. The total number of solutions, , (i.e., the number of purple/grey curves that are intersected by the blue circle) is therefore determined by dividing the radius of the blue circle, , by the range of each solution and using the floor or ceiling functions: In this case there are exactly three solutions, since . and , with the corresponding energies If we want, we can go back and find the values of the constants in the equations now (we also need to impose the normalisation condition). On the right we show the energy levels and wave functions in this case (where ): We note that however small is (however shallow or narrow the well), there is always at least one bound state. Two special cases are worth noting. As the height of the potential becomes large, , the radius of the semicircle gets larger and the roots get closer and closer to the values , and we recover the case of the infinite square well. The other case is that of a very narrow, deep well - specifically the case and with fixed. As it will tend to zero, and so there will only be one bound state. The approximate solution is then , and the energy tends to . But this is just the energy of the bound state of a Delta function potential of strength , as it should be. A simpler graphical solution for the energy levels can be obtained by normalizing the potential and the energy through multiplication by . The normalized quantities are giving directly the relation between the allowed couples as for the even and odd parity wave functions, respectively. In the previous equations only the positive derivative parts of the functions have to be considered. The chart giving directly the allowed couples is reported in the figure. Unbound states If we solve the time-independent Schrödinger equation for an energy , the solutions will be oscillatory both inside and outside the well. Thus, the solution is never square integrable; that is, it is always a non-normalizable state. This does not mean, however, that it is impossible for a quantum particle to have energy greater than , it merely means that the system has continuous spectrum above . The non-normalizable eigenstates are close enough to being square integrable that they still contribute to the spectrum of the Hamiltonian as an unbounded operator. Asymmetrical well