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?


1 Answer 1


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


and the same for $p_y$.


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