Preserving the hermiticity of differential operators when acting on bra

Question

In position basis, $\psi(x_o)=\langle x_o|\psi\rangle$ where $|\psi\rangle=\int dx|x\rangle\psi(x)$
So, we have defined $\langle x_o|x\rangle=\delta(x-x_o)$
Thus, $\langle x_o|\psi\rangle=\langle x_o|\Big(\int dx|x\rangle\psi(x)\Big)=\int dx\langle x_o|x\rangle\psi(x)=\int dx\delta(x-x_o)\psi(x)=\psi(x_o)$

Now, suppose we have an operator $\hat A=\frac{d}{dx}$
$\hat A\psi(x) = \psi'(x)$
So, $\hat A\psi(x)\Big|_{x_o} = \psi'(x)\Big|_{x_o}=\psi'(x_o)$

By the definition of differentiation we get, $\psi'(x_o)=\lim\limits_{\epsilon\to 0}\frac{\psi(x_o+\epsilon)-\psi(x_o)}{\epsilon}\tag{1}$
$(1)$ can be written as $\langle x_o|\hat A\psi\rangle=\lim\limits_{\epsilon\to 0}\frac{\langle x_o+\epsilon|\psi\rangle-\langle x_o|\psi\rangle}{\epsilon}\tag{2}$
$\langle x_o|\hat A\psi\rangle=\Big(\lim\limits_{\epsilon\to 0}\frac{\langle x_o+\epsilon|-\langle x_o| }{\epsilon}\Big)|\psi\rangle=\frac{d}{dx}\langle x_o|\psi\rangle=\hat A\langle x_o|\psi\rangle\tag{3}$
This manipulation has also been used to find the expression wavefunction in position basis as the superposition of the momentum eigenkets because $(3)$ gives us a differential equation.
I want to know in the RHS of $(3)$ whether we should have $\hat A$ or $\hat A^\dagger$. Because for the given $\hat A$ (which is not hermitian), $\hat A^{\dagger}=-\hat A$

I have tried to write $(3)$ in the form of functions instead of vectors.
$(2)$ can be written as
$\int\delta(x-x_o)\hat A\psi(x)dx = \int\delta(x-x_o)\psi'(x)dx$
Using integration by parts we get,
$\int\delta(x-x_o)\hat A\psi(x)dx = -\int\delta'(x-x_o)\psi(x)dx = \int-\hat A\Big(\delta(x-x_o)\Big)\psi(x)dx\tag{4}$

The RHS of $(3)$ suggests that $\int\delta(x-x_o)\hat A\psi(x)dx = \hat A\int\delta(x-x_o)\psi(x)dx\tag{5}$

Doubts
(i) In the RHS of $(3)$ would we have $\hat A$ or $\hat A^\dagger=-\hat A$. I feel that it would be $\hat A^\dagger$ because we are sort of transferring the derivative, but the $(3)$ suggests that it is $\hat A$.

(ii) While proving $(3)$ using the functions instead of vectors, I got stuck in $(4)$. I am not able to justify how we can take $\hat A$ out of the integral to get $(5)$.

Answer 1

WP sets down the rules for the Dirac bra-ket notation, which you are abusing/muffing: $$ \langle x|A|\psi\rangle\equiv A_x \langle x| \psi\rangle= A_x \psi(x), $$ so that, for your $$ A= \int \!\!dx~ |x\rangle \partial_x \langle x| , $$ it reduces to $$ \langle x|A|\psi\rangle =\int \!\!dy~\langle x| y\rangle \partial_y \langle y |\psi\rangle= \int \!\!dy~ \delta (x-y)\partial_y \psi (y)= \partial_x \psi(x)= A_x \psi(x). $$

Your crucial error is not contrasting an operator, A, acting on kets to produce kets, to the coordinate representation of this operator, $A_x$, acting on bracket scalars, that most people often drop the subscript x of, confident there can be no confusion. (It's the "inside" between the ket and the bra!) You should not do that, until you are confident you cannot be in confusion about it.

WP refers you to the better text of Sakurai & Napolitano that cares to avoid that confusion.

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Edit in response to comments

The matrix elements of this A are evidently $$\langle x|A|y\rangle= \partial_x \delta(x-y), $$ which jibes naturally with the coordinate-space representation of $A_x$. Select your $\psi$ to be along y.

Or else, more generally, $$ \langle x|A|\psi\rangle=\int\!\!dy ~ \langle x|A| y\rangle \langle y|\psi \rangle = \int\!\!dy ~ \partial_x\delta(x-y)~ \psi(y) = \partial_x \psi(x). $$

Answer 2

Cosmas Zachos gave a somewhat nice solution that carries the crux of your particular problem. I am, however, much more interested in patching the horrible hole in the standard teaching of quantum theory, and so I would like to present a coherent demonstration and introduction to the story, that textbooks should just state clearly instead of throwing at students.

A suitable definition would be the following trio, with $\hat{\mathbb I}$ being the do-nothing, identity operator: $$ \begin{align} \tag1\int\left|x\right>\mathrm dx\left<x\right|&=\hat{\mathbb I}=\int\left|p\right>\frac{\mathrm dp}{2\pi\hslash}\left<p\right|\qquad\qquad\bigwedge\qquad\qquad\left<x|p\right>=e^{ipx/\hslash} \end {align} $$ The reason why these are definitions and cannot be derived, is because of how much convention is inside them: the $2\pi\hslash$ is often sent somewhere else, and another thing that can only be fixed by convention, is the choice to use $+ipx$ rather than $-ipx$.

Once the above are accepted as reasonable starting points, then we can deal with things in a variety of ways. We can define $\hat p\left|p\right>=p\left|p\right>$, or $\left<p\right|\hat p=\left<p\right|p$ or even $\hat p=\int\left|p\right>p\frac{\mathrm dp}{2\pi\hslash}\left<p\right|$, all of which are trivially equivalent. Note that the scalar value $p$ is very free to be moved all over the place: $$ \begin{align} \tag2 \hat p\left|\psi\right> &=\hat{\mathbb I}\hat p\hat{\mathbb I}\left|\psi\right>\\ \tag3&=\int\left|x\right>\mathrm dx\left<x\right|\int\left|p\right>p\frac{\mathrm dp}{2\pi\hslash}\left<p\right|\int\left|y\right>\mathrm dy\left<y|\psi\right>\\ \tag4&=\int\left|x\right>\mathrm dx\iint\frac{\mathrm dp}{2\pi\hslash}\,\mathrm dy\ p\left<x|p\right>\left<p|y\right>\left<y|\psi\right>\\ \tag5&=\int\left|x\right>\mathrm dx\iint\frac{\mathrm dp}{2\pi\hslash}\,\mathrm dy\ p\,e^{ip(x-y)/\hslash}\left<y|\psi\right>\\ \tag6&=\int\left|x\right>\mathrm dx\iint\frac{\mathrm dp}{2\pi\hslash}\,\mathrm dy\ (-i\hslash\nabla_x)\,e^{ip(x-y)/\hslash}\left<y|\psi\right>\\ \tag7&=\int\left|x\right>\mathrm dx(-i\hslash\nabla_x) \iint\frac{\mathrm dp}{2\pi\hslash}\,\mathrm dy\ e^{ip(x-y)/\hslash}\left<y|\psi\right>\\ \tag8&=\int\left|x\right>\mathrm dx(-i\hslash\nabla_x) \int\mathrm dy\ \delta(x-y)\left<y|\psi\right>\\ \tag9&=\int\left|x\right>\mathrm dx(-i\hslash\nabla_x)\left<x|\psi\right> \end {align} $$ Now, if we consider how, in general, things are written in QM, that you have already correctly written, namely: $$ \begin{align} \tag{10}\left|\psi\right>&=\int\left|\vec r\right>\mathrm d^3\vec r\left<\vec r|\psi\right>=\int\left|\vec r\right>\mathrm d^3\vec r\ \psi(\vec r)\\ \tag{11}\text{Let unnormalised!}\qquad\left|\phi\right>&=\vec{\hat p}\left|\psi\right>\qquad\text{so that we can identify}\\ \tag{12}\text{from Equation (9)}\qquad\left<\vec r|\phi\right> &=\langle\vec r|\vec{\hat p}|\psi\rangle=-i\hslash\vec\nabla\left<\vec r|\psi\right>\\ \tag{13}\text{i.e.}\qquad\left<x\right|\hat p&=-i\hslash\nabla_x\left<x\right|\\ \tag{14}\hat p\left|x\right>&=+i\hslash\nabla_x\left|x\right> \end {align} $$ This last equation, (14), shows that many texts simply thought wrongly on how the older notation $\hat p\psi(x)=-i\hslash\nabla_x\psi(x)$ is meant to be translated into Dirac notation. In fact, the older notation states that $\hat p\left<x\right|=-i\hslash\nabla_x\left<x\right|$, and it is just intolerably subtly wrong all over the place. Dirac notation is the better way to keep things clear.

Luckily, while quite many people have erred here, everybody has agreed to stick with Dirac's choice that $e^{+ipx/\hslash}$ and $\hat p=-i\hslash\nabla_x$ without fail, so there is no profusion of different conventions.

A nice property of this demonstration is that the Hermiticity of the momentum operator is never in question; it is trivial to just glance at the momentum operator's definition that looks like the identity operator, just multiplied by the scalar momentum value, just like the position operator is the identity operator multiplied by the scalar position value, and the Hermiticity is manifest.

Between my answer and Cosmas Zachos's, you should now have everything you need to show that the derivative operator fails to be Hermitian, and other properties.