Noether current in quantum field theory

CSpring432
Messages
1
Reaction score
0
Homework Statement
Finding Noether current for the given action
Relevant Equations
$$J^{\mu}=\frac{\partial L(\phi, \partial (\phi))}{\partial (\partial _{\mu}(\phi)}(\delta_{\alpha}\phi)-F^{\mu}$$
Hi

Have been trying to solve the below question for a while, wondered if anyone could help.

Considering the action

$$S=\int -\frac{1}{2}\sum^2_{n,m=1} (\partial^{\mu}\phi_{nm}\partial_{\mu}\phi_{mn}+m^2 \phi_{nm} \phi_{mn})dx$$
under the transformation

$$\phi'=e^{\alpha}\phi e^{-\alpha}$$

Find the infinitesimal transformation and associated Noether current, where both ##\alpha## and ##\phi## are real 2x2 matrices.

I've managed to find what (I think) is the infinitesimal transformation:

$$e^{\alpha}\phi e^{-\alpha}\approx \phi-\phi \alpha +\alpha\phi+ \mathcal{O}(\alpha^2)$$
$$\therefore \delta_{\alpha}=[\alpha, \phi]$$

I am however, stumped for calculating the Noether constant. I know that I would have to use the formula

$$J^{\mu}=\frac{\partial L(\phi, \partial (\phi))}{\partial (\partial _{\mu}(\phi)}(\delta_{\alpha}\phi)-F^{\mu}$$

The issue, I think, is calculating the covariant derivatives since the phi terms are matrix elements. Any help would be really appreciated.
 
Physics news on Phys.org
I think that ##\phi## represents four real fields ##\phi_{nm}## and the first term in the Noether current is
$$\sum^2_{n,m=1}\frac{\partial L(\phi, \partial (\phi))}{\partial (\partial _{\mu}(\phi_{nm}))}(\delta_{\alpha}\phi_{nm})$$
 
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
It's given a gas of particles all identical which has T fixed and spin S. Let's ##g(\epsilon)## the density of orbital states and ##g(\epsilon) = g_0## for ##\forall \epsilon \in [\epsilon_0, \epsilon_1]##, zero otherwise. How to compute the number of accessible quantum states of one particle? This is my attempt, and I suspect that is not good. Let S=0 and then bosons in a system. Simply, if we have the density of orbitals we have to integrate ##g(\epsilon)## and we have...
Back
Top