A particle is connected to a spring and undergoes one-dimensional motion.

(a) Write an expression for the total (kinetic plus potential) energy of the particle in terms of its position x. its mass m, its momentum p, and the force constantof the spring.

(b) Now treat the particle as a wave. Assume that the product of the uncertainties in position and momentum is governed by an uncertainty relation$\mathbf{∆}\mathbf{p}\mathbf{∆}\mathbf{.}\mathbf{r}\mathbf{\approx }\frac{\mathbf{1}}{\mathbf{2}}\mathbf{h}$. Also assume that because xis on average. the uncertaintyis roughly equal to a typical value of$\left|x\right|$. Similarly, assume that$\mathbf{∆}\mathbf{p}\mathbf{\approx }\left|p\right|$. Eliminate pin favor of xin the energy expression.

(c) Find the minimum possible energy for the wave.

The expression for kinetic energy in terms of momentum is given by,

$\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{p}}^{\mathbf{2}}\mathit{m}$

The expression for the potential energy of the spring is given by,

$\mathit{U}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{k}{\mathit{x}}^{\mathbf{2}}$

The expression for total energy is given by,

T=E+U

a)

The total energy of a l-D harmonic oscillator (HO) consists of kinetic and potential energy (PE) components given as follows,

$$\begin{gathered} \mathrm{E}=\mathrm{KE}+\mathrm{PE} \\ =\frac{{\mathrm{p}}^2}{2\mathrm{m}}+\frac{1}{2}{\mathrm{kx}}^2 \end{gathered}$$

(b)

The expression for momentum and position of the particle is given by,

$$\begin{gathered} \left|{\mathrm{p}}_{\mathrm{x}}\right|\left|\mathrm{x}\right|\approx \frac{\mathrm{h}}{2} \\ {\mathrm{p}}_{\mathrm{x}}\approx \frac{\mathrm{h}}{2\mathrm{x}} \end{gathered}$$

The expression for total energy is given by,

$$\begin{gathered} \mathrm{T}=\frac{{\mathrm{p}}^2}{2\mathrm{m}}+\frac{1}{2}{\mathrm{kx}}^2 \\ =\left(\frac{1}{2\mathrm{m}}\right){\left(\frac{\mathrm{h}}{2\mathrm{x}}\right)}^2+\frac{1}{2}{\mathrm{kx}}^2 \\ =\frac{{\mathrm{h}}^2}{8{\mathrm{mx}}^2}+\frac{1}{2}{\mathrm{kx}}^2 \end{gathered}$$

Therefore, the expression for the total energy of the particles in terms of their position is $\frac{{\mathrm{h}}^2}{8{\mathrm{mx}}^2}+\frac{1}{2}{\mathrm{kx}}^2$.

(c)

Differentiate equation (1) with respect to x and equate to 0.

$$\begin{gathered} \mathrm{T}=\frac{h^2}{8m{x}^2}+\frac{1}{2}{\mathrm{kx}}^2 \\ \frac{dT}{dx}=\frac{h^2}{4m{x}^2}+kx \\ 0=-\frac{h^2}{4m{x}^2}+kx \\ kx=\frac{h^2}{4m{x}^2} \end{gathered}$$

Solve further,

$$\begin{gathered} 4mk{x}^4=h^2 \\ {x}^4=\frac{h^2}{4mk}x \\ =\sqrt[4]{\frac{h^2}{4mk}} \end{gathered}$$

The total minimum energy is calculated as,

$$\begin{gathered} T=\frac{h^2}{8m{k}^2}+\frac{1}{2}k{x}^2 \\ =\frac{h^2}{8m{\left(\sqrt[4]{{\displaystyle \frac{h^2}{4mk}}}\right)}^2}+\frac{1}{2}k{\left(\sqrt[4]{\frac{h^2}{4mk}}\right)}^2 \\ =\frac{h^2}{8m}\left(\sqrt[]{\frac{h^2}{4mk}}\right)+\frac{1}{2}k\left(\sqrt[]{\frac{h^2}{4mk}}\right) \\ =\frac{h^2}{8m}\left[\left(\frac{2}{h}\right)\sqrt{mk}\right]+\frac{1}{2}k\left[\left(\frac{h}{2}\right)\frac{1}{\sqrt{mK}}\right] \end{gathered}$$

Solve further,

$$\begin{gathered} T=\frac{h}{4m}\sqrt{mK}+\frac{hK}{4}\frac{1}{\sqrt{mK}} \\ =\frac{h}{4}\sqrt{\frac{mk}{m^2}}+\frac{h}{4}\sqrt{\frac{k^2}{mk}} \\ =\frac{h}{4}\sqrt{\frac{k}{m}}+\frac{h}{4}\sqrt{\frac{\kappa }{m}} \\ =\frac{h}{2}\sqrt{\frac{k}{m}} \end{gathered}$$

Therefore, the minimum possible energy of the wave is $\frac{h}{2}\sqrt{\frac{x}{m}}$.