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Appendix
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4.-1 Appendix
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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.(
\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.(
\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.)
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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