130_notes.dvi

Note that these states would have a definite parity ifx= 0 were at the center of the box.

Theexpansionof an arbitrary wave function in these eigenfunctions is essentially ouroriginal
Fourier Series. This is a good example of the energy eigenfunctions being orthogonal and covering
the space.

8.5.1 The Same Problem with Parity Symmetry

If we simplyredefine the position of the boxso that−a 2 < x < a 2 , then our problem has
symmetry under the Parity operation.

x→−x

The Hamiltonian remains unchanged if we make the above transformation. The Hamiltonian com-
mutes with the Parity operator.
[H,P] = 0

This means that (Pui) is an eigenfunction ofHwith the same energy eigenvalue.

H(Pui) =P(Hui) =PEiui=Ei(Pui)

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

Pui=cui

The equations says theenergy eigenfunctions are also eigenfunctions of the parity operator.

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

P^2 ui=ui

so theparity eigenvalues must be± 1.

Pui=± 1 ui

The boundary conditions are

ψ(±

a

2

) = 0.

This givestwo types of solutions.

u+n(x) =

√

2

a

cos

(

(2n−1)πx

a

)

u−n(x) =

√

2

a

sin

(

2 nπx

a

)

E+n = (2n−1)^2

π^2 ̄h^2

2 ma^2

En− = (2n)^2

π^2 ̄h^2

2 ma^2