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Cell["Quantum Wells -- An Eigenvalue Problem", "Title",
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Cell["\<\
This notebook follows the lines of the previous notebook on the rectangular \
well. In units where \[HBar] = m = 1 (where m is the mass of the particle), \
Schrodinger's equation is \
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We will find the lowest energy level and corresponding wavefunction for each \
parity. We take the potential\
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the eigenstates are either even or odd functions of x, \[Psi](x) = \
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state. It is shown in the textbooks on quantum mechanics  that (i)  the \
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(ii) the first excited state is odd and has one node, and (iii) for each \
higher energy eigenvalue the the number of nodes increases by oneand the \
parity alternates. \n\nThe method we will use requires that V(x) = 0 for |x| \
> L/2. Now our V(x) is never quite zero but it is very close to zero for |x| \
> 5. Hence we take L=10.  First we define and plot the potential:"
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We also define the Schrodinger equation for region 2, using a delayed \
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We are now in a position to start calculating. First of all we look for an \
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Cell["\<\
This result agrees with the exact value which is -2. This state is actually \
the only even parity bound state.\
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Now we want the eigenfunction coresponding to our eigenvalue. Since we now \
have the eigenvalue, we do not  want to keep recalculating  the wavefunction \
so we define a function \"efunc2\" with immediate assignment, where we input \
the eigenvalue for the energy\
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and then plot it. We first define the plot for \[Psi]norm (but do not yet \
display it because of thesemicolon after the comand). Then we add the labels \
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has the advantage that the current values of the well depth and width, and \
the eigenvalue are automatically printed without needing to modify this \
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We use the \"FindRoot\" command to locate the eigenvalue and give it two \
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This agrees with the exact value of -1/2. This is actually the only odd \
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