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14

No. The total orbital angular momentum is always an integer. Adding the orbital and spin angular momenta for a spin-$\frac{1}{2}$ particle will produce another half-integer value, $j=\frac{1}{2},\frac{3}{2},\frac{5}{2},\ldots$. In contrast, the total angular momentum of two fermions (orbital, plus the first spin-$\frac{1}{2}$, plus the second spin-$\frac{1}...


9

In short (and not very accurately) – a scalar particle is a particle that has no spin and no ‘inner structure’ that requires its field operator to have representation in terms of different components. More accurately – the particles are excitations of fields, and when we construct these fields we have to ask ourselves how do they transform under rotations ...


3

As pointed out by yu-v, the mistake lies in counting the cases with $j=k$. The problem lies in the negligence that as soon as you put $\alpha_j \alpha_k = - \alpha_k\alpha_j$, this implies $j \neq k$. Thus, $\alpha_j\alpha_k\alpha_k\alpha_j=6$. Ofcourse all this trouble could've been avoided using following anticommutation relation which holds for both (...


3

Regardless of energy state, regardless of the choice of isotope, if we have two identical atoms in this molecule (i.e. same isotope of oxygen), this pair must have even numbers of all nucleons and electrons. This then precludes half-integer total spin, which leads to bosonic properties of the system. I initially thought a light (photon) can excite the ...


2

Yes, there is Coulomb interaction, which also leads to correlation in position. As an example, you could look at helium. The binding energy of one electron is 4 Rydberg = 54.4 eV. But the ionization energy of neutral helium is 24.6 eV. Calculating this number is not so easy because it is a three-body problem. One way of taking into account electron-...


2

Loosely speaking, yes. By 'feel each other's Coulomb potential' you mean that the behaviour of one electron is influenced by the presence of the other owing to electrodynamic effects. That is undoubtedly the case. If you were to model the behaviour of an electron in an atom by considering only the potential due to the nucleus and electrons in other orbitals ...


2

The strategy is always the same: to construct states with a given $j$, look for the state $\vert j j\rangle$ as this state is the only state killed by $J_+$ and is an eigenstate of $J_z$ with eigenvalue $j$. Hence, write \begin{align} \bigl\vert \textstyle 1/2, 1/2 \rangle = \alpha \vert 1,1\rangle \vert 1/2,-1/2\rangle + \beta \vert 1,0\rangle \vert 1/2,...


2

The criterion for determining whether mean-field theory is good or not is the Ginzburg criterion. You can estimate how accurate mean-field theory is by computing the leading corrections to it. In particular, the exact Ising model is $$ H = -J \sum_{\langle i j \rangle} s_i s_j, $$ and mean-field theory is done by taking $$ s_i s_j = M^2 + M (s_i + s_j) + (...


1

Elementary particles and the composites from them are defined by spin: if spin is a multiple of 1/2 it is a fermion, if it is a whole number it is a boson. This is incorporated in the spin statistics theorem that describes the difference in their wavefunction, leading to the use of different differential equations. The proof is further in the link. The spin–...


1

I agree with the other answers. The Many Worlds Interpretation has been sensationalised unnecessarily to the degree that some people imagine it to mean the existence of trillions of imaginary universes each as real as ours, and each created every time a wave function collapses. Everett's PhD is available online- there are no such bizarre suggestions in it. ...


1

First, what is he calling the number of excitations for the 2-atom system? There are two atoms coupled to a radiation mode, i.e. to a harmonic oscillator. The states $|n\rangle$ refer to the harmonic oscillator, the states $|S, m_s\rangle$ refer to the two atoms. (Each atom is assumed to have only two energy levels and thus behaves like a spin-1/2. Two spin-...


1

...particles with the same quantum numbers cannot occupy the same space (pauli exclusion principle), while with different numbers such as spin they can. Does that mean the two electrons are invisible to each other<...>? No, this doesn't mean that they are invisible. It's just that Coulomb potential is a "soft" potential: due to Heisenberg uncertainty ...


1

The rule is simple: A combination of particles is a boson is it contains an even number of fermions otherwise it is fermion. Since $\text{O}_2$ is two of the same particle (assuming both are the same isotope), it is necessarily bosonic.


1

You might well be giving the Gell-Mann matrices a bad rap. They are all traceless hermitian, but they are real, except for the three imaginary ones, $\lambda_2,\lambda_5,\lambda_7$ which are imaginary antisymmetric, so, multiplied by i , comprise the three antisymmetric generators of SO(3) in the triplet (spin 1) representation. Behold. So your three σ are ...


1

The Pauli exclusion principle applies to the constituent fermions of the composite bosons. For example, many atoms of helium can be in the same lowest energy state forming a superfluid. However, they cannot be squeezed to a zero volume, because the Pauli exclusion principle holds for protons and neutrons, as well as for their constituent quarks.


1

The Dirac spinor can be interpreted in terms of left/right chiral spinors because the Dirac representation is a reducible representation of the Lorentz group, $$ (\tfrac12,0)\oplus (0,\tfrac12)\tag1 $$ On the other hand, the vector representation of the Lorentz group is irreducible, $$ (\tfrac12,\tfrac12)\tag2 $$ and, as such, no analogous interpretation is ...


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