Podcast #128: We chat with Kent C Dodds about why he loves React and discuss what life was like in the dark days before Git. Listen now.
9

From $\mathbf{L}=\mathbf{Q}\times \mathbf{P}$ we have $L_z=Q_xP_y-Q_yP_x$. Then, introduce the following new operators (assuming units of $\hbar=1$): \begin{align} q_1=\frac{Q_x+P_y}{\sqrt{2}},\\ q_2=\frac{Q_x-P_y}{\sqrt{2}},\\ p_1=\frac{P_x-Q_y}{\sqrt{2}},\\ p_2=\frac{P_x+Q_y}{\sqrt{2}}. \end{align} It is immediate to check that $$ [q_1,q_2]=[p_1,p_2]=0,\...


6

The intuitive proof is straightforward. The group $SU(2)$ is simple, and so any representation is contained in the tensor product of sufficiently many copies of the fundamental (à la Young). More specifically, a field of spin $j$ can be constructed by tensoring $2j$ copies of a spinor: $$ \phi^{\alpha_1\cdots\alpha_{2j}}\sim\psi^{\alpha_1}\otimes\cdots\...


3

"The same rule" is associativity, but everything else is quite different. That is, the group commutator is, in some convention, $$[R_{x}(\alpha),R_{z}(\beta)]= R_{x}(\alpha)R_{z}(\beta) R_{x}(-\alpha)R_{z}(-\beta), $$ where $R_{x}(\alpha)= e^{-i\alpha J_x}$, etc... For notational simplicity, let's call $-i\alpha J_x =A$ and $-i\beta J_z =B$, antihermitean, ...


3

Review the vector model of angular momentum. Absorbing $\hbar$ into L, one has now non-dimensionalized. On the z' axis, the 3 states with eigenvalues -1,0,1 for $\hat L_{z'}$, respectively, are the two back -to back cones and a disc in the middle. You are just projecting the +1 eigenstate cone in your figures. Since $\ell =1$, the length of the side of the ...


3

First on your point 1. There is no contradiction between a two-spin system having an antisymmetric state with respect to exchange of its component parts, yet being symmetric with respect to exchange of that pair with some other pair. The two-spin system is, overall, bosonic, as you say. In particular, the singlet state has the correct properties to be a $S=...


3

There is no contradiction. A particle's spin is not its only attribute. A two-fermion state must be antisymmetric with respect to the exchange of all of their attributes, not just spin. If the state is symmetric with respect to the exchange of their spins, then it is antisymmetric with respect to the excahge of their other attributes (such as location or ...


2

No. Expectation values of observable operators are always real.


2

I was wondering how the stationary flywheel scenario worked. As Adrian said if the flywheel were not to spin, it would fall over (i.e. you would not observe what is called precessional motion). Why? Because there's no angular momentum due to the spin of the flywheel and thus no torque. Why is it that the gyroscopic falls along a "circular path?" This is ...


2

Aside of that, there is another assignment for this problem. I would like to reword the solution of this assignment here in my own words because maybe it can help someone else to understand. Also, if I ever forget how they got those results, I can simply look back at this answer. I'm sure it can be of use to the site and that other users can profit of it. ...


2

He’s going to take a superposition of such solutions at a later step and sum over all values of $m$. That will pick up the parts of the general solution that you’ve identified.


2

ShoutOutAndCalculate, Please take a look on this paper: Proposed physical explanation for the electron spin and related antisymmetry Cetto, A.M., de la Peña, L. & Valdés-Hernández, A. Quantum Stud.: Math. Found. (2019) 6: 45. https://doi.org/10.1007/s40509-017-0152-8 You can read it free here: https://arxiv.org/pdf/1707.08674.pdf The paper presents ...


1

Q1. In the standard theory the total angular momentum of the field is ${\bf r} \times {\bf P} $, where $\bf P$ is the Poynting vector. Note that this depends on whatever origin of the coordinate system you choose. Q2. No, as the Poynting vector vanishes in this case.


1

It's just Euler's $\cos \Phi+i\sin \Phi= e^{i\Phi}$ compbined with the usual forms of Pauli's $\sigma_x$ and $\sigma_y$.


1

Given $$ [ L_z, L_x ] = i \hbar L_y $$ just multiply both sides by $L_x$: $$ L_x \bigg( L_z L_x - L_x L_z \bigg) = i \hbar L_x L_y $$ which gives $$ L_x L_x L_z = L_x L_z L_x - i \hbar L_x L_y $$


1

What you are looking for is a photon OAM switch. The OAM of light is the component of angular momentum of a light beam that is dependent on the field spatial distribution and not the polarization. It can be further split into an internal and an external OAM. The internal OAM is an origin-independent angular momentum of a light beam that can be associated ...


1

I figured this out a long time ago, should probably have deleted this or just added my own answer. Anyway, it is really the angular momentum and I should have never passed to coordinates. Also, I should have kept the mass $m$, for psychological reasons. The point is that $m \det(\gamma(t),\dot{\gamma}(t), v) = \langle m\gamma(t)\times \dot{\gamma}(t), v\...


1

It is well-known that the finite dimensional irreps $V_{\ell}$ of the Lie algebra $so(3)$ are classified by spin $\ell\in\frac{1}{2}\mathbb{N}_0$. To rule out half-integer representations for the orbital angular momentum (OAM) Lie algebra $${\rm span}_{\mathbb{R}}(L_1,L_2,L_3)~\cong~ so(3),$$ one should use the fact the OAM operators $$ L_j~=~\sum_{k,\ell=1}^...


1

In this answer we elaborate on Ballentine's method of finding a canonical transformation (CT) $$(q,p)\quad \longrightarrow\quad (Q,P),\tag{1}$$ cf. NessunDorma's answer. Recall first of all that instead of position and momentum operators, we may equivalently use annihilation & creation operators $$ \begin{align} a^j~=~& \sqrt{\frac{m\omega}{2\hbar}}...


1

The true way of seeing this is that the states are written as a tensor product in that the total Hilbert space is formed of ordered products of states of the form $$\left|j_{1}m_{1}\right> \otimes \left|j_{2}m_{2}\right>.$$ Moreover the so-called addition of angular momentum should really be written as $$J_{T} = J_{1}\otimes \mathbb{I} + \mathbb{I} \...


1

After reviewing the comments I believe KF Gauss is correct in their statement that the atom picks ups angular momentum with respect to the light propagation axis. See Eq. 5.448 in Quantum and Atom Optics by Steck regarding the mechanical force on an atom by an optical field. $$ \langle \boldsymbol{F}\rangle = \frac{i\hbar|\Omega(\boldsymbol{r})|^2}{4\left(\...


1

The key to understanding how orbital angular momentum (OAM) can be conserved when a photon that carries OAM is absorbed by a small particle is to understand OAM itself a little better. Note that OAM in a paraxial light beam is always defined with respect to the propagation axis of that beam. If one would pick some random axis pointing in some random ...


1

This is going to be a bit of a half baked answer but here it is. The usual picture for light with orbital angular momentum is light in a Laguerre-Gaussian Mode mode with non-zero azimuthal index. This sort of field has phase wrapping azimuthally around the propagation axis. We suppose without loss of generality that the atom is on the propagation axis of ...


1

Imagine we have a ball rotating around a pole due to a light string (sort of like a tetherball setup). We can calculate the angular momentum of the ball about the pole. If we imagine drag is minimal, then the ball spins quite freely. In this case, since the only force on the ball is from the string, and that force goes through the axis we are ...


1

These are the fundamental definitions you must know and the rest follow from these. Let's say you are in reference frame $S$ with origin $O$. Assume that there's a particle of mass $m$ located at position $\vec{r}$ (called the position vector) with respect to $O$ and its velocity is $\vec{v}$. Then, $$\vec{L}_{\text{of the point particle with respect to a ...


1

Quantities like the total energy, total momentum, and total angular momentum of the universe are not well defined in general relativity. See http://physics.stackexchange.com/questions/2838/total-energy-of-the-universe It is possible to have an over-all rotation of the universe, but that doesn't convert into an angular momentum, and it pretty strongly ...


1

Your observation is correct. Before I start my answer let me use the picture above to define three axes: Roll axis - the gyroscope wheel spins around the roll axis. Pitch axis - motion of the red frame. Swivel axis - motion of the yellow frame. To understand gyroscopic precession it is key to also recognize how motion of the pitch axis is playing a part. ...


1

My doubt is that, since the wheel is revolving around the vertical axis, there must be a component of angular momentum in the vertical direction. Remember that the wheel is spinning i.e. the wheel is not a stationary disk that is rotating about the vertical axis. If you were to take into account the rotation of the wheel around its own axis as well as its ...


Only top voted, non community-wiki answers of a minimum length are eligible