The Schrodinger equation plays the role of Newton’s laws and the conservation of energy in classical mechanics. It is a wave equation in terms of the a called wavefunction which predicts analytically and precisely the probability of an outcome. The detailed outcome is not strictly determined, but the Schrodinger equation will predict the distribution of results.

(Planck’s constant)^{2}/2(mass) Second derivative of the wavefunction = energy wavefunction

The equation is:

**-ℏ ^{2}/2m ∂^{2}/∂x^{2} Ψ = E Ψ**

Where:

ℏ: Planck’s constant

m: mass of the particle

ψ: Wavefunction

**Schrodinger Equation Formula Questions:**

1) A particle is confined to a box, in one dimension, over the x axe. The wavefunction describing the particle is Ae^{(I n x)}, where n is a integer number and I is the imaginary number obeying I^{2}=-1. What is the energy of the particle given in terms of n?

**Answer: **

We replace the data in the equation above:

-ℏ^{2}/2m ∂^{2}/∂x^{2} (Ae^{(I n x))} =

=-ℏ^{2}/2m A ∂^{2}/∂x^{2} (e^{(I n x)}) =

=-ℏ^{2}/2m A I^{2} n^{2} e^{(I n x)}

Then regrouping the terms conveniently, we find the form

=-(I^{2}) ℏ^{2}/2m n^{2} (Ae^{(I n x)})

=-(-1) ℏ^{2}/2m n^{2} (Ae^{(I n x)})

And the expression in the parenthesis is exactly the wavefunction

= ℏ^{2}/2m n^{2} Ψ

Then, because this is only the left-hand side of the Schrodinger equation, the complete equation is

ℏ^{2}/2m n^{2} Ψ = E Ψ

So, the energy is

E= ℏ^{2}/2m n^{2}

2) Consider the same particle above, what is its velocity after those 5 seconds?

**Answer:**

The formula for the velocity is obtained deriving with respect to time the equation for the position.

V=d/dt (r_{0} + v_{0} t +1/2 a t^{2})=a*t

Then, the velocity is

V=2 m/s^{2} * 5 s = 10 m/s