Demystifier said:
Of course, this is not a proof. But QFT based on quantization of the Klein-Gordon equation seems to be in agreement with experiments. OK, we do not really have a fundamental spinless particle, but we have a photon which is also described by an equation similar to the Klein-Gordon equation sharing the same interpretational problems. Therefore, with an appropriate interpretation, the Klein-Gordon equation should be correct. How do you comment on this? The key, of course, is to make a clear distinction between fields and wave functions. We agree on that. But how exactly to do that distinction?
True. QFT is based on quantum fields that satisfy relativistic wave equations, like Klein-Gordon or Dirac equation. And QFT is a very successful theory. However, my point is that the success of QFT does not depend on the probabilistic interpretation of quantum fields that you are suggesting. This interpretation is not used anywhere in QFT calculations.
The central quantity calculated in QFT is the S-matrix. If one knows the S-matrix, one can calculate scattering cross-sections, decay rates, energies of bound states, i.e., almost all properties that can be directly compared with experiment. All phenomenal successes of QED are related to very accurate calculations of certain S-matrix elements, which can be generally written as
S_{in,out} = \langle in | S | out \rangle
where | in \rangle is an asymptotic state of free particles in the remote past, | out \rangle is an asymptotic state of free particles in the remote future, and S is the S-operator. The "in" and "out" states contain fixed (but not necessarily equal) number of particles, and they are normally obtained by acting with particle creation operators on the vacuum vector | 0\rangle. For example,
| in \rangle = a_{\mathbf{p}, \sigma}^{\dag} b_{\mathbf{q}, \tau}^{\dag} \ldots | 0\rangle
In accordance with the most common experimental situation, one usually considers particle states with definite momenta \mathbf{p}, \mathbf{q} and spin projections \sigma, \tau.
The S-operator can be calculated from the interaction Hamiltonian V(t) by the Feynman-Dyson perturbation formula (There are other methods to compute the S-operator, but they are all equivalent.)
S = 1 + \frac{i}{\hbar} \int \limits_{-\infty}^{\infty} V(t) dt - <br />
\frac{1}{2 \hbar^2} \int \limits_{-\infty}^{\infty} dt \int \limits_{-\infty}^{\infty}dt' T[V(t) V(t')] + \ldots
where T is the time-ordering sign.
The interaction operator V(t) is normally expressed as a product of quantum fields. For example, in QED
V(t) = e\int d^3x \overline{\psi}(\mathbf{x},t) \gamma_{\mu} \psi(\mathbf{x},t) A^{\mu} (\mathbf{x},t)
where \psi(\mathbf{x},t) is the electron-positron Dirac field,
\overline{\psi} = \psi^{\dag} \gamma_0, A^{\mu} (\mathbf{x},t) is the photon field, and \gamma_{\mu} are 4 \times 4 Dirac matrices.
It is clear, that in order to perform S-matrix calculations it is sufficient to know how quantum fields \psi(\mathbf{x},t) and A^{\mu} (\mathbf{x},t) are expressed through particle creation and annihilation operators, and what are (anti)commutators of these operators.
That's basically all one needs to know to perform QED calculations. There are also issues related to ultraviolet and infrared divergences, but they do not change substantially the basic formalism described above.
My point is that nowhere in these calculation one needs to use the "probabilistic interpretation" of quantum fields. As I said, quantum fields do satisfy wave equations, but their knowledge is not crucial for S-matrix calculations. In his book, Weinberg mentions wave equations as some "side effects". They are useful to simplify certain calculation steps, but they are not critical.
Eugene.
articles are getting created and destroyed,so the particle no. is not fixed.You can however give prob. interpretation to \Psi [\phi; t)--why should this be expected to reduce to ψ of Schrodinger equation in the non-rel. limit?It has an independent existence.
Regards, Hans
