reilly said:
meopemuk said:
When a photon leaves its mark on the photographic plate, we measure an observable "photon's position". The probability for such a measurement is determined by the square of photon's wave function in the position space. I'm afraid that photon's quantum field A_{\mu}(x) has nothing to do with this probability.
..........
Then, what does?
Regards,
Reilly Atkinson
Below I will briefly describe how I understand quantum field theory and its relationship to QM. I understand that my views do not look like the mainstream, but I believe they are in full accord with the way QFT presented in S. Weinberg, "The quantum theory of fields", vol. 1. So, here it goes...
Why quantum mechanics is no good enough? In quantum mechanics we normally deal with systems having fixed number of particles (N). The Hilbert space is built as a tensor product of N 1-particle Hilbert spaces where operators of observables, state vectors, dynamics, etc are defined. In relativistic physics energy can be converted to mass due to Einstein's E=mc^2, so we must also take into account the possibility of changing the number of particles (emission, absorption, annihilation, etc.). This can be achieved by switching from fixed-number-of-particles Hilbert spaces to the Fock space, which is simply a direct sum of N-particle Hilbert spaces, where N varies from 0 to infinity. Fundamentally, QFT is nothing else but quantum mechanics in the Fock space.
Operators of observables and wave functions in the Fock space One can define operators of particle observables in the Fock space in exactly the same way as in QM. For example, in each N-particle sector we have well-defined operators (of position, momentum, spin, etc.) for each of the N particles described there. These operators have usual common bases of eigenvectors, and an arbitrary N-particle state vector can be projected on these bases to obtain N-particle wave functions in different representations. The new feature is that one also has state vectors with non-zero projections on sectors with different N. Wave functions of states with such undefined particle numbers are just sets of N-particle wave functions, each with its own coefficient. The square of the coefficient is the probability to find N particles in this general state.
Unitary representation of the Poincare group. The most general way to construct a relativistic quantum theory is by defining an unitary representation of the Poincare group in the Hilbert space (in our case this is the Fock space) of the system. Basically, it is sufficient to construct interaction operators V and \mathbf{W} in the generators of time translations (the Hamiltonian) and boost
H = H_0 + V
\mathbf{K} = \mathbf{K}_0 + \mathbf{W}
so that the Poincare commutation relations between all 10 generators (including the total momentum \mathbf{P}_0 and the total angular momentum \mathbf{J}_0) remain preserved. This is a very non-trivial problem, and currently there is only one known solution which satisfies all requirements, such as the cluster separability, the possibility to describe particle-number-changing interactions, etc. [Please note that the fact that it is the only
known solution does not imply that it is the only
possible solution.] This is where quantum fields come into play:
1. Define particle creation and annihilation operators in the Fock space.
2. For each particle type build a certain linear combination (the free quantum field) \psi(\mathbf{r},t) of the creation and annihilation operators with two major properties
2a. (anti)commutativity at space-like separations
2b. manifestly covariant transformation laws with respect to the non-interacting representation of the Poincare group in the Fock space
3. Then operator V(t) can be constructed as an integral on \mathbf{r} of field products (or polynomials)
V(t) = \int d^3r \psi(\mathbf{r},t) \phi(\mathbf{r},t) A(\mathbf{r},t) \ldots
(for brevity I omit possible indices of operators \psi, \phi, A and summations over these indices). The "boost interaction" \mathbf{W}(t) is given by a similar formula.
It is important to note that we don't need to give any physical interpretation to quantum fields \psi, \phi, A, \ldots. They are just abstract mathematical quantities, whose role is to facilitate the construction of interaction operators V and \mathbf{W}. Once this construction is completed, we have a full interacting quantum theory in the Hilbert space with particle operators, wave functions, the time evolution operator
U(t) = \exp(\frac{i}{\hbar} Ht) [/itex]......(1)<br />
<br />
etc, i.e., everything one would need to solve any kind of physical problem.<br />
<br />
<b>Renormalization and dressing</b> Unfortunately, this nice theory has a serious problem. It appears that for all realistic interaction operators V the scattering (S-) matrix cannot be calculated, because its matrix elements come out infinite. To solve this problem one can add (as Tomonaga, Schwinger, and Feynman did) infinite renormalization counterterms to the Hamiltonian H (and to the boost operator \mathbf{K} as well). Then one can get a very accurate S-matrix, but the Hamiltonian H becomes infinite and useless for calculating the time evolution (1) and for finding bound states via diagonalization. This problem can be solved by applying an "unitary dressing transformation" to the Hamiltonian and all other generators of the Poincare group. Then the theory assumes the form very similar to ordinary quantum mechanics: We have a full description for the system of particles interacting with each other via instantaneous potentials, which, in addition, can change the number of particles in the system. The Poincare commutators, cluster separability and other important physical requirements are exactly satisfied. Quantum fields are not needed for the physical interpretation of this theory..<br />
<br />
Eugene.
