Neues vom Nordkap

11.04.2006 13:44

Swiss elections

Switzerland did very well in the elections on the weekend. First of all they voted for the Centrosinistra. If you bear in mind that Switzerland has a large Italian community (There are more votes in absolute number from Switzerland than from Germany for instance) and that the italiani al estero seem to have a large impact, then you can see how much Switzerland influences international politics. ;-).
And even if there is not much reason for real Freude about the result of the Italian elections, there is at least some reason for Schadenfreude, since Berlusconis law which gives 55% of the seats in the camera to the winning coalition no matter how narrow their advantage is, seems to bite back at himself.
However, the real miracle happened elsewhere, in my very homegrounds. Would you ever have guessed that Bern becomes the 4th canton with a green/red majority in the government. Only after Neuchatel, Geneva and Basel City. Bern is a large canton, with a lot of rural and mountaneous areas. Not exactly the place where the left is strong traditionally.

04.04.2006 17:39

naive fermions vs staggered fermions

If you discretize fermions by just replacing the derivative by an finite difference you get so called naive fermions. The name arises because the discretization scheme is the simplest you could think of. With these fermions you will get 2^d instead of one species in the continuum limit, which is not precisely what you want. Moreover, if you define an infintesimal axial U(1) transformation in the most straightforward way by

$\delta \psi = i\alpha \gamma_5\psi \hspace{2cm} \bar{\psi} = i\alpha \bar{\psi} \gamma_5$

and discretize the axial current by

$J_\mu^5(x)=\frac{1}{2}\left[\bar{\psi}(x +\hat{\mu})U_\mu^\dag(x)\gamma_\mu\gamma_5 \psi(x)+ h.c.\right]$

then there is no axial anomaly term for naive fermions ad the axial current $J_\mu^5(x)$ is exactly preserved in the massless limit. Again this is not what you want. However, if I understood this paper by Peter Weisz et al. you can define an axial transformation by

$\delta \psi = i\alpha \gamma_5\psi(x +\xi) \hspace{2cm} \bar{\psi} = i\alpha \bar{\psi}(x +\xi) \gamma_5$

where $\xi$ is a vector whith $\pm a$ in every component and correspondingly define an axial current
$\tilde{J}_\mu^5(x) = \frac{1}{2}\left[\bar{\psi}(x +\hat{\mu},\Xi)U_\mu^\dag(x)\gamma_\mu\gamma_5 \psi(x)+ h.c.\right].$
There $\psi(x, \Xi) = \frac{1}{2^d}\sum_\xi \prod_y U_\mu(y)\psi(x +\xi).$ The sum is over all $\xi$ vectors and the product over $y$ involves all link variables on a path fron $x$ to $x +\xi$ . This transformation will produce the correct anomaly term in the continuum limit.
There is (among others) a alternative descretization scheme called staggered fermions, where you can reduce the number of Dirac components per lattice sites to 1 through a number of transformations involving multiplications of the original fields $\bar{\psi}(x), \psi(x)$ with an appropriate number of Dirac matrices.
Now, as far as I understood, staggered fermions are equivalent to naive ones, just that there are less species, namely only 4 (in 4d). Then you have to define the axial transformation/current like in the naive case in order to get the anomaly. Do people do this? Or is it not necessary to do it in actual calculations, and it is sufficient that you know there is in principle a way to define the transformation such that there is an anomaly? Furthermore, why do people use staggered and not just naive fermions? Only for practical reasons, because they are maybe cheaper to calculate because you have only one component per site? Or is there a conceptual advantage in using them? To boil it down why would you think there is no problem in taking the 4th root od the 4 taste staggered determinant in order to get one flavour, but there is one in taking the 16th of naive the naive fermion determinant?