Hermitian matrix representation 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?
 A: 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.
A: $\frac{\partial^2}{\partial x^2}$ 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)
