Q/A: What is your favorite number?

What is your favorite number?
The short answer is: $3435$. You may ask: "why?" since this number seems, à priori, pretty boring. But it turns out that it is very interesting: it is the only natural number (along with $1$) which have the following property: $$\begin{align}\color{blue}3\color{red}4\color{green}35&=\color{blue}3^{\color{blue}3}+\color{red}4^{\color{red}4}+\color{green}3^{\color{green}3}+5^5\\&= \color{blue}{27}+\color{red}{256}+\color{green}{27}+3125\\ &={3435}.\end{align}$$ I also like Mills' constant  which is the smallest number $\rm A$ such that $\lfloor \rm A^{3^n} \rfloor$ is a prime number for every $n\in\mathbb N$. Its value is approximately equal to $1.306$ and the primes generated by Mills' constant are known as Mills primes; if the Riemann hypothesis is true, the sequence begins: $$2, 11, 1361, 2521008887, \ldots $$ (sequence A051254 in OEIS).
And what about you? What is your favorite number, and why?

Q/A: What is the math behind the statistical interpretation of Quantum Mechanics?

Srinivisa: What is the math behind the statistical interpretation of Quantum Mechanics?
Physics is all about motion, start with a system $\rm S$ composed of a particle of mass $m$, moving along an axis, let it be subject to a known force, put some physical laws out there, mix well and BAM: $x(t)$ will determine for you the position of the particle at any time! Once you know that, you can find its velocity, its momentum... and a lot of other useful stuff. Of course, this is just under the framework of classical mechanics.
Quantum Mechanics is very different since we can't have any function that determines with absolute certainty the position of a particle at a given time, and thus we can't determine the velocity of the particle without being uncertain. In fact, by Heisenberg's uncertainty principle, the more you know about a particle's position the less you know about its velocity, and the opposite is true. So instead of the well-defined $x(t)$, Quantum Mechanics uses what we call a wavefunction.
As the double-slit experiment shows, light can behave like a particle or like a wave. In fact, even electrons display this same behavior with that famous experiment, which turns out to be very useful, since it will help solving the following problem. If we think of electrons as individual particles orbiting the nucleus of an atom, like the planets in our solar system orbiting our sun, then we get to a serious problem. Indeed, to a very serious one. Since the negatively charged electron is attracted by the positively charged nucleus by the electromagnetic force, then the electron will be continuously accelerating, and would thus radiate away its energy and fall into the nucleus.

That's why quantum mechanics came and proposed that we shouldn't think of the electron not just as a particle, but also as a wave. And this wave is basically described by the wavefunction, just like a 'classical' particle is described by the $x(t)$ function. We get the wavefunction of a particle (denoted as $\psi(x,t)$ or as $\Psi(x,t)$) by solving Schrödinger's equation: $$\underset{\textit{The time dependent Schrödinger equation.}}{\boxed{\displaystyle\,\,
i\hbar\dfrac{\partial \Psi}{\partial t} =
-\dfrac{\hbar^2}{2m}\dfrac{\partial^2\Psi}{\partial x^2}+\mathrm{V}\Psi.\,\,}}$$ The wavefunction is mathematically expressed as: $$\Psi(x,t)=\mathrm{A}e^{\displaystyle i(kx-\omega t)}$$ where $\mathrm{A}$ is the amplitude of the wave, $e$ is a constant which is approximately equal to $2.71$, $i$ is the square root of minus $1$, $k$ is the momentum divided by $\hbar$ which is the same as $h/2\pi$ where $h$ is Planck's constant, and finally $\omega$ denotes the frequency of the wave times $2\pi$.
But isn't a particle, by its nature, located at a single and unique point, whereas a wave is spread out in space? How can we make sense of such object? It's in an attempt to answer those questions that the Born interpretation (or the statistical interpretation) of the wave function was born. The latter propose that $|\Psi(x,t)|^2$ tell us the probability of finding the particle at point $x$, at time $t$. More precisely: $$\int_a^b |\Psi(x,t)|^2\,\mathrm dx=\left\{
\text{probability of finding the particle} \\
\text{between $a$ and $b$, at time $t$.}\\
\right\}.$$ Note: The wavefunction itself is complex, but $|\Psi|^2$ is real and nonnegative. 

Here is a plot of wavefunctions of identical particles, have fun!

Wave Functions of Identical Particles from the Wolfram Demonstrations Project by Michael Trott -
If it doesn't work then install Wolfram CDF Player, it's free!

And as alway thanks for reading!
Best wishes, $\mathcal H$akim.

Do Photons have Mass?

Do Photons have Mass?

As an admin of the Quantum Physics facebook group, I can assure you with great certainty that this is the most popular question we get. In this article, we shall investigate this question to provide a reasonable answer.

One of the ways some people attack this question is by arguing that since gravitational attraction is caused by mass, as Newton's equation shows: $$\mathbf{F}=G\dfrac{m_1\cdot m_2}{r^2}$$ and that since light is bent due to gravity, then light must have mass. But that's actually false, Newton's equation is only an approximation and so: $$\mathbf{F}\approx G\dfrac{m_1\cdot m_2}{r^2}.$$ Furthermore, the cause of gravity is not mass but energy and momentum. And so this argument is false.

From special relativity: $$E^2=(mc^2)^2+(pc)^2$$  which can be written as: $$E^2-(pc)^2=(mc^2)^2\tag{$\star$}$$ where $E$ denotes energy, $m$ rest mass, $p$ is the momentum and $c$ is the speed of light in vacuo which is approximately $3\cdot10^8\rm m/s$.
And we know that the momentum of a photon is given by $\frac{hf}{c}$ and its energy by $E=hf$ where $h$ is Planck's constant (approx $6.62 \times 10^{-34}\rm m^2 kg / s$) and $f$ is its frequency. Therefore, the following holds: $$E=hf\quad p=\frac{hf}c\implies E=pc.$$
Thus, $E^2=(pc)^2\Rightarrow E^2-(pc)^2=0$. We put this result in $\text{($\star$)}$ to get: $$0=E^2-(pc)^2=(mc^2)^2.$$ Solving for $m$ gives $m=0$. So photons are massless. $\square$

to be continued...

Proof that a sum of consecutive odd integers gives rise to a perfect square


This is purely a test

Some matrice: $$\begin{pmatrix}
 1 & 1 & 0 & -2 & 6 & 9 \\
 1 & 3 & 0 & -4 & 6 & 7 \\
 1 & 1 & 0 & -6 & 6 & 9 \\
 4 & 1 & 5 & -4 & 9 & 4 \\
 1 & 1 & 0 & -2 & 6 & 9 \\

Some NT:

\[\lim\limits_{n\to\infty}\dfrac{\pi(x)}{x/\ln(x)}=1\qquad\text{and Ramanujan's famous:}\qquad\frac{2\sqrt{2}}{9801} \sum_{k=0}^\infty \frac{ (4k)! (1103+26390k) }{ (k!)^4 396^{4k} } = \frac1{\pi}.\]

Does mathcal work? $$\mathcal L=-\dfrac14\ldots$$

Maxwell's equations: $$\boxed{\begin{align}\\\,\\\qquad&\qquad\nabla\cdot\mathbf D=\rho&&\text{(1)} \qquad\qquad\text{Gauss' law}&\\\,\\\qquad&\qquad\nabla\cdot\mathbf B=0&&\text{(2)}\quad\text{Gauss' law for magnetism}&\\\,\\\qquad&\,\,\,\nabla\times\mathbf E=-\dfrac{\partial\mathbf B}{\partial t}&&\text{(3)}\qquad\quad\,\text{Faraday's law}&\\\,\\\qquad&\,\nabla\times\mathbf H=\dfrac{\partial\mathbf D}{\partial t}+\mathbf J&&\text{(4)}\qquad\text{Ampère-Maxwell law}\qquad\\\,\\\end{align}}\\\,\\\textit{Maxwell's equations in  vector form}