The expression for the 4-current $j^{\mu}$ in standard QED is $$ e\bar{\Psi}\gamma^\mu\Psi $$ and $$ \frac{e}{2 i}(\psi^\dagger D^\mu \psi - (D^\mu \psi)^\dagger \psi) $$ in scalar QED. I understand the second because in $k$-space it shows explicitely a transfer of momentum, and it contains a part proportional to $A^\mu$ which I am used to. So my first question is :

  1. Why these two currents are fundamentally so different and do not seem to represent the same thing? (e.g. why the second depends on the gauge potential and not the first ?)

In general, the (electrical) current is defined as the Noether current corresponding to (global) gauge invariance of a given Lagrangian. To me, the current on the other hand should be proportional to the velocity which is somehow fixed in the Lagrangian formalism. So the current should have a rather fixed form. My second question is:

  1. Why is this not the case? (or why the current should not be proportional to the velocity?)
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    $\begingroup$ 'the velocity' of what? We're not doing classical mechanics here! I thought your question was good after reading the first paragraph, but the second one didn't really do any good. $\endgroup$
    – Danu
    Commented Mar 9, 2015 at 10:13
  • $\begingroup$ observe that the solutions $\Psi$ to the Dirac equation in momentum space depend on $p^\mu$. The $\gamma$ matrices are then acting on these solutions and the result of $\gamma^\mu\Psi(p)$ of course depends on the point $p$ at which this matrix multiplication is evaluated. $\endgroup$
    – Phoenix87
    Commented Mar 9, 2015 at 10:31
  • $\begingroup$ The velocity of the particle. Do you mean there is no quantum counterpart to the classical velocity ? You're right, there is obviously a p dependence. What about the gauge potential dependence ? $\endgroup$
    – Damlatien
    Commented Mar 9, 2015 at 11:31
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    $\begingroup$ This question doesn't make sense - conserved currents/gauge coupling currents simply need not depend on any "velocity", and I don't see an argument to the contrary here. $\endgroup$
    – ACuriousMind
    Commented Mar 9, 2015 at 12:50
  • 2
    $\begingroup$ @ACuriousMind consider that the current acts as the source term in Maxwell's equations. Since that current is $q\mathbf v$ when the sources are classical particles, a mean-field limit should permit this interpretation. E.g. the EM field should respond to a wavepacket as if it were a classical moving charge distribution. $\endgroup$ Commented Mar 11, 2015 at 14:53

2 Answers 2


I think you bring up an interesting question, I'm not sure why there were some derogatory comments to this...

To see how velocity is incorporated in the Dirac current we need to unwrap some notation. The Dirac field can be expanded in its momentum modes: \begin{equation} \psi = \sum _s \int \frac{ d^3p }{ (2\pi)^3 } \frac{1}{\sqrt{2E_p}}u ^s ( p ) e ^{ i {\mathbf{p}} \cdot {\mathbf{x}} } \end{equation} where $ u ( p ) $ is a momentum dependent spinor which is derived in any QFT textbook and given by, \begin{equation} u ^s ( p ) = \left( \begin{array}{c} \sqrt{ p \cdot \sigma } \xi \\ \sqrt{ p \cdot \bar{\sigma} \xi } \end{array} \right) \end{equation} and a similar expression for the second set of solutions, $ v ^s $.

With these expressions, the Fourier transform of the current is given by, \begin{equation} \vec{ \tilde{ j}} = \frac{1}{2E_p}\bar{u} \vec{ \gamma } u \end{equation}

Here we are working in classical field theory. In general the field is made of a linear combination of all possible solutions, $ u ^{ \pm } _{\mathbf{p}} , v ^{ \pm } _{\mathbf{p}} $. For simplicity we assume the field is primarily composed of the $ u ^+ _{\mathbf{p}} $ component. Its important to keep in mind that here the field does not represent a particle, but simply a solution to the Dirac equation.

Then we can write the solutions at low energies by \begin{equation} u ( p ) \approx \underbrace{ \sqrt{ 2 m } \left( \begin{array}{c} 1 \\ 0 \\ 1 \\ 0 \end{array} \right)} _{ u _0 } + \frac{1}{ 2 \sqrt{ m}} \left( \begin{array}{c} - p _z \\ - p _x - i p _y \\ p _z \\ p _x + i p _y \end{array} \right) \end{equation} The first part is the static solution and the second piece is velocity dependent. If your intuition is correct that currents should depend on velocity then we would expect the pure static contribution to vanish. This is indeed the case, \begin{align} \vec{ \tilde{ j}} _0 & = \frac{1}{2E_p}\bar{u} {\vec \gamma} u \\ & = \frac{1}{2E_p}\bar{u} {\vec \gamma} u\left( \begin{array}{cccc} 1 & 0 & 1 & 0 \end{array} \right) \left( \begin{array}{cc} - {\vec \sigma} & 0 \\ 0 & {\vec \sigma} \end{array} \right) \left( \begin{array}{c} 1 \\ 0 \\ 1 \\ 0 \end{array} \right) \\ &= 0 \end{align} where we have used $E_p \approx m $ (and do so below as well).

Since this vanishes we need to keep terms to first order in the momenta. The lowest order contribution is,\begin{align} \vec{\tilde{j}}_1 & = \frac{1}{ 2\sqrt{2}m } \left( \begin{array}{cccc} 1 & 0 & 1 & 0 \end{array} \right) \left( \begin{array}{cc} - {\vec \sigma} & 0 \\ 0 & {\vec \sigma} \end{array} \right) \left( \begin{array}{c} - p _z \\ - p _x - i p _y \\ p _z \\ p _x + i p _y \end{array} \right) + h.c. \\ & =\frac{1}{2\sqrt{2}m} \left( \begin{array}{cc}1 & 0\end{array} \right) \vec{ \sigma } \left( \begin{array}{c} p _z \\ p _x + i p _y \end{array} \right) +h.c. \end{align} We see that the current is indeed proportional to the momenta as one would expect.

In particular, if $ p _x = p _y = 0 $ then, \begin{equation} \tilde{j} _x = \tilde{j} _y = 0 , \tilde{j} _z = \frac{1}{\sqrt{2}} e v \end{equation} where we have used $p/m = v $. I can't justify the factor of $ \sqrt{2} $ but everything else seems in order (I suspect this factor is due to poor choice of normalization for the fields).


From the Maxwell-Dirac Lagrangian $$ \mathcal L = -\frac{1}{2}F^2 + \overline{\psi}(i\gamma^\mu D_\mu +m) \psi $$ where $D_\mu$ is the gauge covariant derivative it is clear that the 4-current that acts as the source term in Maxwell's equations is $$ j^\mu_D = q\overline{\psi} \gamma^\mu \psi. $$ Using the Dirac equation it can be shown that (see, e.g., Sakurai, Advanced Quantum Mechanics (1967), Section 3.5) $$ j^\mu_D = j^\mu_1 + j^\mu_2 = \frac{iq}{2m}(\overline \psi D^\mu \psi - (D^\mu\overline\psi)\psi) - \frac{q}{2m}\partial_\mu (\overline{\psi}\sigma^{\mu\nu} \psi) $$ where $\sigma^{\mu\nu}$ is the spin tensor. $j_2^\mu$ is clearly conserved (divergenceles) since $\sigma^{\mu\nu}$ is antisymmetric. Therefore $j_1^\mu$ is also conserved, since the sum $j_D^\mu$ is.

The current $j_1^\mu$ is of the same form as what one finds for the Klein-Gordon field, and it is proportional to the probability current of the Dirac field, $\partial_\mu (\overline\psi\psi)$. By taking a plane wave solution you can convince yourself that the time component is the amplitude squared, and the spatial part is proportional to the amplitude times the momentum 3-vector. $j^i_1$ must be the free current.

The current $j_2^\mu$ on the other hand arises entirely from the Dirac field having non-zero spin. Its presence means that in Maxwell's equations we will have as sources not only something like charge times density and flux of particles -- the free charge and current -- but the divergence of an anti-symmetric tensor. Does that sound familiar? Maybe not, but if I call this tensor $D^{\mu\nu}$ and define two 3-vectors by $P^i = D^{0i}$ and $M^i = e^{ijk}M_{jk}$ the system of equations in 3-vector form is \begin{align*} \nabla\cdot \mathbf E & = \rho - \nabla\cdot\mathbf P\\ \nabla\times \mathbf B & = \mathbf j + \nabla\times \mathbf M + \frac{\partial \mathbf P}{\partial t} + \frac{\partial \mathbf E}{\partial t} \end{align*} What is $j_2^\mu$? Evidently it is the bound charge and current!

You can see that this is reasonable by considering a plane wave solution in its rest frame. Then you can work out that $\overline\psi \sigma^{ij} \psi = \epsilon^{ijk}s_k$ where $s_k$ is the spin 3-vector. Thus in Ampére's law will appear a term proportional to $$\partial_i \epsilon^{ijk} s_k= -\nabla\times \mathbf s.$$ Clearly $j_2^i$ is indeed the curl of the magnetization density.

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    $\begingroup$ Thank you very much for your answer. Shouldn't be $j^\mu_2$ rather be called the spin density and spin current ? $\endgroup$
    – Damlatien
    Commented Mar 11, 2015 at 19:04
  • $\begingroup$ And do you understand why the spin current does not depend on the gauge potential ? $\endgroup$
    – Damlatien
    Commented Mar 13, 2015 at 11:29
  • $\begingroup$ Because it does not contain derivatives of the field. To ensure gauge invariance, derivatives that do appear must be covariant derivatives. The gauge potential appears in the free current only to cancel out the gauge dependence in the field. $\endgroup$ Commented Mar 13, 2015 at 12:31

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