Does Electric Charge Have a Definite Value or Does Uncertainty Apply?

[computerphys]

I would like to know if the electric charge of a particle, like electron, has always a definite value, or on the contrary the Heisenberg uncertainty principle should be applied to this magnitude.

Or asking that in another way, what is the observable operator associated to the charge of a particle?

Has the U(1) symmetry something to do with all that?

Thanks in advance!

 

[haael]

I would like to know if the electric charge of a particle, like electron, has always a definite value, or on the contrary the Heisenberg uncertainty principle should be applied to this magnitude.

Or asking that in another way, what is the observable operator associated to the charge of a particle?

Has the U(1) symmetry something to do with all that?

Thanks in advance!

Kaluza-Klein again.

In KK theory the charge is the momentum in fifth (compactified) dimension. Position in fifth dimension is related to charge with uncertainty.

The U(1) group has something to do with this. In KK it's related to the fifth dimension, in standard QED it is unmeasurable. You can't exactly measure EM field - that is, you can always do an U(1) gauge transformation and you will get the same physical result. That means - the exact "position" of EM field in U(1) group can't be measured.

 

[sheaf]

I'm not aware of any uncertainty principle applying to charge (in "conventional" theories anyway). Isn't charge governed by a superselection rule ?

 

[computerphys]

In KK theory the charge is the momentum in fifth (compactified) dimension. Position in fifth dimension is related to charge with uncertainty.

Thanks a lot. This subject is more clear for me now.

Just another question, what about mass and uncertainty? Can we apply HIP to mass? If mass is the charge for the gravity field ...

 

[tom.stoer]

Kaluza-Klein again.

In KK theory the charge is the momentum in fifth (compactified) dimension. Position in fifth dimension is related to charge with uncertainty.

Please be aware that up to now we do not have a reasonable theory describing nature in terms of KK! KK is able to harmonize general relativity and classical electrodynamics, not more. No weak and strong force, no quantum gravity; KK as of today is just wishful thinking.

 

[tom.stoer]

Or asking that in another way, what is the observable operator associated to the charge of a particle?

Has the U(1) symmetry something to do with all that?

You wan't find an answer in quantum mechanics because there is no charge operator (e.g. for an electron in the Schrödinger equation). Charge is related to the charge density which itself is nothing else but the (square of the) wave function. So normalizing the wave function means that for the single particle Schrödinger equation you will always get "1e" for the charge.

You have to look at a theory which allows you to describe a charge operator and where you do not fix the charge of a quantum state in advance. In quantum field theory you can construct a charge operator and a corresponding conservation law. In QED this is related to the Gauss law of the U(1) symmetry and due to consistency (again coming from U(1)) you always get a constraint like

Q|phys> = 0

That means all physical states must have zero charge, otherwise the theory is inconsistent!

In QCD (or in general SU(N) gauge theories) the constraint is generalized to

Q^a|phys> = 0

where ^a labels the different charges. Note that the charge operators generate an su(n) algebra like

[Q^a, Q^b] = if^abcQ^c

which is similar to the angular momentum algebra. So the condition

Q^a|phys> = 0

means that all physical states are charge-singulett states. If you look at angular momentum you know that if the z-comopnent is fixed, the x- and y-component are not. But that does not apply in our case, as in te singulett states all components vanish; this is OK. The singulett state is rather special as it is the only state which is a simultaneous eigenstate of all charge operators!

Now if you look at charges not related to local gauge symmetries there is no Gauss law and therefore no singulett condition. This applies e.g. to isospin. For isospin exactly the same SU(2) symmetry applies as for conventional spin: if there is a proton, which means

I³|proton> = (+1/2)|proton>

where I³ is the 3-component of the isospin, then the 1- and the 2-component are undefined and you can derive an uncertainty relation for these components.

 

[f95toli]

I'm not aware of any uncertainty principle applying to charge (in "conventional" theories anyway). Isn't charge governed by a superselection rule ?

Depends on what you mean by charge. The UP certainly applies to things like charged metallic islands where you have [Q,P]=ie (Q is the charge and P is the phase); devices like Cooper pair boxes, Single electron transistors etc all "obey" this UP.
Generally speaking, charge is the generalized momentum and the phase the generalized position for electronic systems (a trivial example of such a system would be an LC oscillator).

 

[tom.stoer]

Perhaps it helps to look at the uncertainty relation for the angular momentum. I start with

$$\Delta A \cdot \Delta B \ge \frac{1}{2} |\langle\psi|[A,B]|\psi\rangle|$$

For the angular momentum and its eigenstates I get

$$\Delta L_x \cdot \Delta L_y \ge \frac{1}{2} |\langle\ell,m|L_z|\ell,m\rangle| = \frac{m}{2}$$

 

[chrisphd]

I'm not an expert, but I think that in the standard model, in the weak interaction, charge is uncertain due to CKM mixing. That is, weak interaction eigenstates are superpositions of the strong interaction eigenstates. The strong interaction eigenstates have definite charge (are eigenstates of charge), and the weak interaction eigenstates therefore have uncertain charge. Please correct me if I'm mistaken as this is relatively new to me.

 

[tom.stoer]

CKM mixing cannot mix conserved charges, that means it cannot mix states with different electric charge and it does not mix states (quarks) with different color (but that is never mentioned).

CKM mixing can mix states with different masses.

 

[chrisphd]

Ok thanks for clearing that up.