state, expressed as a function of a and m.

2. Homework Equations
$$-\frac{\hbar ^2}{2m} \frac{d^2 u}{dr^2} + \displaystyle \left[ V + \frac{\hbar ^2}{2m} \frac{l (l+1)}{r^2} \right] u = E u$$

if $l=0$
$$E_{n0} = \frac{n^2 \pi^2 \hbar^2}{2ma^2}$$

if $l \neq 0$
$$E_{nl} = \frac{\hbar^2}{2ma^2} \beta^2_{nl}$$
where $\beta_{nl}$ is the nth zero of the lth spherical Bessel function.

3. The Attempt at a Solution

Now the zeroth Spherical bessel function will give me the same energy solution as the l=0 case and so we're consistent. My issue is that I'm not terribly sure which is the proper ground state. Is $l = n =0$ the ground state or is $n = 0, l\neq 0$ the ground state for arbitrary l. I know that l refers to the orbital angular momentum of (in this case) the electron, so I would assume we would need to take arbitrary l into consideration but I'm really not too sure.