# Appendix

Tee Kai Ze

Guan Xin

Jensie Low

Seow Ryan

●

Published 2022-08-27

# 4.-1 Appendix

# Derivation of 2-mass-1-spring simple harmonic oscillator.

The equations of motion of the two masses are

\begin{equation}
m_{1}\ddot{x}_{1}=k(x_{2}-x_{1}-x_{0})
\end{equation}

\begin{equation}
m_{2}\ddot{x}_{2}=-k(x_{2}-x_{1}-x_{0})
\end{equation}

Let x_{c} be the center of mass. By principle of moments,

\begin{equation}
m_{1}(x_{c}-x_{1})=m_{2}(x_{2}-x_{c})
\end{equation}

Let x_{d} be the distance between the two masses

\begin{equation}
x_{d}=x_{2}-x_{1}
\end{equation}

This gives

\begin{align}
m_{1}x_{c}-m_{1}x_{1} & =m_{2}(x_{d}+x_{1})-m_{2}x_{c}\nonumber \\
x_{1}(m_{1}+m_{2}) & =(m_{1}+m_{2})x_{c}-m_{2}x_{d}\nonumber \\
x_{1} & =x_{c}-\frac{m_{2}}{m_{1}+m_{2}}x_{d}
\end{align}

Similarly,

\begin{equation}
x_{2}=x_{c}+\frac{m_{1}}{m_{1}+m_{2}}x_{d}
\end{equation}

We sought to re-write the equations of motion in terms of x_{c}\text{ and }x_{d}. Putting Eq.(4), (5) and (6) into Eq.(1) and (2]), we have

\begin{align*}
m_{1}\left(\ddot{x}_{c}-\frac{m_{2}}{m_{1}+m_{2}}\ddot{x}_{d}\right) & =k(x_{d}-x_{0})\\
m_{2}\left(\ddot{x}_{c}+\frac{m_{1}}{m_{1}+m_{2}}\ddot{x}_{d}\right) & =-k(x_{d}-x_{0})\\
(m_{1}+m_{2})\ddot{x}_{c} & =0\\
\ddot{x}_{c} & =0
\end{align*}

That is a bit of work to show that the center of mass do not acceletate in the vibrational motion! Okay.. but what about the distance of separation? Putting Eq.(1) and (2) into Eq.(4),

\begin{align*}
\ddot{x}_{d} & =\ddot{x}_{2}-\ddot{x}_{1}\\
 & =-\frac{k}{m_{2}}(x_{d}-x_{0})-\frac{k}{m_{1}}(x_{d}-x_{0})\\
 & =-\frac{m_{1}+m_{2}}{m_{1}m_{2}}k(x_{d}-x_{0})\\
 & =-\frac{k}{\mu}(x_{d}-x_{0})
\end{align*}

where {\displaystyle \mu=\frac{m_{1}m_{2}}{m_{1}+m_{2}}} is the effective mass of the system. The above can be made even simpler by letting x=x_{d}-x_{0} to obtain

\ddot{x}=-\frac{k}{\mu}x

To obtain the potential energy of the system, we use the force-potential relationship:

\begin{align*}
F & =-\frac{dV}{dx}\\
\frac{dV}{dx} & =-\mu\ddot{x}=kx\\
V & =\frac{1}{2}kx^{2}
\end{align*}

(We can set the constant of integration to zero without loss of generality.)

# A note on quantum operators

In the previous chapter, we saw that Schrödinger was inspired by the energy conservation equation

\frac{p^{2}}{2m}+V(x)=E

and arrived at the Schrödinger equation

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

In 3 dimensions,

\frac{\mathbf{p}^{2}}{2m}+V(x)=E

where \mathbf{p}=(p_{x},p_{y},p_{z}), leads to

\left(-\frac{\hbar^{2}}{2m}\nabla^{2}+V(\mathbf{r})\right)\psi(\mathbf{r})=E\psi(\mathbf{r})

where \mathbf{r}=(x,y,z) and

\nabla^{2}=\left(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial z^{2}}\right)

In spherical coordinates,

\nabla^{2}=\frac{1}{R^{2}}\frac{\partial}{\partial R}\left(R^{2}\frac{\partial}{\partial R}\right)+\frac{1}{R^{2}\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{R^{2}\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}

It turns out to obtain the Schrödinger equation from classical concepts, one can make the following conversions:

\begin{align*}
x & \rightarrow x\\
p & \rightarrow-i\hbar\frac{\partial}{\partial x}\\
p^{2} & \rightarrow-\hbar^{2}\frac{\partial^{2}}{\partial x^{2}}\\
\mathbf{r} & \rightarrow\mathbf{r}\\
\mathbf{p}^{2} & \rightarrow-\hbar^{2}\nabla^{2}
\end{align*}

The right-hand-side of the above conversions are known as the "operators" in the language of quantum mechanics. For angular momentum L the operator conversions are

\begin{align*}
L_{x} & \rightarrow-i\hbar\left(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}\right)=i\hbar\left(\sin\phi\frac{\partial}{\partial\theta}+\cot\theta\cos\phi\frac{\partial}{\partial\phi}\right)\\
L_{y} & \rightarrow-i\hbar\left(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z}\right)=i\hbar\left(-\cos\phi\frac{\partial}{\partial\theta}+\cot\theta\sin\phi\frac{\partial}{\partial\phi}\right)\\
L_{z} & \rightarrow-i\hbar\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right)=-i\hbar\frac{\partial}{\partial\phi}\\
\mathbf{L}^{2} & \rightarrow-\hbar^{2}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\theta^{2}}\right]
\end{align*}

For the rigid rotor,

T+V=\frac{\mathbf{L}^{2}}{2I}=E

gives the SE

-\frac{\hbar^{2}}{2I}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\theta^{2}}\right]\psi=E\psi