# Origin of charge

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1. May 15, 2015

### driftwood

I understand that the interaction with the Higgs field (Higgs Bosons) confers mass to elementary particles such as the electron. Does that mean that interaction with an electromagnetic field (photons) confers charge to electrons? If not, what is the origin of charge? Also, how is it that the strength a proton's charge is the same as an electron's, considering electrons aren't composed of quarks?

2. May 15, 2015

### Staff: Mentor

No - interaction with the EM field does not confer charge - in fact charge is the origin of the EM field.

The origin of charge is a deep result of QM and gauge symmetry:
http://quantummechanics.ucsd.edu/ph130a/130_notes/node296.html
http://en.wikipedia.org/wiki/Charge_conservation

There is a deep theorem called Noethers theorem that, roughly, says to every symmetry there is a quantity, called the Noether charge, and that quantity is conserved. It turns out the conserved quantity associated with gauge symmetry is charge:

I was in two minds about giving the above link on the detail because the math is advanced and may confuse more than illuminate - but decided to for completeness.

Of course now you need to know why gauge symmetry is true.

Why charge comes in the quantised values it does is a mystery - although I suspect string theory has something to say about it.

Thanks
Bill

Last edited: May 15, 2015
3. May 15, 2015

### atyy

The standard terminology is a bit weird about what a "gauge symmetry" is. I would rather say that charge conservation is not due to a gauge symmetry, but a global symmetry. In the Lagrangian formulation, the global symmetry is a subset of "gauge symmetries", which can be both global and local. However, in the Hamiltonian formulation, one can see the difference between the global symmetry and the gauge symmetry. The conceptual difference is that a gauge symmetry is a redundancy in the description and can never be spontaneously broken. On the other hand, a global symmetry can be spontaneously broken, eg. the BCS state. http://arxiv.org/abs/cond-mat/0503400 (DrDu thinks the article by Greiter is wrong, but I believe it is correct.)

A famous theory about charge quantization is Dirac's monopole theory.
http://www.theory.caltech.edu/~preskill/pubs/preskill-1984-monopoles.pdf

4. May 16, 2015

### haael

In Kaluza-Klein theory and many string theories, charge is a component of momentum in fifth dimension.

In the electroweak theory, electric charge is just a linear combination of other fundamental charges. The specific linear combination is selected by electroweak symmetry breaking by Higgs field. Speaking informally, there are infinite many "charges" taken as linear combinations of more fundamental ones. One of them (electric charge) is the strongest and few of them (weak charges) are the weakest.

5. May 16, 2015

### stevendaryl

Staff Emeritus
Well the standard argument that I've seen goes like this (let me do it nonrelativistically)

Schrodinger's equation for a free particle in 1 spatial dimension looks like this:

$(\frac{-\hbar^2}{2m} \frac{\partial^2}{\partial x^2} - i \hbar \frac{\partial}{\partial t}) \psi = 0$

Since the phase of the wave function is unobservable, there is a global symmetry:

$\psi \rightarrow e^{i \chi} \psi$

where $\chi$ is any real constant. To make this a local symmetry means allowing $\chi$ to vary with position and time; $\chi = \chi(x,t)$. But the equation is not invariant under such a change. Making such a change produces additional terms:

$(\frac{-\hbar^2}{2m} \frac{\partial^2}{\partial x^2} - i \hbar \frac{\partial}{\partial t}) \psi \rightarrow$
$(... + \frac{-i \hbar^2}{2m} \frac{\partial^2}{\partial x^2} \chi + \frac{\hbar^2}{2m} (\frac{\partial}{\partial x} \chi))^2 + \frac{i \hbar^2}{m} \frac{\partial \chi}{\partial x} \frac{\partial}{\partial x}) \psi$

So the free particle equation is not invariant under such a local change. But if we modify the original equation this way:

$(\frac{1}{2m} (-i \hbar \frac{\partial}{\partial x} - e A)^2 + (-i \hbar \frac{\partial}{\partial t} + e\Phi)) \psi = 0$

then this equation is invariant under the following transformation:

$\psi \rightarrow e^{i \chi} \psi$
$A \rightarrow A + \frac{\hbar}{e} \frac{\partial}{\partial x} \chi$
$\Phi \rightarrow \Phi + \frac{-\hbar}{e} \frac{\partial}{\partial t} \chi$

(or something like that).

So this more complicated symmetry, involving the phase $\chi$ and the "gauge fields" $A$ and $\Phi$ is called a "gauge symmetry", and it's local.

6. May 16, 2015

### atyy

This argument is called the "gauge principle", and it is conceptually the same as the "equivalence principle" in general relativity, or minimal coupling. It is not the unique way in which gauge invariance can be satisfied, just as classical general relativity is not the unique diffeomorphism invariant theory.

Anyway, that argument is about the introduction of the EM field which then makes the original particle a "charge". But my comment was about bhobba's additional comments on charge conservation.

7. May 16, 2015

### stevendaryl

Staff Emeritus
Yes, but the issue was whether charge conservation is due to a global or local symmetry. As I understand it, charge is the conserved Noether current corresponding to the local gauge symmetry.

8. May 16, 2015

### atyy

In my understanding it is due to a global symmetry, and not due to a gauge symmetry nor to a global gauge symmetry. (In the path integral language, I don't believe there is a difference between a global symmetry and a global gauge symmetry, but the path integral is only shorthand for the Hamiltonian formulation.)

There is a discussion of the issue in http://arxiv.org/abs/cond-mat/0503400 (DrDu doesn't agree with the article, but I went through it carefully and I believe it is right).

"The difference between the “physical” symmetry (98) and the gauge symmetry (27) can also be appreciated at the level of conservation laws. The former yields particle number (or charge) as a conserved quantity, according to (42), while there is no conservation law associated with the latter. In the literature, (98) is often referred to as a global gauge transformation, and the conservation of charge attributed to gauge invariance. This view, however, is not consistent. If one speaks of a global gauge symmetry, this symmetry has to be a proper subgroup of the local gauge symmetry group. ..."

Last edited: May 16, 2015
9. May 16, 2015

Staff Emeritus
Is there anyone here who thinks this is not above the OP's head?

10. May 16, 2015

### vanhees71

I think atyy is right (also with his opinion on Greiter's article, which I think is a very good one). I forgot, what the debate about it was about, but I also read it pretty carefully, and couldn't find obvious flaws.

On the other hand, since a theory that is invariant under local gauge symmetries, it is also invariant under the global one. This implies the applicability of the Noether theorem and thus the conservation of the corresponding Noether charges, and this in turn implies that the conservation of these charges is a necessary but not sufficient condition for local gauge invariance.

There's a very important difference in spontaneous symmetry breaking of global and local gauge symmetries. If you break the global gauge symmetry that is not a special case of a more comprehensive local gauge symmetry, the QFT necessarily contains massless (zero) modes in the physical spectrum, the famous Nambu-Goldstone bosons related to this spontaneously broken symmetry. How many of such Goldstone bosons occur depends on the group structure of the original group and the unbroken subgroup that acts among the eigenvectors of the (degenerate!) groundstate of the system. One important example related to elementary-particle physics is the light-quark sector of QCD, which has an approximate chiral symmetry. Taking only up and down quarks as "light", we have an $\mathrm{SU}(2)_L \times \mathrm{SU}(2)_R$ global (chiral) gauge symmetry (in the limit of massless quarks). The strong interaction leads, however, to the formation of a quark condensate, i.e., the vacuum expectation value $\langle \overline{\psi} \psi \rangle \neq 0$. This VEV is invariant under the vector transformations, $\psi \rightarrow \exp(-\mathrm{i} \vec{\alpha} \cdot \vec{\tau}) \psi$ but not under the axial-vector transformations $\psi \rightarrow \exp(-\mathrm{i} \vec{\beta} \cdot \vec{\tau} \gamma^5) \psi$. Thus the original symmetry is spontaneously broken to the usual strong-isospin subgroup $\mathrm{SU}(2)_V$. The corresponding Goldstone modes are the pions. Their non-zero mass is only due to the (small) current quark masses of the light quarks u and d.

On the other hand, a chiral symmetry $\mathrm{SU}(2)_{\text{wiso}} \times \mathrm{U}(1)_{\text{Y}}$ is also at work in the electroweak sector of the Standard Model, but this chiral symmetry is also local. In a loose sense you can say, the Higgs sector breaks this symmetry spontaneously, but the implications are totally different from a case of a spontaneously broken symmetry thats "only" local! There are no Nambu-Goldstone modes, but they are in some sense "eaten up" by the gauge fields. In the case of the weak-isospin symmetry the symmetry-breaking pattern is from the original symmetry to a remaining $\mathrm{U}(1)_{\text{em}}$. Thus there would be 3 Goldstone modes, corresponding to the continuous degeneracy of the ground state, if the symmetry was only global. But now in the electroweak theory this symmetry is local, and thus you can parametrize the Goldstone modes (here more accurately dubbed "would-be-Goldstone modes") as a gauge-transformation unitary matrix, and this lumps these degrees of freedom into the gauge fields, when choosing a special gauge fixing, known as "unitary gauge" (for reasons that become obvious) soon. So first of all in this gauge the would-be-Goldstones do not occur as massless particles in the physical spectrum described by the theory but they become the (spatially) longitudinal part of the gauge bosons, and in turn these become massive. The gauge symmetry is still not really broken, and thus the gauge invariance holds in this clever way although the gauge bosons become massive! That came to the rescue of a big puzzle in electroweak theory and earned Higgs and Englert the Nobel Prize after the discovery of a Higgs boson, which is indeed necessarily occuring in the physical particle spectrum of this kind of theory.

This procedure also shows that there is in fact no degeneracy of the ground state, because we have a local gauge transformation, the acting of the gauge group doesn't lead to a different ground state but to another representant of the one ground state that is realized as an equivalence class of vectors modulo local gauge transformations (such as the electromagnetic field in classical electrodynamics is represented by an equivalence class of four-vector potentials with two four-vector potentials representing the same physical em. field if they are connected by a gauge transformation). That's another reason for the fact that there are no massless Goldstone modes in this case. It's better to say the gauge theory is "Higgsed" than to say one spontaneously breaks the local gauge symmetry, which in fact you don't as just explained.

Now the unitary gauge has a drawback. It is called unitary, because it makes the physical particle spectrum manifest on the tree level of the theory (or in the Lagrangian, describing the QFT). That's good, but only on the first glance! Now the propagators of the gauge fields in this gauge have a propagator with a piece $\propto p^{\mu} p^{\nu} /[M^2(p^2-M^2+\mathrm{i} 0^+]$, which has not the nice power counting as a usual bosonic propagator. It now seems as if the Higgsed gauge theory is no longer renormalizable, but that's not right! What should be renormalizable are the physcially relevant S-matrix elements, not necessarily the propagators and vertex functions used to calculate them.

Now fortunately, one can fix the gauge differently from the unitary gauge, and 't Hooft came up with a very clever choice during his PhD work, the socalled renormalizable $\xi$ gauges. They depend on a parameter $\xi$, and for finite values of $\xi$ it leads to nicely renormalizable propagators and vertex functions in the usual sense of perturbative Dyson renormalizability. In this gauge, however you have both, massive gauge fields, would-be Goldstone modes and, in addition, Faddeev-Popov ghosts. This sounds very complicated, and in fact it is! But the would-be Goldstone modes are ghosts similar to the Faddeev-Popov ghosts, but they are represented as usual and not Grassmann fields in the path-integral formalism. Thus, roughly speaking, what happens in this class of gauges is the following: You have one unphysical degree of freedom in the (massive) gauge-boson fields and for each "broken one-parameter symmetry trafo" a would-be Goldstone mode, both of which are compensated in loops of Feynman diagrams by the Faddeev-Popov ghosts. So you are left with the 3 physical gauge-boson degrees of freedom. Now, since one deals with so many "ghost particle degrees of freedom" one might worry about the unitarity of the S matrix, but there's no danger because of local gauge symmetry! The S-matrix elements are independent of the $\xi$ parameter, and the 't Hooft gauges are such that the limit $\xi \rightarrow \infty$ makes the $R_{\xi}$ gauges into the unitary gauge, i.e., the theory becomes manifestly renormalizable in the usual Dyson sense (for the Green's and proper vertex functions) and the S matrix stays unitary! This proof for the renormalizability of local Higgsed gauge theories made up not only a great doctoral thesis for 't Hooft but was the breakthrough for the electroweak standard model (discovered by Glashow, Salam, and Weinberg before) and earned 't Hooft and Veltman their Nobel prize.

11. May 16, 2015

### Staff: Mentor

I think its way above the op's head.

But since its going down that direction I have to mention as far as the explanation of charge goes the key seems the introduction of q in the definition of the covarient derivative. From the paper I linked to before:

Its the q in Du of equation 11.

Guys I simply went down the path of the details for completeness - as I said its likely gibberish to the OP who simply wanted to know its origin, not the technical detail.

That said I did greatly enjoy Vanhees post - as I do many of his.

Thanks
Bill

Last edited: May 16, 2015
12. May 16, 2015

### atyy

In general, charge or mass does not need an "origin" in the sense that there is no mathematical inconsistency whether a particle is charged or not, or whether it is massless or massive, and we just use the data to figure out what the truth is. In the special case of the gauge fields, there will be a mathematical inconsistency if these fields are massive and the Higgs particle does not exist. This is why the Higgs is said to be the "origin" of mass in these special cases.

Interesting question. I think it is just that nature turns out to be like this, but perhaps there is a theoretical constraint?

13. May 17, 2015

### maline

Isn"t it true that protons are created from positrons?

14. May 17, 2015

### Staff: Mentor

No.

They are created from quarks.

Thanks
Bill

15. May 17, 2015

### maline

Ah well. Took GUT too seriously.

16. May 22, 2015

### driftwood

Thanks everyone. Everything was indeed over my head, but at least now I can abandon my misconception about the origin of charge.

17. May 22, 2015