Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Current of Complex scalar field

  1. Nov 1, 2014 #1
    I was trying to derive current for Complex Scalar Field and I ran into the following:


    So we know that the Lagrangian is:

    $$L = (\partial_\mu \phi)(\partial^\mu \phi^*) - m^2 \phi^* \phi$$
    The Lagrangian is invariant under the transformation:
    $$\phi \rightarrow e^{-i\Lambda} \phi $$ and $$\phi^* \rightarrow e^{i\Lambda} \phi^* $$

    Infinitesimal Transformation:
    $$\delta \phi = -i\Lambda \phi$$ and $$\delta \phi^* = i\Lambda \phi^*$$

    So, applying Noether's Theorem and using the Lagrangian above,

    I get

    $$J^{\mu} = \frac{\partial L}{\partial (\partial_\mu \phi} (-i\Lambda \phi) + \frac{\partial L}{\partial (\partial_\mu \phi^*} (i\Lambda \phi^*) =$$
    $$\partial ^\mu \phi^* (-i\Lambda \phi) + \partial _\mu \phi (i\Lambda \phi) = $$
    $$ i(\Lambda \phi^* \partial _\mu \phi - \Lambda \phi \partial ^\mu \phi^*)$$

    but as I googled it there is no $$\Lambda$$ in the final equation of the current. What did I do wrong?
     
  2. jcsd
  3. Nov 1, 2014 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    You did nothing wrong. The current is simply defined as the expression proportional to the infinitesimal group parameter.
     
  4. Nov 1, 2014 #3
    Ah phew! Thanks!!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook