Hermitian matrix representation

Question

I know that a Hermitian matrix is a square matrix with complex entries that is equal to its own conjugate transpose, i.e $H = H^{\dagger}.$ But why is $\frac{\partial^2}{\partial x^2}$ Hermitian?

Answer 1

The inner product in $\mathbb C^n$ is defined by $\displaystyle\langle a,b\rangle=\sum_{k=1}^n \bar a_k b_k$ where $\bar a_k$ is the complex conjugate of $a_k$. This makes $\|a\|^2 =\langle a,a\rangle$ nonnegative. The inner product of $a$ and $Hb$ is

$$ \langle a, Hb \rangle = a^\dagger Hb = a^\dagger H^\dagger b = (Ha)^\dagger b = \langle Ha,b\rangle. $$

It is the case that $\langle a, Hb \rangle = \langle Ha,b\rangle$ if and only if $H=H^\dagger$.

Now define an inner product by

$$ \langle a,b\rangle=\int_{-\infty}^\infty \bar a(x)b(x)\,dx. $$ Now ask whether $$ \left\langle a,\frac{\partial^2}{\partial x^2} b \right\rangle = \left\langle \frac{\partial^2}{\partial x^2}a, b \right\rangle. $$ If so, then $\dfrac{\partial^2}{\partial x^2}$ is Hermitian.

You can show that this holds by integrating by parts twice.

Answer 2

$\frac{\partial^2}{\partial x^2}$ is an operator on some space of functions on a certain interval I. To check that it is Hermitian you need an equality of the form $$\int_I \frac{\partial^2 \phi(x)}{\partial x^2} \cdot\bar \psi(x) = \int_I \phi(x) \frac{\partial^2 \bar\psi(x) }{\partial x^2} $$

(or, perhaps, the conjugations goes on the left, depending on convention)