# Hermiticity of an operator in $2$D [closed]

In $$1$$D, to show an operator $$\hat{A}$$ is Hermitian, all one needs to do is check that

$$(\psi A,\psi) = (\psi, \psi A)$$ for all normalisable wavefunctions $$\psi$$.

Here $$(\cdot,\cdot)$$ is the inner product defined by $$(\psi,\phi) = \int_{-\infty}^\infty \psi^*\phi$$ $$dx$$ $$\color{red}{(\dagger)}$$ (the star denotes complex conjugation).

I'm asked to show that the operators $$-i\hbar \partial/\partial x$$ and $$-i\hbar \partial/\partial y$$ in TWO dimensions are Hermitian.

I'm a bit lost here because so far as I'm aware the analog of $$\color{red}{(\dagger)}$$ in $$2$$D is

$$(\psi,\phi) = \int_{-\infty}^\infty dy \int_{-\infty}^\infty dx$$ $$\psi^*\phi$$

But because $$\hat{p}_x = -i\hbar \partial/\partial x$$ is independent of $$y$$, it looks like the inner product $$(p_x \psi,\psi)$$ will diverge - it doesn't seem to work out.

Where have I got it wrong?

In general, for $$A$$ to be hermitian, $$(A\phi,\psi)=(\phi,A\psi)$$. Let's check for $$p_x=-i\hbar\frac{\partial}{\partial x}$$:
$$(p_x\phi,\psi)=\int_{-\infty}^\infty dy\int_{-\infty}^\infty dx\Big[-i\hbar\frac{\partial\phi}{\partial x}\Big]^*\psi=i\hbar\int_{-\infty}^\infty dy\int_{-\infty}^\infty dx\frac{\partial \phi^*}{\partial x}\psi=\\=i\hbar\int_{-\infty}^\infty dy\int_{-\infty}^\infty dx\Big[\frac{\partial}{\partial x}(\phi^* \psi)-\phi^*\frac{\partial\psi}{\partial x}\Big]=i\hbar\int_{-\infty}^\infty dy\int_{-\infty}^\infty dx\frac{\partial}{\partial x}(\phi^* \psi)+(\phi,p_x\psi).$$
The first integral is $$\int_{-\infty}^\infty dy\int_{-\infty}^\infty dx\frac{\partial}{\partial x}(\phi^* \psi)=\int_{-\infty}^\infty dy\Big[\phi^*\psi\Big]_{x=-\infty}^{x=\infty},$$
but since $$\phi(x,y)$$ and $$\psi(x,y)$$ have to be normalisable, their value at $$x=\pm\infty$$ must be zero (the same for $$y=\pm\infty$$). Then we have
$$(p_x\phi,\psi)=(\phi,p_x\psi),$$
and the same for $$p_y$$.