Quantum Physics Corner

Any state of a single spin is an eigenvector of some component of the spin. In his book Quantum mechanics: The theoretical minimum , Leonard Susskind refers to this statement as the spin-polarization principle. Let us scrutinize the statement by considering the following problem. Problem Consider a spin-\(\tfrac{1}{2}\) particle in a generic spin-state given by \[\Psi = \begin{pmatrix} \psi_1 \\[0.1cm] \psi_2 \end{pmatrix} = \begin{pmatrix} e^{i \alpha} \cos \delta \\[0.1cm] e^{i \beta} \sin \delta \end{pmatrix} \,,\] where \(\alpha\), \(\beta\), \(\delta\) are some real numbers. The state is specified in the \(z\)-representation; that is, the states \[\begin{pmatrix} 1 \\ 0 \end{pmatrix} \quad \text{and} \quad \begin{pmatrix} 0 \\ 1 \end{pmatrix}\] have well-defined \(z\)-components of the spin equal to \(+\hbar/2\) and \(-\hbar/2\), respectively. Find a unit vector \(\boldsymbol{n}\) (in three-dimen

Let \(\boldsymbol{e}_1\), \(\boldsymbol{e}_2\), \(\boldsymbol{e}_3\) be a right-handed triplet of mutually orthogonal unit vectors in a three-dimensional Euclidean space. An operator-valued vector \(\boldsymbol{A}\) is a "vector" whose components along \(\boldsymbol{e}_1\), \(\boldsymbol{e}_2\), \(\boldsymbol{e}_3\) are operators \(A_1\), \(A_2\), \(A_3\), respectively: \[\boldsymbol{A} = \boldsymbol{e}_1 A_1 + \boldsymbol{e}_2 A_2 + \boldsymbol{e}_3 A_3 \,.\] Examples include the position and momentum operators in quantum mechanics: \[\boldsymbol{x} = \boldsymbol{e}_1 x_1 + \boldsymbol{e}_2 x_2 + \boldsymbol{e}_3 x_3 \,,\] \[\begin{align*} \boldsymbol{p} &= \boldsymbol{e}_1 p_1 + \boldsymbol{e}_2 p_2 + \boldsymbol{e}_3 p_3 \\[0.2cm] &= \boldsymbol{e}_1 \frac{\hbar}{i} \frac{\partial}{\partial x_1} + \boldsymbol{e}_2 \frac{\hbar}{i} \frac{\partial}{\partial x_2} + \boldsymbol{e}_3 \frac{\hbar}{i} \frac{\partial}{\partial x_3} \,. \end{align*}\] Let \(\bol

According to the pilot wave theory of de Broglie and Bohm, a single electron is both a wave and a point-like particle. The wave is described by a position-space wave function \(\Psi(\boldsymbol{r}, t)\), whose propagation is governed by the Schrödinger equation \[i \hbar \frac{\partial \Psi(\boldsymbol{r}, t)}{\partial t} = - \frac{\hbar^2}{2 m} \boldsymbol{\nabla}^2 \Psi(\boldsymbol{r}, t) + U(\boldsymbol{r}) \Psi(\boldsymbol{r}, t) \,,\] with \(\hbar\), \(m\) and \(U\) denoting the Planck constant, electron mass, and external potential, respectively. The particle is described by its (well-defined!) position \(\boldsymbol{x}(t)\) and velocity \(\boldsymbol{v}(t)\), that evolve in time according to \[\frac{d \boldsymbol{x}(t)}{d t} = \boldsymbol{v}(t) \,, \qquad \boldsymbol{v}(t) = \boldsymbol{u}\big( \boldsymbol{x}(t), t \big) \,,\] where \[\boldsymbol{u}(\boldsymbol{r}, t) = \frac{\hbar}{m} \, \operatorname{Im} \left( \frac{\boldsymbol{\nabla} \Psi(\boldsymbol{r}, t)}{\

The generalized hypergeometric series is defined as \[\begin{align} &{_pF_q}(a_1, \ldots , a_p; c_1, \ldots, c_q; z) = \sum_{n=0}^{\infty} \frac{(a_1)_n \ldots (a_p)_n}{(c_1)_n \ldots (c_q)_n} \frac{z^n}{n!} \\[0.2cm] &\qquad\qquad = 1 + \frac{a_1 \ldots a_p}{c_1 \ldots c_q} \frac{z}{1} + \frac{a_1 (a_1 + 1) \ldots a_p (a_p + 1)}{c_1 (c_1 + 1) \ldots c_q (c_q + 1)} \frac{z^2}{1 \cdot 2} + \ldots \end{align}\] Show that \[\begin{align} {_0F_0}(z) &= e^z \\[0.3cm] {_1F_0}(-\alpha; -z) &= (1 + z)^{\alpha} \\[0.2cm] {_2F_1}(1, 1; 2; -z) &= \frac{\ln(1 + z)}{z} \\[0.2cm] {_2F_1}\left( \tfrac{1}{2}, \tfrac{1}{2}; \tfrac{3}{2}; z^2 \right) &= \frac{\arcsin z}{z} \\[0.2cm] {_1F_1}\left( \tfrac{1}{2}; \tfrac{3}{2}; -z^2 \right) &= \frac{\sqrt{\pi}}{2 z} \operatorname{erf}(z) \end{align}\] The solution is provided in Issue #4 (premium) - Tutorial: Hypergeometric functions. Part I of Quantum Newsletter .

A weekly problem from Issue #3 - Tension in a ring of Quantum Newsletter . A full solution can be found here . Subscribe to Quantum Newsletter to receive new issues in your inbox.  

A weekly problem from Issue #2 - Scattering at a potential step of Quantum Newsletter . A full solution can be found here . Subscribe to Quantum Newsletter to receive new issues in your inbox.

A weekly problem from Issue #1 - Correspondence principle of Quantum Newsletter . A full solution can be found here . Subscribe to Quantum Newsletter to receive new issues in your inbox.