91影视

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

$1-鈭�唯(x.0{)}^2鈭�dx$

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

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

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

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

(b)

The value are in below

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

For given equation,

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

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

Here,

${蠄}_0\left(x\right)={\left(\frac{mW}{蟺\mathrm{k}}\right)}^{\frac{1}{4}}{e}^{\frac{m蝇}{2{\displaystyle \mathrm{鈩�}}}{x}^2}$.......................(2)

${蠄}_1\left(x\right)=A_1{\left(\frac{mW}{蟺{\displaystyle \mathrm{鈩�}}}\right)}^{\frac{1}{4}}\sqrt{\frac{2m蝇}{{\displaystyle \mathrm{鈩�}}}}x{e}^{-\frac{m_0}{2\xcancel{{\displaystyle \mathrm{鈩�}}}}{x}^2}$ ...........................(3)

$E_n=\left(n+\frac{1}{2}\right)\mathrm{鈩�}蝇$ ........................(4)

$E_0=\frac{{\displaystyle \mathrm{鈩�}}蝇}{2}$

$E_1=\frac{3{\displaystyle \mathrm{鈩�}}蝇}{2}$

And the value of

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

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

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

(c)

The value of will be

$鉄�x鉄�=\frac{1}{25}\left[9鈭�x{蠄}_0^{2}dx+16鈭�x{蠄}_1{}_1{}^{2}dx+24\cos \left(蝇t\right)鈭�x{蠄}_0{蠄}_{1}dx\right]$

But $鈭�x{蠄}_0^{2}dx=鈭�x{蠄}_1^{2}dx=0$while

localid="1660906338752" $$\begin{gathered} 鈭�x{蠄}_0{蠄}_{1}dx=\sqrt{\frac{m蝇}{蟺{\displaystyle \mathrm{鈩�}}}}\sqrt{\frac{2m蝇}{{\displaystyle \mathrm{鈩�}}}}鈭�x{e}^{\frac{m蝇}{蟺{\displaystyle \mathrm{鈩�}}}{x}^2}x{e}^{\frac{m蝇}{蟺{\displaystyle \mathrm{鈩�}}}{x}^2}dx \\ =\sqrt{\frac{2}{蟺}\left(\frac{m蝇}{{\displaystyle \mathrm{鈩�}}}\right)2\sqrt{蟺}2{\left(\frac{1}{2}\sqrt{\frac{{\displaystyle \mathrm{鈩�}}}{m蝇}}\right)}^3} \\ =\sqrt{\frac{{\displaystyle \mathrm{鈩�}}}{2m蝇}} \end{gathered}$$

so,

localid="1660906349104" $鉄�x鉄�=\frac{24}{25}\sqrt{\frac{{\displaystyle \mathrm{鈩�}}}{2m蝇}}\cos \left(蝇t\right);$and

$鉄�p鉄�=m\frac{d}{dt}鉄�x鉄�$

localid="1660906367748" $=-\frac{24}{25}\sqrt{\frac{m蝇{\displaystyle \mathrm{鈩�}}}{2}}\sin \left(蝇t\right).$

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

Ehrenfest's theorem says $\frac{d鉄�p鉄�}{dt}=-\frac{鈭�V}{鈭�x}$. Here

localid="1660906425192" $\frac{d鉄�p鉄�}{dt}=-\frac{24}{25}\sqrt{\frac{m蝇{\displaystyle \mathrm{鈩�}}}{2}}\cos \left(蝇t\right)$

$V=\frac{1}{2}m{蝇}^2{x}^2$

Differentiate with respect to $x$.

$\frac{鈭�V}{鈭�x}-m{蝇}^{2}x$

So,

$-\frac{鈭�V}{鈭�x}=-m{蝇}^2鉄�x鉄�$

localid="1660906435100" $=-m{蝇}^2脳\frac{24}{25}\sqrt{\frac{{\displaystyle \mathrm{鈩�}}}{2m蝇}}\cos \left(蝇t\right)$

localid="1660906449990" $=-\frac{24}{25}\sqrt{\frac{{\displaystyle \mathrm{鈩�}}m蝇}{2}}蝇\cos \left(蝇t\right)$

So, Ehrenfest's theorem holds.

(d)

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