Find the allowed energies of the half harmonic oscillator

$\mathit{V}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{\{}\mathbf{(}\mathbf{1}\mathbf{/}\mathbf{2}\mathbf{)}\mathit{m}\mathit{\omega }\mathbf{2}\mathit{x}\mathbf{2}\mathbf{,}\mathit{x}\mathbf{>}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{,}\mathit{x}\mathbf{<}\mathbf{0}\mathbf{.}$
(This represents, for example, a spring that can be stretched, but not compressed.) Hint: This requires some careful thought, but very little actual calculation.

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

$$\begin{gathered} \mathrm{iħ}\frac{\partial \mathrm{\psi }}{\partial \mathrm{t}}=-\frac{{\mathrm{ħ}}^2}{2\mathrm{m}}\frac{\partial {\mathrm{\psi }}^2}{\partial {\mathrm{x}}^2}+\mathrm{v}\left(\mathrm{x},\mathrm{t}\right)\mathrm{\psi }\left(\mathrm{x},\mathrm{t}\right) \\ \mathrm{iħ}\frac{\partial \mathrm{\psi }}{\partial \mathrm{t}}=-\frac{{\mathrm{ħ}}^2}{2\mathrm{m}}\frac{\partial {\mathrm{\psi }}^2}{\partial {\mathrm{x}}^2}+\infty \mathrm{\psi }\left(\mathrm{x},\mathrm{t}\right) \\ \mathrm{iħ}\frac{\partial \mathrm{\psi }}{\partial \mathrm{t}}=-\frac{{\mathrm{ħ}}^2}{2\mathrm{m}}\frac{\partial {\mathrm{\psi }}^2}{\partial {\mathrm{x}}^2}+\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{x}}^2\mathrm{\psi }\left(\mathrm{x},\mathrm{t}\right) \\ \mathrm{Only}\mathrm{\psi }\left(\mathrm{x},\mathrm{t}\right)=0\mathrm{satisfies}\mathrm{the}\mathrm{one}\mathrm{for}\mathrm{x}<0.\mathrm{Since}\mathrm{the}\mathrm{wave}\mathrm{function}\mathrm{must}\mathrm{be}\mathrm{continuous}, \end{gathered}$$

localid="1658138151187" $$\begin{gathered} iħ\frac{\partial \psi }{\partial t}=-\frac{{ħ}^2}{2m}\frac{{\partial }^2\psi }{\partial x^2}+\frac{1}{2}m{\omega }^2\psi \left(x,t\right), \\ \psi \left(0,t\right)=0 \\ \psi \left(x,0\right)=\psi \left(x\right) \end{gathered}$$

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 $\stackrel{-}{\mathrm{\psi }}$will then be the restriction fo $\stackrel{-}{\mathrm{\psi }}\mathrm{to}\mathrm{x}>0$to .

$iħ\frac{\partial {\displaystyle \stackrel{-}{\psi }}}{\partial t}=-\frac{{ħ}^2}{2m}\frac{\partial {\displaystyle \stackrel{-}{{}^2\psi }}}{\partial x^2}+\frac{1}{2}m{\omega }^2{x}^2\stackrel{-}{\psi }\left(x,t\right)$

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

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

$iħ\frac{\partial }{\partial t}\left[\psi \left(x\right)\varphi \left(t\right)\right]=-\frac{ħ}{2m}\frac{{\partial }^2}{\partial x^2}\left[\psi \left(x\right)\varphi \left(t\right)\right]+\frac{1}{2}m{\omega }^2{x}^2\mathit{\left[}\mathit{\psi }\left(x\right)\mathit{\varphi }\left(t\right)\mathit{\right]}$

role="math" localid="1658135677928" $iħ\left[\psi \left(x\right)\varphi \left(t\right)\right]=-\frac{ħ}{2m}{\psi }^{,,}\left(x\right)\varphi \left(t\right)+\frac{1}{2}m{\omega }^2{x}^2\psi \mathit{\left(}x\mathit{\right)}\varphi \mathit{\left(}t\mathit{\right)}$

$iħ\frac{\varphi \text{'}\left(t\right)}{\varphi \left(t\right)}=-\frac{{ħ}^2}{2m}\frac{{\psi }^{,,}\left(x\right)}{\psi \left(x\right)}\frac{1}{2}m{\omega }^2{x}^2$

Hence,

$$\begin{gathered} iħ\frac{\varphi \text{'}\left(t\right)}{\varphi \left(t\right)}=E \\ -\frac{{ħ}^2}{2m}\frac{\psi \text{'}\text{'}\left(x\right)}{\psi \left(x\right)}+\frac{1}{2}m{\omega }^2{x}^2=E \end{gathered}$$

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

Hence, the energy can be find with the expression:

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

According to the principle of superposition,

$$\begin{gathered} \stackrel{}{\stackrel{-}{\psi }\left(x,t\right)}=\sum _{n=0}^{\infty }{B}_n{\psi }_n\left(x\right)e^{-i{E}_{n}t/h} \\ \stackrel{}{\stackrel{-}{\psi }\left(x,0\right)}=\sum _{n=0}^{\infty }{B}_n{\psi }_n\left(x\right) \\ \stackrel{}{\stackrel{-}{\psi }\left(x,0\right)}={\psi }_{0_{odd}}\left(x\right) \end{gathered}$$

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

${\int }_{-\infty }^{\infty }\sum _{n=0}^{\infty }{B}_n{\psi }_n\left(x\right){\psi }_m\left(x\right)dx={\int }_{-\infty }^{\infty }{\psi }_{0_{odd}}\left(x\right){\psi }_m\left(x\right)dx$

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

$B_n{\int }_{-\infty }^{\infty }{\left[{\psi }_n\left(x\right)\right]}^{2}dx={\int }_{-\infty }^{\infty }{\psi }_{0_{odd}}\left(x\right){\psi }_n\left(x\right)dx$

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

So

$B_n{\int }_{-\infty }^{\infty }{\psi }_{0_{odd}}\left(x\right){\psi }_n\left(x\right)dx$

Eigenstate ${\psi }_n\left(x\right)$ 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 ${\psi }_n\left(x\right)$is an odd function, which means that the integrand is even.

$$\begin{gathered} B_n=2{\int }_0^{\infty }{\psi }_{0_{odd}}\left(x\right){\psi }_n\left(x\right)dx \\ {B}_n=2{\int }_0^{\infty }{\psi }_{0_{}}\left(x\right){\psi }_n\left(x\right)dx \end{gathered}$$

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

localid="1658138598630" $\stackrel{-}{\mathrm{\psi }}\left(x,t\right)=\sum _{q=0}^{\infty }{B}_{2q}{\psi }_{2q}\left(x\right)e^{-i{E}_{2q}t/h}+\sum _{q=0}^{\infty }{B}_{2q-1}{\psi }_{2q-1}\left(x\right)e^{-i{E}_{2q-1}t/h}$

Hence, the solution to Schrödinger’s equation over the whole line,

localid="1658138720669" $\stackrel{-}{\mathrm{\psi }}\left(x,t\right)=\sum _{q=0}^{\infty }{B}_{2q-1}{\psi }_{2q-1}\left(x\right)e^{-i{E}_{2q-1}t/h}-\infty <x<\infty$

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

$\stackrel{-}{\mathrm{\psi }}\left(x,t\right)=\sum _{q=0}^{\infty }{B}_{2q-1}{\psi }_{2q-1}\left(x\right)e^{-i{E}_{2q-1}t/h}$

Therefore, for the half harmonic oscillator,

$$\begin{gathered} \stackrel{}{\mathrm{\psi }}\left(x,t\right)=\left\{\begin{array}{l}0,x<0\\ \sum _{q=0}^{\infty }{B}_{2q-1}{\psi }_{2q-1}\left(x\right)e^{-i{E}_{2q-1}t/h}x>0\end{array}\right. \\ \stackrel{}{\mathrm{\psi }}\left(x,t\right)=\theta \sum _{q=0}^{\infty }{B}_{2q-1}{\psi }_{2q-1}\left(x\right)e^{-i{E}_{2q-1}t/h} \end{gathered}$$

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

i.e.

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