quantum mechanics - Is H=H* sloppy notation or really just incorrect, for Hermitian operators?

I saw it in this pdf, where they state that

$P=P^\dagger$ and thus $P$ is hermitian.

I find this notation confusing, because an operator A is Hermitian if

$\langle \Psi | A \Psi \rangle=\langle A \Psi| \Psi \rangle=\left( \langle \Psi|A\Psi\rangle \right)^\dagger$

or more elaborately

$\int H^\dagger \Psi ^\dagger \Psi dV =\int \Psi^\dagger H \Psi dV $

or in another way ( $\left< X\right>$ being expectation value here, to avoid confusion)

$\left=\left^\dagger$

but surely this does not imply $A=A^\dagger$?

Look at the momentum-operator $\vec{p}=-i\hbar\nabla $. But $\vec{p}\neq \vec{p}^\dagger$

Is it just sloppy notation and are they talking about expectation values, or are Hermitian operators actually operators that are equal to their Hermitian conjugate? My QM course that I've been taught would state otherwise.

Let me try to illustrate this with another (1D )example:

$\langle \Psi | P\Psi \rangle \\ =\int \Psi ^\dagger(-i\hbar \frac{\partial}{\partial x})\Psi dx\\ =\Psi ^\dagger \Psi |^{\infty}_{-\infty}+i\hbar\int \Psi \frac{\partial \Psi ^\dagger}{\partial x}dx\\ =0+\int \Psi (-i\hbar\frac{\partial \Psi )}{\partial x})^\dagger dx\\ =\int (-i\hbar\frac{\partial}{\partial x})^\dagger \Psi ^\dagger \Psi dx\\ =\langle P \Psi | \Psi \rangle$

Now this shows that the expectation values might be the same, but the operator between the parentheses itself is different in the first and last expression (look at the parentheses + $\dagger$ as a whole). I think the reason why I am confused is because I don't understand how $\dagger$ works on something that is not a matrix (if that is even possible).

Answer

The other answers are correct, but the OP still seems confused about how to prove that $\hat{p}$ is hermitian, so I will briefly show how to demonstrate that here using the properties of hermitian operators already given by other posters.

As already stated, the defining property of a hermitian operator is that it is equal to its conjugate transpose, i.e. its matrix elements satisfy $A_{mn} = (A_{nm})^{\ast}$. For the momentum operator you need to use the $L^2$ inner product to find the matrix elements. The OP's confusion stems from the assumption that these can be found from an expectation value, but this is not as general as a matrix element. The expectation value ($\hbar = 1$) $$ \langle\psi|\hat{p}|\psi\rangle = \int\mathrm{d}x\, \psi^{\ast}(x)\left(-i\frac{\partial}{\partial x} \psi (x)\right) $$ only gives a diagonal element in a basis in which $\psi(x)$ is a basis vector.

We want a general (possibly off-diagonal) matrix element in the basis of wave functions $\{\phi_m(x)\}$. This is given by

\begin{eqnarray} p_{mn} = \langle \phi_m|\hat{p}|\phi_n\rangle &=& \int\mathrm{d}x\, \phi^{\ast}_m(x)\left(-i \frac{\partial}{\partial x} \phi_n(x)\right) \\ &=& \int\mathrm{d}x\, \left(+i \frac{\partial}{\partial x} \phi_m^{\ast}(x)\right)\phi_n(x) \\ &=& \left[\int\mathrm{d}x\,\phi^{\ast}_n(x) \left(-i\frac{\partial}{\partial x}\phi_m(x)\right) \right]^{\ast} \\ &=& (p_{nm})^{\ast} \end{eqnarray} The second line follows by integration by parts and dropping boundary terms. The third line is a trivial rearrangement of terms. The final equality showing $p_{mn} = (p_{nm})^{\ast}$ follows by comparison of the third and first lines. Hope that's clear.

Incidentally, the expression for $p_{mn}$ above gives you a recipe for constructing a matrix representation of the operator $\hat{p}$ in some basis, however since the momentum $p$ is a continuous variable the matrix would be infinite-dimensional, so I don't recommend actually doing it!