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Assume there is zero magnetic field, and hence no zeeman splitting and no Mf states Now assume, in a two level system, you pump the medium with a light resonant with hyperfine splitting between the ground state, Fg and excited state Fe. Suppose the atom gets excited from Fg to Fe, it has to fall back to some level, but if that level is same as the level it ...


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The simplest way is to exploit the symmetries here. So, instead of mindlessly going through the algebra, solving equations etc. you just use the fact that you are free to call any direction the x-direction, and set up a right handed coordinate system. In particular, this means that you are free to cyclically permute x, y, z in the equations. The problem of ...


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For a single particle, yes they're parallel. For a system of particles, $$\sum_i \frac{{\bf r}_i\times \dot{\bf r}_i}{\|{\bf r }_i\|^2}\neq \alpha\sum_i{m_i{\bf r}_i\times \dot{\bf r}_i}$$ (you can come up with a specific counterexample but it should be obvious the two sides don't have to be proportional/collinear -- each vector in the sum is weighted ...


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Faced by the same question and a background that includes courses in vector calculus, I have sought a simpler answer. My answer is much that same as to why one can easily balance on a typical bicycle. Bicycles are constructed so that the point where the extension of the front fork pivot would hit the ground is in front of the point where the front tire ...


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The fundamental difference between an electron's spin and that of a baseball is that the electron is (as far as we know) a point particle. It therefore cannot rotate in the usual sense, where individual parts move relative to the center of mass; we say that its angular momentum is intrinsic. The magnitude $\lvert\vec{S}\rvert^2$ of a particle's intrinsic ...


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A simple answer is to look in the Heisenberg picture with the Heisenberg equation of motion. In this picture, operators evolve instead of states as in the Schrodinger picture. Here the evolution of the operator is chosen so that its expectation value has the same time evolution as in the Schrodinger picture. To do so its time evolution is governed by: $ ...


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You must remember that the 'body-frame' (which is the frame 'inside' the rotating body) is a frame and that OTHER bodies (which may, for example, be rotating about the same point with a different speed) will appear in the 'body-frame' to have an angular speed less than if observed in an inertial (stationary) frame. To help visualise, if you stick your arm ...


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When I was asking this question, I didn't understand the relation between the commutativity of two operators and their eigenspaces: If an operator $A$ commutates with another operator $B$, then $A$ leaves the eigenspaces of $B$ invariant: $$ B\psi = \epsilon\psi \implies BA\psi = AB\psi = \epsilon A\psi $$ But this does not imply that $\psi$ is an ...


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My professor defined the angular momentum operators in quantum mechanics as the infinitesimal rotations in the Hilbert space of quantum states. Your professor likely did not mean what you think they meant. Rotations in the Hilbert space are performed by unitary transformations. If you choose the position representation then your Hilbert space is a set ...


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$L_z$ is related to an infinitesimal rotation, and you seem to think you have rotated you coordinates by $90^\circ$. A rotation around the $z$ axis by a small angle $\phi$ is given by $$ R(\phi) = 1 + \imath \phi L_z + O(\phi^2) $$ To see how $L_z$ acts on a wavefunction we can apply $R$ to it, expand and compare terms \begin{align} ...


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In your first paragraph you describe a Stern-Gerlach device as one with a magnetic field in the $\hat z$ direction. And then later you talk about having a large homogeneous component of the magnetic field. I'm not sure you have an accurate physical model of a Stern-Gerlach device. The Hamiltonian for a Stern-Gerlach has magnetic fields components combined ...


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OP is essentially pondering if commutativity is a transitive relation, ie. if three normal operators$^{1}$ $A$, $B$, and $C$ satisfies $$ [A,B]~=~0\quad \wedge\quad [B,C]~=~0 \quad\stackrel{?}{\Rightarrow} \quad [A,C]~=~0 .\tag{T}$$ The answer is No, but OP argues via the existence of a common basis of eigenvectors for two commuting normal operators that ...


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The $L_i$ has many eigenspaces corresponding to many eigenvalues. Each of those eigenspaces is also an eigenspace of the Casimir operator. So they share common eigenspaces in the sense that there are eigenspaces that are eigen to both. But they don't share them in the sense that they are the same. Look at the hydrogen atom. There are energy eigenspaces and ...


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Two operators may be simultaneously diagonalized if and only if they commute. As you can see, $L_z$ commutes neither with $L_x$ nor with $L_y$ – and not with any other linear combinations different from a multiple of $L_z$ – so there's no way to diagonalize two different components of $L_i$ at all. However, $L_z$ (and similarly other components) commutes ...


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The eigenspaces of the quadratic Casimir $L^2 = L_x^2+L_y^2+L_z^2$ of the Lie algebra of infinitesimal rotations $\mathfrak{so}(3)$ are precisely the irreducible representations of $\mathfrak{so}(3)$ - we usually label a representation by its highest weight $l$, which is in this case just a number telling you what the largest possible value for any of the ...


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I think the distinction here is between a free vector and a line vector. A free vector is shared by the entire rigid body regardless of location. Examples are: angular velocity $\mathbf{\omega}$, force $\mathbf{F}$ linear momentum $\mathbf{p}$. line vector is defined to act on a specific location (for example a point A), and along a specific ...


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I don't think this is as dumb a question as everyone downvoting you seems to think. The definitions of angular velocity ($\omega$), angular momentum ($L$), and moment of inertia ($I$) ARE defined in order to perfectly mirror Newton's laws. Angular momentum is the analogue of momentum, angular velocity is the analogue of velocity, and moment of inertia is ...


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You're right, they are mistaken. You can verify this by noticing that (Using their $L_y$): $$L_+=L_x+iL_y=\frac{1}{2}(L_++L_-)+i\frac{i}{2}(L_+-L_-)=L_-$$ Which is obviously wrong. (Do note however that their matrices are correct)


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The quick answer to your question is no, cause nothing can spin that fast. It's an interesting question, though I think you need to take a step back and ask about formation. A spinning object will have less gravity around the equator than on the poles, but if an object spins so fast that it has zero gravity, most likely it would fly apart though there ...


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There are solutions to Einstein's Field equation (General Relativity) where the ratio of angular momentum to mass is so large that the singularity is visible to the outside instead of being shielded by an event horizon. But no known astrophysical black hole has a ratio that high. And it looks like when you try to give more angular momentum to an existing ...


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This is pretty niche notation, and it is indeed not defined in the paper, but the name "vector-coupled product" does seem to be used by a few people beyond Varga and Suzuki. In essence, $$ [\mathcal Y_{l_1}(\mathbf x_1)\mathcal Y_{l_2}(\mathbf x_2)]_{LM} $$ is a coupled wavefunction with total angular momentum $L$ that's made up of the single-particle ...


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If the spin is an actual magnetic moment, then its behavior under time reversal is simply similar to that of classical magnetization, which changes sign. Think of magnetic fields and dipoles as generated by electric currents. Under time reversal the currents reverse direction and so do the corresponding magnetic fields or dipoles. At quantum level, spin ...


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Energy is conserved, but if you ignore some kinds of energy then it will look like it isn't conserved. You can imagine a really big disk with some radial pointing two by fours attached at the one o'clock, two o'clock etcetera positions then attach springs to each two by four with the spring pointing in the clockwise/counter-clockwise directions. Add a nice ...


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As one of the comments mentions, it is simpler to consider a linear case. Dropping a body of mass $m$ on one moving with mass $M$ and velocity $v$ is essentially considered the instantaneous transformation $M \to M + m$. Momentum must be conserved in the collision, but the mass of the system effectively increases, producing a smaller kinetic energy: $$ ...


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The Earth's tilt doesn't change so much as it's position around the sun changes. Notice the North Pole sees more sun in summer than in winter cause it's tilted towards the sun in summer but not in winter. The north pole always points in the same direction into space pretty much. It wobbles slightly, and quite slowly, completing a full wobble every ...


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No difficulties at all. If the net force applied on a body is zero then the center of mass is not going to move accelerate. This leads to the conclusion that the only motion allowed is a rotation about the center of mass. For more details refer to: http://physics.stackexchange.com/a/81078/392 The relevant equations are: $$ \mathbf{F} = m \,\mathbf{a}_C ...


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If you want to define your $J_\pm$ in terms of $J_x$ and $J_y$ then you'll need to know those matrices. If you consider $J_x=\frac{\hbar}{\sqrt 2}\left(\begin{matrix} 0&1&0\\ 1&0&1\\ 0&1&0\end{matrix}\right),$ and $J_y=\frac{\hbar}{\sqrt 2}\left(\begin{matrix} 0&-i&0\\ i&0&-i\\ 0&i&0\end{matrix}\right)$ then ...


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The answer is basically, angular momentum. The collapsing proto-solar nebula has some angular momentum. Whilst dissipative processes can allow the nebula to collapse along the axis of rotation, there is still the problem of how to shed angular momentum in order to allow gas/dust to orbit closer to the rotation axis. This is just a basic application of ...


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Jac's answer excellently lays down how the mechanics of rotating bodies work. Since you still seem a bit confused, I thought I'd try and expand a bit about what this means for gyroscopes in particular. Let's say you have a gyroscope with its angular velocity $\vec \omega$ pointing horizontally, and ignore gravity for now. The set-up will look something ...


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For a body rotating around an axis, we can reformulate the second law of Newton as follows: $$\frac{d\vec L}{dt}=\vec \tau$$ with $\vec L$ the angular momentum and $\vec \tau$ the torque applied to the body, both calculated around the axis of rotation. Furthermore, as $$\vec L=I \vec \omega$$ with $I$ the moment of inertia and $\omega $ the angular ...



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