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I am a little unsure what you are asking. You seem to think that two Majorana fermions can interact in such a way that one gets to Dirac fermions. The Majorana fermion is its own antiparticle. The charge conjugation of $\psi$ is $C\psi~=~i\psi^*$. The appearance of $\psi$ and $C\psi$ in the Lagrangian means that the Majorana field must be electrically ...

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In one of your hang you are already assuming that $P_{ij}|\psi\rangle = |\psi\rangle$, which not true. One way to see that the permutation operator has possible eigenvalues $\pm 1$ is by using that exchanging two identical particles and then exchanging them back should give back the original state, $$P_{ij}P_{ij}|\psi\rangle = P_{ij}\lambda|\psi\rangle = \... 0 Yes. Examples include the superfluid state at 0K of integer spin 0 boson He-4 which itself is composed of several fermions. Other effective composite "bosons" include Cooper pairs of fermions which makes their spin integer thus Pauli Exclusion Principle no longer applies to them. This makes Fermionic Condensate possible. 1 I know that in Full Configurational Interaction Quantum Monte Carlo(FCIQMC), where they start from the Schrödinger equation and sample the full configurational space with integer walkers, there is a spontaneous symmetry breaking between the \Psi and -\Psi after a sufficient number of walkers are spawned into the configurational space. It treats the ... 0 In the following, I will carefully deal with Wick rotation. In the end, I have found that I was confused. The integration is \begin{eqnarray*} & & \int\frac{d^{4}k}{(2\pi)^{4}}\frac{1}{k^{2}+i\epsilon}e^{-ik\cdot x}\\ & = & \frac{1}{(2\pi)^{4}}\int d^{3}ke^{i\mathbf{k}\cdot\mathbf{x}}\int_{-\infty}^{\infty}dk_{0}\frac{1}{k_{0}^{2}-(E_{k}-... 2 As is wellknown, the EFE is a PDE for the metric g_{\mu\nu} / vielbein e^a{}_{\mu}, which plays the role of the dynamical field of GR. OP is apparently pondering whether the EFE's source term (i.e. the matter SEM tensor T^{\mu\nu}) is independent of the unknown fields (i.e. the metric g_{\mu\nu} / vielbein e^a{}_{\mu}) and derivatives thereof? ... 1 As alluded to in the other answer here, the integral can basically be evaluated in several ways. One of them is the one that OP has himself followed. (PS - OP, congratulations on completing that feat!) Let me present here a way of computing this using Schwinger parameterization. We will use$$ \frac{1}{a} = \int_0^\infty d\tau e^{- \tau a} ~, a > 0~. ...

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You shouldn't expect the occupation of the ground state to be one (or zero). That would imply that, for $T>0$, the ground state is ALWAYS (never) occupied. That isn't necessarily true, as you can see from your formula. What is true is that at ZERO temperature states with $E<\mu$ are always occupied, while states with $E>\mu$ are never occupied. At \$...

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Method One: \begin{eqnarray*} & & \int\frac{d^{4}k}{(2\pi)^{4}}\frac{i}{k^{2}+i\epsilon}e^{-ik\cdot x}\\ & = & \frac{i}{(2\pi)^{4}}\int d^{3}ke^{i\mathbf{k}\cdot\mathbf{x}}\int dk_{0}e^{-ik_{0}x_{0}}\frac{1}{[k_{0}+(|\mathbf{k}|-i\epsilon)][k_{0}-(|\mathbf{k}|-i\epsilon)]}\\ & = & \frac{i}{(2\pi)^{4}}\int d^{3}ke^{i\mathbf{k}\...

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When dealing with theories containing both commuting and anti-commuting (even and odd) variables, physicists often use the square bracket notation to denote both commutators and anti-commutators according to the following rule: The brackets are commutators unless both variables are odd, in this case they are anti-commutators, please see footnote no. 3 in ...

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