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Is electric charge truly conserved for bosonic matter?

+ 6 like - 0 dislike
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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?
  • Extra clarifying question: Say we have $10^{23}$ bosons with charge $e$ so that their charge is $e 10^{23}$. Now let us bring these bosons from very far away to very close to each other. As a consequence, they will be in a much stronger field $\Phi$. Does their measured charge change from $e 10^{23}$? If not, how do the bosons compensate in terms of $\phi, B, e, m$? If this is different for bosons rather than fermions, is there an intuitive argument why?


This post imported from StackExchange Physics at 2015-06-09 14:50 (UTC), posted by SE-user Void

asked Sep 24, 2014 in Theoretical Physics by Void (1,505 points) [ revision history ]
edited Jun 9, 2015 by Void
Most voted comments show all comments

By Noether's theorem, Noether currents are  conserved since they are derived from an infinitesimal symmetry; they are observable iff they are gauge invariant. Are you missing something in the answer by Qmechanic?

@ArnoldNeumaier I added an extra clarifying question to what bugs me. I am well aware about the conservation and observability, I mainly wanted to inquire about the deeper physical explanation of these facts.

The charge doesn't change, as it is an integral over the whole space - only the charge density develops a very localized peak. What should need compensation?

Note that bare stuff doesn't matter; it is irrelevant scaffolding removed by renormalization.

Just a dumb idea: Maybe this is somehow related to the fact in the SM, introducing mass-terms for the bosons simply as $\frac{1}{2}m\phi^{*}\phi$ without a higgs field or mechanism breaks the gauge symmetry, and therefore is no conserved current corresponding to the by the mass term broken symmetry?

@Dilaton: Yes, there seems to be something funky about massive or charged elementary bosons. I was just hoping there is an established argument what exactly is the crux of this funkiness -- perhaps through such things as charged pions and their relation to $U(1)$.

Most recent comments show all comments

@drake I just meant that for example Proca mass terms such as $\frac{1}{2}m^2 B^{*\nu} B_{\nu}$ break gauge symmetries such as $U(1)$, and could therefore spoil charge conservation.

@Dilaton I don't get your point... Here the gauge field is $A$, which doesn't have any mass term. In the SM one wants to give mass to the gauge fields. I think you are wrong.

4 Answers

+ 3 like - 0 dislike

Comments to the question (v3):

  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.

This post imported from StackExchange Physics at 2015-06-09 14:50 (UTC), posted by SE-user Qmechanic
answered Sep 24, 2014 by Qmechanic (2,790 points) [ no revision ]
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".

This post imported from StackExchange Physics at 2015-06-09 14:50 (UTC), posted by SE-user Void
${\cal J}^{\mu}$ 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).

This post imported from StackExchange Physics at 2015-06-09 14:50 (UTC), posted by SE-user Trimok
I updated the answer.

This post imported from StackExchange Physics at 2015-06-09 14:50 (UTC), posted by SE-user Qmechanic
+ 1 like - 0 dislike

I have actually taken the time to compute the equations of motion and the situation is more complicate than I previously thought. The Lagrangian in the static situation $\vec{A} = 0, \partial_t \to 0$ reads

$$\mathcal{L} = -\frac{1}{2} |\nabla \phi|^2 - \frac{1}{2} m^2 |\phi|^2 + e^2 |\Phi|^2 |\phi|^2 + \frac{1}{2} |\nabla \Phi|^2 $$

which leads to EOM:
$$(\Delta - m^2 + 2 e^2 |\Phi|^2) \phi = 0$$

$$ (\Delta - 2 e^2 |\phi|^2) \Phi = 0 $$

Amongst other things, this implies that minimally coupled bosons do not act as a usual source of the electromagnetic field at all. As it stands (a more detailed analysis of the non-stationary equations might show otherwise), the bosons actually "easen" their motion (effectively loose mass) in the presence of the electromagnetic field at the cost of weakening (rendering massive and short-range) the electromagnetic field.

The coupling constant $e$ really does not have any reasonable interpretation in terms of a usual charge. For instance, the sign of $e$ is irrelevant and the particles and antiparticles of quantized $\phi$ have the same effect on $\Phi$.  The $U(1)$ charge is just a conserved quantity with no intuitive interpretation in terms of the usual charge. Hence, the original form of the question does not have a proper meaning; $U(1)$ coupling for bosons simply means something totally different than for fermions.

(If you have any more observations or a different view, please contribute, I am interested.)

answered Jun 10, 2015 by Void (1,505 points) [ no revision ]

Are you allowed to simply put $A=0$? It changes the dynamics.

@ArnoldNeumaier: If we still hold $\partial_t \to 0$ a nonzero $\vec{A}$ would only make $|\Phi|^2 \to |\Phi|^2 - |A|^2$ and an extra $\vec{A}$ equation coupled to $\phi$ similarly as in the $\Phi$ case.

+ 0 like - 0 dislike

Dear mods, I am sorry this answer is not graduate-upward level, but I have not been able to come up with a more sophisticated one.

1) Yes, the charge is truly conserved but the respective current depends on the 4-potential A. What it is confusing you, I think, is that the current for a scalar field depends on 4-potential $A$, whereas that of 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 the charge too) 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.

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 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). 

answered Jun 12, 2015 by drake (885 points) [ revision history ]
edited Jun 12, 2015 by drake
+ 0 like - 4 dislike

The charge $e$ introduced into your Lagrangians/equations is a constant in time by definition, no Noether theorem is necessary to "conserve" it: $\frac{de}{dt}=0$.

Another thing is your equations/theory or "charge definition" via equations/solutions (as an integral bla-bla-bla). Here everything depends on your equations. Do not think that equations for bosons are already well established and finalized. For one formulation you get one result, for another you do another. So, there is no 'truly" thing, keep it firmly in your mind!

answered Jun 9, 2015 by Vladimir Kalitvianski (22 points) [ revision history ]
edited Jun 9, 2015 by Vladimir Kalitvianski

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