Eigenfunctions of Hermitian Operators are Orthogonal

We wish to prove that eigenfunctions of Hermitian operators are orthogonal. In fact we will first do this except in the case of equal eigenvalues.

Assume we have a Hermitian operator $A$ and two of its eigenfunctions such that

A\psi_1=a_1\psi_1 \\
A\psi_2=a_2\psi_2 .\\

Now we compute \bgroup\color{black}$\langle\psi_2\vert A\vert\psi_1\rangle$\egroup two ways.

\langle\psi_2\vert A\psi_1\rangle&=&a_1\langle\psi_2\vert\psi_...
...A\psi_2\vert\psi_1\rangle=a_2\langle\psi_2\vert\psi_1\rangle \\

Remember the eigenvalues are real so there's no conjugation needed.

Now we subtract the two equations. The left hand sides are the same so they give zero.

\begin{displaymath}\bgroup\color{black} 0=(a_2-a_1)\langle\psi_2\vert\psi_1\rangle \egroup\end{displaymath}

If \bgroup\color{black}$a_1\neq a_2$\egroup then

\begin{displaymath}\bgroup\color{black} \langle\psi_2\vert\psi_1\rangle=0 .\egroup\end{displaymath}

The eigenfunctions are orthogonal.

What if two of the eigenfunctions have the same eigenvalue? Then, our proof doesn't work. Assume \bgroup\color{black}$\langle\psi_2\vert\psi_1\rangle$\egroup is real, since we can always adjust a phase to make it so. Since any linear combination of \bgroup\color{black}$\psi_1$\egroup and \bgroup\color{black}$\psi_2$\egroup has the same eigenvalue, we can use any linear combination. Our aim will be to choose two linear combinations which are orthogonal. Lets try

\psi_+&=&{1\over\sqrt{2}}\left(\psi_1 + \psi_2\right) \\
\psi_-&=&{1\over\sqrt{2}}\left(\psi_1 - \psi_2\right) \\


&=&{1\over 2}\left(1-1+(\lang...
...\vert\psi_2\rangle-\langle\psi_2\vert\psi_1\rangle\right)=0 .\\

This is zero under the assumption that the dot product is real.

We have thus found an orthogonal set of eigenfunctions even in the case that some of the eigenvalues are equal (degenerate). From now on we will just assume that we are working with an orthogonal set of eigenfunctions.

Jim Branson 2013-04-22