- #1
latentcorpse
- 1,411
- 0
Given the Lagrangian
[itex]L=i \psi^\star \dot{\psi} - \frac{1}{2m} \nabla \psi^\star \nabla \psi[/itex]
which has an internal symmetry [itex]\psi \rightarrow e^{i \alpha} \psi[/itex] so [itex]\delta \psi = i \psi[/itex] (am I correct in saying that we omit the infinitesimal parameter [itex]\alpha[/itex] here because we don't want it appearing in the Noether current?)
Anyway I want to show the Noether current is given by
[itex]j^\mu = \left( \psi^\star \psi , \frac{i}{2m} ( \psi^star \nabla \psi - \psi \nabla \psi^star ) \right)[/itex]
so my formula for the noether current is
[itex]j^\mu = \frac{\partial L}{\partial ( \partial_\mu \psi)} \delta \psi - F^\mu[/itex] where [itex]F^\mu satisfies [itex]\delta L = \partial_\mu F^\mu[/itex]<br /> <br /> now if i set [itex]\mu =0[/itex] and work through i get [itex]j^0=- \psi^star \psi[/itex] does anyone know how to get rid of this minus sign?<br /> <br /> and if i work on the [itex]\mu=i[/itex] components i get it to work out using the formula for [itex]j=i[/itex] \bove added to the formula for [itex]j=i[/itex] where I replace [itex]\psi[/itex] with [itex]\psi^\star[/itex]<br /> <br /> so my only real problems are <br /> a) how to get rid of the minus sign in the 0th component<br /> b)why do we get rid of the infinitesimal parameter when writing out [itex]\delta \psi[/itex] - is it so it doesn't appear in [itex]j^\mu[/itex] and if so, why do we not want it in j^\mu[/itex]
[itex]L=i \psi^\star \dot{\psi} - \frac{1}{2m} \nabla \psi^\star \nabla \psi[/itex]
which has an internal symmetry [itex]\psi \rightarrow e^{i \alpha} \psi[/itex] so [itex]\delta \psi = i \psi[/itex] (am I correct in saying that we omit the infinitesimal parameter [itex]\alpha[/itex] here because we don't want it appearing in the Noether current?)
Anyway I want to show the Noether current is given by
[itex]j^\mu = \left( \psi^\star \psi , \frac{i}{2m} ( \psi^star \nabla \psi - \psi \nabla \psi^star ) \right)[/itex]
so my formula for the noether current is
[itex]j^\mu = \frac{\partial L}{\partial ( \partial_\mu \psi)} \delta \psi - F^\mu[/itex] where [itex]F^\mu satisfies [itex]\delta L = \partial_\mu F^\mu[/itex]<br /> <br /> now if i set [itex]\mu =0[/itex] and work through i get [itex]j^0=- \psi^star \psi[/itex] does anyone know how to get rid of this minus sign?<br /> <br /> and if i work on the [itex]\mu=i[/itex] components i get it to work out using the formula for [itex]j=i[/itex] \bove added to the formula for [itex]j=i[/itex] where I replace [itex]\psi[/itex] with [itex]\psi^\star[/itex]<br /> <br /> so my only real problems are <br /> a) how to get rid of the minus sign in the 0th component<br /> b)why do we get rid of the infinitesimal parameter when writing out [itex]\delta \psi[/itex] - is it so it doesn't appear in [itex]j^\mu[/itex] and if so, why do we not want it in j^\mu[/itex]