A particle of mass m in the harmonic oscillator potential (Equation 2.44) starts $\mathit{\psi }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{=}\mathit{A}{\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{2}\sqrt{\frac{\mathbf{m}\mathbf{\omega }}{\mathbf{ħ}}}\mathbf{x}\mathbf{)}}^{\mathbf{2}}{\mathit{e}}^{\mathbf{-}\frac{\mathbf{m}\mathbf{\omega }}{\mathbf{2}\mathbf{ħ}}{\mathbf{x}}^{\mathbf{2}}}$out in the state for some constant A.
(a) What is the expectation value of the energy?
(c) At a later time T the wave function islocalid="1658123604154" $\mathit{\psi }\mathbf{(}\mathit{x}\mathbf{,}\mathit{T}\mathbf{)}\mathbf{=}\mathit{B}{\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{2}\sqrt{\frac{\mathit{m}\mathit{\omega }}{\mathit{ħ}}}\mathit{x}\mathbf{)}}^{\mathbf{2}}{\mathit{e}}^{\mathbf{-}\frac{\mathit{m}\mathit{\omega }}{\mathbf{2}\mathit{ħ}}{\mathit{x}}^{\mathbf{2}}}$
for some constant B. What is the smallest possible value of T ?

The harmonic oscillator is an essential concept of quantum mechanics. The oscillator oscillates about an equilibrium point, and the system's total energy is always constant. The potential that can represent the system is harmonic oscillator potential.

a)

Since,

$$\begin{gathered} \xi =\sqrt{m\omega /ħx} \\ \alpha ={\left(m\omega /\pi ħ\right)}^{1/4} \end{gathered}$$

The wave function is:

$$\begin{gathered} \psi \left(x,0\right)=A{\left(1-2\xi \right)}^2{e}^{-{\xi }^2/2} \\ \psi \left(x,0\right)=A{\left(1-4\xi \right)}^2{e}^{-{\xi }^2/2} \end{gathered}$$

……………(1)

Now,

${\psi }_0\left(x\right)=\alpha e^{-{\xi }^2/2}$

localid="1658127827170" $$\begin{gathered} {\psi }_1\left(x\right)=\sqrt{2}\alpha \xi e^{-{\xi }^2/2} \\ {\psi }_1\left(x\right)=\frac{\alpha }{\sqrt{2}}\left(2{\xi }^2-1\right)e^{-{\xi }^2/2} \\ So, \\ \psi \left(x,0\right)=c_0{\psi }_0+c_1{\psi }_1+c_2{\psi }_2 \end{gathered}$$

Substitute values in the above expression, and we get,$\psi \left(x,0\right)=\alpha \left(c_0+\sqrt{2}\xi c_1+\sqrt{2}{\xi }^2{c}_2-\frac{1}{\sqrt{2}}{c}_2\right)e^{-{\xi }^2/2}$ ……………..(2)

Comparing equations 1 and 2 as:

$$\begin{gathered} \alpha \sqrt{2}{c}_2=4A \\ {c}_2=2\sqrt{2}A/\alpha \\ \alpha \sqrt{2}{c}_1=-4A \\ {c}_1=-2\sqrt{2}A/\alpha \\ \alpha \left(c_0=\frac{c_2}{\sqrt{2}}\right)=A \\ {c}_0=\left(\frac{A}{\alpha }\right)+\frac{c_2}{\sqrt{2}} \end{gathered}$$

$$\begin{gathered} \left(1+2\right)\frac{A}{\alpha }=\frac{3A}{\alpha } \\ 1={\left|c_0\right|}^2+{\left|c_1\right|}^2+{\left|c_2\right|}^2 \end{gathered}$$

Substitute values in the above expression, and we get,

$$\begin{gathered} 1=\left(8+8+9\right){\left(\frac{A}{\alpha }\right)}^2 \\ A=\frac{\alpha }{5} \end{gathered}$$

Therefore, the values will be:

$c_0=\frac{3}{5}$

Substitute values in the above expression, and we get,

role="math" localid="1658127161820"

Thus, the expectation value of energy can be written as: .

$$\begin{gathered} \psi \left(x,t\right)=\frac{3}{5}{\psi }_0{e}^{-i\omega t/2}-\frac{2\sqrt{2}}{5}{\psi }_1{e}^{-3i\omega t/2}+\frac{2\sqrt{2}}{5}{\psi }_1{e}^{-5i\omega t/2} \\ \psi \left(x,t\right)=e^{-i\omega t/2}\left[\frac{3}{5}{\psi }_0-\frac{2\sqrt{2}}{5}{\psi }_1{e}^{-i\omega t/2}+\frac{2\sqrt{2}}{5}{\psi }_2{e}^{-2i\omega t/2}\right] \end{gathered}$$

Since for the lowest value, we need,

$$\begin{gathered} e^{-i\omega T}=-1 \\ {e}^{-2i\omega T}=1 \\ \omega T=\pi \\ T=\frac{\pi }{\omega } \end{gathered}$$

The smallest possible value of T can be written as: $T=\frac{\pi }{\omega }$.