Quantum Harmonic Oscillator - Operator Method - 3

From our previous post harmonic-oscillator-operator-method-2,
we got the differential equation to solve for the ground state of the harmonic oscillator problem given by, $$ \frac{1}{\sqrt{2}}\left(\rho +\frac{\partial}{\partial\rho}\right)u_0 = 0 $$ with corresponding scale factors for non-dimensionalization. 
Solving this we get, $$ u_0 = Ae^{\frac{-\rho^2}{2}}$$ with the normalizing constant as, $$ A = \frac{\sqrt{\alpha}}{\pi^{\frac{1}{4}}}$$ All other excited states are obtained from the ground state using creation operator as, $$ u_n = A_n a{u_0}$$ and Energy eigenvalues are obtained as $$ E_n = \hbar\omega\left(n+\frac{1}{2}\right)$$ Now, we define an operator named as Number operator and defined as $$ \hat{N} = aa^\dagger \\~\\ \hat{H} = \hbar\omega\left(\hat{N} + \frac{1}{2}\right) \\~\\ \hat{N}u_n = nu_n \,\,\,\,\,\,where \,\,\,n=0,1,2,..$$ from the analogy $$ E_n = \hbar\omega\left(n+\frac{1}{2}\right)\,\,\,\, n = 0,1,2,..$$ Then we can work out for the normalizing constants of all eigen functions as follows, 
Let us start with defining, $$ a{u_n} = c_n u_{n+1} \\~\\ a^\dagger{u_n} = d_n u_{n-1}$$ Now, we need to normalize this new eigenfunction $u_{n-1} \,\,or\,\,u_{n+1}$ from the fact that $u_n$ is normalized. So, we get, $$ \langle{u_{n-1}}\vert{u_{n-1}}\rangle = 1 = \langle \frac{a^\dagger{u_n}}{d_n}\vert\frac{a^\dagger{u_n}}{d_n}\rangle \\ \rightarrow |d_n|^2 = \langle{aa^\dagger{u_n}}\vert{u_n}\rangle \\~\\ = \langle\hat{N}u_n\vert{u_n}\rangle = ||u_n||^2 n = n $$ 
where we know $a,a^\dagger$ are adjoint to each other and $\hat{N}$ is hermitian $$ \left(aa^\dagger\right)^\dagger = aa^\dagger$$ and Remember $$\langle {au}\vert{au}\rangle = \langle {a^\dagger{a}u}\vert{u}\rangle = \langle{u}\vert{a^\dagger{a}u}\rangle \\~\\\neq \langle{aa^\dagger{u}}\vert{u}\rangle \neq \langle{aua^\dagger}\vert{u}\rangle $$ Similarly, $$ \langle{u_{n+1}}\vert{u_{n+1}}\rangle = 1 = \langle\frac{a{u_n}}{c_n}\vert\frac{a{u_n}}{c_n}\rangle \\ \rightarrow \,\,\,\, |c_n|^2 = \langle{u_n}\vert{a^\dagger{a}}u_n\rangle $$ where $$ a^\dagger {a}= \hat{N} + 1$$ and $$ |c_n|^2 = \langle{u_n}\vert\left(\hat{N}+1\right)u_n\rangle = (n+1)u_n\\\rightarrow \,\,\,\, c_n = \sqrt{(n+1)}$$ Then, we get$$ u_n = \left[\prod_{n=0}^{n-1} \frac{1}{\sqrt{n+1}}\right](a^n) u_0 = \frac{1}{\sqrt{n!}}(a^n)u_0$$  In general, $$ u_n = \frac{A}{\sqrt{n!2^n}} \left(\rho - \frac{\partial}{\partial\rho}\right)^nu_0 \propto H_n(\rho)e^{-\frac{\rho^2}{2}} $$ where $H_n(\rho) $ is the Hermite polynomials. And thus we get the general solution of a Simple Linear Harmonic Oscillator as, $$ u_n(\rho) = \sqrt{\frac{\alpha}{\sqrt{\pi}2^nn!}} H_n(\rho) e^{-\frac{\rho^2}{2}} $$ 

Quantum Harmonic Oscillator - Operator Method - 2

We have from Harmonic oscillator operator method - 1, $$a = \sqrt{\frac{m\omega}{2\hbar}} x - \sqrt{\frac{1}{2m\omega\hbar}} i p \\~\\ a^\dagger = \sqrt{\frac{m\omega}{2\hbar}} x + \sqrt{\frac{1}{2m\omega\hbar}} i p\\~\\ [a,a^\dagger] = -1\\~\\ aa^\dagger = \frac{1}{\hbar\omega}\left[H - \frac{\hbar\omega}{2}\right] \\~\\ a^\dagger{a}u = \lambda{u} \\~\\ a^\dagger{a} (au) = (\lambda+1) (au) \\~\\ a^\dagger{a} (a^\dagger{u}) = (\lambda - 1)(a^\dagger{u}) $$
Now, we can derive the result of Hamiltonian operator on this eigen vector, $$ H (a^\dagger{u}) = \hbar\omega \left(a^\dagger{a} - \frac{1}{2}\right)(a^\dagger{u}) = \hbar\omega \left( a^\dagger{a}a^\dagger{u} - \frac{a^\dagger{u}}{2}\right) \\~\\ = \hbar\omega\left[(\lambda -1) - \frac{1}{2} \right] (a^\dagger{u}) $$
Using the relation, $$ Hu = \hbar\omega (a^\dagger{a} - \frac{1}{2})u = \hbar\omega\left[\lambda-\frac{1}{2}\right]u = Eu $$
With its correspondence we can rewrite,
$$ H (a^\dagger{u}) = (E - \hbar\omega)(a^\dagger{u}) $$
Similarly, $$ H (au) = (E+\hbar\omega) (au)$$ 
or the general proof, $$ H (a^\dagger{u}) = Ha^\dagger{u} - a^\dagger{H}u + a^\dagger{H}u = [H,a^\dagger]u + a^\dagger{H}u \\~\\= [-\hbar\omega{a^\dagger} + E {a^\dagger}]u = (E - \hbar\omega)(a^\dagger{u}) $$
where, $$ [H, a^\dagger] = \hbar\omega[{aa^\dagger}, a^\dagger] +\hbar\omega[\frac{1}{2}, {a^\dagger}] \\~\\=\hbar\omega \left([ a,a^\dagger ] a^\dagger + a [ a^\dagger, a^\dagger] + [1/2, a^\dagger]\right) \\~\\ =\hbar\omega\left([a,a^\dagger]a^\dagger +0\right) = -\hbar\omega{a^\dagger} $$
As we can see, the repeated application of $"a^\dagger"$ continuously on any eigen function will give negative values for Energy. But, we have assumed the harmonic oscillator has positive energy, which restricts the negative energy states. 
So, we demand a wave function such that it gets annihilated when operated by operator $"a^\dagger"$ on its lower state. Let us say, the lowest ground state is $u_0$, then we have, $$ a^\dagger{u_0} = 0 $$ or $$ \sqrt{\frac{m\omega}{2\hbar}}x{u_0} + \sqrt{\frac{1}{2m\omega\hbar}} i p{u_0} = 0 $$ Non-dimensionalization gives, $$ \frac{1}{\sqrt{2}}(\rho +\frac{\partial}{\partial{\rho}})u_0 = 0 $$ where $\rho = \alpha{x} $ and $ \alpha = \sqrt{\frac{m\omega}{\hbar}}$ 

By solving this, the ground state wave function can be obtained. Once we get the ground state wave function, using creation operator we can derive all other eigen states.