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I would say it's because at its core quantum mechanics is the study of atomic transitions, and the knowledge of the internal states must be inferred. What we see from that perspective is that all angular momentum transitions are integer multiples of $\hbar$, and are quite frequently exactly $\hbar$. There are sets of related states in some atoms which are ...


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The "spin" tells us how the wavefunction changes when we rotate space (or spacetime). Just because I double all charges by convention, the behaviour of the wavefunction will not be any different. What will happen is that the "doubling" or charges will lead to the "halving" of your definition of angles such that the physical results (which depends on angle ...


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It doesn't matter that the ladder operators are Hermitian or not. If they commute with some other operator, then you can have simultaneous eigenvalues. The relation between simultaneous eigenstates and commutation relations is a result of mathematics, not physics.


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In quantum mechanics, one can form a solid understanding of these matters. The conceptually simplest example which demonstrates the essential physics is, of course, the hydrogen atom. In solving it, we find that the (spatial part of the) allowed eigenstates of the Hamiltonian are parametrized by three quantum numbers, $n$, $\ell$ and $m$. So, we have ...


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Spin is an intrinsic property of quantum objects that, unlike a particle's orbital momentum, does not depend on the frame of reference you are considering. Another intrinsic quantity of that kind would be charge, which is also just a fundamental number you assign to a particle, no matter its state of motion. One possible source of confusion when talking ...


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L, angular momentum, is not like linear momentum ($p$) that means that the object is actually moving in that direction. Angular momentum and torques are defined via the "cross-product", which means mathematically that angular momentum and torque vectors point 90$^\circ$ from the radius and force that produce the torque and angular momentum. The radius is ...


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This is because there are just two possible values to the spin in any direction, $-\frac{\hbar}{2}$ and $\frac{\hbar}{2}$, the just differ in a sign, so when you square it you get a single value $\frac{\hbar^2}{4}$. Think about this, the only possible value when you measure the square of $S_z$ is $\frac{\hbar^2}{4}$ for any state, so $$ ...


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OP asks: Is there any physical meaning to this? Yes, the Pauli matrix $\sigma_j$ represents (up to a proportionality factor) the spin in the $j$th direction of a spin $\frac{1}{2}$ system. Such system has only two spin states: $\uparrow$ and $\downarrow$, with opposite eigenvalues. The square $\sigma_j^2$ can no longer see the sign, so it only has one ...


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Physically, how can it be that tidal friction on Earth makes the Moon do something? I know it is because conservation of angular momentum, No, conservation of angular momentum alone can't predict that one object will lose angular momentum and another will gain. It would be equally consistent with conservation of angular momentum if both stayed the same. ...


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The orbital wavefunctions of the hydrogen atom, which obey the eigenvalue equation $$ \left[-\frac{1}{2\mu}\nabla^2-\frac{e^2}{r^2}\right]\psi_{nlm}=E_{nl}\psi_{nlm}, $$ are functions of the separation vector $\mathbf r$ which points from the proton towards the electron. This is a standard trick in the two-body problem and it is done in both the classical ...


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The Earth is currently rotating at one revolution per sidereal day. Converting to radians, this is an angular velocity of $\omega_0 = \frac {2\pi}{\text{sidereal day}}$. You want to have 360 solar days per year, or 361 sidereal days per year. That means a rotation rate of one revolution per 1/361 tropical year, or an angular velocity of $\omega_1 = \frac ...


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Yes the days would be longer. Since the ice that was at the South Pole is now distributed around the world, there is a change in the distribution of mass. Like an ice skater opening their arms. Since the mass is further from the axis of rotation, the rotation slows; which means the days are longer.


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L(angular momentum) = (moment of Inertia) x (angular velocity)


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The angular momentum $L_{A/B}$ of a rigid body $A/B$ about its center of mass is $$L_{A/B} = I_{A/B} \omega_{A/B},$$ where $I_{A/B}$ is the inertia matrix of $A/B$ about its center of mass in the world frame and $\omega_{A/B}$ is the angular velocity of $A/B$. The angular momentum $L_{A/B}^0$ of a rigid body $A/B$ about the origin of the world frame is ...


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Some bits and pieces on angular momentum: Angular momentum is that which is conserved in rotationally invariant systems, just like energy is that which is conserved in time translation invariant systems and momentum is that which is conserved in space translation invariant system. This is the essence of Noether's theorem. The analogue in QFTs are ...


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If you are looking for quantitative values, look to further than the equations of motion $$ \begin{align} \sum_{i} \left( \vec{F_i} \right) & = m \vec{a}_{cm} \\ \sum_i \left( \vec{\tau}_i + (\vec{r}_i-\vec{r}_{cm}) \times \vec{F}_i \right) & = I_{cm} \vec{\alpha} + \vec{\omega} \times I_{cm} \vec{\omega}\end{align} $$ where the left hand side is ...


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The angular equivalent of the impulse-momentum theorem states The change in angular momentum of a system is equal to the product of the (average) external torque time the time it is applies In math that is (finite version): $$ \langle \vec{\tau}_{ext} \rangle \,\Delta t = \Delta \vec{L} \,,$$ or (infinitesimal version): $$ \vec{\tau} \, \mathrm{d}t = ...


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Some conservation laws are related to conservation of angular momentum. There is a famous example (from Feynman if I recall correctly), where you assume an infinite flat space and conservation of angular momentum about any point, and then you get conservation of linear momentum for free. Intuitively, to get say the $x$ component of linear momentum is ...


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Answer to first question: At the instant being considered, the space and body axes are identical, so at that moment the matrix $a$ that relates the two sets of axes is simply the identity matrix. $dG'$ is a vector, so $a_{ji}dG_j' = dG_i'$ is simply equivalent to the statement that with $I$ the identity matrix and $V$ an arbitrary vector, $IV=V$. Answer ...


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Electromagnetic field of light has two kind of angular momentum first spin angular momentum (SAM) and secondly orbital angular momentum (OAM). former one represent the dynamical rotation of electric (or magnetic) field of around propagation direction and indicate the polarization of beam. Later one represent the rotation of light around beam axes. The ...


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By no means not a complete answer, more a criticism of @luksen’s one. It is posted here because the text is too long to fit in the comment field. First of all, the spin is not a well-defined concept for composite particles. More precisely, whether the spin of a particle is defined depends on how the “particle” is defined. Look at an atom: it has the nucleus ...



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