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 $\int\delta(x-x_o)\hat A\psi(x)dx = \hat A\int\delta(x-x_o)\psi(x)dx\tag{5}$

(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 $(5)$.

  • 1
    $\begingroup$ Your initial notation choice is already in trouble. When you wrote down $\hat A\psi(x)=\psi^\prime(x)$, you left no way for you to connect that with $\langle x_0\vert\hat A\psi\rangle$ that is written in Equations (2) and (3); this scheme is simply not likely to get you what you want. $\endgroup$ Oct 17, 2023 at 19:21
  • $\begingroup$ @naturallyInconsistent, thanks a lot for the reply. I am not able to understand why the notation you mentioned is incorrect. I have just used the way we use to represent derivative. A derivative operator acting on a function giving another function $f\to f'$. $\endgroup$
    – Iti
    Oct 17, 2023 at 19:26
  • $\begingroup$ The former is $\hat A\left<x|\psi\right>$ whereas the latter is $\langle x|\hat A\psi\rangle$ and are thus different. $\endgroup$ Oct 17, 2023 at 19:28
  • $\begingroup$ But I have seen that same manipulation with momentum operator. $\langle x_o|\hat p\psi\rangle=\hat p\langle x_o|\psi\rangle$. Using that they show that if $\psi$ is momentum eigenstate of $\hat p$ with momentum p, then $p\langle x_o|\psi\rangle=-i\bar h\frac{d}{dx}\langle x_o|\psi\rangle$ Thus, giving, $\langle x_o|\psi\rangle=Nexp(ipx_o/\bar h)$. So, how this is incorrect in the above post. $\endgroup$
    – Iti
    Oct 17, 2023 at 19:33
  • $\begingroup$ That is just them doing it wrong. And if $\psi$ is a momentum eigenstate, then the momentum operator acting on it will just bring down a scalar number multiple, which is then free to be commuted anywhere. $\endgroup$ Oct 17, 2023 at 19:35

2 Answers 2


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.

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

  • $\begingroup$ I think OP wants to know how $\langle x|\hat A\psi\rangle\equiv\hat A_x\langle x|\psi\rangle$, this definition comes. And then OP might get confused between $\hat A_x\langle x|\psi\rangle$ and $\langle x\hat A^\dagger|\psi\rangle$ $\endgroup$
    – Manu
    Oct 17, 2023 at 20:12
  • $\begingroup$ @Manu: I explained it: $A_x$ mimics the operator inside the ket and the bra. It is a reasonable definition. His latter confusion is completely unwarranted in the Dirac bracket notation. He simply failed to learn the rules. $\endgroup$ Oct 17, 2023 at 21:20
  • $\begingroup$ My sense/advice is that he should never use $𝐴_𝑥$, and he should never skip | demarcators in kets and bras, both points of confusion and meaninglessness. $\endgroup$ Oct 17, 2023 at 21:41
  • $\begingroup$ @CosmasZachos , thanks a lot for the reply. I have read your replies and the references you mentioned. So, if we are working in position basis and the operator $\hat A=\hat A(\hat x)$ then the definition you mentioned is obvious. All the problem started when $\hat A=\hat A(\frac{d}{dx})$. So, what Sakurai did is that he defined translational operator in the form of momentum operator. Using translation operator he finds the action of momentum operator on ket and thus deduced the first expression in your answer. With that we can generalise it for higher order derivatives also. $\endgroup$
    – Iti
    Oct 18, 2023 at 9:46
  • $\begingroup$ So, the definition you mentioned is the result is due to the fact that translation operator can be written sing momentum operator. Is it so? $\endgroup$
    – Iti
    Oct 18, 2023 at 9:48

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.

  • $\begingroup$ Thanks a lot for the answer. I was reading your reply and Sakurai. So you have assumed that momentum eigenstate is known in position basis and then you move forward first to find the representation of momentum operator in position basis and other results. What Sakurai did is that after defining the position basis, he defined translational operator (coming out to be unitary after imposing the required conditions) and write it using the momentum operator. $\endgroup$
    – Iti
    Oct 18, 2023 at 10:10
  • $\begingroup$ After that he analyzed the action of momentum operator on wavefunction in position basis and then he get the form of momentum operator and also the identity $(13)$ also. And then further solving to get the $\langle x|p\rangle$. So, which approach is more natural, defining translational operator or assuming the form of momentum eigenket in position basis? $\endgroup$
    – Iti
    Oct 18, 2023 at 10:12
  • $\begingroup$ I would say that translational is more natural, i.e. Sakurai is better (I read it long before.) However, then you cannot see that there is a lot of choice of conventions involved in this. My presentation here is specifically to highlight that, as opposed to finding a most natural derivation. This is important because part of the insanity is in the confusion coming from old classical notation, and the incomplete translation into Dirac notation. I wanted to clarify this maximally. $\endgroup$ Oct 18, 2023 at 15:51

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