Schrodinger's equation

Assume a particle-in-a-box scenario. x-axis is the position, y-axis is energy (which we already know must be discrete, so it’s basically n), two infinitely tall walls on two sides (at x=0 and x=L). The particle cannot be at or beyond the walls.
Manipulate the Hamiltonian (expanded & substituted into the equation) such that only the second derivative of phi (wave function) is on the left side.
The differential equation says that the wave function, when differentiated twice, will have the same form but multiplied by a constant ( $E$ ).
The wave function can be $A \cdot sin k x$ .
After twice-differentiating the wave function in sine form and rearranging the Schrodinger’s equation, we get a formula for $E$ in terms of $k$ .
Square of wave function = probability of appearance
definite integral of wave function sqaured (e.g. from 0 to box size) = 1 (total probability of apperance)
$E = \frac{n ^{2} h ^{2}}{8 m L ^{2}}$
Energy lower than the ground state energy (n = 1, a.k.a. zero-point energy). Since energy is discrete, the only energy level possible that is lower than the ground state energy is 0, and a energy of zero is impossible because the wave function will be 0 at all values (since n = 0), which violates the boundary condition

📖 Notes