Life with Physics

I occasionally browsed a youtube video on "2023 MIT Integration Bee". I thought it is a competition between students taking the course of calculus. I took a try and it turns out that I was not able to even solve the first problem in the allowed 4 mins.  The first problem is to solve \begin{equation} I \equiv \int_0^{\infty} \frac{\sqrt[3]{\tan x}}{(\sin x + \cos x)^2}dx\,.\tag{1}\end{equation} It is obvious that \begin{eqnarray}I &=&  \int_0^{\infty} \frac{\sqrt[3]{\tan x}}{(\tan x + 1)^2} \frac{dx}{\cos^2 x} =  \int_0^{\infty} \frac{\sqrt[3]{x}}{(x+1)^2}dx\\&=&-\left. \frac{\sqrt[3]{x}}{x+1}\right|_0^{\infty} +  \int_0^{\infty} \frac{d\,\sqrt[3]{x}}{x+1}\\&=&  \int_0^{\infty}\frac{dx}{x^3+1}\,.\tag{2}\end{eqnarray} I know how to solve the indefinite integral $\int\frac{dx}{x^3+1}$, but I was not able to accomplish the entire calculation within 4 mins. Here is another youtube video  that provides step-by-step calculat...

What is a particle? S. Weinberg described a particle simply as "a physical system that has no continuous degrees of freedom except for its total momentum".  Recall that the spacetime admits the so-called Poincare symmetry \begin{equation} {x'}^{\mu} = \Lambda^{\mu}_{\,\,\nu}\,x^{\nu} + a^{\mu}\,.\end{equation} By Noether's theorem , each continuous symmetry indicates a conservation. Thus, the total momentum $p^{\mu}$ is a good quantum because of the spacetime translation invariance. As we shall see below, there are other discrete degrees of freedom, denoted by $\sigma$, that are associated with the Lorentz invariance. So the quantum state of a particle is described as $|p, \sigma\rangle$.  Under the Lorentz transformation $\Lambda$, the one-particle state $|p, \sigma\rangle$ changes to a new state $U(\Lambda)|p,\sigma\rangle$ by a unitary operator $U(\Lambda)$. Note that the Lorentz transformation only changes the reference frames, elementary particles should remain...

The sum of all natural numbers One may see the following mysterious identity that the sum of all natural numbers equals to $-\frac{1}{12}$:   \begin{equation} 1 + 2 + 3 + \cdots = -\frac{1}{12}\,.\tag{1}\end{equation} The rigorous mathematical interpretation relies on the so-called Riemann zeta function $\zeta(s)$ and its analytic continuation. This is not the focus of this post, so we just highlight the main conclusions: \begin{equation}\zeta(s) = \sum_{n=1}^{\infty}\frac{1}{n^s}\tag{2}\end{equation} is only valid when $\mathfrak{Re} (s) > 1$. After analytic continuation, we can compute $\zeta(-1)=-\frac{1}{12}$. Therefore, mathematically, Eq.(1) should be interpreted in the language of analytic continuation. In this post, we explain Eq. (1) by applying a scheme that is commonly used in physics when treating a divergent series $\sum_{n=1}^{\infty}a_n$:  Multiply each term $a_n$ by some regulator such as $e^{-n\epsilon}$ to make the series converge.  Compute the r...

When I was an undergraduate and took the course of advanced quantum mechanics, I did not fully grasp the concept of second quantization. In my impression, second quantization was summarized as the following recipe: Expand the wave function in some basis: $\psi(x, t)=\sum_k a_k(t)\phi_k(x)$.  Impose the commutation relation like $[a_k, a^{\dagger}_{k'}]=\delta(k-k')$ (second quantization). Replace operators $\hat{\cal{O}}$ by their expectation values $\int dx \psi(x)^{\dagger} \hat{\cal{O}} \psi(x)$. I don't remember clearly whether the above summary was from my crash study for the final exam or exactly what I was taught on class. So I asked one of my best friends Joking for confirmation since we had that class together. Without any surprise, Joking told me that he never understand second quantization either.  Now let me try to explain what second quantization actually is in this post. Before start, I have to say second quantization is really a misleading terminology. As sho...