Calculating Energy Difference of Quantum Dot in Spherical Well

[Kreizhn]

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.

Homework Equations

The Radial Schrodinger equation:
$$-\frac{\hbar ^2}{2m} \frac{d^2 u}{dr^2} + \displaystyle \left[ V + \frac{\hbar ^2}{2m} \frac{l (l+1)}{r^2} \right] u = E u$$

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.

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.

 

[Jhero]

Thank you for your question about the energy difference between the ground state and first excited state in a quantum dot. As you correctly stated, the energy levels in a quantum dot can be described by the radial Schrodinger equation, which takes into account the confinement of the electron in an infinite spherical well.

In this case, the ground state is defined as the lowest energy state in which the electron can exist. The ground state will always have l=0, since this corresponds to the lowest possible value of the orbital angular momentum. Therefore, the ground state energy for a quantum dot is given by E_{00} = \frac{\hbar ^2}{2m_e a^2}.

The first excited state, on the other hand, can have a non-zero value of l. However, it will still have the same principle quantum number n=1, since it is the first excited state. Therefore, the energy of the first excited state is given by E_{10} = \frac{\hbar ^2}{2m_e a^2} \beta_{10}^2.

The energy difference between these two states can be calculated by subtracting the ground state energy from the first excited state energy: \Delta E = E_{10} - E_{00} = \frac{\hbar ^2}{2m_e a^2} \left(\beta_{10}^2 - 1\right).

I hope this helps to clarify the energy levels in a quantum dot. Please let me know if you have any further questions.

Scientist