As a general rule, bound states are considered to be orthogonal. Once again, the solutions for the infinite well are an exception to this rule. Although not all states are orthogonal, pairs of solutions representing orthogonal states are present.

In this section, I examine why the standard proof for orthogonality of two solutions does not apply to the infinite well.  However, it can be shown that when the two solutions differ by an even quantum number, the solutions are orthogonal.

Orthogonality

Keeping in mind that the inner product $⟨f|g⟩$ of two functions $f(x)$ and $g(x)$ is defined as

$$⟨f|g⟩=\int f^{\ast }(x)g(x)dx$$

The rule for orthogonality between normalized states $f_n={\psi }_n(x)$ and $f_m={\psi }_m(x)$ is given by

$$⟨f_m|f_n⟩={\delta }_{mn}$$

where ${\delta }_{mn}$ is the Kronecker delta defined as

$$\begin{array}{}& {\delta }_{mn}=\{\begin{array}{l}0,\text{ if }m\ne n\\ 1,\text{ if }m=n\end{array}\end{array}$$

Standard Proof of Orthogonality

The proof for orthogonality of two distinct wave functions is usually given as follows:

Let $f={\psi }_a$ and $g={\psi }_b$. When acted upon by a hermitian operator $\hat{Q}$, eigenfunctions $f$ and $g$ give real eigenvalues of $q$ and $q^{\prime }$ respectively. Since $q\in \mathbb{R}$, $q^{\ast }=q$. The general rule is that there are no degenerate states in 1-dimension which implies that $q^{\prime }\ne q^{\ast }$ Then,

$$\begin{array}{}& ⟨f|\hat{Q}g⟩=⟨\hat{Q}f|g⟩\end{array}$$

$$\begin{array}{}& q^{\prime }⟨f|g⟩=q^{\ast }⟨f|g⟩\end{array}$$

$$⟨f|g⟩=0\text{since}{q}^{\prime }\ne q^{\ast }$$

The Exception for the Infinite Well

However, as show in section on Degneracy, the infinite well does allow degenerate solutions. Since it is possible for $q^{\prime }=q^{\ast }$, the proof given above fails and the rule that all bound quantum states are orthogonal doesn’t apply.

Orthogonality when the principal quantum numbers differ by an even integer

For the particle in the box, the solutions (see The Solutions page) are as follows:

$${\psi }_{1_n}(x)=A{e}^{ıkx}|A|=\sqrt{\frac{1}{L}}$$

or

$${\psi }_{2_n}(x)=B^{-ıkx}|B|=\sqrt{\frac{1}{L}}$$

with

$$k_n=\frac{n\pi }{L}n=\pm 1,\pm 2,\pm 3,\dots$$

Orthogonality does exist if the principal quantum number $n$ for solution ${\mathrm{\Psi }}_{1_n}$ differs from the principal quantum number $m$ for state ${\mathrm{\Psi }}_{1_m}$ by an even integer.

Keeping in mind that $A^{\ast }A=1/L$,

$$\begin{array}{}& ⟨{\psi }_{1_m}|{\psi }_{1_n}⟩={\int }_{x_0}^{x_0+L}{\psi }_{1_m}^{\ast }{\psi }_{1_n}dx\end{array}$$

$$\begin{array}{}& ={\int }_{x_0}^{x_0+L}{A}^{\ast }{e}^{-ık_{m}x}A{e}^{ık_{n}x}dx\end{array}$$

$$\begin{array}{}& =A^{\ast }A\cdot \frac{1}{(k_n-k_m)}\cdot e^{ı(k_n-k_m)x}{|}_{x_0}^{x_0+L}\end{array}$$

$$\begin{array}{}\text{(1)}& =\frac{1}{L}\cdot \frac{1}{(k_n-k_m)}\cdot \{[\cos (k_n-k_m)(x_0+L)+ı\sin (k_n-k_m)(x_0+L)]–[\cos (k_n-k_m)x_0+ı\sin (k_n-k_m)x_0]\}\end{array}$$

Noting that

$$\begin{array}{}& k_m=\frac{m\pi }{L}m=\pm 1,\pm 2,\pm 3,\dots \end{array}$$

$$\begin{array}{}& k_n=\frac{n\pi }{L}n=\pm 1,\pm 2,\pm 3,\dots \end{array}$$

For $n\ne m$, Eq.$\text{(1)}$ reduces to zero if $|m\pm n|$ is even and the two states are therefore orthogonal. A similar analysis yields the same results for the states ${\psi }_{2_m}$ and ${\psi }_{2_n}$.

Each state is also orthonormal since

$$\begin{array}{}& ⟨{\psi }_{1_n}|{\psi }_{1_n}⟩={\int }_{x_0}^{x_0+L}{\psi }_{1_n}^{\ast }{\psi }_{1_n}dx\end{array}$$

$$\begin{array}{}& ={\int }_{x_0}^{x_0+L}{A}^{\ast }{e}^{-ık_{n}x}\cdot A{e}^{ık_{n}x}dx\end{array}$$

$$\begin{array}{}& =A^{\ast }A(L)\end{array}$$

$$=1$$