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 # The schrodinger equation
 
-## Mathematical prerequisites
+## Preface
 
-To understand this article, you will need a thorough understanding of complex numbers.
+This article will cover the schrodinger equation, which represents a non-relativistic differential equation governing
+quantum particles. Relativistic effects are not taken into account here.
 
-I will also use Euler's formula $e^{i*x} = \cos(x) + i*\sin(x)$
+## Deriviation
 
-$$e^{ix} = \cos(x) + i*\sin(x)$$
+It should be mentioned that the schrodinger equation cannot be derived traditionally but rather represents a
+semi-logical conclusion from a number of assumptions/axioms.
+
+### Known properties of traditional waves
+
+In quantum mechanics, particles are assumed to be governed by a wave function. [Louis-
+de-Broglie](https://de.wikipedia.org/wiki/Louis_de_Broglie) postulated that matter waves should have the same wavelength
+as photons, whose wavelength depends on their respective impulse:
+
+$$\lambda = \frac{h}{p} = \frac{h}{m*v} $$
+
+A basic wave function for a travelling monochromatic wave can be written as:
+
+$$ \Psi = \cos(k*x - \omega * t) = e^{i*(k*x - \omega * t)} $$
+
+Schrodinger assumed that a matter wave should have the same wave form. From this, we can derive $k =
+\frac{2*\pi}{\lambda} = \frac{2 * \pi * p}{h} $, using the de-Broglie-wavelength as mentioned above. With $ v = f *
+\lambda $ for each wave, one can also conclude that $\omega = 2 * \pi * f = 2 * \pi * \frac{v}{\lambda} = \frac{2 * \pi
+* v * p}{h} $. We can rewrite the wave function as follows:
+
+$$ \Psi = e^{i*(\frac{2 * \pi * p}{h}*x - \frac{2 * \pi * v * p}{h} * t)} = e^{\frac{i*p}{\hbar}*(x - v*t)} $$
+
+### Obtaining the Schrodinger equation through energy conversion
+
+The total sum of energy should be a constant for physical particles. For this,
+the sum of the kinetic Energy $E_{kin} = \frac{p^{2}}{2*m}$ and potential energy $E_{pot} = V$, where V describes some
+arbitrary potential, which in practice will need to be multiplied by its respective "charge" of a particle (this could
+be its mass for a gravitational potential or the charge for electric fields) needs to stay constant. Thusly, one can
+write:
+
+$$ E = E_{kin} + E_{pot} = \frac{p^{2}}{2*m} + V $$
+$$ \hat{E} * \Psi = \frac{\hat{p}^{2}}{2*m} * \Psi + V * \Psi $$
+
+$\hat{E}$ and $\hat{p}$ are some (yet unknown) operators on $\Psi$ which respectively should return the impulse and
+total energy for the wave function. For wave function to be valid, this equation must be true. On the other hand, we can
+obtain these operators using the basic wave function mentioned above.
+
+Noting the factor $p$ in the exponent of $\Psi$, you might write $\frac{\partial{\Psi}}{\partial{x}} = \frac{i *
+p}{\hbar}$ and obtain $p = -i*\hbar*\frac{\partial{\Psi}}{\partial{x}}$. Plugging this back into the energy conservation
+equation, we obtain the time-independent schrodinger equation:
+
+$$ \hat{E} * \Psi = -\frac{\hbar^2}{2*m} * \frac{\partial^2 \Psi}{\partial^2 x} + V*\Psi$$