I came across a derivation in my book which proves that any two eigenfunctions with different eigenvalues are orthogonal.
The derivation goes like this: $$\hat{A}u_1(x)=a_1u_1(x)\tag{i}$$ $$\hat{A}u_2(x)=a_2u_2(x)\tag{ii}$$
Here, $u_1(x), u_2(x)$ are eigenfunctions of a Hermitian operator $\hat{A}$ with eigenvalues $a_1$ and $a_2$.
Multiplying $u_2(x)$ with the complex conjugate of eq. (i) then integrating over all space:
$$\int_\text{all space}u_2(x)\hat{A}^*u_1^*(x)dV=a_1\int_\text{allspace}u_2(x)u_1^*(x)dV\tag{iii}$$
With eq. (ii) we multiply complex conjugate of $u_1(x)$,
$$\int_\text{all space}u_1^*(x)\hat{A}u_2(x)dV=a_2\int_\text{allspace}u_1^*(x)u_2(x)dV\tag{iv}$$
Since, $\hat{A}$ is a Hermitian operator, the LHS of eq. (iii) is equal to the LHS of eq. (iv) and $a_1$ and $a_2$ are not equal so $u_1(x)$ and $u_2(x)$ are orthogonal.
Why are they equal? Am I missing something in the definition of Hermitian operator?
By definition I know that if $\hat{A}$ is a Hermitian operator, then $$\int u^*(x)\hat{A}u(x)dx=\int u(x)\hat{A}^*u^*(x)dx$$
But in the scenario of our derivation $u_1(x)$ and $u_2(x)$ are different.
Please explain what I am missing.
Best Answer
The following uses the notation $\langle u,v\rangle = \int u^\ast(x)v(x)$, since nothing here relies on the specific detail that the inner product on our space is given in integral form.
Your definition of Hermiticity is unusual. The standard definition of a Hermitian operator $A$ on a Hilbert space $H$ is that $\langle u,Av \rangle = \langle Au,v\rangle$ for all vectors $u,v\in H$.
Your definition of Hermiticity does, however, imply the standard one: Let $u = u_1 + c u_2, c\in\mathbb{C}, u_1,u_2\in H$, then \begin{align} \langle u,Au\rangle & = \langle u_1,Au_1\rangle + c\langle u_1, Au_2\rangle + c^\ast\langle u_2, Au_1\rangle + \lvert c\rvert^2\langle u_2, Au_2\rangle \\ \langle Au,u\rangle & = \langle Au_1,u_1\rangle + c\langle Au_1, u_2\rangle + c^\ast\langle Au_2, u_1\rangle + \lvert c\rvert^2\langle Au_2, u_2\rangle \end{align} and since $A$ has $\langle u,Au\rangle = \langle Au,u\rangle$ for all $u\in H$ by your definition, the difference between these two expressions is zero: $$ c(\langle Au_1, u_2\rangle - \langle u_1, Au_2\rangle) + c^\ast(\langle Au_2,u_1\rangle - \langle u_2, Au_1\rangle) = 0,$$ but the equation $cx + c^\ast y= 0$ for all $c\in\mathbb{C}$ is only solved by $x=y=0$ since for $c=1$ we have $x+y = 0$ but for $c=\mathrm{i}$ we have $x-y = 0$, and so $$ \langle Au_1,u_2\rangle = \langle u_1,Au_2\rangle, $$ i.e. $A$ is Hermitian in the standard sense.
It's notable that this implication works only over the complex numbers, but not over e.g. the reals, and this is the reason that the standard definition of Hermiticity (which is for inner product spaces over arbitrary fields) directly uses the variant with two different vectors.