Find F automatically
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@ -2,7 +2,6 @@ import numpy as np
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import pylab as pl
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a = 5
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F = 4000
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D = 1
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hbar = 1
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m = 1
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@ -20,6 +19,7 @@ x_pos = x[x > 0]
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z0 = (a / hbar) * np.sqrt(2 * m * V0)
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print(f'z0 = {z0}')
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k = -l * cot(l * a)
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F = np.sin(l * a) / np.exp(-k * a)
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psi_conditions = [
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(0 < x_pos) & (x_pos <= a),
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