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Schrodinger equation governs the time evolution of the wave function $\mathit{蠄}\left(x,t\right)$

$$\begin{gathered} \mathrm{颈魔}\frac{鈭�\mathrm{蠄}}{鈭�\mathrm{t}}=-\frac{{\mathrm{魔}}^2}{2\mathrm{m}}\frac{鈭�{\mathrm{蠄}}^2}{鈭�{\mathrm{x}}^2}+\mathrm{v}\left(\mathrm{x},\mathrm{t}\right)\mathrm{蠄}\left(\mathrm{x},\mathrm{t}\right) \\ \mathrm{颈魔}\frac{鈭�\mathrm{蠄}}{鈭�\mathrm{t}}=-\frac{{\mathrm{魔}}^2}{2\mathrm{m}}\frac{鈭�{\mathrm{蠄}}^2}{鈭�{\mathrm{x}}^2}+鈭�\mathrm{蠄}\left(\mathrm{x},\mathrm{t}\right) \\ \mathrm{颈魔}\frac{鈭�\mathrm{蠄}}{鈭�\mathrm{t}}=-\frac{{\mathrm{魔}}^2}{2\mathrm{m}}\frac{鈭�{\mathrm{蠄}}^2}{鈭�{\mathrm{x}}^2}+\frac{1}{2}{\mathrm{m蝇}}^2{\mathrm{x}}^2\mathrm{蠄}\left(\mathrm{x},\mathrm{t}\right) \\ \mathrm{Only}\mathrm{蠄}\left(\mathrm{x},\mathrm{t}\right)=0\mathrm{satisfies}\mathrm{the}\mathrm{one}\mathrm{for}\mathrm{x}<0.\mathrm{Since}\mathrm{the}\mathrm{wave}\mathrm{function}\mathrm{must}\mathrm{be}\mathrm{continuous}, \end{gathered}$$

localid="1658138151187" $$\begin{gathered} i魔\frac{鈭�蠄}{鈭�t}=-\frac{{魔}^2}{2m}\frac{{鈭�}^2蠄}{鈭�x^2}+\frac{1}{2}m{蝇}^2蠄\left(x,t\right), \\ 蠄\left(0,t\right)=0 \\ 蠄\left(x,0\right)=蠄\left(x\right) \end{gathered}$$

Consider the corresponding problem over the whole line, using the odd extension of the initial condition. Doing so automatically satisfies the boundary condition at $x=0$. The solution for $\stackrel{-}{\mathrm{蠄}}$will then be the restriction fo $\stackrel{-}{\mathrm{蠄}}\mathrm{to}\mathrm{x}>0$to .

$i魔\frac{鈭�{\displaystyle \stackrel{-}{蠄}}}{鈭�t}=-\frac{{魔}^2}{2m}\frac{鈭�{\displaystyle \stackrel{-}{{}^2蠄}}}{鈭�x^2}+\frac{1}{2}m{蝇}^2{x}^2\stackrel{-}{蠄}\left(x,t\right)$

$\stackrel{-}{蠄}\left(x,0\right)={蠄}_{0_{odd}}\left(x\right)=\left\{\begin{array}{ll}{蠄}_0& \left(x\right)\\ -{蠄}_0& \left(-x\right),\end{array}\right.\begin{array}{c}x>0\\ x<0\end{array}$

Assuming a product solution of the form role="math" localid="1658135063605" $\stackrel{-}{蠄}\left(x,t\right)=蠄\left(x\right)蠒\left(t\right)$and plugging it into the partial differential equation.

$i魔\frac{鈭�}{鈭�t}\left[蠄\left(x\right)蠒\left(t\right)\right]=-\frac{魔}{2m}\frac{{鈭�}^2}{鈭�x^2}\left[蠄\left(x\right)蠒\left(t\right)\right]+\frac{1}{2}m{蝇}^2{x}^2\mathit{\left[}\mathit{蠄}\left(x\right)\mathit{蠒}\left(t\right)\mathit{\right]}$

role="math" localid="1658135677928" $i魔\left[蠄\left(x\right)蠒\left(t\right)\right]=-\frac{魔}{2m}{蠄}^{,,}\left(x\right)蠒\left(t\right)+\frac{1}{2}m{蝇}^2{x}^2蠄\mathit{\left(}x\mathit{\right)}蠒\mathit{\left(}t\mathit{\right)}$

$i魔\frac{蠒\text{'}\left(t\right)}{蠒\left(t\right)}=-\frac{{魔}^2}{2m}\frac{{蠄}^{,,}\left(x\right)}{蠄\left(x\right)}\frac{1}{2}m{蝇}^2{x}^2$

Hence,

$$\begin{gathered} i魔\frac{蠒\text{'}\left(t\right)}{蠒\left(t\right)}=E \\ -\frac{{魔}^2}{2m}\frac{蠄\text{'}\text{'}\left(x\right)}{蠄\left(x\right)}+\frac{1}{2}m{蝇}^2{x}^2=E \end{gathered}$$

The system was solved using the method of operator factorization, refer to problem 2.10

Hence, the energy can be find with the expression:

$E_n=\left(n+\frac{1}{2}\right)魔蝇$; where, n=0,1,2,...

According to the principle of superposition,

$$\begin{gathered} \stackrel{}{\stackrel{-}{蠄}\left(x,t\right)}=\underset{n=0}{\overset{鈭�}{鈭�}}{B}_n{蠄}_n\left(x\right)e^{-i{E}_{n}t/h} \\ \stackrel{}{\stackrel{-}{蠄}\left(x,0\right)}=\underset{n=0}{\overset{鈭�}{鈭�}}{B}_n{蠄}_n\left(x\right) \\ \stackrel{}{\stackrel{-}{蠄}\left(x,0\right)}={蠄}_{0_{odd}}\left(x\right) \end{gathered}$$

Multiplying by ${蠄}_m\left(x\right)$and Integrating both sides,

${鈭�}_{-鈭�}^{鈭�}\underset{n=0}{\overset{鈭�}{鈭�}}{B}_n{蠄}_n\left(x\right){蠄}_m\left(x\right)dx={鈭�}_{-鈭�}^{鈭�}{蠄}_{0_{odd}}\left(x\right){蠄}_m\left(x\right)dx$

Since, the eigenstates are orthogonal, this integral on the left is zero for all $n鈮�m$. The infinite series consequently yields one term,$n=m$one.

$B_n{鈭�}_{-鈭�}^{鈭�}{\left[{蠄}_n\left(x\right)\right]}^{2}dx={鈭�}_{-鈭�}^{鈭�}{蠄}_{0_{odd}}\left(x\right){蠄}_n\left(x\right)dx$

Integral on the left side would be one, since the eigenstates are normalized.

So

$B_n{鈭�}_{-鈭�}^{鈭�}{蠄}_{0_{odd}}\left(x\right){蠄}_n\left(x\right)dx$

Eigenstate ${蠄}_n\left(x\right)$ is an even function of x if n is even, which means thatis zero because the integrand is odd and the integration interval is symmetric. And if n is odd, then the eigenstate ${蠄}_n\left(x\right)$is an odd function, which means that the integrand is even.

$$\begin{gathered} B_n=2{鈭�}_0^{鈭�}{蠄}_{0_{odd}}\left(x\right){蠄}_n\left(x\right)dx \\ {B}_n=2{鈭�}_0^{鈭�}{蠄}_{0_{}}\left(x\right){蠄}_n\left(x\right)dx \end{gathered}$$

Writing the general solution for$\stackrel{-}{\mathrm{蠄}}$for the even and odd integers separately.

localid="1658138598630" $\stackrel{-}{\mathrm{蠄}}\left(x,t\right)=\underset{q=0}{\overset{鈭�}{鈭�}}{B}_{2q}{蠄}_{2q}\left(x\right)e^{-i{E}_{2q}t/h}+\underset{q=0}{\overset{鈭�}{鈭�}}{B}_{2q-1}{蠄}_{2q-1}\left(x\right)e^{-i{E}_{2q-1}t/h}$

Hence, the solution to Schr枚dinger鈥檚 equation over the whole line,

localid="1658138720669" $\stackrel{-}{\mathrm{蠄}}\left(x,t\right)=\underset{q=0}{\overset{鈭�}{鈭�}}{B}_{2q-1}{蠄}_{2q-1}\left(x\right)e^{-i{E}_{2q-1}t/h}-鈭�<x<鈭�$

Now, take the restriction of $\stackrel{-}{\mathrm{蠄}}$to x > 0

$\stackrel{-}{\mathrm{蠄}}\left(x,t\right)=\underset{q=0}{\overset{鈭�}{鈭�}}{B}_{2q-1}{蠄}_{2q-1}\left(x\right)e^{-i{E}_{2q-1}t/h}$

Therefore, for the half harmonic oscillator,

$$\begin{gathered} \stackrel{}{\mathrm{蠄}}\left(x,t\right)=\left\{\begin{array}{l}0,x<0\\ \underset{q=0}{\overset{鈭�}{鈭�}}{B}_{2q-1}{蠄}_{2q-1}\left(x\right)e^{-i{E}_{2q-1}t/h}x>0\end{array}\right. \\ \stackrel{}{\mathrm{蠄}}\left(x,t\right)=胃\underset{q=0}{\overset{鈭�}{鈭�}}{B}_{2q-1}{蠄}_{2q-1}\left(x\right)e^{-i{E}_{2q-1}t/h} \end{gathered}$$

Where, only the odd energies of the harmonic oscillator is allowed.

i.e.

$E_{2q-1}\left(2q-\frac{1}{2}\right)魔蝇,\mathrm{where},\mathrm{q}=1,2,3,...$ where,