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-2 votes

What are the actual interactions between the Higgs field and fermions?

I understand the higgs field as the average of the electromagnetic spectrum at rest from which the big bang created noise, subsequent thought & the electromagnetic signal of intent (phi?=psi?). ...
John Gomez's user avatar
0 votes

Fermi-Dirac Distribution for Multiple Species

The sum, $n_1 + n_2$, of two Fermi probabilities, as you call it, can exceed $1$ and therefore cannot be considered the correct probability. Instead, for the case of two independent ideal Fermi gases ...
Gec's user avatar
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1 vote

A confusion: Why are composite bosons possible?

This is a great question and something that confused me while studying superconductivity, where we often think of bound states of two electrons as a boson. I think the other answer and comments by ...
Nandagopal Manoj's user avatar
1 vote

A confusion: Why are composite bosons possible?

Consider a boson comprised of two fermions A and B. The boson has spin $0$. One way this can happen is one fermion is spin up and the other is spin down. Another way is A and B are mixtures of spin up ...
mmesser314's user avatar
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0 votes

Calculation of canonical partition function for fermion system with degenerate energy levels

I think LPZ already gave a very good answer, so here I'll just add a comment and a general picture. In some sense, you can say that this kind of problem is exactly one of the reasons you would like to ...
Jun_Gitef17's user avatar
0 votes

Calculation of canonical partition function for fermion system with degenerate energy levels

In general it is easier to compute the canonical partition function from the grand canonical one. For a direct computation of the canonical ensemble, it is also easier to think in terms of orbitals ...
LPZ's user avatar
  • 11.7k
2 votes

Proving a Grassmann integral identity

Using the hint in @Qmechanic's answer I am able to prove the identity Method 1: $$ \begin{align} & \frac{\partial}{\partial \eta_{1}}\frac{\partial}{\partial \bar\eta_{1}} \left(1+a \left(\bar{\...
Faber Bosch's user avatar
4 votes
Accepted

Proving a Grassmann integral identity

Hints: Method 1: Use that Berezin integration is the same as differentiation, and that the Taylor series for the exponentials truncate. Method 2: Use the WKB/saddle point/stationary phase ...
Qmechanic's user avatar
  • 203k
5 votes
Accepted

The dimension of the Clifford algebra for the Dirac equation

The $V$ as in $\mathcal{C}\!\ell(V,Q)$ is 4 dimensional, while the $V$ as in $\circ:V\times V\to V$ is 16 dimensional. These are two different $V$s. Unfortunately your are conflating the two $V$s.
MadMax's user avatar
  • 3,842
3 votes

The dimension of the Clifford algebra for the Dirac equation

The anticommutator is not the algebra product in the Clifford algebra. The algebra product is the product $\cdot$ that relates to the anticommutator through $\gamma^\mu \cdot \gamma^\nu + \gamma^\nu \...
Níckolas Alves's user avatar
1 vote

Why can't we insist that the strong interactions must preserve $CP$?

The answer is pretty simple. To make $$ \bar{\theta} = \theta + Arg det M = 0 $$ you have to chirally rotate the quarks so that $$ Arg det M = -\theta $$ The chiral rotation of quarks such as $$ \psi \...
MadMax's user avatar
  • 3,842
0 votes

Propagator in massive QED/Schwinger model

Inserting unity $\gamma_5^2=1$, we will obtain $\mathrm{det}(i\not D+m)=\mathrm{det}(\gamma_5(i\not D+m)\gamma_5) $. Therefore, the fermion determinant can be rewritten as follows $$\mathrm{det}(i\not ...
Siam's user avatar
  • 1,363
1 vote

Is there a probability distribution associated with fermionic Gaussian states

This is probably only a partial answer. Firstly, although there are exceptions, most fermion phase space representations are defined with the aid of Grassmann functions (or more precisely, linear ...
flippiefanus's user avatar
  • 14.6k
0 votes

Grassmann variables and orthogonality of coherent fermionic states

The reason that some lecture notes say that the inner product of those two states is $1 + \phi \phi'$ and others say it is $\phi - \phi'$ is because they are using different conventions. From memory, ...
user1379857's user avatar
  • 11.5k
3 votes
Accepted

Grassmann variables and orthogonality of coherent fermionic states

Bosonic (Grassmann-even) and fermionic (Grassmann-odd) coherent states are overcomplete bases, that are not orthogonal. One may show that the fermionic definition of coherent states$$ |\eta \rangle~:...
Qmechanic's user avatar
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