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$\require{cancel} \newcommand{\Ket}[1]{\left|{#1}\right\rangle} \newcommand{\Bra}[1]{\left\langle{#1}\right|} \newcommand{\Braket}[1]{\left\langle{#1}\right\rangle} \newcommand{\Rsr}[1]{\frac{1}{\sqrt{#1}}} \newcommand{\RSR}[1]{1/\sqrt{#1}} \newcommand{\Verti}{\rvert} \newcommand{\HAT}[1]{\hat{\,#1~}} \DeclareMathOperator{\Tr}{Tr}$
First created in April 2019
$\displaystyle i\hbar\frac{\partial\psi}{\partial t} =-\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}+V\psi .$
Let us find the expression of $\displaystyle p(x)=\frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar},$ as in https://en.wikipedia.org/wiki/Wave_function.
$\displaystyle \int_{-\infty}^{+\infty}\delta(x-x_0)~g(x)~dx=g(x_0).~~ \text{When }g(x)=1, \int_{-\infty}^{+\infty}\delta(x-x_0)~dx=1 .$
Fourier Transform:
$\displaystyle F(k) =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}f(x)~e^{-ikx}~dx .$
$\displaystyle f(x) =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}F(k)~e^{ikx}~dk .$
Let $~f(x)=\delta(x-x_0).$
$\displaystyle F(k) =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\delta(x-x_0)~e^{-ikx}~dx =\frac{1}{\sqrt{2\pi}}e^{-ikx_0} .$
$\displaystyle \delta(x-x_0) =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\left(\frac{1}{\sqrt{2\pi}}e^{-ikx_0}\right)~e^{ikx}~dk =\frac{1}{2\pi}\int_{-\infty}^{+\infty}e^{ik(x-x_0)}~dk .$
The expectation value of position is given by $\displaystyle \Braket x =\int_{-\infty}^{+\infty}x\lvert\psi(x)\rvert^2~dx =\int_{-\infty}^{+\infty}\psi^*(x)~x~\psi(x)~dx .$
Similarly, its counterpart in momentum is $\displaystyle \Braket p =\int_{-\infty}^{+\infty}p\lvert\Phi(p)\rvert^2~dp =\int_{-\infty}^{+\infty}\Phi^*(p)~p~\Phi(p)~dp .$
The difference between $\psi$ and $\Phi$ can be viewed as different basis of $\Ket x$ and $\Ket p$.
Please note that while $\lvert\psi(x)\rvert^2$ is probability density function over the position space, $\lvert\Phi(p)\rvert^2$ is probability density function over the momentum space. i.e. The probability of momentum measured in $p\pm\epsilon.$
Let us derive $\Phi$ from $\psi$.
Fourier Transform linking position and momentum:
$\displaystyle\psi(x) =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\phi(k)~e^{ikx}~dk .$
$\displaystyle\phi(k) =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\psi(x)~e^{-ikx}~dx .$
Dimensional Analysis: $[x]=[L],~~[p]=[LMT^{-1}],~~[k]=[L^{-1}],~~[\hbar]=[L^2MT^{-1}] .$
$[\psi(x)]=[L^{-1/2}],~~[\Phi(p)]=[L^{-1/2}M^{-1/2}T^{1/2}],~~[\phi(k)]=[L^{1/2}] .$
Let us define $\Phi(p)$ such that $\phi(k)=\sqrt\hbar~\Phi(p),$ which is consistent with dimensional analysis.
$\displaystyle \frac{\partial\psi(x)}{\partial x} =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\phi(k)~ik~e^{ikx}~dk .$
Inverse Fourier Transform: $\displaystyle \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\frac{\partial\psi(x)}{\partial x}e^{-ikx}~dx =ik~\phi(k) =ik~\sqrt\hbar~\Phi(p) =i\frac{p}{\sqrt\hbar}~\Phi(p) .$
$\displaystyle p~\Phi(p) =\sqrt{\frac{\hbar}{2\pi}}\int_{-\infty}^{+\infty}-i\frac{\partial}{\partial x}\psi(x)~e^{-ikx}~dx \ldots(1) .$
Recalling the original form $\phi(k) =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\psi(x)~e^{-ikx}~dx$, substituting $\phi(k)=\sqrt\hbar~\Phi(p)$ and conjugating both sides:
$\displaystyle\Phi(p)^* =\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{+\infty}\psi(x)^*~e^{ikx}~dx \ldots(2) .$
From $(1)$ and $(2):$
$\displaystyle \Braket p =\int_{-\infty}^{+\infty}\Phi(p)^*p~\Phi(p)~dp$
$\displaystyle =\int_{-\infty}^{+\infty}\left(\frac{1}{\sqrt{2\pi\cancel{\hbar}}}\int_{-\infty}^{+\infty}\psi(x)^*~e^{ikx}~dx\right) \left(\sqrt{\frac{\cancel{\hbar}}{2\pi}}\int_{-\infty}^{+\infty}-i \frac{\partial}{\partial y}\psi(y)~e^{-iky}~dy\right)\hbar~dk$
$\displaystyle =\frac{-i\hbar}{2\pi}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} \psi(x)^*\frac{\partial}{\partial y}\psi(y)~e^{ik(x-y)}~dk~dy~dx$
$\displaystyle =-i\hbar\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} \psi(x)^*\frac{\partial}{\partial y}\psi(y)\left(\frac{1}{2\pi}\int_{-\infty}^{+\infty}e^{ik(x-y)}~dk\right)dy~dx$
$\displaystyle =-i\hbar\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} \psi(x)^*\frac{\partial}{\partial y}\psi(y)~\delta(x-y)~dy~dx~~~\text{(see Preparation)}$
$\displaystyle =-i\hbar\int_{-\infty}^{+\infty}\psi(x)^*\left(\int_{-\infty}^{+\infty} \frac{\partial}{\partial y}\psi(y)~\delta(x-y)~dy\right)dx$
$\displaystyle =-i\hbar\int_{-\infty}^{+\infty}\psi(x)^*\left(\frac{\partial}{\partial x}\psi(x)\right)dx$
$\displaystyle =\int_{-\infty}^{+\infty}\psi(x)^*\left(-i\hbar\frac{\partial}{\partial x}\right)\psi(x)~dx .$
This proved that $\displaystyle\boxed{\HAT p=-i\hbar\frac{\partial}{\partial x}.}$
This section is loose. Needs more rigorous proof:
As $\displaystyle \Braket p =\int_{-\infty}^{+\infty}p~\lvert\psi(x)\rvert^2dx =\int_{-\infty}^{+\infty}\psi(x)^*p~\psi(x)~dx .$
$\HAT p\psi(x)=p~\psi(x).~~$ i.e. $p$ is the eigenvalue of operator $\displaystyle\HAT p=-i\hbar\frac{\partial}{\partial x}$.
Attempt to prove $\psi(x)$ is an eigenvector of operator $\frac{\partial}{\partial x}$ with eigenvalue $ik.~$
It would make $\displaystyle \psi(x) =\frac{1}{ik}\frac{\partial\psi(x)}{\partial x} =\frac{-i\hbar}{p}\frac{\partial\psi(x)}{\partial x}.~~ p~\psi(x)=\left(-i\hbar\frac{\partial}{\partial x}\right)\psi(x) .$
From $(1):$
$\displaystyle p~\psi(x) =p\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\phi(k')~e^{ik'x}~dk' =p\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\phi(k')~\frac{1}{ik'}\frac{\partial}{\partial x}e^{ik'x}~dk' .$
$\displaystyle =-ip\frac{\partial}{\partial x} \left(\int_{-\infty}^{+\infty}\frac{1}{k'}~\frac{1}{\sqrt{2\pi}}\phi(k')~e^{ik'x}~dk'\right) =-ip\frac{\partial}{\partial x} \left(\cancelto{0}{\Big[\frac{1}{k'}\psi(x)\Big]_{-\infty}^{+\infty}} -\int_{-\infty}^{+\infty}\psi(x)\frac{-1}{k'^2}dk' \right) =-ip\frac{\partial}{\partial x}\psi(x)\int_{-\infty}^{+\infty}\frac{1}{k'^2}dk' =\ldots .$
$\displaystyle -i\hbar\frac{\partial}{\partial x}\psi(x) =-i\hbar\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\phi(k)~ik~e^{ikx}~dk =-i\hbar\left( \int_{-\infty}^{+\infty}ik~\left(\frac{1}{\sqrt{2\pi}}\phi(k)~e^{ikx}\right)~dk \right) .$
$\displaystyle =-i\hbar\left( \cancelto{0}{\Big[ik~\psi(x)\Big]_{-\infty}^{+\infty}} -\int_{-\infty}^{+\infty}\psi(x)~i\cancelto{1}{\frac{dk}{dk}}~\frac{dp}{\hbar} \right) =-\psi(x)\int_{-\infty}^{+\infty}dp .$
??
$\displaystyle \frac{\partial\psi(x)}{\partial x} =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\phi(k)\frac{\partial}{\partial x}e^{ikx}~dk =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\phi(k)~ik~e^{ikx}~dk$
$\displaystyle =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\psi(y)~e^{-iky}~dy\right)ik~e^{ikx}~dk$
$\displaystyle =\int_{-\infty}^{+\infty}\psi(y)\left(\int_{-\infty}^{+\infty}ik~\frac{1}{2\pi}~e^{ik(x-y)}~dk\right)dy$
$\displaystyle ?=\int_{-\infty}^{+\infty}\psi(y)~ik~\delta(x-y)dy$
$\displaystyle =ik~\psi(x) .$
$\displaystyle \psi(x) =\frac{1}{ik}\frac{\partial\psi(x)}{\partial x} =\frac{-i\hbar}{p}\frac{\partial\psi(x)}{\partial x}.~~ p~\psi(x)=\left(-i\hbar\frac{\partial}{\partial x}\right)\psi(x) .$
To do it with the continuous basis:
$\displaystyle \Bra{\psi}:\Ket\zeta:\mapsto\Braket{\psi\Verti\zeta}\equiv\int\psi(q)^*\zeta(q)~dq .$
Let $\displaystyle\zeta(x) =-i\hbar\frac{\partial}{\partial x}\psi(x).~$
?? As stated before, $\psi(x)$ is an eigenvector of operator $\frac{\partial}{\partial x}$ with eigenvalue $ik,~ \zeta(x)=-i\hbar~ik~\psi(x)=p~\psi(x) .$
$\displaystyle \because\zeta(x)=\left(-i\hbar\frac{\partial}{\partial x}\right)\Ket\psi=p\Ket\psi,~ \therefore\HAT p=-i\hbar\frac{\partial}{\partial x}$
$\displaystyle \Braket{p} =\Braket{\psi\Verti\HAT p\Verti\psi} =\int_{-\infty}^{+\infty}\psi(x)^*\left(-i\hbar\frac{\partial}{\partial x}\right)\psi(x)~dx .$
Schrödinger Equation: $\displaystyle i\hbar\frac{\partial\psi}{\partial t} =-\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}+V\psi .$
$\displaystyle \frac{\partial\psi}{\partial t} =\frac{i\hbar}{2m}\frac{\partial^2\psi}{\partial x^2}-\frac{i}{\hbar}V\psi .$
$\displaystyle \frac{\partial\psi^*}{\partial t} =-\frac{i\hbar}{2m}\frac{\partial^2\psi^*}{\partial x^2}+\frac{i}{\hbar}V\psi^* .$
An interesting relation:
$\displaystyle \frac{\partial}{\partial t}\left(\psi^*\psi\right) =\psi^*\frac{\partial\psi}{\partial t}+\psi\frac{\partial\psi^*}{\partial t} =\psi^*\left(\frac{i\hbar}{2m}\frac{\partial^2\psi}{\partial x^2}\cancel{-\frac{i}{\hbar}V\psi}\right) +\psi\left(-\frac{i\hbar}{2m}\frac{\partial^2\psi^*}{\partial x^2}\cancel{+\frac{i}{\hbar}V\psi^*}\right)\\ \displaystyle =\frac{i\hbar}{2m}\left(\psi^*\frac{\partial^2\psi}{\partial x^2}+\frac{\partial\psi^*}{\partial x}\frac{\partial\psi}{\partial x}\right) -\left(\psi\frac{\partial^2\psi^*}{\partial x^2}+\frac{\partial\psi}{\partial x}\frac{\partial\psi^*}{\partial x}\right)\\ \displaystyle =\frac{i\hbar}{2m}\frac{\partial}{\partial x} \left(\psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x}\right) .$
$\displaystyle \int\frac{\partial}{\partial t}\left(\psi^*\psi\right)dx =\frac{i\hbar}{2m}\left(\psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x}\right) .$
A side-benefit of the above relation is the proof of normalisation being time invariance - Once normalised, forever normalised.
$\displaystyle \frac{\partial}{\partial t}\int_{-\infty}^{+\infty}\lvert\psi\rvert^2dx =\int_{-\infty}^{+\infty}\frac{\partial}{\partial t}\left(\psi^*\psi\right)dx \displaystyle =\frac{i\hbar}{2m}\int_{-\infty}^{+\infty}\frac{\partial}{\partial x} \left(\psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x}\right)dx =\frac{i\hbar}{2m}\left[\psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x}\right]_{-\infty}^{+\infty} =0,~$
as $\psi$ and $\psi^*$ are both square integrable and therefore must vanish at $-\infty$ and $+\infty.$
Review of integral by parts: $\displaystyle \int_a^budv=\Big[uv\Big]_a^b-\int_a^bvdu .~$ Alternatively, $\displaystyle \int_a^bu\left(\frac{dv}{dx}\right)dx =\Big[u\left(\int\frac{dv}{dx}dx\right)\Big]_a^b -\int_a^b\frac{du}{dx}\left(\int\frac{dv}{dx}dx\right)dx .~$
i.e. First term is to "take $u$ out", and the second term is to differentiate $u$ and integrate it again.
In the previous section, we "labeled" the eigenvalue of position operator $\HAT x$ with its measurement $x$. With momentum, we need to derive the operator $\HAT p$, so that $\Braket{\HAT p}$ is related to $\Braket{\HAT x}$ according to physical law.
To simplify, let us make $\varphi$ implicit.
$\displaystyle \frac{\partial}{\partial t}\Braket{\HAT x} =\frac{\partial}{\partial t}\int_{-\infty}^{+\infty}x~\psi^*\psi~dx =\int_{-\infty}^{+\infty}x~\frac{\partial}{\partial t}\left(\psi^*\psi\right)dx =\frac{i\hbar}{2m}\int_{-\infty}^{+\infty}x~\frac{\partial}{\partial x} \left(\psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x}\right)dx\\ \displaystyle =\cancelto{0}{\frac{i\hbar}{2m}\left[x\left(\psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x}\right)\right]_{-\infty}^{+\infty}} -\frac{i\hbar}{2m}\int_{-\infty}^{+\infty}\cancelto{1}{\frac{\partial x}{\partial x}}\left(\psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x}\right)dx .$
The first term approaching $0$ is questionable, with the bracket approaching $0$ but $x$ approaching $\pm\infty$.
$\displaystyle \int_{-\infty}^{+\infty}\psi\frac{\partial\psi^*}{\partial x}dx =\cancelto{0}{\left[\psi\psi^*\right]_{-\infty}^{+\infty}} -\int_{-\infty}^{+\infty}\psi^*\frac{\partial\psi}{\partial x}dx .$
$\displaystyle \frac{\partial}{\partial t}\Braket{\HAT x} =\frac{-i\hbar}{2m}\int_{-\infty}^{+\infty}2\psi^*\frac{\partial\psi}{\partial x}dx .$
$\displaystyle \Braket{\HAT p} =m\frac{\partial}{\partial t}\Braket{\HAT x} =\int_{-\infty}^{+\infty}\psi^*\frac{\hbar}{i}\frac{\partial}{\partial x}~\psi~dx .$
i.e. $\displaystyle\HAT p=\frac{\hbar}{i}\frac{\partial}{\partial x}.$
*** WORKING ***
$=\frac{i\hbar}{2m}\int_{-\infty}^{+\infty}x\frac{\partial}{\partial x} \left(\psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x}\right)dx .$
$=\left[x\int\frac{\partial}{\partial t}\left(\psi^*\psi\right)dx\right]_{-\infty}^{+\infty} -\int_{-\infty}^{+\infty}\cancelto{1}{\frac{\partial x}{\partial x}} \left(\int\frac{\partial}{\partial t}\left(\psi^*\psi\right)dx\right)dx\\ \displaystyle =\left[x\frac{i\hbar}{2m}\left(\psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x}\right)\right]_{-\infty}^{+\infty} -\int_{-\infty}^{+\infty}\frac{i\hbar}{2m}\left(\psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x}\right)dx .$
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