# What is the status of E=mc^2 in QM/QFT

1. Oct 2, 2013

### ftr

I mean is the formula taken as a postulate or can we derive it from QFT at least?

2. Oct 2, 2013

### SteamKing

Staff Emeritus
3. Oct 2, 2013

### ftr

Yes, I know it was derived in SR but I meant with regards to QFT. In other words, shouldn't QFT be the right place to make the derivation(it is about particles after all) instead of using it from a semi-classical concept like SR. Why not?

4. Oct 2, 2013

### SteamKing

Staff Emeritus
Einstein derived it from SR first because he discovered/developed SR before QFT was developed.

There is a certain historical order in which discoveries have been made, which is separate from what we may consider to be the logical order in which things should be developed.

For instance, Archimedes played around with the concepts of the infinitesimal calculus centuries before number systems and algebra were developed, when only arithmetic was crudely understood.

This is not to say that energy-mass equivalence may not be developed using QFT (I can't say because I am not a quantum mechanic), but that's not how it happened first.

5. Oct 2, 2013

### ftr

Yes, I understand. My question was more toward a QFT guru I guess. before QFT people accepted the 1/r law as a fact, then QFT derived the law(this is basically QED). Zee claimed that to be the greatest discovery of all time. So I am wondering about E=mc^2 in the same context.

6. Oct 2, 2013

### dauto

QFT is usually started by postulating some field Lagrangian which is assured to be relativistic by design. So we cannot really be surprised when we find out that the particles obtained by quantizing the fields also turn out to obey fully relativistic relations, can we. In other words, we feed our knowledge that the world is relativistic into the theory from the get go.

7. Oct 2, 2013

### ftr

Exactly. Usually newer theories include the older ones as a limit. But in this case it is curious that supposedly QFT our most advanced tool that probes nature depend on a semi-classical theory for its survival. You would expect the opposite wouldn't you?

8. Oct 2, 2013

### dauto

No, because neither relativity nor quantum mechanics are theories. They are meta-theories. Theories about theories.

9. Oct 2, 2013

### rubi

Special relativity is nothing that is tied to classical physics. And quantum field theory doesn't need to be relativistic either. It's wrong to think about special relativity as something classical that has to be carried over to QFT. Instead, it is a basic insight about space and time that applies equally to classical physics and to quantum physics.

10. Oct 6, 2013

### ftr

Thank you for your input. Of course I know that SR is used in QFT. But I guess I was just trying to clarify something in my mind. For instance, why is it then that we cannot apply GR to QFT(Noble prize) even so GR is considered to be the generalization of SR. Second, many theories (almost all) of QG try to consider GR as a quantum theory which implies that the origin of SR is indeed in QM system somehow.

You get very fancy theories these days. for example

Gravitational Dynamics From Entanglement "Thermodynamics"

http://arxiv.org/abs/1308.3716

also

11. Oct 6, 2013

### tom.stoer

I'll try to provide some new insights.

1) Usually a QFT is derived from a classical field theory via a procedure called quantization. This translates fields into field operators. Classical structures of the Hamiltonian formulation, especially the Poisson algebra, are translated into structures on Hilbert space, especially commutation relations. In a classical, relativistic field theory one can construct generators of the Poincare algebra, i.e.

$P^\mu = (H, P^i), L^i, K^i$

In the QFT context the commutators generate the Poincare algebra, just as the classical Poisson brackets do in the classical field theory context. The Poincare algebra is represented on the Hilbert space, which means that physical states members of irreducable representations of the Poincare algebra, and that there are two Casimir operators

$C^{(1)} = P_\mu P^\mu$
$C^{(2)} = W_\mu W^\mu$

which means that physical states with 4-momentum p have invariant mass

$m^2 = p_\mu p^\mu$

(Wigner classification)

This is nothing else but a translation of relations known from classical field theories into QFT language. So SR cannot be derived from QFT. SR is a general framework which fixes certain symmetry aspects for both classical and quantum field theories.

2) I don't think that it's correct to say ...
First we do not want to apply GR to QFT (or formulate QFT on curved but classical spacetime), but we want to apply the above mentioned quantization to GR. I agree that this is still work in progress, but it is not true that this is impossible. It is correct that certain methods (especially perturbation theory and perturbative renormalization) do fail when quantizing GR, but this is not due to a fundamental incompatibility of GR with quantization, but only an indication of the limited applicability of these methods. It is true that we do not have a complete (and testable) theory of quantum gravity (= quantization of GR), but we have a couple of candidate theories passing many consistency tests (Loop Quantum Gravity, Asymptotic Safety approach, Causal Dynamical Triangulation, several Supergravity models, ...)

So my conclusion is that we (or our descendants) will be able to quantize GR.

3) The conclusion in
is not correct. Yes, many theories (all candidates I have listed above) try to quantize GR. But the conclusion that this quantization results in SR (in some limit) is not correct. SR is already present in classical GR, but only as a local (instead of a global) symmetry. Spacetime in GR need not have any global symmetry at all, but is always has local Lorentz invariance by construction. This is nothing else but the famous equivalence principle. A free-falling observer cannot detect spacetime curvature locally; to her spacetime always looks flat. So for local experiments a free-falling observer can always use SR calculations.

So the origin of SR in QG is not the Q, it's the G ;-)

And it's the same idea that I already indicated above: Quantization respects certain classical symmetries, in our case local Lorentz invariance.

Last edited: Oct 6, 2013