Application of Noether's theorem in Classical Field Theory Peskin's book I am currently reading 
Peskin's book about quantum field theory but really
struggling with Noether's theorem. I am trying to understand the following example
I have given the following Lagrange density
$$
\mathcal{L}=\left | \partial_{\mu} \varphi\right  |^2 - m^{2}\left | \varphi \right|^2,
$$
where $\partial_{\mu}$ is a 4-gradient, $m \in \mathbb{R}$ and $\varphi:\mathbb{R}^{3,1}\rightarrow \mathbb{C}$. Now the book tells me "This Lagrangian is invariant under the transformation 
$\varphi \rightarrow e^{i\alpha}\varphi$". The Noether current is given by
$$
j^{\mu}=\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\varphi)}\delta\varphi-\mathcal{J}^{\mu}
$$
where $\mathcal{J}^{\mu}$ is some vector field chosen such that
$$
\partial_{\mu}j^{\mu}=0.
$$
Now I would start by first computing $\delta \varphi$. For this I would define a function $\hat{\varphi}(\alpha,x)=e^{i\alpha}\varphi(x)$ and compute
$$
\delta \varphi = \frac{\partial\hat{\varphi}(\alpha,x)}{\partial \alpha}.{\huge{\mid}}_{\alpha=0}=i\varphi(x)
$$
The book also tells me I should consider $\varphi$ and $\varphi^{*}$(complex conjugate) as two independent fields.
Therefore I get two equations for $\delta\varphi$
$$
\delta \varphi=i\varphi \text{  and   }\delta \varphi^{*}=-i\varphi^{*}.
$$
Now I compute 
$$
\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\varphi)}=\partial_{\mu}\varphi^{*}
$$
therefore I obtain for the Noether current
$$
j^{\mu}=i(\partial_{\mu}\varphi^{*})\varphi-\mathcal{J}^{\mu}.
$$
And here I am completely stuck because I don't know how to determine $\mathcal{J}^{\mu}$.
The book tells me the answer has to be
$$
j^{\mu}=i\left[(\partial_{\mu}\varphi^{*})\varphi-(\partial_{\mu}\varphi)\varphi^{*}\right]
$$
Any help or hint is very much appreciated
 A: Perhaps it helps if one explains where the $\mathcal{J}^\mu$ comes from. In general, Noether's theorem only requires that such a symmetry (of the action) changes the Lagrangian by
$$\mathcal{L} \mapsto \mathcal{L} + \partial_\mu \mathcal{J}^\mu$$
By expanding $\mathcal{L}$ about a general variation of the fields $\delta \phi$, one can show that (a more complete discussion of what follows is available here)
$$\delta \mathcal{L} = \partial_\mu \left( \frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi_i)} \delta \phi_i \right)$$
where we sum over the indices on the fields $i$. But we also know $\delta \mathcal{L} = \partial_\mu \mathcal{J}^\mu$, so we have
$$0 = \delta \mathcal{L} - \partial_\mu \mathcal{J}^\mu = \partial_\mu \left[ \frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi_i)} \delta \phi_i - \mathcal{J}^\mu\right]$$
i.e. $j^\mu = \frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi_i)} \delta \phi_i - \mathcal{J}^\mu$ is conserved.
Note that in your case, we have two fields $$\phi_i = \begin{cases} \varphi & i = 0 \\ \varphi^* & i = 1 \end{cases}$$
so, we sum over the variations we found for both fields $\varphi$ and $\varphi^*$ (as you correctly found for $\varphi$) to get
$$j^\mu + \mathcal{J}^\mu = i (\partial^\mu \varphi^*) \varphi - i (\partial^\mu \varphi) \varphi^*$$
But in this case, our transformation leaves $\mathcal{L}$ invariant, i.e. there is no change at all in $\mathcal{L}$! That is, we have $\mathcal{J}^\mu = 0$, and this gives the desired result for $j^\mu$.
