Question

Show that the constant b used in the quantum-harmonic-oscillator wave functions (a) has units of length and (b) is the classical turning point of an oscillator in the $\mathrm{n}=1$ground state.

Short answer

All the statements are proved.

Step-by-step solution

Part (a) Step 1: Given information 

We have given,

System of quantum harmonic oscillator.

We have to find the unit of constant b.

Simplify

Since we know that the value of b is given by,

$\mathrm{b}=\sqrt{\frac{\mathrm{h}}{2\mathrm{\pi m\omega }}}$

Where

h has unit J.s and m has unit of kg and w is in per second then,

$$\begin{gathered} \mathrm{b}\mathrm{unit}={\left(\frac{{\mathrm{ML}}^2{\mathrm{T}}^{-1}}{{\mathrm{MT}}^{-1}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ \mathrm{b}\mathrm{unit}=\mathrm{L} \end{gathered}$$

Hence proved.

Part (b) Step 1: Given information

We have to show that b is classical turning point for ground state.

Simplify

To show that the classical turning point is b is ,

$$\begin{gathered} E=\frac{1}{2}k{x}^2=\frac{1}{2}m{\omega }^2{A}^2 \\ A=\sqrt{\frac{2E}{m{\omega }^2}} \\ A=\sqrt{\frac{2}{m{\omega }^2}(n+1)\frac{h\omega }{2\pi }} \\ A=\sqrt{(2n+1)\frac{h}{2\pi m\omega }} \\ for,n=0 \\ A=\sqrt{\frac{h}{2\pi m\omega }}=b \\ b=x \end{gathered}$$

Hence proved.