# Tag Info

## New answers tagged bosons

0

If we identify a force as a scattering process, i.e. with a mediator of some interactions, then this need not be a vector boson of course. One can speak of "Higgs" force for instance if the process under consideration is mediated by the the Higgs (which is a scalar). There are also numerous cases where the interaction is mediated by a fermion. Therefore, ...

4

Well, $1/2\otimes1/2=0\oplus1$, so a system with two fermions has integer spin. But it is still a two fermion system, and therefore its wavefunction must be antisymmetric, as usual. This is not specific to Cooper pairs, but is basic Quantum Mechanics... [what is specific to Cooper pairs is that their size is $\gg a_0$, which means they are highly ...

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Let's say we could perform the experiment with $W^\pm$ bosons. These particles are similar to electrons, but the possible spin states are $-1,0,+1$, that is, three different possibilities. The magnetic moment of these bosons is, therefore, $$\mu_z=\begin{cases} -\mu_W\\\phantom{+}0\\+\mu_W\end{cases}$$ where $\mu_W=6\ 10^{-6}\ \mu_B$ is the $W$ magneton. ...

3

It is not true that all superconductors are gapped. For example, d-wave superconductors in cuprates are gapless. The energy gap in the superconductor arises from the fact that breaking the Cooper pair requires finite energy. The low-lying quasi-particle excitations are all pair breaking excitations, so they are gapped from the ground state by the amount of ...

0

Other answers paraphrase it well in technical terms. It might be easier to see if you remember that when two particles interact they must do so in a way so that the momentum, energy, spin, etc. are conserved. After the interaction the two particles still remain in a superposition state but if you measure one of them after an interaction you can find out ...

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In general in Feynman diagrams an incoming particle can be read as an outgoing antiparticle and W+ is the antiparticle of W- and vice verso. Quantum number conservation holds at the vertices. (charge , lepton number..) The reaction studied in 11 is a change of a proton to a neutron through the weak interaction. The charge of the proton has to go to the ...

0

Let $\phi(z)=\sum_{k\geq 1} a_k z^k$ be absolutely convergent and invertible in a neighborhood of $z=0$, with $a_1\neq 0$. Let us denote its (compositional) inverse by $\phi^{-1}(u)=\sum_{k\geq 1} b_k u^k$. It is not difficult to check that the following relations hold (see below): $$\sum_{n=1}^m a_n\sum_{\substack{k_1,\dots,k_n\geq 1\\ k_1+\cdots +k_n=m}} ... 0 Ok. After experimenting a bit I think I answered the question, although confirmation would be useful. I will write N\lambda^3/V = \alpha and 1/2 \sqrt{2} = \gamma so that we have$$ \alpha = z + \gamma z^2 $$or that$$ z = \frac{-1 + \sqrt{1+ 4\gamma \alpha}}{2\gamma} $$Taylor expanding to second order,$$ -\frac{1}{2 \gamma} + \frac{1}{2 ...

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The situation is not symmetric at all: This diagram describes a force between two fermions, but a diagram such as just doesn't exist (in the Standard Model). Fermions can in fact mediate a force between bosons, like in: Such diagrams are highly suppressed loop diagrams though, and the one above would after renormalization be seen as just one ...

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I think that this is more about the historical construction of the theory than about the actual interactions. In a lagrangian, two fields A, B interact when there is a product term of both such as AB. So, I see no real fundamental distinction there, even with more complicated expressions. But when one introduces the interaction bosons, it's by the mean of ...

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Its just a consequence of garlac experiment the particles on the have possibly two eigen states up |z>and down ie half of particles go up nd rest half down |_z> that's y they are called spin half. And note that spin is nothing to do with rotation it is purely intrinsic and quantum mechanical

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