Is electric charge truly conserved for bosonic matter?

Even before quantization, charged bosonic fields exhibit a certain "self-interaction". The body of this post demonstrates this fact, and the last paragraph asks the question.

Notation/ Lagrangians

Let me first provide the respective Lagrangians and elucidate the notation.

I am talking about complex scalar QED with the Lagrangian $$\mathcal{L} = \frac{1}{2} D_\mu \phi^* D^\mu \phi - \frac{1}{2} m^2 \phi^* \phi - \frac{1}{4} F^{\mu \nu} F_{\mu \nu}$$ Where $D_\mu \phi = (\partial_\mu + ie A_\mu) \phi$, $D_\mu \phi^* = (\partial_\mu - ie A_\mu) \phi^*$ and $F^{\mu \nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$. I am also mentioning usual QED with the Lagrangian $$\mathcal{L} = \bar{\psi}(iD_\mu \gamma^\mu-m) \psi - \frac{1}{4} F^{\mu \nu} F_{\mu \nu}$$ and "vector QED" (U(1) coupling to the Proca field) $$\mathcal{L} = - \frac{1}{4} (D^\mu B^{* \nu} - D^\nu B^{* \mu})(D_\mu B_\nu-D_\nu B_\mu) + \frac{1}{2} m^2 B^{* \nu}B_\nu - \frac{1}{4} F^{\mu \nu} F_{\mu \nu}$$

The four-currents are obtained from Noether's theorem. Natural units $c=\hbar=1$ are used. $\Im$ means imaginary part.

Noether currents of particles

Consider the Noether current of the complex scalar $\phi$ $$j^\mu = \frac{e}{m} \Im(\phi^* \partial^\mu\phi)$$ Introducing local $U(1)$ gauge we have $\partial_\mu \to D_\mu=\partial_\mu + ie A_\mu$ (with $-ieA_\mu$ for the complex conjugate). The new Noether current is $$\mathcal{J}^\mu = \frac{e}{m} \Im(\phi^* D^\mu\phi) = \frac{e}{m} \Im(\phi^* \partial^\mu\phi) + \frac{e^2}{m} |\phi|^2 A^\mu$$ Similarly for a Proca field $B^\mu$ (massive spin 1 boson) we have $$j^\mu = \frac{e}{m} \Im(B^*_\mu(\partial^\mu B^\nu-\partial^\nu B^\mu))$$ Which by the same procedure leads to $$\mathcal{J}^\mu = \frac{e}{m} \Im(B^*_\mu(\partial^\mu B^\nu-\partial^\nu B^\mu))+ \frac{e^2}{m} |B|^2 A^\mu$$

Similar $e^2$ terms also appear in the Lagrangian itself as $e^2 A^2 |\phi|^2$. On the other hand, for a bispinor $\psi$ (spin 1/2 massive fermion) we have the current $$j^\mu = \mathcal{J}^\mu = e \bar{\psi} \gamma^\mu \psi$$ Since it does not have any $\partial_\mu$ included.

"Self-charge"

Now consider very slowly moving or even static particles, we have $\partial_0 \phi, \partial_0 B \to \pm im\phi, \pm im B$ and the current is essentially $(\rho,0,0,0)$. For $\phi$ we have thus approximately $$\rho = e (|\phi^+|^2-|\phi^-|^2) + \frac{e^2}{m} (|\phi^+|^2 + |\phi^-|^2) \Phi$$ Where $A^0 = \Phi$ is the electrostatic potential and $\phi^\pm$ are the "positive and negative frequency parts" of $\phi$ defined by $\partial_0 \phi^\pm = \pm im \phi^\pm$. A similar term appears for the Proca field.

For the interpretation let us pass back to SI units, in this case we only get a $1/c^2$ factor. The "extra density" is $$\Delta \rho = e\cdot \frac{e \Phi}{mc^2}\cdot |\phi|^2$$ That is, there is an extra density proportional to the ratio of the energy of the electrostatic field $e \Phi$ and the rest mass of the particle $mc^2$. The sign of this extra density is dependent only on the sign of the electrostatic potential and both frequency parts contribute with the same sign (which is superweird). This would mean that classicaly, the "bare" charge of bosons in strong electromagnetic fields is not conserved, only this generalized charge is.

After all, it seems a bad convention to call $\mathcal{J}^\mu$ the electric charge current. By multiplying it by $m(c^2)/e$ it becomes a matter density current with the extra term corresponding to mass gained by electrostatic energy. However, that does not change the fact that the "bare charge density" $j^0$ seems not to be conserved for bosons.

Now to the questions:

• On a theoretical level, is charge conservation at least temporarily or virtually violated for bosons in strong electromagnetic fields? (Charge conservation will quite obviously not be violated in the final S-matrix, and as an $\mathcal{O}(e^2)$ effect it will probably not be reflected in first order processes.) Is there an intuitive physical reason why such a violation is not true for fermions even on a classical level?
• Charged bosons do not have a high abundance in fundamental theories, but they do often appear in effective field theories. Is this "bare charge" non-conservation anyhow reflected in them and does it have associated experimental phenomena?
• I have answered all your questions in full detail. What do you not understand? Jun 16 '15 at 16:32

1. In contrast to QED with fermionic matter, in QED with bosonic matter, the full Noether current ${\cal J}^{\mu}$ (for global gauge transformations) tends to depend explicitly on the gauge potential $A^{\mu}$, see e.g. Refs. 1-2 and this Phys.SE post.

2. The reason for this difference is because the QED Lagrangian for fermionic (bosonic) matter typically contains one (two) spacetime derivative(s) $\partial_{\mu}$, which after minimal coupling $\partial_{\mu}\to D_{\mu}$ leads to e.g. no (a) quartic matter-matter-photon-photon coupling term, respectively.

3. The full Noether current ${\cal J}^{\mu}$ is a gauge-invariant and conserved quantity, $d_{\mu }{\cal J}^{\mu} \approx 0$. [Here $d_{\mu}\equiv\frac{d}{dx^{\mu}}$ means a total spacetime derivative, and the $\approx$ symbol means equality modulo eom.] The electric charge $Q=\int \! d^3x ~{\cal J}^{0}$ is a conserved quantity.

4. The only physical observables in a gauge theory are gauge-invariant quantities. The quantity $j^{\mu}$, which OP calls the "bare current", is not gauge-invariant, and hence not a consistent physical observable to consider.

5. As Trimok mentions in a comment, the situation for non-Abelian (as opposed to Abelian) Yang-Mills is radically different. The full Noether current ${\cal J}^{\mu a}$ (for global gauge transformations) is a conserved $d_{\mu }{\cal J}^{\mu a} \approx 0$, but ${\cal J}^{\mu a}$ is not gauge-invariant (or even gauge covariant), and hence not a consistent physical observable to consider. There is not a well-defined observable for color charge that one can measure. This follows also from Weinberg-Witten theorem (for spin 1): A theory with a global non-Abelian symmetry under which massless spin-1 particles are charged does not admit a gauge- and Lorentz-invariant conserved current, cf. Ref. 3.

References:

1. M. Srednicki, QFT, Chapter 61.

2. M.D. Schwartz, QFT and the Standard Model, Section 8.3 and Chapter 9.

3. M.D. Schwartz, QFT and the Standard Model, Section 25.3.

• Yes, some of these are the observations which lead me to this question. But say we have a macroscopic material with bosonic charged particles, object it to a very strong electrostatic field and measure it's charge. Would we have to be measuring $\mathcal{J}^0$ under all conditions? I guess 3. implies yes, and that means we would measure the object to have a charge different from the zero field situation. The extra "non-bare" charge obviously comes from the field, but this is a very different notion from the usual intuition of "charge".
– Void
Sep 24 '14 at 22:17
• ${\cal J}^{\mu}$ is a covariant quantity, then it should verify $D_\mu {\cal J}^{\mu}=0$, but a conserved quantity corresponds to $\partial_\mu {\cal J}^{\mu}=0$. So, here, are covariant and conserved current compatible notions ? (for instance, this is not the case in Yang-Mills theories). Sep 25 '14 at 11:50
• I updated the answer. Sep 28 '14 at 17:08
• 1. I don't see how this answers Trimok's question. Could you answer it more explicitly? 2. Also, your notation is nonstandard - is your "total spacetime derivative" $d_\mu$ different from the usual partial derivative $\partial_\mu$? Jul 23 '17 at 14:28
• 1. See point 5 in my answer. 2. Yes, the symbol $d_{\mu}$ is used to stress that it is a total derivative, i.e. it includes both implicit & explicit spacetime differentiations. Jul 23 '17 at 15:12

1) Yes, the charge is truly and exactly conserved. What it is confusing you, I think, is that the current for a scalar field explicitly depends on 4-potential $A$, whereas that for a spin-1/2 does not. This is obviously related to the number of derivatives in the Lagrangian kinetic term and, likewise, to the number of derivatives in the current. It can help you understand what it's going on to adopt the canonical formalism (also known as the language of gentlemen), in which in both cases the density (and also the charge) involves the product of the canonical momentum and the field, as it could not be otherwise because the charge is nothing else but the infinitesimal generator of $U(1)$ transformations for both the field and the canonical momentum.

Let's see the expressions. For spin-1/2 the charge density is: $$\rho_{1/2}\propto\psi ^{*}\, \psi$$ where $\psi ^{*}$ is the canonical conjugated momentum of $\psi$ (replace stars with daggers for operators).

For the complex scalar field: $$\rho_0\propto \phi^*\,D^0\phi - \phi\,D^0\phi^* =\phi^*\, \pi - \phi\,\pi^*$$

where $\pi^*$ is likewise the canonical momentum for $\phi$. Note that among the omitted prefactors there is a the imaginary unit $i$, so that the charge density is real.

It is obvios that the charge, defined as the integral over all space of the charge density, generates $U(1)$ phase transformations when acting on fields and momenta of both spins.

Now consider the Hamiltonian densities (the subscript refers to the spin) of these fields in an electromagnetic field:

$$\mathcal H_{1/2} = \psi^*(-\vec{\gamma}\cdot\vec D+ m )\,\psi + q\,A_0\,\rho_{1/2}$$ $$\mathcal H_{0} = \pi^*\pi+(\vec D\phi)^*\cdot(\vec D\phi) + m^2\phi^*\phi + q\,A_0\,\rho_{0}$$

Now we want to compute how the charge changes when the electromagnetic field changes. We can, for example, think of a situation where the electromagnetic field is an external one that depends on time (we can switch it on and off, vary its intensity, etc.) and the fields are connected with test particles. (Of course we know that the charge is a conserved quantity by Noether theorem, but we want to compute its variation explicitly). In both cases the variation of the density will be given by its commutator (or poisson bracket we are in the classical realm) with the Hamiltonian. Then it is easy to check by making use the canonical relations between conjugate pairs that for both spins, surprise, surprise, we get: $$\dot\rho = - \vec{\nabla} \cdot \vec{J}$$ And therefore, the charge is constant as long as the matter fields go to zero fast enough at infinity. This might be seen much faster from $U(1)$ invariance of the Hamiltonian, but I wanted to reproduced the continuity equation.

What the OP is perhaps missing is that besides the change in $A$, one has to considerer the change in the matter fields. That is, $A$ changes and the matter fields change in such a way that the charge doesn't change. It is not magic, it is like this by construction (the interaction is respecting the $U(1)$ symmetry). If rather than external electromagnetic field, one desires to considerer the internal one——that produced by the charged matter fields, then by simply making use of Maxwell equation, one gets the continuity equation again.

2) What you call the "bare charge", which probably is not a good name since this term is reserved for something else, lacks of physical content before fixing a gauge, as it is not a gauge invariant quantity. Note however that one can always choose one's favorite gauge. And if one picks the temporal gauge ($A_0=0$), the charge does not depend on the 4-potential ($D_0\phi=\partial _0 \phi$ in this gauge) and the form is the same as your "bare charge", which is conserved in this gauge.

3) The only difference in the movement of spin-one-half particles and spin zero-particles in an electromagnetic field is a term proportional to

$$\sigma_{\mu\nu}\, F^{\mu\nu}$$

in the equation for spin-1/2 particles. This term gives rise to the term

$$\bf{S}\cdot \bf{B}$$

in the non-relativistic limit, that is, the interaction between the spin of the particle and the magnetic field.

4) It can help you to get the equation in your answer to first think of the equation of motion in the non-relativist limit, which is the Schrödinger equation in an electromagnetic field, that is, the Schrödinger equation replacing partial derivatives with gauge-covariant ones (for scalar particles, for spin-1/2 there is the additional term I wrote above).

• 1) Consider a fixed number $N$ of $\phi$ particles of charge $e$. The EM field is coupled as $\partial_\mu F^{\mu \nu} = \mathcal{J}^\nu$ and the asymptotics of the field are thus $\sim Q/r$ far away from the bosons. Now increase the EM field, say by a homogeneous electrostatic field. So since the charge is exactly conserved, does the EM field increase imply that the $\phi$ particles vanish? Is it through $N$ or $e$ that the particles vanish? Why should this be different from the fermionic case?
– Void
Jun 16 '15 at 17:40
• The Hamiltonian formalism seems to simply cover up things in this case. If you take the charge operator in terms of creation and annihilation operators, there is simply something extra which has an unclear meaning. The quantization of charge and conservation of numbers of (particles-antiparticles) seems to be somehow in conflict with charge conservation.
– Void
Jun 16 '15 at 17:44
• 2) Yes, "bare charge" is not a good name. But temporal gauge is in conflict with putting all time derivatives equal to zero, since you have to have $\vec{E} \sim - \partial_0 \vec{A}$ etc. Furthermore, the charge would be simply zero in the case $\partial_0 \phi = 0$ and $\Phi =0$.
– Void
Jun 16 '15 at 17:47
• 3) We actually do not do the non-relativistic limit of $U(1)$ coupled bosons, we take the non-relativistic limit of Yukawa coupled bosons for which the Yukawa coupling is a consequence of the particles being really composite particles (such as pions) with the fermionic $U(1)$ coupling to EM fields.
– Void
Jun 16 '15 at 17:51
• @Void 1) When you switch on the external homog electrostatic field, the dynamics of the matter fields $\phi$ change in such a way that they compensate the change in $A_0$. This is the gauge covariant version in configuration space (no canonical momentum). Another way to see it is fixing the temporal gauge, where there is no explicit dependence on $A$. So the external homog field is just $A_0=0$, $\vec A=\vec E_0 \, t$. Neither particles disappear nor their charge vanish. They just change their motion. Jun 16 '15 at 18:09