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The spin of the electron can change as long as angular momentum conservation is not violated. If a photon leaves carrying 1+ then an electron can flip spin. – anna v That is correct but oversimplified. Spin-flip transitions are dipole-forbidden, and they can only happen via magnetic-dipole interactions, which means that they are generally very weak and ...


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For the (c) case, you have parallel spins so total spin is $S = 1$. Then we have orbital angular momentum (quantum number) $L = 1$. Our total nuclear spin will then have any of the values $I = |L-S|, ..., L+S$. This means $I = 0, 1, 2$. So, this combination of $L$ and $S$ does give us a total $I$ of $1$, even though $I=1$ is not the only possibility! The ...


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First of all, try to imagine a simple MRI experiment with an perfectly homogeneous main magnetic field in the order of 1-3 T and an object to image that only contains protons of the same kind (i.e. the signal-baring protons are chemically all of the same kind in the object, i.e. there is only water in your object). When the object is brought into the main ...


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It is true that two electrons can't have identical quantum numbers, but spin itself is a quantum number. That is, the state with quantum numbers 111 can hold two electrons: one of spin up and one of spin down. So when you are finding the ground state, for example, find the four lowest energy eigenstates (ignoring spin), and the ground state of eight ...


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You just have to find an eigenvalues of this operator. Suppose you have the case of spin $\frac{1}{2}$ particle. Then $$ \vec{S} = \frac{1}{2}\left( \sigma_{x}, \sigma_{y}, \sigma_{z}\right), $$ where $\sigma_{i}$ is Pauli matrices. Now you have that $$ \hat{h}\equiv \vec{S}\cdot \vec{n} =\sum_{\sigma = \pm \frac{1}{2}}\sigma| v_{\sigma}\rangle\langle ...


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It is straightforward matrix multiplication and the operators enter through from either side. $exp(iS_zϕ/ℏ)$ would act from the left on the kets while $exp(−iS_zϕ/ℏ)$ would act from the right on the bras. You can see it more clearly if you do it explicitly. Use the basis $|+⟩$ and $|−⟩$ to represent the operators as matrices $|+⟩$ = ...


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I fear your question is not very precise. Unfortunately my reputation is not high enough to allow me to comment instead of answer. What exactly do you mean by the term "exciton"? The way it is normaly understood, is a pair of one electron and one hole. So there are no multiple electrons whose spins could line up parallel oder antiparallel. In a biexciton ...


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I think the answer to my first question is no: Let's say A and B are detectors. Each has two states: |detected> and |not detected>. Then we will have: Ψ′up∣up⟩|A detected>|B not detected> + Ψ′down∣down⟩|A not detected>|B detected> Both spatially separated parts of the wave function entangle with both detectors. (Normalization is suppressed for ...


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Spin is determined from the representation of the Lorentz group the quantum field transforms in. The projective finite-dimensional representations of the Lorentz group are labeled by two half-integers $(s_1,s_2)$. The spin of a field is the sum $s = s_1+s_2$. For example, a scalar transforms in $(0,0)$, a vector field in $(\frac{1}{2},\frac{1}{2})$, a Dirac ...


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It all depends on definitions used of course, but overwhelmingly I have found that when people talk about "unpolarized" or "depolarized" they mean a classical mixture of pure quantum states. So they definitely do not mean a superposition state. Another way to understand that a superposition is not "unpolarized", or at least that this is not a good ...


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Let us try to clear up what light means and what a photon means. Light is a solution of classical Maxwell's equations , an electromagnetic wave. As a wave it can have ploarizations that go through filters etc. A photon is an elementary particle which has as energy h*nu, nu the frequency of an electromagnetic wave solution, and is described by a wave ...


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First, you are right about the unpolarized light. Second, whether a photon will be absorbed or not is according to the so-called selection rule which is determined by the conservation law (In this case it is the conservation of angular momentum). Thus you may think that the polaroid is "trying to" absorb each photon. However, the possibility for the ...


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The important thing here is that a magnetic dipole, like a permanent magnetic or induced magnetism in ferrous material, produces a nonuniform field. The potential energy of a magnetic dipole $\vec\mu$ in a magnetic field $\vec B$ is $$ U = -\nabla( \vec \mu \cdot \vec B).$$ Most frequently (as in anna v's answer) this is used to explain the torque which ...


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It is an experimental fact that atoms and molecules have magnetic moments. Also elementary particles with spin have magnetic moments as well as protons and neutrons. It is not classical mechanics that will describe the behavior of atoms and molecules, but Quantum mechanics. Quantum mechanical theoretical models of magnetism exist, and early on in the ...


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Not at all. The split of the 6 component electromagnetic field into a 3 component electric field and a three component magnetic field is an artifact of choosing an inertial frame. A different frame will break the same electromagnetic field into different parts. Just like if you have combinations of jumping to the future and teleporting in space then when ...


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Since you know about $SU(2)$ characters, this is a doable exercise. Let me remind you that the character of spin-$j$ is $$\chi_j(t) = \sum_{m=-j}^j e^{2imt} = \frac{\sin (2j+1)t}{\sin t}.$$ The crucial property is that the character of a tensor product is the product of characters, i.e. $\chi_{j \otimes j'}(t) = \chi_j(t) \chi_{j'}(t).$ In your case, ...


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The shapes can be predicted if you know the azimuthal quantum number. L=0 s orbital and hence, spherical. L=1 p orbital dumbell shape. L=2 d orbital, double dumbell shape, or biconical shape. L=3 would be f orbital. The second number written corresponds to the azimuthal quantum number. Each shape shows you the electron density about the nucleus. Principal ...


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Why, yes, they do! Probably the most decisive way to show this involves "handedness" of neutrinos. I won't get too much into the details here, but neutrinos are nearly massless and travel at nearly the speed of light; and when something travels at the speed of light you can describe its spin as either "right-handed" (momentum and spin are in-line) or ...


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There are two kinds of operators which have an intuitive meaning when you apply them to a state: Evolution operators like $e^{-i\hat H\Delta t/\hbar}$ simply take the state at time $t$ and give you the state at $t+\Delta t$ Projectors tell you what the state will be after you take a measurement. For example, supose you measure an observable $\hat A$ on ...


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$\langle z\rangle=\int_0^\infty r^3dr[\int_0^{2\pi}d\phi\int_0^\pi \sin \theta \cos \theta d\theta]=0$ As the $\theta $ integration is zero. But the symmetry argument is clear if the integration is written is Cartesian coordinates.In that case $$\langle z\rangle=\int_{-\infty}^\infty \int_{-\infty}^\infty \int_{-\infty}^\infty z dz |\psi|^2 dx dy$$ As you ...


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When you have two levels separated by energy $\Delta E$, you can induce a transition between them by introducing a time-dependent term to the hamiltonian (for example exciting it with a periodic electromagnetic field). This term must have a time dependence of the form $\cos(\omega t)$ with $\omega=\frac{\Delta E}{\hbar}$. The angular frequency being referred ...


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This is not my answer, it's one of the answers you can find here Is there a reason why the spin of particles is integer or half integer instead of even and odd? I just wrote here (and re-posted) the work of @Siva which I found a very good answer. However, follow the link to read more interesting useful answers The "spin" tells us how the wavefunction ...



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