I am trying to understand the subsection on Noether's Theorem in the second chapter of Peskin's as Schroeder's book. I am trying to understand the simplest example they give as follows (I am going to try to be as detailed as possible in order to "debug" my thoughts easily):
Consider the Lagrangian $\mathcal{L}(\phi,\partial_\mu\phi)=\frac{1}{2} (\partial_\mu\phi)^2$ and the transformation $T$ given by $T(\phi)=\phi'$ where $\phi'(x)=\phi(x)+\alpha$ with $\alpha\in\mathbb{R}$.
Notice that $$ \mathcal{L}'(\phi,\partial_\mu\phi):=\mathcal{L}(\phi',\partial_\mu\phi')= \frac{1}{2} (\partial_\mu\phi')^2=\frac{1}{2} (\partial_\mu(\phi+\alpha))^2=\frac{1}{2} (\partial_\mu\phi)^2 $$
So $\mathcal{L}'=\mathcal{L}+\alpha\partial_\mu \mathcal{J}^\mu$ with $\partial_\mu \mathcal{J}^\mu=0$. Notice that any "divergenceless" $4$-vector field $\mathcal{J}$ will do.
Using Noether's Theorem directly as stated in Peskin's and Schroeder's book, we see that $$ \partial_\mu j^\mu=0 \,\,\,\,\text{ where }\,\,\,\, j^\mu=\partial_\mu \phi\Delta\phi-\mathcal{J}^\mu $$ because $\frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)}=\partial_\mu \phi$
Now, the conclusion of the authors is that $j^\mu=\partial_\mu\phi$, but I do not see why: it seems to me that any $\mathcal{J}$ will do, the job, and even if one chooses $\mathcal{J}=0$, one gets $j^\mu=\partial_\mu\phi\Delta\phi$.
The definition given in the Peskin's and Schroeder's book for a general continuous transformation on the field $\phi$ is: $$ T(\phi)=\phi′\:\:\:\:\:\: where \:\:\:\:\:\:\phi′(x)=\phi(x)+\alphaΔ\phi(x)$$
In this example, as you said, the transformation is $$ϕ′(x)=ϕ(x)+α$$ so the term $Δϕ(x)=1$.