We can solve the Schrodinger problem for a particle in 2 - dimension hole:
H^Ψn,m=En,mΨn,m\widehat H \Psi_{n,m} = E_{n,m}\Psi_{n,m}HΨn,m=En,mΨn,m
where :
H^→−ℏ22m(∂2∂x2+∂2∂y2)+U(x,y)\widehat H \rightarrow -\frac{\hbar^{2}}{2m}\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right) + U(x,y)H→−2mℏ2(∂x2∂2+∂y2∂2)+U(x,y)
if we solve it for a hole with rectangle as a shape, then we get the level of energy fo such system:
En,m=π2ℏ22m(n2a2+m2b2)E_{n,m} = \frac{\pi^2\hbar^2}{2m}\left(\frac{n^2}{a^2} + \frac{m^2}{b^2}\right)En,m=2mπ2ℏ2(a2n2+b2m2)
n,m - quantum numbers
a,b - sides of the rectangle
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments