6

As a hint, try to show that: $$\frac{1}{e^{E/T}+1} = \frac{1}{e^{E/T}-1} - \frac{2}{e^{2E/T}-1}.$$


5

"Tastes" is the name for the additional fermions produced by fermion doubling when putting actions with fermions on a lattice. "Taste symmetry" is a symmetry exchanging these additional fermions with each other. These "tastes" are unphysical and purely an artifact of the lattice theory - they have no relation to flavor except ...


3

I think the confusion arises out of a misunderstanding of "complex conjugation" in the Grassmann formalism. Grassmann "variables" are simply elements of an exterior algebra. If $V$ is a vector space, and $\xi_1,...,\xi_n$ some vectors, then $\xi_1 \cdots \xi_n$ is simply a way of writing the exterior product $\xi_1 \wedge \cdots \wedge \...


2

So I guess I'm ultimately asking why the formula for the partition function of N non interacting indistinguishable particles isn't working for bosons and fermions. What am I missing? You're not missing anything. $N!$ is a naive overcounting factor which is wrong for precisely the reasons you've worked out, both for bosons and for fermions. For fermions, $N!...


2

Experiments in high energy physics use statistical definitions in describing their data, and deciding how a fit of data to a hypothesis predicted can be compared with the theory. The usual standard deviation measure for all statistical quantities. This is usually done using monte carlo methods: generating a large number of artificial events that fit the ...


2

The masses come from, and are proportional to, the yukawa couplings to the Higgs field. So part of the question is, how to obtain yukawas that differ by so many orders of magnitude? I may have overlooked something, but all the examples I can think of, fall into one of two categories. Either the yukawa is an exponential function of something else - in which ...


1

If we swap particles $a$ and $b$ so that $a$ is in state $\psi_a ( \mathbf{r}_2)$ and $b$ is in state $\psi_b (\mathbf{r}_1)$ then the system of two particles is now in state $\psi_a (\mathbf{r}_2) \psi_b (\mathbf{r}_1)$ Since in general, $\psi_a$ and $\psi_b$ are different functions of $\mathbf{r}$, then $\psi_a (\mathbf{r}_2) \psi_b (\mathbf{r}_1) \ne \...


1

Allow me to use an analogy. Let me draw two imaginary chess boards. In the first, two pawns of the same color occupy opposite corners, whereas in the second one, it's one white and one black pawn. In the second board, the pawns clearly have an "identity". One is black, the other is white. You will not mix them at any time. If they switch positions, ...


1

The energy $E$ is related to the moment $p$ in the following way (with $c=1$) $$E=\sqrt{p^2+m^2}\rightarrow p=\sqrt{E^2-m^2}\tag{1}.$$ $d^3p$ is a infinitesimal volume in the $p$-space. Since $f(p)$ depends only in the magnitude of $\vec{p}$, we can use spherical coordinates, so $$d^3p=p^2\sin\theta \,dp\,d\theta\,d\phi$$ $$n=\frac{g}{2\pi^3}\int f(p)d^3p=\...


1

If we take, assuming isotropy on p: $$d^{3}p = 4\pi p^{2}dp, $$ and taking $c = 1$ in the energy-momentum relation: $$E^{2} = p^{2} + m^{2}. $$ We have: $$|\frac{dp}{dE}| = \frac{E}{\sqrt{E^{2} - m^{2}}}. $$ Then: $$d^{3}p = 4\pi p^{2}dp = 4\pi \frac{(E^{2} - m^{2})}{\sqrt{E^{2} - m^{2}}}EdE.$$ So, finally: $$d^{3}p = 4\pi E\sqrt{E^{2} - m^{2}}dE.$$


1

1. Are these state eigenstates? How can i calculate their energy? Act Hamiltonian on the states $\vert 00,00\rangle $ and $\vert 11,11\rangle $. The 4 number denotes occupation for site 1 spin-up, site 1 spin-down, site 2 spin-up, and site 2 spin-down: $$ H=\sum_{\sigma=\uparrow,\downarrow}[\epsilon_1 c_{1\sigma}^\dagger\,c_{1\sigma} + \epsilon_2 c_{2\...


1

You are not alone in this. I tend to regard the Grassmann integral as tool for combinatorics, but if you want a deeper view and a discussion of analytic subtleties you might like to read Martin R. Zirnbauer, Riemannian symmetric superspaces and their origin in random-matrix theory, J. Math. Phys.\ 37 (1996) 4986; arXiv:math-ph/9808012.


1

Term decay is rather rare in condensed matter physics, so I will assume that what you really mean is the finite lifetime. Condensed matter studies complex many-paryicle systems. For example, description of a crystal in terms of electron Bloch waves is valid only under assumption of a a rigid lattice and absence of Coulomb interaction. Once the interactions ...


Only top voted, non community-wiki answers of a minimum length are eligible