I have frequently read in articles that the energy level of spherical quantum dots depends on radius. We simply match the wave functions inside and outside the particle and find the intersection of the curves which give us the appropriate $k$:

$$ V(r)=0 \hspace{2cm} for \hspace{2mm}r<R $$ $$ \hspace{1.cm}V_0 \hspace{1.8cm} for\hspace{2mm} r>R $$ The wave functions inside and outside the dot of radius $R$ are $\phi_1$ \begin{equation} =N \frac{\sin(kr)}{r} \hspace{1cm} r<=R \\ =N\sin(kR)\exp(KR)\frac{\exp(-Kr)}{r}\hspace{1cm} r>=R \end{equation} The transcendental equation $$ kR+KRtankR=0 $$ was solved to obtain the lowest ‘subband’ energy $E_1$.In $ r=R$ by matching the wave function we earn a $k$. $ k^2=\frac{2mE}{\hbar^2}$ inside and $K^2=\frac{2m(V_0-E)}{\hbar^2}$ outside the particle. I don't know how can we insert the radius dependence.

enter image description here

Any help is appreciated.

  • $\begingroup$ Figure from: S. Davatolhagh "Oscillator strengths of the intersubband electronic transitions in the hydrogenic nano-antidots" , Superlattices and Microstructures 51 (2012) 62–72 $\endgroup$
    – Abolfazl
    Jul 13, 2014 at 8:34


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.