Energy of Electon and Energy of the Field

[ghotra]

The energy of an electron at rest is mc^2.

1) Can an electron even be at rest? It seems that the answer is "no" by the uncertainty principle. Thus, it would seem that every electron has energy greater than mc^2. Is this a correct statement?

2) Is this a classical quantity? That is, if I were to determine the electric and magnetic fields of an electron quantum mechanically, and if I integrated the square of the electric field to determine the total energy stored in those fields, would I get E = mc^2 as an answer.

Basically, I am wondering which (if any) of these statements is true.

E = E_rest + E_fields

E = E_rest = E_fields

Can we say: An electron has no energy---rather, it's fields due.

Anyway, this wasn't super organized, but I hope my question is clear. I am trying to resolve (if it needs to be) E = mc^2 with quantum mechanics and I don't understand how the energy of a particle relates to the field that it creates.

 

[MalleusScientiarum]

The way that you resolve special relativity with quantum mechanics is a rather complicated process (well, not horridly complicated, but unpleasant to say the least).

Essentially, the energy of any particle is frame-dependent. At the moment, my computer has no kinetic energy because we aren't moving relative to each other. Once I dash out the door for a beer run I could say that the computer has kinetic energy relative to my frame of reference.

You can have non-violation of the Heisenberg principle by having a particle in a momentum eigenstate. Only, in special relativity, we contend with four-vectors, such as
\mathbf{p} = (p^0, p^1, p^2, p^3)
where p^0 represents the energy component of the momentum four-vector. So let's suppose that we start out with a particle in a four-momentum eigenstate. The Poincare Group clearly defines how this state transforms under translations, boosts, and rotations. In our particular case a translation has no effect on the momentum (had it been in a position eigenstate that would be different). However, the four-momentum that defines the state vector transforms according to the Lorentz transformations. Thus, out pops a new momentum value and the state measured will be different between two reference frames that are moving relative to each other, call them \mathcal{O} and \mathcal{O}'.

As for your other questions, the electric field energy is distinct from the kinetic energy of the electron, even though the source is different. Energy is a frame-dependent quantity (as you can see up above), and so the energy of the electron will be different in differing frames. Furthermore, if I'm moving relative to the electron, I will detect a new magnetic field that has formed and which has energy. This arises out of the Lorentz invariance of charge. So I'll have an electric field (which will in general be time-varying) and a magnetic field that an observer in the rest frame of the electron won't see.

I guess to sum up my answer, energy depends largely on what frame of reference you are in.

 

[ghotra]

That all sounds good. So let me state this:

1) An electon and I are in the same inertial frame.
2) The electron is in a momentum eigenstate.

Thus,

1) The electon is a "rest".
2) The energy of the electron is E = mc^2
3) The energy stored is the fields does not contribute to the energy of the electron.

Is this correct?

 

[MalleusScientiarum]

I believe so, yes. Maybe someone else might check my work. I just note that
E^2 = p^2 + m^2
in naturalized units (i.e. c = 1). Just keep that in mind.

 

[reilly]

Yes, QM allows an electron to be at rest, that is it can be in a sate wih p=0. Note, however, that a localized state, say x=0, requires all momenum eigenstates to be present -- due to the Fourier expansion of the delta function.

The energy in the electron's field is certainly non-zero; classically the local energy density of E*E + B*B does not vanish, also true in QM. Frankly we don't have a clue about a "bare" electron -- is the electron's mass due entirely to its fields, or does it have some residual mass due to who knows what. (See any advanced E&M book, and any QED or QFT dealing wih renormalization; or Dirac's QM book)

Regards,
Reilly Atkinson