# Tag Info

## New answers tagged fermions

0

WSC - Not all matter is made of fermions. Things such as helium-4 and carbon-12 are bosons, and there are are many other composite particles and molecules that are actually composite bosons. Any composite particle with an even number of fermions, and thus with an integer value of spin, is boronic, which to me seems a little moronic. I imagine that you are ...

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You only need to use the anticommutations relations for dirac creation/annihilation operators and their linearity. Expand the definition of $P^\mu$ or $Q$ and use those relations on each term. Only one term per case will be non-zero, the one in the answers.

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If q is the momentum for positron, then the propagator for it is still $i\frac{\not{q}+m}{\not{q}^2 +m^2}$.

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For reference, the fermion propagator is $$\left\langle 0 \right| T\psi(x)\overline\psi(y) \left|0\right \rangle= S(x-y) = \int \frac{d^4k}{(2\pi)^4} \frac{i}{\not k-m}e^{-ik\cdot(x-y)}$$ Depending on the time ordering, this describes a particle moving from $y$ to $x$, or an antiparticle moving from $x$ to $y$. Now, consider a one-loop diagram in which ...

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You are correct in assuming that there should be no more than $L$ Fourier modes. In book you specified considered two different sets of boundary conditions. For periodic boundary conditions it is the usual problem and you get multiples of $2\pi/L$ which they wrote as $q=j\pi/L$ with $j=-L,\dots,-2,0,2,\dots,L$ (note that only even values of $j$ appear). ...

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Let's start from $$H = \hbar \omega \left(f^\dagger f - \frac{1}{2}\right),$$ with $\{f, f^\dagger\}=1$, $\{f, f\} = 0$ and define fermionic position and momentum coordinates by $$\psi_1 = \sqrt{\frac{\hbar}{2}} \left(f + f^\dagger\right) \\ \psi_2 = i\sqrt{\frac{\hbar}{2}} \left(f - f^\dagger\right)$$ with the following anticommutation relations: $$... 0 Fermions are strange beasts in many ways. The first problem you will encounter, and which will make it impossible to write an harmonic oscillator for fermions is the following: The fermion ladder operators f and f^\dagger require that \{f,f^\dagger\}=1. Translated to X and P this means that \{X,P\}=i\hbar. But is also means that \{X,X\}=0 and ... 1 Assuming that X=X^\dagger, P=P^\dagger and [X,P] = i\hbar, let me try$$f = \sqrt{\frac{m\omega}{2\hbar}}\left( \alpha X + \frac{\beta}{m\ \omega } P \right) $$where \alpha and \beta are complex numbers of modulus one. From this follows that$$ \hbar \omega \left( f^\dagger f - \frac{1}{2} \right) = \frac{P^2}{2m}+ \frac{1}{2} m \omega^2 X^2 + ...

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I would say that the question is still not very well defined, as it is important if in the high-temperature (classical) limit spin is conserved or not. If spin is conserved (think strong magnetic fields $B\gg T$), then in the classical limit spinful quantum gases become a mixture of classical gases, as coherent superpositions between spin states are not ...

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It was pointed out by @Peter Anderson in the comment that you forgot the transformation of the derivative, which in infinitesimal form should read $$\delta \partial_n = - g^{lm} \Lambda_{mn}\partial_l$$ which comes from the Lorentz transformation $$\partial_n \to g^{lm}(L^{-1})_{mn} \partial_l$$ (the metric is there to keep the indices in agreement with OP's ...

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Chemistry as we know it would not exist if electrons were Bosons, because many of the details of the chemical "bonding" of atoms are strongly affected by the requirement that the overall wavefunction of all the electrons involved be anti-symmetric. This anti-symmetrization principle (which is the basis of the "Pauli Exclusion Principle") along with the fact ...

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It seems to be referring to the fact that bosons don't have to obey the Pauli exclusion principle, whereas fermions do. With fermionic electrons, only one can occupy each atomic energy state at a time, so as you add more electrons, they have to form sequential shells, which makes different atoms behave differently chemically. If electrons were bosons, then ...

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