because positive and negative are completely the same

because we are working in a positive potential we only keep positive normalisations

because we cannot have negative energy, we ignore negative normalisations.

because positive and negative are the same and differ only in phase

hyperbolic trigonometric functions

cosines

sines

tangents

both sines and cosines

V(x) = V₀ for -L/2 ≤ x ≤ L/2 and 3V₀ elsewhere

V(x) = 0 for -L/2 ≤ x ≤ L/2 and V₀ elsewhere

V(x) = 0 for -L ≤ x ≤ L and V₀ elsewhere

V(x) = V₀ for -L/2 ≤ x ≤ L/2 and 0 elsewhere

V(x) = 0 for -L ≤ x ≤ L and -V₀ elsewhere

solving the Schrödinger Equation for an infinite square well and a finite well only.

solving the Schrödinger Equation for an infinite square well only.

solving the Schrödinger Equation for a harmonic oscillator only.

solving the Schrödinger Equation for an infinite square well, a finite well and for a wall (i.e. quantum tunnelling).

It applies the DSolve function to dissolve the equation and separate into terms of y[x] and x alone.

DSolve tests if the sum y’[x] + y[x] is equal to a*Sin[x] and returns y[x] if so or x if not.

It solves a first order differential equation for the function y[x] but no boundary condition has yet been specified.

It solves a first order differential equation in y[x] using the boundary condition mySolve = a*Sin[x]

Solution[x_] := DSolve[{y'[X] + y[X] == a Sin[X]}, y[X], X]  /.C[1]->1/.X->x

solution[x] = DSolve[{y'[x_] + y[x_] == a Sin[x_], y[0] == -a⁄2}, y[x_], x_ ]

solution[x_] := DSolve[y'[X] + y[X] == a Sin[X], y[X], X]/.X->x

solution[x_] := y[X] /. DSolve[{y'[X] + y[X] == a Sin[X], y[0] == -a/2}, y[X], X][[1]]/.X->x

The input cells define lists of transformation rules for k, V₀ and L.  Some lists contain only single elements.

The cells don’t do anything useful, as this is not valid Mathematica syntax.

The input cells test whether variables such as defk are equal to variables such as k, once the potential and total energies are evaluated and the square roots are calculated.