The Same Problem with Parity Symmetry

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 energy eigenfunctions are also eigenfunctions of the parity operator.

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

so the parity eigenvalues must be .

The boundary conditions are

This gives two types of solutions.

Together, these are exactly equivalent to the set of solutions we had with the box defined to be from 0 to . The have eigenvalue +1 under the parity operator. The have eigenvalue -1 under the parity operator.

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