Relating Fock states to eigenfunctions in space domain

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 $hbaromega$ and moreover whose units were chosen so that $sqrt{frac{hbar}{momega}}=1$. Then the eigen functions of the Schroedinger equation of the harmonic oscillator $Hpsi_n =E_n psi_n$ are:

$$psi_n(x) = langle x|nrangle = frac{pi^{-1/4}}{sqrt{2^n n!}}exp(-x^2/2) H_n(x) quad quad text{(1)}$$

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:

$$ a =frac{1}{sqrt{2}}left( x + frac{d}{dx}right) equiv frac{1}{sqrt{2}}left( x +frac{i}{hbar} pright) $$ and $$ a^dagger =frac{1}{sqrt{2}}left( x - frac{d}{dx}right) equiv frac{1}{sqrt{2}}left( x -frac{i}{hbar} pright) $$

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:

$$ Hpsi = (a^dagger a + frac{1}{2})psi equiv ( -frac{1}{2}frac{d^2}{dx^2} + frac{1}{2} x^2)psi$$

Furthermore an annihilation operator acting on the ground should be zero:

$langle x| a |0rangle =0$

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)):

$$langle x| a^dagger |0rangle =sqrt{1}langle x | 1rangle =psi_1(x)$$

or the same in differential language (knowing the correct normalization constant $C =pi^{-1/4}$)

$$frac{1}{sqrt{2}}left( x - frac{d}{dx}right) psi_0(x) equivfrac{1}{sqrt{2}}left( x - frac{d}{dx}right) pi^{-1/4}e^{-x^2/2} = frac{pi^{-1/4} }{sqrt{2}} 2x e^{-x^2/2}= psi_1(x) $$

and so on. As it is wellknown, the ladder operators fulfill:

$$a|nrangle =sqrt{n}|n-1rangle quad text{and} quad a^dagger|nrangle =sqrt{n+1}|n+1rangle$$

These relations can be checked with their differential form given above in the following way:

$$langle x | a| nrangle =sqrt{n}langle x|n-1rangle quad text{respectively}quad a psi_n(x) =sqrt{n}psi_{n-1}(x) $$ and $$langle x | a^dagger| nrangle =sqrt{n+1}langle x|n+1rangle quad text{respectively}quad a^dagger psi_n(x) = sqrt{n+1}psi_{n+1}(x) $$

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.