If we simply **redefine the position of the box** so that
,
then our problem has **symmetry under the Parity operation**.

The Hamiltonian remains unchanged if we make the above transformation. The Hamiltonian commutes with the Parity operator.

This means that
is an eigenfunction of
with the same energy eigenvalue.

Thus, it must be a constant times the same energy eigenfunction.

The equations says the

If we operate twice with parity, we get back to the original function,

so the

The boundary conditions are

This gives **two types of solutions**.

Together, these are

This is an **example of a symmetry** of the problem, causing an operator to commute
with the Hamiltonian.
We can then have simultaneous eigenfunctions of that operator and
.
In this case all the energy eigenfunctions are also eigenstates of parity.
Parity is conserved.

An arbitrary wave function can be written as a sum of the energy eigenfunctions
recovering the **Fourier series in its standard form**.

Jim Branson 2013-04-22