The Same Problem with Parity Symmetry

If we simply redefine the position of the box so that $-{a\over 2}<x<{a\over 2}$, then our problem has symmetry under the Parity operation.

$$\bgroup\color{black}x\rightarrow -x \egroup$$

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

$$\bgroup\color{black}[H,P]=0 \egroup$$

This means that $(Pu_i)$ is an eigenfunction of $H$ with the same energy eigenvalue.

$$\bgroup\color{black}H(Pu_i)=P(Hu_i)=PE_iu_i=E_i(Pu_i) \egroup$$

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

$$\bgroup\color{black}Pu_i=cu_i \egroup$$

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,

$$\bgroup\color{black}P^2u_i=u_i \egroup$$

so the parity eigenvalues must be $\pm 1$.

$$\bgroup\color{black}Pu_i=\pm 1 u_i \egroup$$

The boundary conditions are

$$\bgroup\color{black}\psi(\pm {a\over 2})=0 .\egroup$$

This gives two types of solutions.

$$\begin{eqnarray*} u_n^+(x)&=&\sqrt{2\over a}\cos\left({(2n-1)\pi x\over a}\right... ...ar^2\over 2ma^2} \\ E_n^-&=&(2n)^2{\pi^2\hbar^2\over 2ma^2} \\ \end{eqnarray*}$$

Together, these are exactly equivalent to the set of solutions we had with the box defined to be from 0 to $a$. The $u_n^+(x)$ have eigenvalue +1 under the parity operator. The $u_n^-(x)$ 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 $H$. 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.

$$\bgroup\color{black}\psi(x)=\sum\limits_{n=1}^\infty[A_n^+ u_n^+(x)+A_n^-u_n^-(x)] \egroup$$