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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
Note: Here we use $\varphi$ instead of $x$ to avoid confusion between the amplitude $\varphi$ and the position measurement $x$. It is different from the dummy index $j=1,2,\ldots,n.$ Although $\varphi$ has the same range, it is not used as a free index.
The basis vector for a "sharply" defined discrete measurement is $\Ket{\varphi}=[0,0,\ldots,1,\ldots,0]^T.$
i.e. $\Ket{\varphi} =[\delta_{\varphi 1},\delta_{\varphi 2},\ldots,\delta_{\varphi n}]^T =[\delta_{\varphi j}]^T.$
More generally, a state vector $\Ket\psi$ is not "sharply" defined but in a superposition.
$\displaystyle\Ket\psi=\sum_{j=1}^n a_j\Ket{j}$, where $\displaystyle a_j\in\mathbb{C}\text{ and }\sum_{j=1}^n\lvert a_j\rvert^2=1.$
We denote $a_j$ as $\psi_j\in\mathbb{C},~$ so $\boxed{\displaystyle\Ket\psi=\sum_{j=1}^n\psi_j\Ket{j}}$, the superposition form.
i.e. $\displaystyle\boxed{\Ket\psi=[\psi_1,\psi_2,\ldots,\psi_n]^T,\text{ where } \psi_j\in\mathbb C.}$
Its dual space $\Bra\psi$ is therefore defined as $\Bra\psi\equiv[\psi_1^*,\psi_2^*,\ldots,\psi_n^*].$
The inner product $\displaystyle \Braket{\psi\Verti\psi} =[\psi_1^*,\psi_2^*,\ldots,\psi_n^*]~[\psi_1,\psi_2,\ldots,\psi_n]^T =\sum_{j=1}^n\psi_j^*\psi_j =\sum_{j=1}^n\lvert\psi_j\rvert^2 =1.$
As a result, $\Bra{\varphi}=[\delta_{\varphi j}^*]=[\delta_{\varphi j}]$ makes $\Braket{\varphi\Verti\psi}=\psi_\varphi.$
$\Ket{\varphi}\Braket{\varphi\Verti\psi}=\psi_\varphi\Ket{\varphi}.~$ i.e. $\Ket{\varphi}\Bra{\varphi}$ projects $\Ket\psi$ onto $\Ket{\varphi}$.
An observable can be expressed as $\displaystyle \HAT x= \begin{bmatrix} x_1&0&\ldots&0\\ 0&x_2&\ldots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\ldots&x_n \end{bmatrix}.$
i.e. $\HAT x=\sum_{j=1}^nx_j\Ket{j}\Bra{j}.~$
Please note that it is not a density matrix, although its form is similar to $\sum_{k=1}^nP_k\Ket{\psi_k}\Bra{\psi_k}.$ With a density matrix, $\Ket{\psi_k}$ is the state of particle $k$ in a mixed state and $P_k$ is the probability distribution of $\psi_k$ in the ensemble, while with the observable, $\Ket{j}$ is an orthonormal basis vector and $x_j$ is a measurement with a physical unit.
$\Braket{\HAT x}_\psi \equiv\Braket{\psi\big|\HAT x\big|\psi} =\Braket{\psi\big|\left(\sum_{j=1}^nx_j\Ket{j}\Bra{j}\right)\big|\psi} =\sum_{j=1}^nx_j\Braket{\psi\Verti j}\Braket{j\Verti\psi} =\sum_{j=1}^nx_j\psi_j^*\psi_j =\sum_{j=1}^nx_j\lvert\psi_j\rvert^2 ,$ which is called the expectation value of $\HAT x$ when measuring a system of state $\Ket\psi$.
Again, although the form is similar, the following is not related to expectation value:
$\Braket{\psi\Verti\psi} =\Braket{\psi\Big|\HAT I\Big|\psi} =\Braket{\psi\big|\left(\sum_{j=1}^n\Ket{j}\Bra{j}\right)\big|\psi} =\sum_{j=1}^n\Braket{\psi\Verti j}\Braket{j\Verti\psi} =\sum_{j=1}^n\psi_j^*\psi_j =\sum_{j=1}^n\lvert\psi_j\rvert^2 =1 .$
But it makes sense to see how the probability distribution (above) relates to the expectation value, which is simply possible measurements time their corresponding probability.
Let us track back the discrete relations and write down their continuous counterparts. Here we use $q$ as the dummy variable.
The "sharply" defined basis vector has infinite number of components: $\Ket{\varphi}\approx~_{-\infty}[\ldots,0,1,0,\ldots]_{+\infty}^T.$
Let us use $q\in\mathbb R$ as the index instead of a discrete one. The infinitely tall column is now a distribution: $\varphi:q\mapsto\varphi(q).$
The function $\varphi(q)$ is in fact $\delta(\varphi-q)$. If we define $\delta_\varphi(q)\equiv\delta(\varphi-q)$, it effectively means $\varphi=\delta_\varphi.~$ i.e. $\boxed{\varphi:q\mapsto\delta(\varphi-q).}$
For a general state $\Ket\psi\approx[\psi(-\infty),\ldots,\psi(q),\ldots,\psi(+\infty)]^T,$ we can simply express it as $\boxed{\Ket\psi:q\mapsto\psi(q).}$
Please note $\psi(q)\in\mathbb C$ is an amplitude, not a probability distribution.
The dual vector $\Bra\phi\approx[\phi(-\infty)^*,\ldots,\phi(q)^*,\ldots,\phi(+\infty)^*]$ cannot be defined as $\phi(q)^*$, as it would mean $\Ket{\phi^*}$.
Therefore $\Bra\phi$ needs to be expressed as a transformation from $\Ket\psi$ to $\displaystyle\sum_{q=-\infty}^{+\infty}\phi(q)^*\psi(q).$
Formally, $\displaystyle\Bra\phi:~\Ket\psi\mapsto\Braket{\phi\Verti\psi}\equiv\int\phi(q)^*~\psi(q)~dq .~$
$\displaystyle\Bra\varphi:~\Ket\psi\mapsto\int\delta(\varphi-q)~\psi(q)~dq=\psi(\varphi),~ \therefore\boxed{\Braket{\varphi\Verti\psi}=\psi(\varphi)},~$ which is the component of $\Ket\psi$ on $\Ket\varphi.$
(A reminder: $\varphi$ is a basis vector and $q$ is dummy, and $\delta(q-\varphi)=\delta(\varphi-q)\in\mathbb R$ so its conjugate equals to itself.)
The normalisation condition is therefore $\displaystyle \Braket{\psi\Verti\psi} =\int_{-\infty}^{+\infty}\psi(q)^*\psi(q)~dq =\int_{-\infty}^{+\infty}\lvert\psi(q)\rvert^2dq =1.$
This condition can be satisfied if $\psi$ is a square integrable function, which means $\int_{-\infty}^{+\infty}\lvert\psi(q)\rvert^2dq$ is finite and therefore normalisable.
This also means that $\lim_{q\to\pm\infty}\psi(q)=0.$
Sometimes the function notation $\psi(q)$, the vector $\Ket\psi$, and the superposition $\left(\text{loosely expressed as } \int_{-\infty}^{+\infty}\psi(\varphi)\Ket\varphi d\varphi\right)$ are used interchangably.
The infinitely tall column vector and the mapping of a function are conceptually the same. They both capture the idea of superposition on standard basis. For non-standard basis, the superposition would require more complex integrals.
To project $\Ket\psi$ onto basis vector $\Ket\varphi$ means $\Ket\varphi\Braket{\varphi\Verti\psi}=\psi(\varphi)\Ket\varphi.$ Given $\Ket\varphi=\delta(\varphi-q),~~\psi(\varphi)\Ket\varphi:q\mapsto\psi(\varphi)~\delta(\varphi-q).$
This is not a unit length vector as $\int_{-\infty}^{+\infty}\lvert\psi(\varphi)\delta(\varphi-q)\rvert^2dq\le 1.$
In terms of measurement, if we have an apparatus that clicks if $\Ket\psi$ is aligned with $\Ket\varphi$, the probability it clicks is $\lvert\psi(\varphi)\rvert^2$. (The posterior state after collapsing is $\Ket\varphi$.)
$\big|\Ket\varphi\Braket{\varphi\Verti\psi}\big|^2 =\left(\Ket\varphi\Braket{\varphi\Verti\psi}\right)^* \left(\Ket\varphi\Braket{\varphi\Verti\psi}\right) =\Braket{\psi\Verti\varphi}\Braket{\varphi\Verti\varphi}\Braket{\varphi\Verti\psi} =\psi(\varphi)^*\psi(\varphi) =\lvert\psi(\varphi)\rvert^2.$
Still have troubles with these but proceed anyway: $\Braket{\varphi\Verti\varphi} =\int_{-\infty}^{+\infty}\lvert\delta(\varphi-q)\rvert^2~dq =1,$ or $\Braket{\varphi\Verti\varphi}\to+\infty\text{ as}\Braket{q\Verti\varphi}=\delta(q-\varphi)$? Would it be that $\Braket{\varphi\Verti\varphi}$ being a probability density and it needs to be considered within a domain $\Braket{\varphi\Verti\varphi}dx\to 1.$
If we align all such $\Ket\varphi\Bra\varphi$, each assigned an "eigenvalue" as a label (explained later), we have an observable
$\displaystyle \HAT x \approx\sum_{q=-\infty}^{+\infty}x_q\Ket{\varphi_q}\Bra{\varphi_q} \approx\int_{-\infty}^{+\infty}x(\varphi)\Ket{\varphi}\Bra{\varphi}~d\varphi .$
$\displaystyle \Braket{\HAT x}_\psi \equiv\Braket{\psi\big|\HAT x\big|\psi} =\Bra\psi\left(\int_{-\infty}^{+\infty}x(\varphi)\Ket{\varphi}\Bra{\varphi}~d\varphi\right)\Ket\psi$
$\displaystyle =\int_{-\infty}^{+\infty}x(\varphi)\Braket{\psi\Verti\varphi}\Braket{\varphi\Verti\psi}~d\varphi =\int_{-\infty}^{+\infty}x(\varphi)~\psi(\varphi)^*~\psi(\varphi)~d\varphi =\int_{-\infty}^{+\infty}x(\varphi)\lvert\psi(\varphi)\rvert^2~d\varphi .$
Let us define projector $P_\varphi\equiv\Ket\varphi\Bra\varphi,~$ the eigenvector is obviously $\Ket\varphi$ with eigenvalue $1$, as $P_\varphi\Ket\varphi=\Ket\varphi\Braket{\varphi\Verti\varphi}=\Ket\varphi.$
If we attach a physical "label" to the projector, i.e. $x_\varphi P_\varphi\Ket\varphi=x_\varphi \Ket\varphi,~$ the operator $\left(x_\varphi P_\varphi\right)$ has an eigenvector $\Ket\varphi$ with eigenvalue $x_\varphi$.
If we lined up many detectors, each of which would give a click when a particular measurement is detected, we would know the "label". Such apparatus effectively becomes a measuring device.
In practice, all devices are discrete, just the level of granularity.
$\displaystyle \HAT x =\sum_{q=-\infty}^{+\infty}x_qP_q =\sum_{q=-\infty}^{+\infty}x_q\Ket q\Bra q =\begin{bmatrix} x_{-\infty}&0&\ldots&\ldots&0\\ 0&\ddots&\ldots&\ldots&\vdots\\ \vdots&\vdots&x_q&\ldots&\vdots\\ \vdots&\vdots&\vdots&\ddots&0\\ 0&\ldots&\ldots&0&x_{+\infty} \end{bmatrix} .$
We have $\HAT x\Ket q=x_q\Ket q,~$ so $\HAT x$ has infinite number of eigenvectors $\Ket q$ each with eigenvalue $x_q$.
The conceptual continuous observable is therefore $\displaystyle \HAT x =\int_{-\infty}^{+\infty}x(q)\Ket{q}\Bra{q}~dq .$
Again, $\HAT x\Ket\varphi =x(\varphi)\Ket\varphi =\int_{-\infty}^{+\infty}x(q)\Ket{q}\Braket{q\Verti\varphi}~dq =\int_{-\infty}^{+\infty}x(q)\Ket{q}\delta(q-\varphi)~dq =x(\varphi)\Ket\varphi .$
That means $\HAT x$ has infinite number of eigenvectors $\Ket\varphi$ each with eigenvalue $x(\varphi)$. For short, $x(\varphi)$ is simplified to just $x$, indicating the position measurement corresponding to $\Ket\varphi$.
$\boxed{\HAT x\Ket\varphi=x\Ket\varphi.}$
Similarly, $\HAT x\Ket\varphi =x(\varphi)\Ket\varphi =\int_{-\infty}^{+\infty}x(q)\Ket{q}\Braket{q\Verti\varphi}~dq =\int_{-\infty}^{+\infty}x(q)\Ket{q}\delta(q-\varphi)~dq =x(\varphi)\Ket\varphi .$
That means $\HAT x$ has infinite number of eigenvectors $\Ket\varphi$ each with eigenvalue $x(\varphi)$.
Based on the (loose) superposition form $\displaystyle \Ket\psi=\int_{-\infty}^{+\infty}\psi(\varphi)\Ket{\varphi}~d\varphi .$
$\displaystyle \HAT x\Ket\psi =\int_{-\infty}^{+\infty}\psi(\varphi)\HAT x\Ket{\varphi}~d\varphi =\int_{-\infty}^{+\infty}\psi(\varphi)~x(\varphi)\Ket{\varphi}~d\varphi =\int_{-\infty}^{+\infty}x(\varphi)~\psi(\varphi)\Ket{\varphi}~d\varphi .$
However, $\HAT x\Ket\psi$ by itself does not mean much. It needs to be put into the context of an expectation value via a "bra transformation".
$\displaystyle \Bra\psi:\HAT x\Ket\psi\mapsto\int_{-\infty}^{+\infty}\psi(\varphi)^*x(\varphi)~\psi(\varphi)~d\varphi .$
Recall $\displaystyle \Braket{\HAT x}_\psi \equiv\Braket{\psi\big|\HAT x\big|\psi} =\int_{-\infty}^{+\infty}\psi(\varphi)^*x(\varphi)~\psi(\varphi)~d\varphi =\int_{-\infty}^{+\infty}x(\varphi)\lvert\psi(\varphi)\rvert^2~d\varphi ,~$ same conclusion as before.
Based on the (loose) superposition form $\displaystyle \Ket\psi=\int_{-\infty}^{+\infty}\psi(\varphi)\Ket{\varphi}~d\varphi .$
Recall $\displaystyle \Braket{\HAT x}_\psi \equiv\Braket{\psi\big|\HAT x\big|\psi} =\Bra\psi\left(\HAT x\Ket\psi\right) =\Bra\psi\left(\int_{-\infty}^{+\infty}\psi(\varphi)x(\varphi)\Ket{\varphi}~d\varphi\right)$
$\displaystyle =\int_{-\infty}^{+\infty}x(\varphi)~\psi(\varphi)^*~\psi(\varphi)~d\varphi =\int_{-\infty}^{+\infty}x(\varphi)\lvert\psi(\varphi)\rvert^2~d\varphi ,~$ same conclusion as before.
*** WORKING ***
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.
$\displaystyle \frac{\partial}{\partial t}\Braket{\HAT x} =\frac{\partial}{\partial t}\int_{-\infty}^{+\infty}x~\psi^*\psi~dx .$
Schrodinger Equation: $\displaystyle i\hbar\frac{\partial\psi}{\partial t} =-\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}+V\psi .$
Let us discuss $\displaystyle~~ \eta(\xi,\epsilon)=\frac{1}{\epsilon\sqrt{\pi}}e^{-(\xi/\epsilon)^2}.~~ \int_{-\infty}^{+\infty}\eta(\xi,\epsilon)~d\xi =\frac{1}{\sqrt\pi}\int_{-\infty}^{+\infty}e^{-\xi'^2}~d\xi' =\frac{1}{\sqrt\pi}\cdot\sqrt\pi =1.$
This result is independent of $\epsilon$. Therefore, $\delta(\xi)=\lim_{\epsilon\leftarrow 0}\eta(\xi,\epsilon),~ \int_{-\infty}^{+\infty}\delta(\xi)~d\xi=1.$
An observable can be expressed as $\displaystyle \HAT x= \begin{bmatrix} x_1&0&\ldots&0\\ 0&x_2&\ldots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\ldots&x_n \end{bmatrix}.$
i.e. $\HAT x=\sum_{j=1}^nx_j\Ket{j}\Bra{j}.~$
Please note that it is not a density matrix, although its form is similar to $\sum_{k=1}^nP_k\Ket{\psi_k}\Bra{\psi_k}.$ With a density matrix, $\Ket{\psi_k}$ is the state of particle $k$ in a mixed state and $P_k$ is the probability distribution of $\psi_k$ in the ensemble, while with the observable, $\Ket{j}$ is an orthonormal basis vector and $x_j$ is a measurement with a physical unit.
Now if we want $\lvert\Ket\psi\rvert^2=\int\lvert\psi(p)\rvert^2dp=1,~$ we can set $\lvert\psi(p)\rvert^2=\eta(p,\epsilon).$
The Probability Density Function (PDF) of Normal Distribution $\displaystyle\mathcal{N}(\mu,\sigma^2)$ is $\displaystyle\frac{1}{\sqrt{2\pi\sigma^2}}~\exp\left[-\frac{(x-\mu)^2}{2\sigma^2}\right],$ which also integrates to unity.
That means $\eta(x-\mu,\sqrt{2\sigma^2})\in\mathcal{N}(\mu,\sigma^2),$ being a normal distribution centred on $\mu$ with standard deviation of $\epsilon^2/2.$
If $\psi(x)=\sqrt{\eta}~e^{-ikx},$ we can define $\Ket{\psi}=\int dx~\psi(x),$ with $\int\lvert\Ket{\psi}\rvert^2dx =\int\eta dx =1.$
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