# Quantum Mechanics

By

Ervin Chia

,

Chua Wei Hong

,

Hillson Hung

,

Nemo Chen

●

Published 2022-08-15

We shall now dive a little deeper into modern quantum mechanics¹. From 1925-1927, three versions of quantum mechanics was independently developed by Erwin Schrödinger, Werner Heisenerg and Paul Dirac. They differed drastically from their initial formulism, but were later shown to be equivalent. Here we will adopt the Schrödinger "picture", as out of the three, it is the least chim².

Schrödinger postulated that a quantum system can be specified completely by a wavefunction:

\Psi

If one knows the wavefunction, then one will know all the mechanical information such as energy, position and momentum of the system. Schrödinger sought to find a differential equation that relates how the derivatives (change) of the wavefunction of a particle is affected by the energy and boundary conditions it is in.

He starts by assuming that the wavefunction can take on a simple but general wave-like expression

\begin{equation}
\Psi(x,t)=\psi_{0}e^{i(kx-\omega t)}
\end{equation}

where \psi_{0} is the amplitude³ of the wave, k=2\pi/\lambda is the wavenumber and \omega=2\pi f is the angular frequency (see Eq. Light(5) and Light(6)).

The derivatives of \Psi are

\begin{align*}
\frac{\partial\Psi}{\partial t} & =-i\omega\psi_{0}e^{i(kx-\omega t)}=-i\omega\Psi\\
\frac{\partial^{2}\Psi}{\partial t^{2}} & =-i\omega(-i\omega\Psi)=-\omega^{2}\Psi\\
\frac{\partial\Psi}{\partial x} & =ik\psi_{0}e^{i(kx-\omega t)}=ik\Psi\\
\frac{\partial^{2}\Psi}{\partial x^{2}} & =-k^{2}\Psi
\end{align*}

Borrowing de Broglie's idea of matter waves where we have

\begin{align*}
E & =hf=\hbar\omega\\
p & =\frac{h}{\lambda}=\hbar k
\end{align*}

The above derivatives can be rewritten as

\begin{align}
\frac{\partial\Psi}{\partial t} & =-i\frac{E}{\hbar}\Psi\\
\frac{\partial^{2}\Psi}{\partial t^{2}} & =-\frac{E^{2}}{\hbar^{2}}\Psi\\
\frac{\partial\Psi}{\partial x} & =i\frac{p}{\hbar}\Psi\\
\frac{\partial^{2}\Psi}{\partial x^{2}} & =-\frac{p^{2}}{\hbar^{2}}\Psi
\end{align}

Using a total energy consideration

\begin{align*}
\text{Total energy} & =\text{Kinetic energy }+\text{Potential energy}\\
E & =\frac{p^{2}}{2m}+V(x)
\end{align*}

Multiply throughout by \Psi and using Eq.(2) and (5), we have

\begin{align}
E\Psi & =\frac{p^{2}}{2m}\Psi+V(x)\,\Psi\nonumber \\
i\hbar\frac{\partial\Psi}{\partial t} & =-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}}+V(x)\,\Psi
\end{align}

The differential equation Eq.(6) is known as the Schrödinger's equation (SE). Different systems will have a different potential energy landscape V(x) and a different set of boundary conditions. By finding the solution of the SE, ie \Psi(x,t), one will essentially be able to determine everything that can be known about the system.

The SE however is a difficult equation to solve as it has derivatives in both time t and position x. If one is interested to find the "stationary solutions" such as the standing waves mentioned in the previous section, then it is beneficial to remove time from the SE.

To do so, we go back to the assumed general wavefunction in Eq.(1) and rewrite it as

\begin{align*}
\Psi(x,t) & =\psi_{0}e^{ikx}e^{-i\omega t}\\
 & =\psi(x)e^{-i\omega t}
\end{align*}

Put this into the SE,

\begin{align}
i\hbar\frac{\partial\psi(x)e^{-i\omega t}}{\partial t} & =-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi(x)e^{-i\omega t}}{\partial x^{2}}+V(x)\,\psi(x)e^{-i\omega t}\nonumber \\
i\hbar(-i\omega)e^{-i\omega t}\psi(x) & =-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi(x)}{\partial x^{2}}e^{-i\omega t}+V(x)\,\psi(x)e^{-i\omega t}\nonumber \\
E\psi(x) & =-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi(x)}{\partial x^{2}}+V(x)\,\psi(x)\nonumber \\
\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V(x)\right)\,\psi(x) & =E\,\psi(x)
\end{align}

Eq.(7) is known as the time-independent Schrödinger equation. This equation is the one that will allow us to calculate the allowed discrete energy states of matter (such as electrons) in the atom or other systems.

For now, we understand a wavefunction as something that contains a particle's mechanical information, and the SE as the equation to solve to obtain the wavefunction as well as allowed energy states. But...

The wavefunction \Psi(x,t) is a complex (as oppose to real) function. Complex quantities cannot be measured in the real world. Is there actually a physical meaning associated with this all-important wavefunction?

No.

But Max Born offered a "working" statistical interpretation to the wavefunction:

\left|\Psi\right|^{2}\equiv\Psi^{*}\Psi\text{ plays the role of a probability density function.}

This interpretation allows one to use the wavefunction to find physically relevant stuffs. For example:

\int_{a}^{b}\left|\Psi(x,t)\right|^{2}dx=\text{probability of finding the particle between $a$ and $b$ at time $t$}

With the intepretation of \left|\Psi\right|^{2} as a probability density function, we have the property that

\begin{equation}
\int_{-\infty}^{\infty}\left|\Psi(x,t)\right|^{2}dx=1,
\end{equation}

Eq.(8) is a statement that says that \left|\Psi\right|^{2} is normalised, which simply means that all probabilities add up to 1.

One can go further into the statistical interpretation by looking the expected values and varriances of observable physical quantities, but it is out of the scope of the course and we will leave the interpretation and use of \Psi as is for now.

We can now try and explore how different systems give rise to different wavefunctions and allowed energy states..

The particle in a box problem is a thought experiment made by Schrödinger to illustrate the basics of 'motion' in the quantum world. In this section, we will solve the time-independent SE for the particle in box problem⁴.

The formal way to state the problem is to ask what happens to the wavefunction of a single particle along a 1D line when subjected to the following boundary conditions:

V(x)=\begin{cases}
0\text{ if }0<x<L,\\
\infty\text{ elsewhere}
\end{cases}

where V(x) is the potential at any point in space. For simplicity, we shall assume that the particle's wavefunction does not change with time.

We will first deal with the case of infinite potential as it is easier. Later we will see what happens when the barriers are finite.

Recall that Schrödinger's equation states that

\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V(x)\right)\,\psi(x)=E\psi(x)

Inside the box 0<x<L where V=0, we have

\begin{equation}
-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi(x)}{\partial x^{2}}=E\psi(x)
\end{equation}

and the wavefunction outside the box must obey

\begin{equation}
-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi(x)}{\partial x^{2}}+\infty\,\psi(x)=E\psi(x)
\end{equation}

We can make an educated assumption that the wavefunction outside the box must be zero such that Schrodinger's equation still holds⁵ (because 0=0). This also gives us the boundary conditions of the problem:

\begin{align}
\psi(0) & =0\\
\psi(L) & =0
\end{align}

But now, what about the wavefunction inside the box?

We can guess the solution must resemble something of form (recall the restoring force case in Chapter 2 and see the similarity of the equations)

\begin{equation}
\psi(x)=ce^{\lambda x}
\end{equation}

Putting the guess solution into the SE, we obtain the general solution to be

\begin{equation}
\psi(x)=c_{1}e^{ikx}+c_{2}e^{-ikx}
\end{equation}

where

k={\displaystyle \sqrt{\frac{2mE}{\hbar^{2}}}}

Applying the boundary conditions Eq.(11) and (12), we will find that

c_{1}=-c_{2}

and

kL=n\pi

where n takes positive integer values. Consequently

\psi(x)=A\sin kx

and

\begin{equation}
E=\frac{\hbar^{2}n^{2}\pi^{2}}{2mL^{2}}
\end{equation}

A can be determined by normalising the wavefunction with Eq.(8). Finally the wavefunction of the particle in a box is

\begin{equation}
\psi(x)=\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi}{L}x\right)
\end{equation}

and it's allowed energy levels are given by Eq.(15).

Also important is to recall that n takes only positive integer values, thus effectively quantising the energy states and wavefunction. For this reason, n is known as the quantum number for particle in a box.

Welcome back to the Hydrogen atom. The hydrogen atom consists of a proton and an electron separated by some distance r. The potential of the system is

\begin{equation}
V(r)=-\frac{1}{4\pi\epsilon_{0}}\frac{e^{2}}{r}=-\frac{1}{4\pi\epsilon_{0}}\frac{e^{2}}{\left(x^{2}+y^{2}+z^{2}\right)^{1/2}}
\end{equation}

As the system is 3 dimensional, the time independent SE is

\left[-\frac{\hbar^{2}}{2\mu}\left(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial x^{2}}\right)-\frac{1}{4\pi\epsilon_{0}}\frac{e^{2}}{\left(x^{2}+y^{2}+z^{2}\right)^{1/2}}\right]\psi=E\,\psi

where

\mu=\frac{m_{e}m_{p}}{m_{e}+m_{p}}

is so called the reduced mass of the system. We will talk more about the reduced mass in the next chapter. Meanwhile, in here, since the mass of the proton m_{p} is much larger than the mass of the electron m_{e,} we have \mu\approx m_{e}. We can concise the notation and state that the SE of the hydrogen atom is

\begin{equation}
\left[-\frac{\hbar^{2}}{2m_{e}}\nabla^{2}-\frac{1}{4\pi\epsilon_{0}}\frac{e^{2}}{r}\right]\psi=E\,\psi
\end{equation}

The boundary conditions are

\begin{align*}
\psi(0) & =0\\
\psi(\infty) & =0
\end{align*}

[Translates as: you will not find the electron at the nucleus (r=0) and at somewhere infinitely far away (r\rightarrow\infty)].

The procedure to solve the SE is same as that of particle in a box: Solve the SE with the above bounday conditions. The workings are much more complicated since the system is three dimensional. The higher dimensionality also means that more quantum numbers to required to describe the allowed states of the hydrogen atom. The quantum numbers are

1. n, the principal quantum number. n=1,2,3,\dots
2. l, orbital quantum number. l=0,1,2,\dots,n-1.
3. m_{l}, magnetic quantum number. m_{l}=-l,\dots,-1,0,1,\dots l.

The allowed energies of the system is given by

E=-\frac{e^{4}m_{e}}{32\pi^{2}\epsilon_{0}^{2}\hbar^{2}n^{2}}

The wavefunctions for the hydrogen atom are

\begin{align*}
\psi_{n=1,l=0,m_{l}=0} & =\frac{1}{\sqrt{\pi}a_{0}^{3/2}}e^{-r/a_{0}}\\
\psi_{n=2,l=0,m_{l}=0} & =\frac{1}{4\sqrt{2\pi}a_{0}^{3/2}}\left(2-\frac{r}{a_{0}}\right)e^{-r/2a_{0}}\\
\psi_{n=2,l=1,m_{l}=0} & =\frac{1}{4\sqrt{2\pi}a_{0}^{3/2}}\frac{r}{a_{0}}e^{-r/2a_{0}}\cos\theta\\
\psi_{n=2,l=1,m_{l}=+1} & =\frac{1}{8\sqrt{\pi}a_{0}^{3/2}}\frac{r}{a_{0}}e^{-r/2a_{0}}\sin\theta e^{i\phi}\\
\psi_{n=2,l=1,m_{l}=-1} & =\frac{1}{8\sqrt{\pi}a_{0}^{3/2}}\frac{r}{a_{0}}e^{-r/2a_{0}}\sin\theta e^{-i\phi}
\end{align*}

Here is 2D visualization of the probabilites of the electron's position around the nucleus.

The solutions to the hydrogen atom not only opens a pathway to many new rich areas of physics such a molecular physics which is critical to the development of physical chemistry. Let us begin to dive into the how these orbitals play a role in composite systems such as molecules and their necessary rammifications.