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The Dirac Lagrangian reads: \begin{equation*} \mathcal{L} = \bar\psi(i\not\partial-m)\psi.\tag{1} \end{equation*} It's invariant under the transformation $\psi(x)\rightarrow e^{i\alpha}\psi(x)$. Now suppose we perform the transformation $\psi(x)\rightarrow e^{i\alpha(x)}\psi(x)$ instead. We can find \begin{equation*} \mathcal{L}\rightarrow \bar\psi e^{-i\alpha(x)}(i\not\partial-m) e^{i\alpha(x)}\psi\\ = \mathcal{L}+\bar\psi e^{-i\alpha(x)}(i\not\partial e^{i\alpha(x)})\psi\\=\mathcal{L}-(\bar\psi\gamma^\mu\psi)\partial_\mu \alpha.\tag{2} \end{equation*} I'm trying to compare the above derivation to equation 2.10 on Peskin and Schroeder: $$ \mathcal{L(x)}\rightarrow\mathcal{L(x)}+\alpha\partial_\mu\mathcal{J^\mu(x)}.\tag{2.10} $$ Thus it seems like we have the relation

$$ \mathcal{J^\mu(x)} = -\bar\psi(x)\gamma^\mu\psi(x)\tag{3} $$ Is that correct? How do I find $$j^\mu(x) = \bar\psi(x)\gamma^\mu\psi(x)\tag{4}$$ for Dirac equation? Is there a way we can show that $$\partial_\mu j^\mu(x) = 0~?\tag{5}$$ I think I got confused about how to transform from $\mathcal{J^\mu(x)}$ to $j^\mu(x)$.

Qmechanic
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IGY
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3 Answers3

3

If $j^{\mu}$ is conserved, so is $- j^{\mu}$, and hence the sign is meaningless.

As to prove that $\partial_\mu j^\mu = 0$, notice that this is necessary for the action to be invariant under the transformation. If the current is not conserved, the transformation will change the action. If you prefer, a more detailed proof of Noether's theorem in the general case can be found in Nivaldo Lemos's Analytical Mechanics, specifically in the chapter on field theory.

Níck Aguiar Alves
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How do I find $j^\mu(x) = \bar\psi(x)\gamma^\mu\psi(x)$ for Dirac equation?

The Dirac Equation is: $$ \frac{1}{c}\frac{\partial \psi}{\partial t} + \vec{\alpha}\cdot \vec\nabla\psi + \frac{imc}{\hbar}\beta\psi = 0\;, $$ where $\vec\alpha^\dagger = \vec \alpha$ and $\beta^\dagger = \beta$ are Hermitian 4x4 matrices.

The probability density is: $$ \rho = \psi^\dagger\psi\;. $$

Taking the time derivative and applying the Dirac equation gives: $$ \frac{\partial \rho}{\partial t} = c\vec\nabla\cdot\psi^\dagger\vec\alpha\psi\;, $$ which shows that $$ \vec j = c\psi^\dagger\vec \alpha \psi $$ and that the four-current density is: $$ j^\mu = (c\rho, \vec j) = c\bar\psi\gamma^\mu\psi\;, $$ where $\gamma^0=\beta$ and $\vec\gamma = \beta\vec\alpha$.

hft
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1

  1. First of all, it should be stressed that different authors have different sign conventions

    • for the Noether current,

    • and independently for the gauge parameter $\alpha$.

  2. Be aware that there are at least 2 different methods to find the Noether current for a global symmetry:

    • On one hand, OP is using the trick of an $x$-dependent $\alpha$ parameter to infer the full Noether current.

    • On the other hand, P&S is keeping the $\alpha$ parameter $x$-independent. Therefore ${\cal J}^{\mu}=0$ vanishes for the Dirac theory (1) according to P&S. The full Noether current is later given in P&S as $$ j^{\mu}(x)~=~{\cal L}\frac{\stackrel{\leftarrow}{\partial} }{\partial \partial_{\mu}\psi}\Delta\psi - {\cal J}^{\mu}.\tag{2.12}$$

  3. If one does try to compare eqs. (2) and (2.10), don't forget there's a minus sign from integration by parts.

  4. See also this related Phys.SE post.

Qmechanic
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