$\require{cancel} \newcommand{\Ket}[1]{\left|{#1}\right\rangle} \newcommand{\Bra}[1]{\left\langle{#1}\right|} \newcommand{\Braket}[1]{\left\langle{#1}\right\rangle} \newcommand{\Rsr}[1]{\frac{1}{\sqrt{#1}}} \newcommand{\RSR}[1]{1/\sqrt{#1}} \newcommand{\Verti}{\rvert} \newcommand{\HAT}[1]{\hat{\,#1~}} \DeclareMathOperator{\Tr}{Tr}$

Fourier Transform as Change of Basis

^{First created in March 2019}

Ref:

• https://www.youtube.com/watch?v=myxx2uaqPLM

Background

---

Let us define a Hilbert Space by $\displaystyle\Ket\psi=\int dx~\psi(p)~,$ with inner product $\Braket{\phi\Verti\psi}=\int\phi(p)^*\psi(p)~dp.$

$\Bra\phi$ is a transformation from $\int dp~\psi(p)$ to $\int\phi(p)^*\psi(p)~dp\in\mathbb{C}.$

$\big|\Ket\psi\big|^2 =\Braket{\psi\Verti\psi} =\int\psi(p)^*\psi(p)~dp=\int\lvert\psi(p)\rvert^2~dp\in\mathbb{R}$ and is positive-definite.

---

Basis $\{\Ket\varphi\}$ is defined as $\displaystyle\Ket{\varphi_\alpha}=\int dp~\varphi_\alpha\in\mathbb{F},$ where $\varphi_\alpha(p)\equiv\delta(p-\alpha) =\lim_{\epsilon\rightarrow 0}\frac{1}{\epsilon\sqrt{\pi}}e^{-((p-\alpha)/\epsilon)^2}$ (Dirac Delta function).

Commonly written as $\displaystyle\Ket{\varphi}=\int dp~\delta(p-\varphi).$

Fourier Transform

---

Let us look at the Fourier Transform between position $x$ and momentum $k$.

Firstly, what is the basis for position? i.e. what does $\Ket x=\int dp~x$ mean?

Here $x$ is a probability density function $P_x$ and $\displaystyle\int_{x_0-\epsilon}^{x_0+\epsilon}P_x(x)~dx$ is the probability that the particle is found to be within the neighbourhood of $x_0$.

---

Let us define $\displaystyle\Ket{x_0}=\int_{-\infty}^{+\infty}\delta(x-x_0)~dx.$

---

Continuous Form:

$\psi(x)=\frac{1}{\sqrt{2\pi}}\int\phi(k)~e^{ikx}dk~,~$ and

$\phi(k)=\frac{1}{\sqrt{2\pi}}\int\psi(x)~e^{-ikx}dx~.$

---

Discrete Form:

$\displaystyle\psi_n=\Rsr{N}\sum_{k=0}^{N-1}\phi_k~e^{+ink2\pi/N}~,~$ and

$\displaystyle\phi_k=\Rsr{N}\sum_{n=0}^{N-1}\psi_n~e^{-ink2\pi/N}~.~$

Notes

Given quantum state $\Ket\psi$ and the state of the particle at $x$ is $\Ket x$, the amplitude of finding the particle at $x$ is $\Braket{x\Verti\psi}=\psi(x).$

If the state of the particle having momentum $\hbar k$ is $\Ket k$, the amplitude is $\Braket{k\Verti\psi}=\phi(k).$

So $\Braket{x\Verti k}$ is the amplitude of knowing that the momentum is exactly $\hbar k$, what is the amplitude of finding the particle at $x$.

$\displaystyle\Braket{x\Verti k}=\frac{1}{\sqrt{2\pi}}e^{ikx}~,~$ and $\displaystyle\Braket{k\Verti x}=\frac{1}{\sqrt{2\pi}}e^{-ikx}.$

---

More rigorously, given quantum state $\Ket\psi$ and the state of the particle at $x$ is $\Ket x$, the amplitude of finding the particle at $x$ is $\Braket{x\Verti\psi}=\psi(x).$

If the state of the particle having momentum $\hbar k$ is $\Ket k$, the amplitude is $\Braket{k\Verti\psi}=\phi(k).$

---

Suppose we know the exact position of $x$ being at $x_0$, from the above, the $k$ distribution is

$\displaystyle F(k) =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\delta(x-x_0)~e^{-ikx}~dx =\frac{1}{\sqrt{2\pi}}e^{-ikx_0} .$ This means $k$ is completely undetermined.

We write $\Braket{k\Verti x_0}=\frac{1}{\sqrt{2\pi}}e^{-ikx_0}.$

---

Suppose we know the exact momentum of $p$ being $p_0=\hbar k_0$, the $x$ distribution is

$\displaystyle f(x) =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\delta(k-k_0)~e^{ikx}~dk =\frac{1}{\sqrt{2\pi}}e^{ik_0x} .$ This means $x$ is completely undetermined.

We write $\Braket{x\Verti k_0}=\frac{1}{\sqrt{2\pi}}e^{ik_0x}.$

---

Generally, $\Braket{x\Verti k}$ is the amplitude of finding the particle at $x$, knowing that the momentum is $\hbar k$ (removing the word exactly).

$\displaystyle\Braket{x\Verti k}=\frac{1}{\sqrt{2\pi}}e^{ikx}~,~$ and $\displaystyle\Braket{k\Verti x}=\frac{1}{\sqrt{2\pi}}e^{-ikx}.~$ They are multually conjugated, which is consistent with notation.

Change of Basis

Let us define two bases: $\{\Ket i,\Ket j\}$ and $\{\Ket u,\Ket v\}$. For example, $\{\Ket 0,\Ket 1\}$ and $\{\Ket +,\Ket -\}$.

The identity operator $\HAT I=\Ket i\Bra i+\Ket j\Bra j=\Ket u\Bra u+\Ket v\Bra v.$ i.e. $\HAT I=\sum_k\Ket k\Bra k.$

---

For continuous case, $\displaystyle\HAT I=\int\Ket k\Bra kdk.$

---

$\displaystyle\psi(x) =\Braket{x\Verti\psi} =\Braket{x\big\lvert\HAT I\big\rvert\psi} =\Bra{x}\left(\int\Ket k\Bra kdk\right)\Ket\psi =\left(\int\Braket{x\Verti k}\Braket{k\Verti\psi}dk\right) =\frac{1}{\sqrt{2\pi}}\int e^{ikx}\phi(k)~dk .$

---

$\displaystyle\phi(k) =\Braket{k\Verti\psi} =\Braket{k\big\lvert\HAT I\big\rvert\psi} =\Bra{k}\left(\int\Ket x\Bra xdx\right)\Ket\psi =\left(\int\Braket{k\Verti x}\Braket{x\Verti\psi}dx\right) =\frac{1}{\sqrt{2\pi}}\int e^{-ikx}\psi(x)~dx .$

---

KEPT

---

To prove, substitute $\phi_k$ into the first equation.

$\displaystyle\Rsr{N}\sum_{k=0}^{N-1}\phi_k~e^{+ink2\pi/N} =\Rsr{N}\sum_{k=0}^{N-1}\left(\Rsr{N}\sum_{m=0}^{N-1}\psi_m~e^{-imk2\pi/N}\right)~e^{+ink2\pi/N} =\frac{1}{N}\sum_{m=0}^{N-1}\psi_m\sum_{k=0}^{N-1}e^{-imk2\pi/N}~e^{+ink2\pi/N} =\frac{1}{N}\sum_{n=0}^{N-1}\psi_n~N =\psi_n .$