What are quantum fields?
Relativistic QFT is a fine theory. If you hold your nose and follow textbook recipes without asking too many questions, you can finally calculate a lot of stuff (scattering cross-sections, Lamb shifts, etc.) in a remarkable agreement with experiment.
However, on this blog we are going to ask questions. Why not? For example:
- Q: What are quantum fields?
- A: They are operator functions on the Minkowski space-time.
- Q: What is their physical meaning?
- A: They are obtained by second quantization of wave functions, at least in the non-interacting theory.
- Q: Wave functions have probabilistic interpretation, and in a relativistic theory the probabilities must be invariant with respect to Poincare transformations. Is it true?
- A: Yes.
- Q: Then particle wave functions must transform by a unitary representation of the Poincare group. Right?
- A: Right.
- Q: So, if quantum fields (even non-interacting) are somehow related to particle wave functions, they also must transform unitarily. Right?
- A: That sounds plausible.
- Q: However, this is not true. Take, for example, Dirac field for electrons. Lorentz transformations of this field are represented by 4x4 matrices. However, it is well-known that there are no finite-dimensional unitary representations of the non-compact Lorentz group. So, Lorentz transformations of the Dirac field are non-unitary.
- A: So what?
- Q: This means that there is no direct connection between the electron's quantum field and electron's wave function?
- A: I am not sure.
- Q: In the quantum field ψ(x,t) the argument x is physical position and t is time. Isn't it?
- A: Yes.
- Q: Is it possible to think (as we do in quantum mechanics) that x as an eigenvalue of the position operator?
- A: I guess so.
- Q: Then we should accept the existence of the time operator as well, which is impossible .
- A: Why did you make this conclusion?
- Q: Because the Lorentz transformation of the field ψ(x,t) involves intermixing of the arguments x and t, which is only possible if x and t are eigenvalues of operators X and T that do not commute with the boost generator.
- A: I guess you may be right.
- Q: So, there is no connection between quantum field and quantum-mechanical wavefunctions. And quantum fields are just abstract operator functions on an abstract 4-dimensional space whose coordinates are not necessarily related to experimentally observed positions and times.
- A: Why do you call Minkowski space-time "abstract"?
- Q: We will talk about Minkowski space-time later. However, arguments x and t of quantum fields are just dummy integration variables without any particular physical meaning.
- A: Why it is so?
- Q: Because, as we discussed here, the only physical thing predicted by relativistic QFT is S-matrix. And in the expression for the S-matrix quantum fields enter integrated over x and t. All dependence on x and t is lost.
P.S. Everything said above refers only to relativistic quantum fields used in theories like QED. They should be separated from non-relativistic quantum fields used, for example, in condensed matter physics. The quantum field of phonons is a well-defined operator which is related to displacements of real atoms from lattice positions.
4 Comments:
If you have a Lorentz vector |μ> then the norm <μ|ν> ∝ η_{μν} which is obviously not positive definite and therefore the representation is not unitary. This is not unexpected, there being no finite-dimensional unitary representations of the Lorentz group. But one may use finite-dimensional representations of the Lorentz group within the context of infinite-dimensional unitary representations of the Poincare group. For example
<μ|ν> ∝ η_{μν} - p_μ p_ν/m²
for massive vector particles works OK because, within the constraint that contraction with p vanishes, the normalization tensor is positive semi-definite, as may be seen by examining in the particle's rest frame. Similar considerations apply to Dirac spinors.
Chris,
sorry, I can't comment on your post, because I don't understand your notation and terminology. What is "Lorentz vector |μ>"? Is it a state in the Hilbert space? Then why do you label it by index μ= 0,1,2,3 rather than by usual momenta and spins of particles?
What is the reason behind your desire to treat quantum field as some king of "quantized" wave function? In my view, quantum fields and particle wave functions are completely different things. There are few similarities between them, but they are rather superficial. The role of wave functions is to give us probability distributions for measurements of observables. The role of quantum fields is to give us "building blocks" for constructing interacting Hamiltonians (or Lagrangians) in the Fock space. This view on QFT is well explained in Weinberg's book [1]. If you accept that quantum fields have nothing to do with probability distributions and wave functions, then there is nothing wrong in having non-unitary transformation laws of fields.
[1] S. Weinberg, The Quantum Theory of Fields, Vol. 1 , (University Press, Cambridge, 1995)
Sorry - I did not mean specifically a Hilbert space with |μ>: I just meant anything that transforms as the four vector under the Lorentz group. An invariant norm of the vector, however defined, cannot be positive definite if the vector space is of finite dimension.
Further down, yes, this is a spin 1 massive particle, and I could, if I wanted, label it with m_s = -1, 0, 1 as everyone else does. But I prefer covariant notation.
Non-second-quantized wave functions AFAIC are just devices for learning the subject. With QFT - if done correctly - one should be starting afresh. Of course, the connection with probabilities motivates the unitarity requirement, but personally I suspect that it is deeper than that. Unitarity, in the sense of positive definite norms for quantum states is, I believe, is an absolute requirement.
Chris,
An invariant norm of the vector, however defined, cannot be positive definite if the vector space is of finite dimension.
Why do you say that? Isn't it true that the norm AμAμ of a 4-vector Aμ is a Lorentz invariant which can be positive, negative, or zero?
Unitarity, in the sense of positive definite norms for quantum states is, I believe, is an absolute requirement.
Yes, I agree with you 100%. The unitarity means that the Hamiltonian and other generators of the Poincare group representation in the Fock space must be Hermitian. However, it does not impose any restrictions on quantum fields. Quantum fields may be non-Hermitian (and they often are), and they may transform non-unitarily (and they often do). The only important thing is that the Hamiltonian (constructed as a polynomial of fields) is Hermitian and satisfies correct Poincare algebra commutators with other generators.
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