Relating Fock states to eigenfunctions in space domain How can I relate the eigenvalues of $H=\hbar\omega(a^\dagger a+1/2)$ to the eigenfunctions of $H=\frac{p^2}{2m}+\frac{1}{2}m\omega^2 x^2$, with $p=-i\hbar\nabla$? I mean, how the analytical approach to the solution of this problem can be related to the algebraic one, obtaining a 1:1 correspondence between the solutions?
 A: The ladder operators $a$ and $a^\dagger$ can perfectly be defined as differential operators. One starts off from the Hamiltonian
$$H = -\frac{1}{2}\frac{d^2}{dx^2} +  \frac{1}{2} x^2$$
that for simplicity is normalized on units of $\hbar\omega$ and moreover whose units were chosen so that $\sqrt{\frac{\hbar}{m\omega}}=1$. Then the eigen functions of the Schroedinger equation of the harmonic oscillator $H\psi_n =E_n \psi_n$ are:
$$\psi_n(x) = \langle x|n\rangle = \frac{\pi^{-1/4}}{\sqrt{2^n n!}}\exp(-x^2/2) H_n(x)  \quad \quad \text{(1)}$$
with the corresponding eigenvalues $E_n = n+1/2$. $H_n(x)$ is an abreviation for the Hermite polynomials ($H_0(x)=1,\, H_1(x)=2x\,$ etc.).
Then the ladder operators:
$$ a =\frac{1}{\sqrt{2}}\left( x + \frac{d}{dx}\right) \equiv \frac{1}{\sqrt{2}}\left( x +\frac{i}{\hbar} p\right) $$
and
$$ a^\dagger =\frac{1}{\sqrt{2}}\left( x - \frac{d}{dx}\right) \equiv \frac{1}{\sqrt{2}}\left( x -\frac{i}{\hbar} p\right) $$
can be defined. They do exactly the job that you expect from them:
First of all the Hamiltonian written in $a^\dagger$ and $a$ is written as expected:
$$ H\psi = (a^\dagger a + \frac{1}{2})\psi  \equiv ( -\frac{1}{2}\frac{d^2}{dx^2} +  \frac{1}{2} x^2)\psi$$
Furthermore an annihilation operator acting on the ground should be zero:
$\langle x| a |0\rangle =0$
leads to the following differential equation
$$\left( x + \frac{d}{dx}\right)\psi_0(x) =0$$
that once solved yields the wave function of the ground state (apart from the normalization):
$$\psi_0(x) = C e^{-\frac{x^2}{2}}$$
Furthermore one can find (look up $\psi_0$ and $\psi_1$ in formula (1)):
$$\langle x| a^\dagger |0\rangle  =\sqrt{1}\langle x | 1\rangle   =\psi_1(x)$$
or the same in differential language (knowing  the correct normalization constant $C =\pi^{-1/4}$)
$$\frac{1}{\sqrt{2}}\left( x - \frac{d}{dx}\right) \psi_0(x) \equiv\frac{1}{\sqrt{2}}\left( x - \frac{d}{dx}\right) \pi^{-1/4}e^{-x^2/2}  = \frac{\pi^{-1/4} }{\sqrt{2}} 2x e^{-x^2/2}=  \psi_1(x) $$
and so on. As it is wellknown, the ladder operators fulfill:
$$a|n\rangle =\sqrt{n}|n-1\rangle \quad \text{and} \quad a^\dagger|n\rangle =\sqrt{n+1}|n+1\rangle$$
These relations can be checked with their differential form given above in the following way:
$$\langle x | a| n\rangle =\sqrt{n}\langle x|n-1\rangle \quad \text{respectively}\quad    a \psi_n(x) =\sqrt{n}\psi_{n-1}(x) $$
and
$$\langle x | a^\dagger| n\rangle =\sqrt{n+1}\langle x|n+1\rangle \quad \text{respectively}\quad    a^\dagger \psi_n(x) =  \sqrt{n+1}\psi_{n+1}(x)  $$
where the $\psi_n(x)$ are given by the formula (1) indicated above respectively the $\psi_n$ are the eigenfunctions of the differential operator $H = -\frac{1}{2}\frac{d^2}{dx^2} +  \frac{1}{2} x^2$.
Actually, the eigenvalues of the Hamiltonian of the quantum harmonic oscillator can be determined completely algebraically.
