Question 3.4: An asymmetric one-dimensional potential well of width L has ......

In this exercise, an asymmetric one-dimensional potential well of width L has an infinite potential energy barrier on the left-hand side and a finite constant potential of energy V_{0} on the right-hand side. We aim to find the minimum value of L for which a particle of mass m has at least one bound state. A quick way to a solution is to compare with results for the symmetric potential considered in Section 3.7. The first figure sketches four bound-state wave functions for a particle of mass m in a symmetric potential for which V_{0} L^{2} =25\hbar ^{2}/2m, so that LK_{0} =L\sqrt{2mV_{0} } /\hbar =5 . As expected, the wave functions have alternating even and odd parity, with even parity for the lowest energy state, \psi _{1}. The lower part of the above figure shows the asymmetric potential that arises when the potential for x < 0 is infinite. Because the wave function must be zero for x < 0, the only allowed solutions for bound states are of the form \psi _{2} and \psi _{4} previously obtained for the symmetric case and x > 0. These correspond to the odd-parity wave functions that are found using the graphical method sketched below when V_{0} L^{2} =25\hbar ^{2}/2m, so that LK_{0} =L\sqrt{2mV_{0} } /\hbar =5 .

The lowest energy solution exists where the radius of the arc LK_{0} =L\sqrt{2mV_{0} } /\hbar  intersects the function kLcot(kL)=−\kappa L. Vanishing binding energy will occur when the particle is in a potential so that LK_{0} \rightarrow \pi /2 and \kappa \ll k . This situation is sketched below.

Since LK_{0} =L\sqrt{2mV_{0} } /\hbar  , in the limit of vanishing binding energy L\sqrt{2mV_{0} } /\hbar\geq \pi /2, so that

To find the minimum value of L, we need to know the value of V_{0} and m. For the case in which the step potential energy is V_{0} = 1  eV and m=m_{0}, the minimum value of L for an electron is L_{min} \geq \pi \hbar /2(2m_{0} V_{0} )^{1/2} \sim 0.3 nm.