# Does the weak force violate conservation of charge?

Dear Physics Forum,

I am a little confused about conservation of charge vs. charge conjugation. I'm reading David Griffith's book "Intro. to Elementary Particles"...

pg. 81 "All three interactions (weak, strong, EM), of course, conserve electric charge. In the case of weak interactions, the lepton (or quark) that comes out may not have the same charge as the one going in, but if so, the difference is carried by the W."

pg. 144 "The weak force is not invariant under C" (charge conjugation, the laws of physics remain the same if all particles are exchanged for their antiparticles).

Am I right in thinking that total electric charge is conserved at the weak vertex, but, if you exchange all + and - signs (more specifically, particles for antiparticles), this will effect the laws of physics (the C operator acting on certain weak force interactions predicts a left-handed particle to be produced but only right handed particles are produced).

I thought conservation laws resulted from symmetries, so if charge is not symmetrical, why is overall charge nonetheless conserved? Are there two symmetries, one for overall charge and a second for particle/antiparticle?

Sorry to ramble...Mark

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tom.stoer
C-invariance says that that a process with particles X,Y,Z has exactly the same properties (matrix elements) as the process with antiparticles CX, CY, CZ where C acts as a charge conjugation operator. The processes are different (electron-photon scattering vs. positron-photon scattering) but in both processes charge is conserved.

Charge conservation i.e. dQ/dt=0 is not due to charge conjugation but due to the Noether theorem associated with a continuous symmetry which guarantuees the existence of a conserved current j and a conserved charge Q. In QFT the Noether theorem for current conservation is replaced by the Ward-Takahashi identities.

C-invariance says that that a process with particles X,Y,Z has exactly the same properties (matrix elements) as the process with antiparticles CX, CY, CZ where C acts as a charge conjugation operator. The processes are different (electron-photon scattering vs. positron-photon scattering) but in both processes charge is conserved.

Charge conservation i.e. dQ/dt=0 is not due to charge conjugation but due to the Noether theorem associated with a continuous symmetry which guarantuees the existence of a conserved current j and a conserved charge Q. In QFT the Noether theorem for current conservation is replaced by the Ward-Takahashi identities.
Charge conjugation is a discrete symmetry, but can't discrete symmmetries still give rise to conservation laws? So does charge conjugation symmetry lead to any conservation laws?

tom.stoer
I don't think so.

The only conservation laws not related to continuous symmetries which I am aware of are topological conservation laws - but they do not play any role here.

I don't think so.

The only conservation laws not related to continuous symmetries which I am aware of are topological conservation laws - but they do not play any role here.
Don't parity operators have associated conservation laws?

tom.stoer
As I said; I am not aware of such a conservation law.

Bill_K
Well! That's because no sooner has a conserved quantity been proposed, bingo somebody comes along and invents a continuous symmetry to 'explain' it!

Such as R parity.

As I said; I am not aware of such a conservation law.
I'm a bit confused.
I know Noether theorem only applies on continuos symmetries, but I used to think discrete symmetries still has conserved quantity associated, as for example parity eigenvalues for parity invariant systems.

Ilm

tom.stoer
Can you please explain how to define a conservation law for parity in a simple (perhaps classical) system? Perhaps then I understand better what you mean.

Can you please explain how to define a conservation law for parity in a simple (perhaps classical) system? Perhaps then I understand better what you mean.
If we denote with $\Pi$ the parity operator and with $H$ the Hamiltonian of our system, then if $\,\Pi\,|\psi(t=0)>= \,\pi\,|psi(t=0)> ,\; \pi\in\{0,1\}$:

$[\Pi,H]=0 \;\Rightarrow \; \Pi|\psi(t)>= \,\pi\,|\psi(t)>$.

I always tought of it as a conservation law for parity, am I wrong?

Ilm

tom.stoer
I see. Yes you are right, that makes sense.

Matterwave
Gold Member
Noether's theorem applies to a continuous symmetry which implies the existence of a conserved current/momentum.

You may get conserved quantities which correspond to discrete symmetries, but I know of no general theorem which proves this.

In other words, I know of no theorem that says for EVERY discrete symmetry there is some conserved quantity.

I think it's pretty intuitive in nonrelativistic quantum mechanics that for any symmetry of the system, discrete or continuous, there exists a self-adjoint operator such that the fact that it commutes with the Hamiltonian operator follows from the symmetry. Am I right about this?

tom.stoer
In QM a conserved quantity is described by an observable Q commuting with the Hamiltonian [H,Q]=0. And of course one can construct a unitary operator U[θ] = exp(iθQ) describing a rotation in Hilbert space, i.e. a symmetry. I think for continuous symmetries this is rather obvious in the modern quantum mechanical formalism.

The Noether theorem works the other way round: it starts with the symmetry and gives us a conserved quantity. And it does not start with a symmetry in Hilbert space but with a symmetry in position space. Here I think it's not so obvious that this way does always work well.

In QM a conserved quantity is described by an observable Q commuting with the Hamiltonian [H,Q]=0. And of course one can construct a unitary operator U[θ] = exp(iθQ) describing a rotation in Hilbert space, i.e. a symmetry. I think for continuous symmetries this is rather obvious in the modern quantum mechanical formalism.

The Noether theorem works the other way round: it starts with the symmetry and gives us a conserved quantity. And it does not start with a symmetry in Hilbert space but with a symmetry in position space. Here I think it's not so obvious that this way does always work well.
A symmetry in QM isn't necessarily in the form

$\exp{i\theta Q}, \quad [Q,H]=0$,

this is only true for continuos connected group. In general it suffice

$[g,H]=0 \quad\forall g \in G$

to have a symmetry (G is a group representation on the Hilbert space).

Moreover Noether Theorem applies to every continuos symmetry of the Lagrangian, even if it doesn't involve spacetime transformation, i.e. for internal symmetries too.

Charge conservation is obtained via Noether theorem from the $U(1)_{em}$ (internal) symmetry of the Lagrangian, while charge coniugation symmetry in QED (broken if we add weak interaction) is discrete and it's a consequence of $[L,C]=0$ ($L$ is the Lagrangian density, $C$ the charge coniugation operator).

I never thought about wich quantum number we associate to $C$ operator, maybe none because it's anti-hermitian and so there aren't eigenvalue?

I have now the same problem for the time-reversal symmetry in QED

Ilm

A symmetry in QM isn't necessarily in the form

$\exp{i\theta Q}, \quad [Q,H]=0$,

this is only true for continuos connected group. In general it suffice

$[g,H]=0 \quad\forall g \in G$

to have a symmetry (G is a group representation on the Hilbert space).
Yes, that was the kind of thing I was trying to intuitively describe in my last post.
Charge conservation is obtained via Noether theorem from the $U(1)_{em}$ (internal) symmetry of the Lagrangian
No, it's technically not a consequence of Noether's theorem, rather as tom.stoer said it comes from the Ward-Takahashi identity, which are a field-theoretic counterpart to Noether's theorem.
I never thought about wich quantum number we associate to $C$ operator, maybe none because it's anti-hermitian and so there aren't eigenvalue?
I don't think that's an issue. Just multiply $C$ by $i$ to make it Hermitian.
I have now the same problem for the time-reversal symmetry in QED
I'm interested in hearing the answer to this as well.

tom.stoer
A symmetry in QM isn't necessarily in the form

$\exp{i\theta Q}, \quad [Q,H]=0$,

this is only true for continuos connected group.
Exactly; but when we want to compare to Noether's theorem we need this form.

In general it suffice

$[g,H]=0 \quad\forall g \in G$

to have a symmetry.
I don't think so. Here g is a conserved quantity, but in which sense would you call it a symmetry? (OK, b/c it acts as a rep. of a group G on the Hilbert space).

Moreover Noether Theorem applies to every continuos symmetry of the Lagrangian, even if it doesn't involve spacetime transformation, i.e. for internal symmetries too.
Correct, but that was not my focus. I only wanted to point out that there's a difference whether a) you start with a conserved quantity which commutes with H or whether b) you start with as spacetime (opr internal) symmetry and then construct a conserved quantity. It's definitly not the same.

I never thought about wich quantum number we associate to $C$ operator, maybe none because it's anti-hermitian and so there aren't eigenvalue?

I have now the same problem for the time-reversal symmetry in QED
Perhaps it's enough to have an operator O commuting with H, [H,O] = 0, and allowing for the rep. of a group on the Hilbert space.

No, it's technically not a consequence of Noether's theorem, rather as tom.stoer said it comes from the Ward-Takahashi identity, which are a field-theoretic counterpart to Noether's theorem.
I'd certainly need to read again my QFT notes, but I thought Ward-Takahashi identity was used to show that gauge invariance preserves the mass of photons to vary along the renormalization group flow, while Noether theorem shows the conservation of the em current...

I don't think that's an issue. Just multiply $C$ by $i$ to make it Hermitian.
$C$ is anti-linear too (like time-reversal), could this be the issue?

Exactly; but when we want to compare to Noether's theorem we need this form. [...]

Correct, but that was not my focus. I only wanted to point out that there's a difference whether a) you start with a conserved quantity which commutes with H or whether b) you start with as spacetime (opr internal) symmetry and then construct a conserved quantity. It's definitely not the same.
Once we represent a discrete group $G$ on the Hilbert space can we define a Lie algebra with it (using matrix sum and product with scalars) and then construct a corresponding Lie group of symmetry for the theory?
If so can't we then apply Noether theorem to this Lie group to find conserved quantities?
Should eventually this mechanism result in the same conserved quantities we define from $[H,G]=0\;$?
I'm now a lot confused

Perhaps it's enough to have an operator O commuting with H, [H,O] = 0, and allowing for the rep. of a group on the Hilbert space.
I don't understand what you're suggesting here

Ilm

tom.stoer
Once we represent a discrete group $G$ on the Hilbert space can we define a Lie algebra with it (using matrix sum and product with scalars) and then construct a corresponding Lie group of symmetry for the theory?
If so can't we then apply Noether theorem to this Lie group to find conserved quantities?
Should eventually this mechanism result in the same conserved quantities we define from $[H,G]=0\;$?
I don't understand the idea of constructing a continuous Lie group from a discrete group. What's the Lie group for parity?

I don't understand what you're suggesting here
I try to define "conservation law" as an operator O commuting with H, [H,O] and "symmetry" as some representation acting on the Hilbert space.

I don't understand the idea of constructing a continuous Lie group from a discrete group. What's the Lie group for parity?

I try to define "conservation law" as an operator O commuting with H, [H,O] and "symmetry" as some representation acting on the Hilbert space.
As I said before I'm a bit confused now, but I was thinking that, for example, if $\Pi$ is the parity operator then $exp{(i\theta\Pi)}\, ,\; \theta\inℝ$ would be an abelian Lie group, with conserved charge $\Pi$...

Ilm

tom.stoer
But \Pi^2 = 1 so nobody would consider the exponential

But \Pi^2 = 1 so nobody would consider the exponential
Yes, maybe it was not the best example :)

What I was thinking about was that given a group $G$ of order $n$, such that

$[H,g]=0 \; \forall g \in G \, ,$

then, if we construct an associated algebra $\tilde{G}$ ,

$\exp{(i \sum_{i=1 ... n}\theta_i \tilde{g}_i)}\;,\quad \tilde{g}_i \in G$

would be a continuous group of symmetry for the system.

If this could be done, then we could then regard conserved quantities as they would all come from Noether theorem, even those we usually think as consequences of discrete symmetries.

Ilm