A **system** is some collection of particles, we describe it using
**observables**. To know the **state** of a system we must know the values of
the observables. Both quantum and classical mechanics are **dynamical** theories
in that they describe states; however, while classical is **deterministic** (we
can predict the future using it), quantum mechanics can only describe
probability. Mathematically, this is done with **wave functions** ($\Psi(\vec{r}
,t)$) which are defined as the **probability density amplitude**. To derive the
probability density from this we use:

Here the notation $\Psi^\ast$ refers to the complex conjugate of the wave function. A wave function must meet the following conditions:

- Must be
**normalized**($\int_{-\infty}^{\infty}\Psi(\vec{r},t)d\vec{r}=1$) - Must match the boundary conditions. In a well $\Psi=0$ at the walls, whereas when not confined $\lim_{|x|\to\infty}\Psi=0$
- Arbitrary phases are allowed e.g. the phase $\phi$ in $\Psi=f(x,t)e^{i\phi}$ since $P(x,t)=\Psi^\ast\Psi=f^\ast e^{-i\phi}fe^{i\phi}=f^\ast f$ (the phase does not affect observables)
- Must be a function
- Must be continuous
- The derivative of $\Psi$ must be continuous (such that $\Psi$ is a smooth function)
- Must satisfy the
**Schrödinger Equation**:

For a 1-Dimensional **free particle** (a particle with no potential energy) this
can be simplified to:

While this cannot be derived, we can feel like we derived it by proving that $\Psi=Ae^{i(kx-\omega t)}$ is a solution to it. We start this by expanding the left hand side of Eq.3 with the assumption that $E=hf=\hbar\omega$:

then, given that $KE=\frac{p^2}{2m}$ where $p=\frac{h}{\lambda}\cdot\frac{2\pi} {2\pi}=\hbar k$ such that $KE=\frac{\hbar^2k^2}{2m}$, for a free particle $V=0$, and that $E=KE+V$, we say that

On the right hand side of Eq.3, we can simply evaluate the partial derivative to find that $\frac{\partial^2}{\partial x^2}\Psi=(ik)^2\Psi=-k^2\Psi$ so that

Since the right hand sides of Eq.6 and Eq.7 are equal, we can then set their left hand sides to be equal, giving us the final form of

thus proving that $\Psi=Ae^{i(kx-\omega t)}$ is a solution… probably.

For further exploration of the Schrödinger Equation, we can derive a time
independent form. Where Eq.3 will hereafter be referred to as the **Time
Dependent Schrödinger Equation** or “TDSE,” this will be the **Time Independent
Schrödinger Equation** or “TISE.” This is done by the process of separation of
variables, beginning with defining some function $\Psi(x)$ and $T(t)$ such that
$\Psi(x,t)=\Psi(x)T(t)$ and proceeding to substitute them into the TDSE:

For the sake of laziness we’ll just refer to $T(t)$ as $T$ and $\Psi(x)$ as $\Psi_E$ from hereon. Moving each function to its own side of the equation yields

This equation does describe the energy of a system, so we can say that both sides are equal to the constant value $E$ (I don’t have a more thorough explanation for this, sorry). First, the left hand side:

Haha, like the alien movie. Anyway, moving on to the right hand side:

This last equation (Eq.12b) is the TISE. If we guess a function for the time dependence to be $T=e^{-i\omega t}$ and evaluate $\frac{\partial T}{\partial t}$ to be $-i\omega e^{-i\omega t}=-i\omega T$ which can be substituted into Eq.11b:

showing that $T=e^{-i\omega t}$ is a solution for time dependence, and so is the time dependence for any $\Psi_E$.

Finally, some discussion of $k^2$. As previously mentioned, $KE=\frac{\hbar^2 k^2}{2m}$, and since $E=KE+V$ we can say that $KE=E-V$. Earlier $0$ was used for $V$ but in this instance it will be an arbitrary constant $V_0$:

here Eq.14c describes a dispersion relation between $k$ and $\omega$. We can also consider velocity $v$:

With this we can revisit Eq.14 to find a value for the velocity of the particle: