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Quantum Dot

  • Thread starter Kreizhn
  • Start date
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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 [itex] m = f m_e[/itex], where f is some real number and [itex]m_e[/itex] 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
The Radial Schrodinger equation:
[tex] -\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 [/tex]

if [itex] l=0[/itex]
[tex]E_{n0} = \frac{n^2 \pi^2 \hbar^2}{2ma^2}[/tex]

if [itex] l \neq 0 [/itex]
[tex] E_{nl} = \frac{\hbar^2}{2ma^2} \beta^2_{nl} [/tex]
where [itex] \beta_{nl} [/itex] 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 [itex] l = n =0 [/itex] the ground state or is [itex] n = 0, l\neq 0 [/itex] 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.
 

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