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) =
Φ_{E}(x)
is an eigenfunction of the operator H, and the energy of the system is
certain. We find the eigenfunction of H by solving
HΦ_{E}(x) = EΦ_{E}(x).

**Regions that do not contain a well
**

There exists an eigenfunction for every E > U_{min}. 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 > E_{rim}.
These eigenfunctions are not square integrable.

For E < E_{rim} eigenfunctions exist only for selected
eigenvalues.

Confinement leads to energy quantization.

We can solve HΦ_{E}(x) = EΦ_{E}(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}/∂x^{2})Φ(x)
+ (2m/ħ^{2})(E
- U)Φ(x)
= 0.

i. Let E > U: (∂^{2}/∂x^{2})Φ(x)
+ k^{2}Φ(x)
= 0. E - U = ħ^{2}k^{2}/(2m).

The most general solution is
Φ(x) = Aexp(ikx) + A'exp(-ikx), with A and A’
complex constants.

ii. Let E < U: (∂^{2}/∂x^{2})Φ(x)
- ρ^{2}Φ(x) = 0.
U - E = ħ^{2}ρ^{2}/(2m).

The most general solution is
Φ(x) = Bexp(ρx)
+ B'exp(-ρx),
with B and B’ complex constants.

iii. Let E = U: (∂^{2}/∂x^{2})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)Φ_{e}(x)|_{ε-->0
}is discontinuous, but
it has a finite discontinuity. Therefore
Φ_{e}(x) remains continuous as
ε -->
0.

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

H = -(ħ^{2}/(2m))(∂^{2}/∂x^{2})
+ ½mω^{2}x^{2}
is the harmonic oscillator Hamiltonian.

(∂^{2}/∂x^{2})Φ(x)
+ (2m/ħ^{2})(E
- ½mω^{2}x^{2})Φ(x)
= 0 is the eigenvalue equation.

The eigenvalues are E_{n}
= (n + ½)ħω,
n = 0, 1, 2, ... .

The normalized ground state wave
function of the 1-dimensional harmonic oscillator is

Φ_{0}(x)
= (mω/(πħ))^{¼}exp(-½mωx^{2}/ħ).

The wave functions of the excited states are

Φ_{n}(x) = (n! 2^{n})^{-½}(β/√π)^{½}H_{n}(η)^{ }
exp(-½η^{2}),

where η = (mω/ħ)^{½} x = βx.

Specifically,

Φ_{1}(x) = ((4/π)(mω/ħ)^{3})^{¼} x exp(-½mωx^{2}/ħ),

Φ_{2}(x) = (mω/(4πħ))^{¼}[2mωx^{2}/ħ
- 1] exp(-½mωx^{2}/ħ).

Φ_{n}(x) is the product of exp(-½mωx^{2}/ħ) and a polynomial of degree n
and parity (-1)^{n} called a
Hermite
polynomial.

Defining scaled operators
X_{s} = (mω/ħ)^{½}X, P_{s} = (mωħ)^{-½}P,
and new operators

a = (2)^{-½}(X_{s} + iP_{s}), and its adjoint, a^{†}
= (2)^{-½}(X_{s} - iP_{s}), we can write

H = ħω(a^{†}a + ½) = ħω(aa^{†} - ½) =
½ħω(aa^{†} + a^{†}a).

The operators a and a^{†} do not commute.

[a,a^{†}] = 1.

H and a^{†}a have the same eigenstates {|n>}.

a^{†}a|n> = n|n>, H|n> = (n + ½)ħω|n>.

a|n> = √(n) |n>, a^{†}|n> = √(n+1) |n>.