A particle in the harmonic oscillator potential starts out in the state$\mathit{\Psi }\mathbf{(}\mathit{x}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{=}\mathit{A}\mathbf{\left[}\mathbf{3}{\mathbf{\psi }}_{\mathbf{0}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathbf{4}{\mathbf{\psi }}_{\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\right]}$

a) Find $\mathit{A}$.

b) Construct $\mathit{\Psi }\mathbf{(}\mathit{x}\mathbf{,}\mathit{t}\mathbf{)}$and$\left|\Psi (x,t{)}^2\right|$

c) Find $\mathbf{⟨}\mathit{x}\mathbf{⟩}$and $\mathbf{⟨}\mathit{p}\mathbf{⟩}$. Don't get too excited if they oscillate at the classical frequency; what would it have been had I specified ${\mathit{\psi }}_{\mathbf{2}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$, instead of ${\mathit{\psi }}_{\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$?Check that Ehrenfest's theorem holds for this wave function.

d) If you measured the energy of this particle, what values might you get, and with what probabilities?

The starting state of the particle in the harmonic oscillator potential is, $\Psi (x,0)=A\left[3w_0\left(x\right)+4w_1\left(x\right)\right]$......................(1)

A system is said to be a harmonic oscillator if it experiences a restoring force F proportionate to the displacement x, where k is a positive constant, when it is moved from its equilibrium position.

(a)

The value of $A$will be :

$1-\int \Psi (x.0{)}^2\mid dx$

$=|A{|}^2\int \left(9{\left|{\psi }_0\right|}^2+12{\psi }_0^{*}{\psi }_1+12{\psi }_1^{*}{\psi }_0+16{\left|{\psi }_1\right|}^2\right)dx$

$$\begin{gathered} ={\left|A\right|}^2(9+0+0+16) \\ =25{\left|A\right|}^2 \end{gathered}$$

$A=\frac{1}{5}$

Thus, the value of $A=\frac{1}{5}$

(b)

The value are in below

$\left|\Psi \right(x,t\left)\right|=\frac{1}{5}\left[3{\psi }_0\left(x\right)e^{-E_{0}t/\hslash }+4{\psi }_1\left(x\right)e^{-i{E}_{t}t/U\hslash }\right]$

For given equation,

Using the equation 2, 3 and 4 , then the above equation is expressed as,

$\left|\Psi \right(x,t\left)\right|=\frac{1}{5}\left[3{\psi }_0\left(x\right)e^{-i\omega t/2}+4{\psi }_1\left(x\right)e^{-3\text{iωt/2}}\right]$

Here,

${\psi }_0\left(x\right)={\left(\frac{mW}{\pi \mathrm{k}}\right)}^{\frac{1}{4}}{e}^{\frac{m\omega }{2{\displaystyle \hslash }}{x}^2}$.......................(2)

${\psi }_1\left(x\right)=A_1{\left(\frac{mW}{\pi {\displaystyle \hslash }}\right)}^{\frac{1}{4}}\sqrt{\frac{2m\omega }{{\displaystyle \hslash }}}x{e}^{-\frac{m_0}{2\xcancel{{\displaystyle \hslash }}}{x}^2}$ ...........................(3)

$E_n=\left(n+\frac{1}{2}\right)\hslash \omega$ ........................(4)

$E_0=\frac{{\displaystyle \hslash }\omega }{2}$

$E_1=\frac{3{\displaystyle \hslash }\omega }{2}$

And the value of

$\left|\Psi \right(x,t){|}^2=\frac{1}{25}\left[9{\psi }_0^2+12{\psi }_0{\psi }_1{e}^{i\omega t/2}{e}^{-3i\omega t/2}+12{\psi }_0{\psi }_1{e}^{-i\omega t/2}{e}^{3i\omega t/2}+16{\psi }_1^2\right]$

$\left|\Psi \right(x,t){|}^2=\frac{1}{25}\left[9{\psi }_0^2+16{\psi }_1^2+24{\psi }_0{\psi }_1\cos \left(\omega t\right)\right]$

The value of $\left|\Psi \right(x,t\left)\right|$is $\frac{1}{5}\left[3{\psi }_0\left(x\right)e^{-\text{iωt}/2}+4{\psi }_1\left(x\right)e^{-i\text{iωt/2}]}\right.$and the value of $\left|\Psi \right(x,t){|}^2$islocalid="1657966352148" $\frac{1}{25}\left[9{\psi }_0^2+16{\psi }_1^2+24{\psi }_0{\psi }_1\cos \left(\omega t\right)\right]$

(c)

The value of will be

$⟨x⟩=\frac{1}{25}\left[9\int x{\psi }_0^{2}dx+16\int x{\psi }_1{}_1{}^{2}dx+24\cos \left(\omega t\right)\int x{\psi }_0{\psi }_{1}dx\right]$

But $\int x{\psi }_0^{2}dx=\int x{\psi }_1^{2}dx=0$while

localid="1660906338752" $$\begin{gathered} \int x{\psi }_0{\psi }_{1}dx=\sqrt{\frac{m\omega }{\pi {\displaystyle \hslash }}}\sqrt{\frac{2m\omega }{{\displaystyle \hslash }}}\int x{e}^{\frac{m\omega }{\pi {\displaystyle \hslash }}{x}^2}x{e}^{\frac{m\omega }{\pi {\displaystyle \hslash }}{x}^2}dx \\ =\sqrt{\frac{2}{\pi }\left(\frac{m\omega }{{\displaystyle \hslash }}\right)2\sqrt{\pi }2{\left(\frac{1}{2}\sqrt{\frac{{\displaystyle \hslash }}{m\omega }}\right)}^3} \\ =\sqrt{\frac{{\displaystyle \hslash }}{2m\omega }} \end{gathered}$$

so,

localid="1660906349104" $⟨x⟩=\frac{24}{25}\sqrt{\frac{{\displaystyle \hslash }}{2m\omega }}\cos \left(\omega t\right);$and

$⟨p⟩=m\frac{d}{dt}⟨x⟩$

localid="1660906367748" $=-\frac{24}{25}\sqrt{\frac{m\omega {\displaystyle \hslash }}{2}}\sin \left(\omega t\right).$

(With ${\psi }_1$and ${\psi }_2$in place of ${\psi }_1$the frequency would be localid="1660906408776" $\frac{\left(E_2-E_0\right)}{{\displaystyle \hslash }}=\frac{\left[{\displaystyle \underset{2}{\overset{5}{}}}{\displaystyle \hslash }\omega {\displaystyle \underset{2}{\overset{-1}{}}}{\displaystyle \hslash }\omega \right]}{{\displaystyle \hslash }}$$=2\omega )$

Ehrenfest's theorem says $\frac{d⟨p⟩}{dt}=-\frac{\partial V}{\partial x}$. Here

localid="1660906425192" $\frac{d⟨p⟩}{dt}=-\frac{24}{25}\sqrt{\frac{m\omega {\displaystyle \hslash }}{2}}\cos \left(\omega t\right)$

$V=\frac{1}{2}m{\omega }^2{x}^2$

Differentiate with respect to $x$.

$\frac{\partial V}{\partial x}-m{\omega }^{2}x$

So,

$-\frac{\partial V}{\partial x}=-m{\omega }^2⟨x⟩$

localid="1660906435100" $=-m{\omega }^2\times \frac{24}{25}\sqrt{\frac{{\displaystyle \hslash }}{2m\omega }}\cos \left(\omega t\right)$

localid="1660906449990" $=-\frac{24}{25}\sqrt{\frac{{\displaystyle \hslash }m\omega }{2}}\omega \cos \left(\omega t\right)$

So, Ehrenfest's theorem holds.

(d)

The value can be got $E_0-\frac{1}{2}\hslash \omega$, with probability ,or ${\left|c_0\right|}^2=\frac{9}{25}$, or localid="1660906537134" $E_1=\frac{3}{2}\hslash \omega$, with probability ${\left|c_1\right|}^2=\frac{16}{25}$.