

For the problems in which you are sketching states for potentials with several identical wells, the approach is to: 1) work with the ground states for the single wells, and 2) combine them, using the +- freedom, in such a way as to increase the number of nodes by one each time. The ground state has zero nodes, 1st-excited state has one node, 2nd excited state has 2, etc. That principle can guide you in creating solutions to all the problems except 1e, which goes one state beyond what you can create with the ground state. There you have to use the single-well 1st-excited state for each well (you can't mix ground and excited states; that is related to momentum "conservation"). In that case you get 4 nodes in the well centers (none between the wells). There is a pretty big jump in energy between state 4 and 5 in that case (see picture and graph). Conceptually, that state will belong to a different band than the four below it.
Problem 2 involves a graph of state energy vs "q", which has units of inverse length. The E vs q function has a minimum at q=0 (long wavelength and is biggest at the largest values of q (+-pi/a). For this to make sense, 2t should be pretty small compared to E_0.
Problem 9 gives us a rough intuitive sense of the nature of the bound state that forms around an impurity, including the effect of dielectric screening, and carrier effective mass, in making this state only very weakly bound.
We will discuss q and effective mass more when we consider transport properties, e.g., the movement of electrons in the conduction band.
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