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Question 2.62: Suppose the bottom of the infinite square well is not flat (...

Suppose the bottom of the infinite square well is not flat  (V(x)= 0 ) , but rather

V(x)=500 V_{0} \sin \left(\frac{\pi x}{a}\right), \text { where } \quad V_{0} \equiv \frac{\hbar^{2}}{2 m a^{2}}

Use the method of Problem 2.61 to find the three lowest allowed energies numerically, and plot the associated wave functions (use N= 100).

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\lambda=\frac{\hbar^{2}}{2 m a^{2}}(N+1)^{2}=(N+1)^{2} V_{0} \quad(\text { here } N=100) ;

\left.V_{j}=b V_{0} \sin \left(\frac{\pi x_{j}}{a}\right)=b V_{0} \sin \left(\frac{\pi j \Delta x}{a}\right)=b V_{0} \sin \left(\frac{\pi j}{N+1}\right) \quad \text { (here } b=500\right) ;

v_{j}=\frac{V_{j}}{\lambda}=\frac{b}{(N+1)^{2}} \sin \left(\frac{\pi j}{N+1}\right) .

Factoring out λ, the diagonal elements of H are

2+v_{j}=2+\frac{b}{(N+1)^{2}} \sin \left(\frac{\pi j}{N+1}\right)=2+\frac{500}{10201} \sin \left(\frac{\pi j}{101}\right) .

h =\text { Table }[\operatorname{If}[ i == j , 2+(500 / 10201) \sin [\pi j / 101], 0],\{ i , 100\},\{j, 100\}] .

k =\text { Table }[I f[i=j+1,-1,0],\{i, 100\},\{j, 100\}] .

m =\text { Table }[I f[i=j-1,-1,0],\{i, 100\},\{j, 100\}] .

p=\operatorname{Table}[h[[i, j]]+k[[i, j]]+m[[i, j]],\{i, 100\},\{j, 100\}] .

EIG = Eigenvalues[N[p]]

10 201 * EIG

{841 255., 41 158.1, 41 063.4, 40 968.8, 40 869.4, 40 758.8, 40 632., 40 486.7, 40 322.1,

40 138.4, 39 935.6, 39 714.1, 39 474., 39 215.7, 38 939.5, 38 645.5, 38 334.2,

38 005.7, 37 660.6, 37 299., 36 921.3, 36 528., 36 119.3, 35 695.8, 35 257.7,

34 805.6, 34 339.9, 33 860.9, 33 369.3, 32 865.4, 32 349.7, 31 822.8, 31 285.1,

30 737.3, 30 179.7, 29 613., 29 037.7, 28 454.3, 27 863.4, 27 265.6, 26 661.5,

26 051.7, 25 436.7, 24 817.1, 24 193.6, 23 566.7, 22 937., 22 305.2, 21 671.9,

21 037.6, 20 403.1, 19 768.8, 19 135.5, 18 503.7, 17 874.1, 17 247.2, 16 623.6,

16 004., 15 389., 14 779.2, 14 175.1, 13 577.3, 12 986.5, 12 403.1, 11 827.8,

11 261.1, 10 703.5, 10 155.6, 9617.98, 9091.08, 8575.44, 8071.55, 7579.91,

7100.98, 6635.24, 6183.13, 5745.1, 5321.57, 4912.95, 4519.64, 4142.02, 3780.47,

3435.34, 3106.97, 2795.68, 2501.8, 2225.62, 1967.43, 1727.51, 1506.15, 1303.63,

1120.24, 956.36, 812.457, 689.191, 590.237, 499.854, 476.163, 304.8, 304.66}.

So the lowest three energies are 304.66 V _{0}, 304.8 V _{0}, \text { and } 476.163 V _{0} . Notice that the ground state is almost degenerate-essentially we have two separated wells with a huge barrier in between them, and the particle can be either in the left one or in the right one (or the even and odd linear combinations thereof).

EVE = Eigenvectors[N[p]].

ListLinePlot[EVE[[100]], PlotRange → \left\{0,0.2\right\} ].

ListLinePlot[EVE[[99]], PlotRange → \left\{-.2,0.2\right\} ].

ListLinePlot[EVE[[98]], PlotRange → \left\{-.1,0.2\right\} ].

Notice that the central barrier pushes the wave function out to the wings.

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