Current of Complex scalar field

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The discussion focuses on deriving the current for a complex scalar field using Noether's Theorem. The Lagrangian provided is invariant under specific transformations of the scalar field and its conjugate. The user initially derived the current but was confused about the absence of the infinitesimal transformation parameter, Lambda, in the final expression. It was clarified that the current is defined as the expression proportional to this infinitesimal group parameter, confirming the user's derivation was correct. This highlights the importance of understanding the role of transformation parameters in field theory.
PhyAmateur
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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?
 
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You did nothing wrong. The current is simply defined as the expression proportional to the infinitesimal group parameter.
 
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Ah phew! Thanks!
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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