[Physics] Eigenstates of momentum and energy of a free particle

momentumobservablesoperatorsquantum mechanics

Given the momentum operator $\hat{P}:= \frac{\hbar}{i}\frac{d}{dx}$, as I understand, the eigenvalue equations are $$\hat{P}f_{p}(x)= \frac{\hbar}{i}\frac{d}{dx}f_{p}(x) = p f_{p}(x)$$ and the eigenfunctions which correspond to this are $$f_{p}(x) = A e^{\frac{ipx}{\hbar}}.$$ Given then that the square of this operator $\hat{P}^2$ commutes with $\hat{P}$, so $[\hat{P}, \hat{P}^2] = 0$, it follows that the two operators share a common set of eigenstates which form a basis (apparently in something called a rigged Hilbert space) such that any quantum state, represented as a member of some Hilbert space can be written as a linear combination of elements of these basis vectors (this is a postulate of QM).

Question: I know that all eigenvectors of $\hat{P}$ are eigenvectors of $\hat{P}^2$ (which are of the form $f_{p}(x) = A e^{\frac{ipx}{\hbar}}$) this is easy to show. But what are the eigenvectors of $\hat{P}^2$ which are not eigenvectors of $\hat{P}$ and can you form a basis of such vectors? Thanks.

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the eigenvectors of $\hat{p}^2$ which are not eigenvectors of $\hat{p}$.

These do exist: each plane wave $|p⟩$ is an eigenvector of $\hat p$ with eigenvalue $p$, and an eigenvector of $\hat{p}^2$ with eigenvalue $p^2$, which means that the plane wave $\left|-p\right>$ has the same eigenvalue of $\hat{p}^2$. This means, in turn, that any linear combination of the two will still be an eigenvector of $\hat{p}^2$.

As a quick example of how you turn these into a basis, you can simply consider the basis $$ \left\{\frac{\left|p\right>+\left|-p\right>}{\sqrt{2}}\middle|\, p\geq0\right\} \bigcup \left\{\frac{\left|p\right>-\left|-p\right>}{\sqrt{2}}\middle|\, p<0\right\} , $$ which consists of the real-valued plane waves with position-representation wavefunctions $$ \left<x\right|\frac{\left|p\right>+\left|-p\right>}{\sqrt{2}} = \frac{1}{\sqrt{\pi}}\cos(px) \quad\text{and}\quad \left<x\right|\frac{\left|p\right>-\left|-p\right>}{\sqrt{2}} = \frac{i}{\sqrt{\pi}}\sin(px) , $$ respectively. It should be a relatively easy exercise for you to transform the completeness relation over the $|p⟩$ states to show that these are a basis. (While doing that, by the way, never mind the zero: all your integrals are Lebesgue integrals, and pointwise values do not matter.)

In three dimensions, of course, it becomes a bit more complicated (you need to change the condition $p>0$ for a set that includes half of all $p$ space, for this example) but you also have more freedom, as there is a much bigger set of $\mathbf p$s that share the same $p^2$.

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[Physics] Complex conjugate of momentum operator

The short story is, you can't distribute the * onto the operator like that. You need to keep it as $\left(\hat{p}e^{ipx}\right)^* = {p}\,e^{-ipx}$ because $\hat{p}$ is an operator on a complex vector space which looks like a derivative in the $x$ basis.

$\hat{p} =-i{\partial \over \partial x} $ is the representation of the momentum operator in the $x$ basis, which is to say $\langle x \left| \hat p \right|\psi\rangle = -i{\partial \over \partial x}\langle x\left|\psi\right\rangle$. Complex conjugation is an operation we know how to do on complex numbers, so let's make sure that the objects we're working with are complex numbers first. The complex conjugate of an inner product is $\langle x | \psi \rangle^* = \langle\psi|x\rangle$. The operator $\hat{p}$ changes the vector $|\psi\rangle$ into some other vector $\hat{p}|\psi\rangle = |p\psi\rangle$. Now taking the complex conjugate looks like:

$$ \langle x \left| \hat p \right|\psi\rangle^* = \langle x |p\psi\rangle^* = \langle p\psi|x\rangle $$

The question now is what is $\langle p\psi|$? In order to have $\langle\psi|\psi\rangle = |\psi|^2$ then $\langle \psi|$ must be hermitian adjoint of $|\psi\rangle$. If you think of kets as column vectors, bras as row vectors, and operators as matrices, then this operation is taking the transpose of the operator matrix in addition to complex conjugation.

Quantum Mechanics – Understanding the Hermiticity of Momentum Operator Represented in Position Basis

The following is a copy-paste from Dirac⁽¹⁾

This gives \begin{equation} \bigg< \phi\dfrac{\mathrm d \hphantom{q}}{\mathrm d q}\bigg| q'\bigg>=\boldsymbol{-}\dfrac{\mathrm d \phi\left(q'\right)}{\mathrm d q'}, \nonumber \end{equation} showing that \begin{equation} \bigg< \phi\dfrac{\mathrm d \hphantom{q}}{\mathrm d q}=\boldsymbol{-}\bigg<\dfrac{\mathrm d\phi}{\mathrm dq}. \tag{016}\label{eq016} \end{equation} Thus $\:\mathrm d /\mathrm d q\:$ operating to the left on the conjugate complex of a wave function has the meaning of $\mathrm{minus}$ differentiation with respect to q.

And two paragraphs later

The conjugate complex of the linear operator $\:\mathrm d /\mathrm d q\:$ can be evaluated by noting that the conjugate imaginary of $\:\mathrm d /\mathrm d q.\psi\rangle\:$ or $\:\mathrm d\psi /\mathrm d q\rangle\:$ is $\:\langle\mathrm d \overset{\boldsymbol{\_\!\_}}{\psi}/\mathrm d q$, or $\:\boldsymbol{-}\langle\overset{\boldsymbol{\_\!\_}}{\psi}\,\mathrm d /\mathrm d q\;$ from \eqref{eq016}. Thus the conjugate complex of $\:\mathrm d /\mathrm d q\:$ is $\:\boldsymbol{-}\mathrm d /\mathrm d q\:$, so $\:\mathrm d /\mathrm d q\:$ is a pure imaginary linear operator.

^{(1)
The Principles of Quantum Mechanics, P.A.M.Dirac, 4th edition 1967-
$\S\;$22. Schröndiger's representation.}

ADDENDUM

Let $\;p\;$ an operator in a finite $n-$dimensional Hilbert space represented by the finite $n\times n$ matrix $\mathrm{p}_{\mu\nu}$. The hermitian conjugate operator $\;p^{\boldsymbol{\dagger}}\;$ is represented by the matrix $\mathrm{p}^{\boldsymbol{\dagger}}_{\mu\nu}$, the hermitian conjugate of $\mathrm{p}_{\mu\nu}$ which is produced by transposing and complex conjugating \begin{equation} \mathrm{p}^{\boldsymbol{\dagger}}_{\mu\nu}=\overset{\boldsymbol{\_\!\_}}{\mathrm{p}}_{\nu\mu} \tag{01}\label{eq01} \end{equation} The operator $\;p\;$ is hermitian if its matrix $\mathrm{p}_{\mu\nu}$ is hermitian \begin{equation} p^{\boldsymbol{\dagger}}=p \quad \Longleftarrow\!\Longrightarrow \quad \mathrm{p}^{\boldsymbol{\dagger}}_{\mu\nu}=\mathrm{p}_{\mu\nu} \tag{02}\label{eq02} \end{equation}

Now, if $\;q\;$ is the position operator for the $x-$axis and $\;p\;$ the operator of the $x-$component of the linear momentum in the $\;q-$basis, that is with respect to the complete set of the eigenstates of $\;q$, then
\begin{equation} p=-i\hbar\dfrac{\mathrm d\hphantom{q}}{\mathrm dq} \tag{03}\label{eq03} \end{equation} If we want to check the hermiticity of this momentum operator by transposing and complex conjugating the elements of a matrix then equation \eqref{eq03} is not convenient, as it is, since it's far away from being an equation between matrices.

In the following we'll see a matrix representation of \eqref{eq03} by which we'll prove the hermiticity of $\;p\;$ exactly by transposing and complex conjugating of the elements of a matrix. This is achieved by the basic quantum-mechanical equation \begin{equation} qp-pq =i\hbar\, I \tag{04}\label{eq04} \end{equation} and the Dirac $\delta-$function. It is known that there don't exist matrices $\;\mathrm{q}_{\mu\nu},\mathrm{p}_{\mu\nu},\mathrm{I}_{\mu\nu}\;$ representations of $\;q,p,I\;$ satisfying equation \eqref{eq04} with $\mu,\nu$ taking values from a finite or even countable set of values (any effort to succeed this would convince us). So these matrices must be of infinite uncountable dimensions.

Now, our first step would be to find a representation for the operator $\;q\;$ since we must express the momentum operator $\;p\;$ in the $\;q-$basis. This problem is equivalent to the determination of the eigenstates of $\;q$.

Before examine the infinite continuous case we'll try to build an example for a finite discrete case and see the correspondence. So, suppose that a particle could be only on a finite set of points on the $\;x-$axes, let $x_1=1,x_2=2, \cdots,x_n=n$. The space of states is the $\;n-$dimensional Hilbert space $\;\mathbb{C}^{n\;}$ so it's reasonable to take as eigenstates of the position operator the $n$ state vectors
\begin{equation} \left(\mathbf{e}_\mu\right)_\nu=\delta_{\mu\nu}\quad \text{that is : } \mathbf{e}_1= \begin{bmatrix} 1\\ 0\\ \vdots\\ 0\\ 0 \end{bmatrix} \quad \mathbf{e}_2= \begin{bmatrix} 0\\ 1\\ \vdots\\ 0\\ 0 \end{bmatrix} \quad \cdots\cdots\cdots \quad \mathbf{e}_n= \begin{bmatrix} 0\\ 0\\ \vdots\\ 0\\ n \end{bmatrix} \tag{05}\label{eq05} \end{equation} and since the operator $\;q\;$ has eigenvalues $\;1,2,\cdots,n\;$ it is represented by the diagonal matrix \begin{equation} \mathrm q= \begin{bmatrix} 1 & \hphantom{n} &\hphantom{n} &\hphantom{n}&\hphantom{n} \\ \hphantom{n} & 2 &\hphantom{n} &\hphantom{n}&\hphantom{n} \\ \hphantom{n} & \hphantom{n} & \ddots &\hphantom{n}&\hphantom{n}\\ \hphantom{n} & \hphantom{n} &\hphantom{n} &\ddots&\hphantom{n}\\ \hphantom{n} & \hphantom{n} &\hphantom{n} &\hphantom{n} & n \end{bmatrix} \tag{06}\label{eq06} \end{equation} expressed via the Kronecker $\delta$ \begin{equation} \mathrm q_{\mu\nu}=\mu\delta_{\mu\nu} \tag{07}\label{eq07} \end{equation} In \eqref{eq07} the Einstein's convention for repeated indices is not valid.

A general state $\;f\;$ is a $\;n-$dimensional complex vector represented with respect to the basis \eqref{eq05} by the one column matrix \begin{equation} \mathrm f= \begin{bmatrix} f(1)\\ f(2)\\ \vdots\\ \vdots\\ f(n) \end{bmatrix} \tag{08}\label{eq08} \end{equation}

Application of the $\;q\;$ operator on $\;f\;$ produces a new state $\;g\;$ \begin{equation} g=qf \tag{09}\label{eq09} \end{equation} represented by the matrix \begin{equation} \mathrm g= \begin{bmatrix} 1\cdot f(1)\\ 2\cdot f(2)\\ \vdots\\ \vdots\\ n\cdot f(n) \end{bmatrix} \tag{10}\label{eq10} \end{equation} that is \begin{equation} g(\mu)=\sum\limits_{\nu}\mathrm q_{\mu\nu}f(\nu)=\sum\limits_{\nu}\mu\delta_{\mu\nu}f(\nu)=\mu f(\mu) \tag{11}\label{eq11} \end{equation}

Now, returning to the infinite continuous case, a particle could be at any point on the real $\;x-$axis so we expect the $\;q\;$ to be represented in the basis of its own eigenstates by a diagonal $^\prime$improper$^\prime$ matrix with all real numbers as eigenvalues on the main diagonal as below
\begin{equation} \mathrm q= \begin{bmatrix} \boldsymbol{-\infty} & \hphantom{n} &\hphantom{n} &\hphantom{n}&\hphantom{n}\vphantom{\dfrac{a}{b}} \\ \hphantom{n} & \nwarrow &\hphantom{n} &\hphantom{n}&\hphantom{n}\vphantom{\dfrac{a}{b}} \\ \hphantom{n} & \hphantom{n} & \mathbb{R} &\hphantom{n}&\hphantom{n}\vphantom{\dfrac{a}{b}}\\ \hphantom{n} & \hphantom{n} &\hphantom{n} &\searrow&\hphantom{n}\vphantom{\dfrac{a}{b}}\\ \hphantom{n} & \hphantom{n} &\hphantom{n} &\hphantom{n} & \boldsymbol{+\infty}\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{12}\label{eq12} \end{equation} But the row and column indices $\;\mu,\nu\;$ which take discrete values in \eqref{eq07} must be replaced by the indices $\;x_{\boldsymbol{r}},x_{\boldsymbol{c}}\;$ which take continuous values in $\;\mathbb{R}$ as shown below \begin{align} &\begin{matrix} \hphantom{\cdots\cdots} &\hphantom{\!\!\! \cdots\cdots} &\hphantom{\!\!\! \cdots\cdots} & \:\:\:x_{\boldsymbol{c}} &\hphantom{\!\!\!\! \cdots\cdots} \end{matrix} \nonumber\\ &\begin{matrix} \hphantom{\cdots\cdots} &\hphantom{\!\!\! \cdots\cdots} &\hphantom{\!\!\! \cdots\cdots} & \:\:\:\:\downarrow &\hphantom{\!\!\!\! \cdots\cdots} \end{matrix} \nonumber\\ \mathrm q \left( x_{\boldsymbol{r}}, x_{\boldsymbol{c}}\right)\boldsymbol{=}\quad \begin{matrix} \vphantom{\dfrac{a}{b}} \\ x_{\boldsymbol{r}} \longrightarrow\vphantom{\dfrac{a}{b}} \\ \vphantom{\dfrac{a}{b}} \\ \vphantom{\dfrac{a}{b}} \\ \vphantom{\dfrac{a}{b}} \end{matrix}\!\! &\begin{bmatrix} \hphantom{n} & \hphantom{n} &\hphantom{n} &\vdots&\hphantom{n} \\ \cdots\cdots\!\!\!\! & \cdots\cdots\!\!\!\! &\cdots\cdots\!\!\!\! &\cdots\vdots\cdots &\!\!\!\! \cdots\cdots \vphantom{\dfrac{a}{b}}\\ \hphantom{n} & \hphantom{n} &\hphantom{n} &\vdots&\hphantom{n}\vphantom{\dfrac{a}{b}} \\ \hphantom{n} & \hphantom{n} &\hphantom{n} &\vdots&\hphantom{n} \\ \hphantom{n} & \hphantom{n} &\hphantom{n} &\vdots&\hphantom{n} \end{bmatrix} \tag{13}\label{eq13} \end{align} For notation reasons in \eqref{eq13} the row and column indices $\;x_{\boldsymbol{r}},x_{\boldsymbol{c}}\;$ are not placed as subscripts as $\;\mu,\nu\;$ do in $\;\mathrm q_{\mu\nu}\;$ but are placed in parentheses so that $\;\mathrm q \left( x_{\boldsymbol{r}}, x_{\boldsymbol{c}}\right)\;$ looks like a function of two real variables. Here the eigenstates of the position operator $\;q\;$ that correspond to those of the finite case, see equation \eqref{eq05}, are \begin{equation} \left(\mathbf{e}_{ x_{\boldsymbol{r}}}\right)_ {x_{\boldsymbol{c}}}\boldsymbol{=}\delta\left( x_{\boldsymbol{r}}\!\boldsymbol{-}\!x_{\boldsymbol{c}}\right)\qquad x_{\boldsymbol{r}},x_{\boldsymbol{c}} \in \mathbb{R} \tag{14}\label{eq14} \end{equation} where $\;\delta\left(x\right)\;$ the Dirac $\;\delta-$function.

The position operator $\;q\;$ is represented by the $^\prime$improper$^\prime$ matrix \begin{equation} \mathrm q \left( x_{\boldsymbol{r}}, x_{\boldsymbol{c}}\right)\boldsymbol{=} x_{\boldsymbol{r}}\delta\left( x_{\boldsymbol{r}}\!\boldsymbol{-}\!x_{\boldsymbol{c}}\right) \tag{15}\label{eq15} \end{equation} an expression that corresponds to equation \eqref{eq07} of the finite case.

The matrix \eqref{eq15} is hermitian due to the fact that the Dirac $\;\delta-$function is real and even, that is transposing and complex conjugating \begin{equation} \mathrm{q}^{\boldsymbol{\dagger}}\left( x_{\boldsymbol{r}}, x_{\boldsymbol{c}}\right)\boldsymbol{=}\overset{\boldsymbol{\_\!\_}}{\mathrm{q}}\left( x_{\boldsymbol{c}}, x_{\boldsymbol{r}}\right)\boldsymbol{=} x_{\boldsymbol{r}}\delta\left( x_{\boldsymbol{c}}\!\boldsymbol{-}\!x_{\boldsymbol{r}}\right)\boldsymbol{=} x_{\boldsymbol{r}}\delta\left( x_{\boldsymbol{r}}\!\boldsymbol{-}\!x_{\boldsymbol{c}}\right)\boldsymbol{=}\mathrm q \left( x_{\boldsymbol{r}}, x_{\boldsymbol{c}}\right) \tag{16}\label{eq16} \end{equation} Note that the identity operator $\;I\;$ is represented by \begin{equation} \mathrm I \left( x_{\boldsymbol{r}}, x_{\boldsymbol{c}}\right)\boldsymbol{=} \delta\left( x_{\boldsymbol{r}}\!\boldsymbol{-}\!x_{\boldsymbol{c}}\right) \tag{17}\label{eq17} \end{equation} A general state $\;f\;$ is a function $\;f\left(x\right)\;$ of the $\;x-$coordinate and application of the position operator $\;q\;$ produces an other state $\;g\;$ expressed in matrix form as \begin{equation} g\boldsymbol{=}qf\boldsymbol{=}\!\!\!\!\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\!\!\mathrm q \left( x_{\boldsymbol{r}}, x_{\boldsymbol{c}}\right)f\left(x_{\boldsymbol{c}}\right)\mathrm dx_{\boldsymbol{c}}\boldsymbol{=}\!\!\!\!\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\!\!x_{\boldsymbol{r}}\delta\left( x_{\boldsymbol{r}}\!\boldsymbol{-}\!x_{\boldsymbol{c}}\right)f\left(x_{\boldsymbol{c}}\right)\mathrm dx_{\boldsymbol{c}}\boldsymbol{=}x_{\boldsymbol{r}}f\left(x_{\boldsymbol{r}}\right) \tag{18}\label{eq18} \end{equation} that is \begin{equation} g\left(x\right)\boldsymbol{=}xf\left(x\right) \tag{19}\label{eq19} \end{equation} Equations \eqref{eq18}-\eqref{eq19} correspond to \eqref{eq11} of the finite case where the summation $\;\sum_\nu\;$ with respect to the index $\;\nu\;$ is replaced by the integral $\;\int\mathrm dx_{\boldsymbol{c}}\;$ with respect to the real index $\;x_{\boldsymbol{c}}$.

The final step will be to find the matrix representation of the linear momentum operator $\;p$, let it be $\;\mathrm{p}\left( x_{\boldsymbol{r}}, x_{\boldsymbol{c}}\right)$. So, let a general state $\;f$, that is a function $\;f\left(x\right)$. Application of the basic quantum-mechanical equation \eqref{eq04} on $\;f\;$ yields \begin{align} &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\left(qp-pq\right)f =i\hbar\, I f \quad \Longrightarrow \nonumber\\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\!\!\mathrm q \left( x_{\boldsymbol{r}},u\right)\left(\:\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\!\!\mathrm p \left( u,x_{\boldsymbol{c}}\right)f\left(x_{\boldsymbol{c}}\right)\mathrm dx_{\boldsymbol{c}}\right)\mathrm du\boldsymbol{-}\!\!\!\!\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\!\!\mathrm p \left( x_{\boldsymbol{r}},v\right)\left(\:\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\!\!\mathrm q \left( v,x_{\boldsymbol{c}}\right)f\left(x_{\boldsymbol{c}}\right)\mathrm dx_{\boldsymbol{c}}\right)\mathrm dv \nonumber\\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\boldsymbol{=}i\hbar\!\!\!\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\!\!\mathrm I \left( x_{\boldsymbol{r}}, x_{\boldsymbol{c}}\right)f\left(x_{\boldsymbol{c}}\right)\mathrm dx_{\boldsymbol{c}} \quad \stackrel{\eqref{eq15},\eqref{eq17}}{=\!=\!=\!\Longrightarrow} \nonumber\\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\!\!x_{\boldsymbol{r}}\delta\left( x_{\boldsymbol{r}}\!\boldsymbol{-}\!u\right)\left(\:\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\!\!\mathrm p\left( u,x_{\boldsymbol{c}}\right)f\left(x_{\boldsymbol{c}}\right)\mathrm dx_{\boldsymbol{c}}\right)\mathrm du\boldsymbol{-}\!\!\!\!\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\!\!\mathrm p \left( x_{\boldsymbol{r}},v\right)\left(\:\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\!\!v\delta\left( v\!\boldsymbol{-}\!x_{\boldsymbol{c}}\right)f\left(x_{\boldsymbol{c}}\right)\mathrm dx_{\boldsymbol{c}}\right)\mathrm dv \nonumber\\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\boldsymbol{=}i\hbar\!\!\!\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\!\!\delta\left( x_{\boldsymbol{r}}\!\boldsymbol{-}\!x_{\boldsymbol{c}}\right)f\left(x_{\boldsymbol{c}}\right)\mathrm dx_{\boldsymbol{c}} \quad \Longrightarrow \nonumber\\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! x_{\boldsymbol{r}}\!\!\!\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\!\!\mathrm p \left( x_{\boldsymbol{r}}, x_{\boldsymbol{c}}\right)f\left(x_{\boldsymbol{c}}\right)\mathrm dx_{\boldsymbol{c}}\boldsymbol{-}\!\!\!\!\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\!\!\mathrm p \left( x_{\boldsymbol{r}}, x_{\boldsymbol{c}}\right)x_{\boldsymbol{c}}f\left(x_{\boldsymbol{c}}\right)\mathrm dx_{\boldsymbol{c}}\boldsymbol{=}i\hbar\!\!\!\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\!\!\delta\left( x_{\boldsymbol{r}}\!\boldsymbol{-}\!x_{\boldsymbol{c}}\right)f\left(x_{\boldsymbol{c}}\right)\mathrm dx_{\boldsymbol{c}} \quad \Longrightarrow \nonumber\\ &\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\!\!\Bigl[\left( x_{\boldsymbol{r}}\!\boldsymbol{-}\!x_{\boldsymbol{c}}\right)\mathrm p \left( x_{\boldsymbol{r}}, x_{\boldsymbol{c}}\right)\!\boldsymbol{-}\!i\hbar\delta\left( x_{\boldsymbol{r}}\!\boldsymbol{-}\!x_{\boldsymbol{c}}\right)\Bigr]f\left(x_{\boldsymbol{c}}\right)\mathrm dx_{\boldsymbol{c}}=0 \tag{20}\label{eq20} \end{align} Since \eqref{eq20} must be valid for any state-function $\;f\left(x\right)\;$ the expression in brackets under the integral must be identically zero, so \begin{equation} \bbox[yellow] {\:\:\:\mathrm p \left( x_{\boldsymbol{r}}, x_{\boldsymbol{c}}\right)\boldsymbol{=} i\hbar \dfrac{\delta\left( x_{\boldsymbol{r}}\!\boldsymbol{-}\!x_{\boldsymbol{c}}\right)}{\left( x_{\boldsymbol{r}}\!\boldsymbol{-}\!x_{\boldsymbol{c}}\right)}\boldsymbol{=}\boldsymbol{-}i\hbar\delta^{\,\prime}\!\left( x_{\boldsymbol{r}}\!\boldsymbol{-}\!x_{\boldsymbol{c}}\right)\boldsymbol{=} \boldsymbol{-}i\hbar \dfrac{\mathrm d\delta\left( x_{\boldsymbol{r}}\!\boldsymbol{-}\!x_{\boldsymbol{c}}\right)}{\mathrm d\left( x_{\boldsymbol{r}}\!\boldsymbol{-}\!x_{\boldsymbol{c}}\right)}\:\:\:\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}} } \tag{21}\label{eq21} \end{equation} where \begin{equation} \delta^{\,\prime}\!\left( x\right)\boldsymbol{=} \dfrac{\mathrm d\delta \left(x\right)}{\mathrm dx}\boldsymbol{=} \boldsymbol{-}\dfrac{\delta \left(x\right)}{x} \tag{22}\label{eq22} \end{equation} the first derivative of the $\;\delta-$function.

Now, the hermiticity of the momentum operator is proved by transposing and complex conjugating the matrix in \eqref{eq21}

\begin{equation} \mathrm{p}^{\boldsymbol{\dagger}}\left( x_{\boldsymbol{r}}, x_{\boldsymbol{c}}\right)\boldsymbol{=}\overset{\boldsymbol{\_\!\_}}{\mathrm{p}}\left( x_{\boldsymbol{c}}, x_{\boldsymbol{r}}\right)\boldsymbol{=} \boldsymbol{-}i\hbar \dfrac{\delta\left( x_{\boldsymbol{c}}\!\boldsymbol{-}\!x_{\boldsymbol{r}}\right)}{\left( x_{\boldsymbol{c}}\!\boldsymbol{-}\!x_{\boldsymbol{r}}\right)}\boldsymbol{=}i\hbar \dfrac{\delta\left( x_{\boldsymbol{r}}\!\boldsymbol{-}\!x_{\boldsymbol{c}}\right)}{\left( x_{\boldsymbol{r}}\!\boldsymbol{-}\!x_{\boldsymbol{c}}\right)}\boldsymbol{=}\mathrm p \left( x_{\boldsymbol{r}}, x_{\boldsymbol{c}}\right) \tag{23}\label{eq23} \end{equation} The matrix \eqref{eq21} is hermitian due to the fact that the first derivative $\;\delta^{\,\prime}\!\left(x\right)\;$ of the Dirac $\;\delta-$function is real and odd and to the antihermiticity of the imaginary unit $\;i$.

Note also that equation \eqref{eq21} is a matrix representation of \eqref{eq03} since application of the momentum operator $\;p\;$ on a state $\;f\;$ yields \begin{equation} pf=\!\!\!\!\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\!\!\mathrm p \left( x_{\boldsymbol{r}}, x_{\boldsymbol{c}}\right)f\left(x_{\boldsymbol{c}}\right)\mathrm dx_{\boldsymbol{c}}\boldsymbol{=}\boldsymbol{-}i\hbar\!\!\!\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\!\!\delta^{\,\prime}\!\left( x_{\boldsymbol{r}}\!\boldsymbol{-}\!x_{\boldsymbol{c}}\right)f\left(x_{\boldsymbol{c}}\right)\mathrm dx_{\boldsymbol{c}}\boldsymbol{=}\boldsymbol{-}i\hbar\dfrac{\mathrm df}{\mathrm dx} \tag{24}\label{eq24} \end{equation}

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Quantum Mechanics – Understanding the Hermiticity of Momentum Operator Represented in Position Basis

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