The integral of the axial current conservation law

Question

In the section 19.1 "The Axial Current in Two Dimensions" of Peskin & Shroeder.
On page 657,

In free fermion theory, the integral of the axial current conservation law gives
$$ \int d^2x\, \partial_{\mu}j^{\,\mu 5} =N_R -N_L \quad (=0) \tag{19.30} $$

What's this?
If (19.30) is correct, it follows that in the ordinary four-dimensional case,

\begin{cases} \displaystyle N_R=\int d^4x \, \partial_{\mu} \bar{\psi} \gamma^{\mu}\left(\frac{1+\gamma^5}{2}\right)\psi =\int d^4x \, \partial_{\mu}j_R^{\,\mu}, \\ \displaystyle N_L=\int d^4x \, \partial_{\mu} \bar{\psi} \gamma^{\mu}\left(\frac{1-\gamma^5}{2}\right)\psi =\int d^4x \, \partial_{\mu}j_L^{\,\mu}, \\ \displaystyle N=\int d^4 \, \partial_{\mu} \bar{\psi} \gamma^{\mu}\psi = \int d^4x \, \partial_{\mu}j^{\,\mu}. \end{cases}

These look weird to me.

Answer 1

One correcrion: these definitions are the time integrals over the time derivatives of the quantities $$ N(t)=\int d^{3}\mathbf r J^{0}(\mathbf r, t) $$ under an assumption that the spatial currents vanishes at infinity, $$ \int d^{4}\mathbf r \nabla \cdot \mathbf J =\oint d\mathbf S \cdot \mathbf J \to 0 $$ With these assumptions, the integral $$ \int d^{4}x \partial_{\mu}J^{\mu} = \int \limits_{t_{0}}^{t_{1}} dt \frac{dN}{dt} $$ of a time derivative of $N$ is reduced to the values of $N$ at the boundaries $t_1, t_0$ of the integration, namely, $$ N(t_{1})-N(t_0) = \Delta N $$

Answer 2

In four dimensions, The correct definition of $N_R ,\,N_L \, \text{and}\, N$ is \begin{cases} \displaystyle N_R=\int d^3 \mathbf{x} \, \bar{\psi} \gamma^{0}P_R \psi =\int d^3 \mathbf{x} \, \bar{\psi} \gamma^{0}\left(\frac{1+\gamma^5}{2}\right)\psi =\int d^3 \mathbf{x} \, j_R^{\,0}\,, \\ \displaystyle N_L=\int d^3 \mathbf{x} \, \bar{\psi} \gamma^{0}P_L \psi =\int d^3 \mathbf{x} \, \bar{\psi} \gamma^{0}\left(\frac{1-\gamma^5}{2}\right)\psi =\int d^3 \mathbf{x} \, j_L^{\,0}\,, \\ \displaystyle N=\int d^3 \mathbf{x} \, \bar{\psi} \gamma^{0}\psi = \int d^3 \mathbf{x} \, j^{\,0}\,. \end{cases} $N$ is the Noether charge $Q$ associated with the vector current conservation law.
Let $Q^5$ be the Noether charge associated with the axial current conservation law.
$$ Q^5= \int d^3 \mathbf{x} \, j^{\,05} =\int d^3 \mathbf{x} \, \bar{\psi} \gamma^{0}\gamma^{5}\psi = \int d^3 \mathbf{x} \,\left( \bar{\psi} \gamma^{0}P_R\psi - \, \bar{\psi} \gamma^{0}P_L\psi \right) =N_R-N_L $$ $$ \partial_0 Q^5=\partial_0 N_R - \partial_0 N_L= \int d^3 \mathbf{x} \, \partial_0 \bar{\psi} \gamma^{0}\gamma^{5}\psi = \int d^3 \mathbf{x} \, \partial_0 j^{\,05} = \int d^3 \mathbf{x} \, \partial_{\mu} j^{\,\mu 5}=0 $$ This is under an assumption that the spatial axial currents vanishes at infinity. $$ \int d^3 \mathbf{x} \,\partial_i j^{\,i5} =\oint d\mathbf S \cdot \mathbf j^5 \to 0 $$ $$ \therefore \Delta N_R - \Delta N_L =0 = \int d^4 x \, \partial_0 \bar{\psi} \gamma^{0}\gamma^{5}\psi = \int d^4 x \, \partial_{\mu} j^{\,\mu 5} $$

In two dimensions,
$$ Q^5= \int d x^1 \, \bar{\psi} \gamma^{0}\gamma^{5}\psi = \int d x^1 \,\left( \bar{\psi} \gamma^{0}P_R\psi - \bar{\psi} \gamma^{0}P_L\psi \right) =N_R-N_L $$ $$ \partial_0 Q^5=\partial_0 N_R - \partial_0 N_L= \int d x^1 \, \partial_0 \bar{\psi} \gamma^{0}\gamma^{5}\psi = \int d x^1 \,\partial_0 j^{\,05}=\int d x^1 \,\partial_{\mu} j^{\,\mu 5} =0 $$

Here the authors assume the spatial axial current vnishes at infinity.
$$ \int dx^1 \, \partial_1 j^{\,15} =\left[ j^{\,15}\right]_{-\infty}^{\infty} \to 0 $$

(19.30) should be replaced by $$ \int d^2x\, \partial_{\mu}j^{\,\mu 5} =\Delta N_R -\Delta N_L =0 .\tag{19.30} $$ So the sentence just below (19.30)

This relation implies that the difference in the number of right-moving and left-moving fermions cannot be changed in any possible process.

should be replaced by

This relation implies that the difference in the variation of right-moving and left-moving fermion numbers cannot be changed in any possible process.

The LHS of (19.39) should be also $$ \Delta N_R -\Delta N_L .$$