Notes on Quantum Mechanics I

Abridged Edition

By Una Ada, April 26, 2018

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:

$|\Psi(\vec{r},t)|^2=\Psi^\ast(\vec{r},t)\Psi(\vec{r},t).\tag1$

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

1. Must be normalized ($\int_{-\infty}^{\infty}\Psi(\vec{r},t)d\vec{r}=1$)
2. Must match the boundary conditions. In a well $\Psi=0$ at the walls, whereas when not confined $\lim_{|x|\to\infty}\Psi=0$
3. 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)
4. Must be a function
5. Must be continuous
6. The derivative of $\Psi$ must be continuous (such that $\Psi$ is a smooth function)
7. Must satisfy the Schrödinger Equation:
$i\hbar\frac{\partial\Psi}{\partial t}=\frac{-\hbar^2}{2m}\nabla^2\Psi +V(\vec{r})\Psi.\tag2$

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

$i\hbar\frac{\partial\Psi}{\partial t}=\frac{-\hbar^2}{2m}\frac{\partial^2\Psi} {\partial x^2}.\tag3\label3$

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:

$i\hbar\Psi(x)\frac{\partial}{\partial t}T(t)=\frac{-\hbar^2}{2m}T(t) \frac{\partial^2}{\partial x^2}\Psi(x).\tag9$

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}