# Tag Info

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There exists a very simple and concise semi-classical explanation of the electron's spin angular momentum, without the notion of rotation of any material object: Qualitatively speaking, the electron’s spin angular momentum is the electromagnetic field's angular momentum resulting from the combined electromagnetic field surrounding an electron just in such a ...

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I almost voted to close your question as a duplicate of How do you rotate spin of an electron?. This would be controversial because the question looks related at a first glance, but ACuriousMind's answer to that question also (indirectly) answers this question. When we talk about rotating an electron, or any fermion, we are not talking about a physical ...

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I assume that in the quadrupole transition there is a 2 photon interaction, since photons only have spin 1. Hence why it is a less probable transition.

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Madelung's Rule: Orbitals fill from lowest energy to highest energy, so for example $\mathrm{1s,2s,2p,3s,3p,4s,}$ etc. Pauli's Exclusion Principle: No two electrons can have the same set of four quantum numbers, in the same orbital. In practice this means atomic (or molecular) orbitals can only accomodate 2 electrons, one with spin quantum number ...

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It depends if we look at particle as classical ones or quantum ones. In the first case, particles are usually following a Boltzmann statistics. However, things become more interesting when entering the quantum world. Here, the spin of the particles become crucial. We have that particles with integer spin follow a Bose-Einstein statistics. Whereas particles ...

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If you want to reduce the "spaghetti of algebra" you can reorient the coordinates. If $\hat{z}'=\hat{m}$ then in spherical coordinates you have $$\sigma_\hat{m}= \left[ \begin{array}{cc} \cos(\theta) & \sin(\theta) \\ \sin(\theta) & -\cos(\theta) \\ \end{array} \right]$$ where $\cos(\theta)=\hat{n} \cdot \hat{m}$. It's easier to find the ...

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So the matrix element I tried to calculate is indeed zero for the dipole moment. In order to find the hyperfine splitting, one must calculate $|\langle a_f|\mu|a_i \rangle|^2$.

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The expectation value is a single number, it is the sum of all the possible values with a weight based on how often you get the result. So $\langle S_x\rangle=P(+\frac \hbar 2)\frac{\hbar}{2}+P(-\frac \hbar 2)\frac{-\hbar}{2}$ And you can get the probabilities by projecting the original spinor onto the eigenspaces of the operator and comparing the $L^2$ ...

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The options 1,2 are actually physically identical because the electrons are identical particles. Once we have two electrons, we can't say which of them is "Paul" and which of them is "Peter". When the addition is slow etc., the option 1=2 violates the conservation law for the angular momentum. So it is indeed 3 that has to happen: the ion will refuse to ...

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Spin and orbital angular momentum are two different things, as already pointed out in Aniket's answer, but there is a good reason why we still call spin a "spin". This is because the Einstein-de Haas-Richardson experiment shows that electron spin is indeed of the nature of an angular momentum, although not exactly due to a "spinning electron". In fact, ...

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In quantum mechanics and particle physics, spin is an intrinsic form of angular momentum carried by elementary particles, composite particles (hadrons), and atomic nuclei. Spin is one of two types of angular momentum in quantum mechanics, the other being orbital angular momentum. The orbital angular momentum operator is the quantum-mechanical counterpart to ...

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There is a very interesting Am J Phys paper by Ohanian, titled "What is spin?". You can find free PDF copies on google, in case you don't have academic access. He points out that ALL forms of angular momentum, even spin, arise from linear momentum via the relationship $\vec r \times \vec p$. In other words, even spin is orbital angular momentum. This is ...

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I tried to ask you to clarify the problem so I'll assume a strong external classical field. The Hamiltonian for a magnetic moment $\vec \mu=\gamma \vec S$ (where $\gamma$ is the gyromagnetic ratio) in an external magnetic field is $$H=(P_x^2+P_y^2+P_z^2)/2m-\vec \mu \cdot \vec B$$ Which assumes no other interactions, such as electric charge or electric ...

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