## One-dimensional eigenvalue problems

The operator representing the energy of a system is H.  The eigenvalues of H are E.  If the potential energy function U(x) is independent of time, then separation of variables is possible, and we can write ψ(x,t) = Φ(x)Χ(t).  If the wave function is of this form, then Φ(x) = ΦE(x) is an eigenfunction of the operator H, and the energy of the system is certain.  We find the eigenfunction of H by solving HΦE(x) = EΦE(x).

Regions that do not contain a well
There exists an eigenfunction for every E > Umin.  These eigenfunctions, however, are plane waves and are not square integrable.  They cannot represent a single particle, but can represent a constant flux of particles.  We calculate transmission and reflection coefficients by comparing fluxes.  (Flux = k|Φ(x)|2.)

Regions that do contain a well
There exists an eigenfunction for every E > Erim.  These eigenfunctions are not square integrable.
For E < Erim eigenfunctions exist only for selected eigenvalues.

### "Square" potentials

We can solve HΦE(x) = EΦE(x) in regions of piecewise constant potentials.
Assume U(x) = U = constant in certain regions of space.  In such a region the Schroedinger equation yields
(∂2/∂x2)Φ(x) + (2m/ħ2)(E - U)Φ(x) = 0.

i.  Let E > U: (∂2/∂x2)Φ(x) + k2Φ(x) = 0.   E - U = ħ2k2/(2m).
The most general solution is Φ(x) = Aexp(ikx) + A'exp(-ikx), with A and A' complex constants.
ii.  Let E < U: (∂2/∂x2)Φ(x) - ρ2Φ(x) = 0.   U - E = ħ2ρ2/(2m).
The most general solution is Φ(x) = Bexp(ρx) + B'exp(-ρx), with B and B' complex constants.
iii.  Let E = U: (∂2/∂x2)Φ(x) = 0.  Φ(x) = Cx + C', with C and C' complex constants.

(Note: A solution exists in the classically forbidden regions.)

How does the wave function behave at a point where U is discontinuous, i.e. at a step?
(a)  At a finite step the boundary conditions are that Φ(x) and (∂/∂x)Φ(x) are continuous.
(b)  At an infinite step (∂/∂x)Φe(x)|ε-->0 is discontinuous, but it has a finite discontinuity.  Therefore Φe(x) remains continuous as ε --> 0.

We can solve for the bound states in a square-well potential using a graphical solution.

### The Harmonic Oscillator

H = -(ħ2/(2m))(∂2/∂x2) + ½mω2x2 is the harmonic oscillator Hamiltonian.
(∂2/∂x2)Φ(x) + (2m/ħ2)(E - ½mω2x2)Φ(x) = 0 is the eigenvalue equation.
The eigenvalues are En = (n + ½)ħω, n = 0, 1, 2, ... .

The normalized ground state wave function of the 1-dimensional harmonic oscillator is
Φ0(x) = (mω/(πħ))¼exp(-½mωx2/ħ).
The wave functions of the excited states are
Φn(x) = (n! 2n)(β/√π)½Hn(η) exp(-½η2),
where η = (mω/ħ)½ x = βx.

Specifically,
Φ1(x) = ((4/π)(mω/ħ)3)¼ x exp(-½mωx2/ħ),
Φ2(x) = (mω/(4πħ))¼[2mωx2/ħ - 1] exp(-½mωx2/ħ).
Φn(x) is the product of exp(-½mωx2/ħ) and a polynomial of degree n and parity (-1)n called a Hermite polynomial.

Defining scaled operators Xs = (mω/ħ)½X,  Ps =  (mωħ)P, and new operators
a = (2)(Xs + iPs), and its adjoint, a = (2)(Xs - iPs), we can write
H = ħω(aa + ½) = ħω(aa - ½) = ½ħω(aa + aa).
The operators a and a do not commute.
[a,a] = 1.
H and aa  have the same eigenstates {|n>}.
aa|n> = n|n>, H|n> = (n + ½)ħω|n>.
a|n> = √(n) |n>,  a|n> = √(n+1) |n>.