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 $\eqref3$ with the assumption that $E=hf=\hbar\omega$:

\[i\hbar\frac{\partial\Psi}{\partial t}=i\hbar(-i\omega)\Psi=\hbar\omega\Psi= E\Psi,\tag5\]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

\[i\hbar\frac{\partial\Psi}{\partial t}=E\Psi=\frac{\hbar^2k^2}{2m}\Psi\tag6 \label6\]On the right hand side of $\eqref3$, 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

\[\frac{-\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}=\frac{\hbar^2k^2}{2m} \Psi.\tag7\label7\]Since the right hand sides of $\eqref6$ and $\eqref7$ are equal, we can then set their left hand sides to be equal, giving us the final form of

\[i\hbar\frac{\partial\Psi}{\partial t}=\frac{-\hbar^2}{2m}\frac{\partial^2\Psi} {\partial x^2},\tag8\]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 $\eqref3$ will hereafter be referred to as the **Time
Dependent Schrödinger Equation** (TDSE), this will be the **Time Independent
Schrödinger Equation** (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

\[i\hbar T^{-1}\frac{\partial T}{\partial t}=\frac{-\hbar^2}{2m}\Psi_E^{-1} \frac{\partial^2\Psi_E}{\partial x^2}.\tag{10}\]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:

\[\begin{align} i\hbar T^{-1}\frac{\partial T}{\partial t}&=E,\tag{11a} \\ i\hbar\frac{\partial T}{\partial t}&=ET.\tag{11b}\label{11b} \end{align}\]Haha, like the alien movie. Anyway, moving on to the right hand side:

\[\begin{align} \frac{-\hbar^2}{2m}\Psi_E^{-1}\frac{\partial^2\Psi_E}{\partial x^2}&=E, \tag{12a} \\ \frac{-\hbar^2}{2m}\frac{\partial^2\Psi_E}{\partial x^2} & = E\Psi_E. \tag{12b}\label{12b} \end{align}\]This last equation, $\eqref{12b}$, 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 $\eqref{11b}$:

\[\begin{align} i\hbar(\partial T/\partial t) & = i\hbar(-i\omega T)\tag{13a} \\ & = -i^2\hbar\omega T\tag{13b} \\ & = \hbar\omega T\tag{13c} \\ & = ET,\tag{13d} \end{align}\]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, $KE=\frac{\hbar^2k^2}{2m}$, 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 $V_0$:

\[\begin{align} \frac{\hbar^2k^2}{2m}&=E-V_0,\tag{14a}\label{14} \\ k^2&=\frac{2m}{\hbar^2}(E-V_0)\tag{14b} \\ &=\frac{2m}{\hbar^2}(\hbar\omega-V_0),\tag{14c}\label{14c} \end{align}\]here $\eqref{14c}$ describes a dispersion relation between $k$ and $\omega$. We can also consider velocity $v$:

\[\begin{align} v&=f\lambda\cdot\frac{2\pi}{2\pi}\tag{15a} \\ &=2\pi f\cdot\frac\lambda{2\pi}\tag{15b} \\ &=\frac\omega k.\tag{15c} \end{align}\]With this we can revisit $\eqref{14}$ to find a value for the velocity of the particle:

\[\begin{align} E-V&=KE,\tag{16a} \\ \hbar\omega-V_0&=\frac{k^2\hbar^2}{2m},\tag{16b} \\ \omega&=\frac{k^2\hbar}{2m}+\frac{V_0}{\hbar},\tag{16c} \\ v&=\frac{1}{k}\left(\frac{k^2\hbar}{2m}+\frac{V_0}{\hbar}\right)\tag{16d} \\ v&=\frac{k\hbar}{2m}+\frac{V_0}{k\hbar}.\tag{16e} \end{align}\]