Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Does the renormalization of QFT contradict canonical quantization?

In canonical quantization, you take the classical fields and canonical momenta and turn them into operators, and you require that the commutators ( or anti-commutators) equal the Poisson brackets of the original fields. Also, you need to quantize the classical Hamiltonian (of course, since classical fields commute but quantum operators don't, you need to quantize the Hamiltonian according to a well defined prescription, such as normal-ordering). The form of the quantum Hamiltonian will resemble the form of the classical one, but it will be an operator rather than a number. In particular, if a physical constant appears in the classical Hamiltonian, it will also appear in the Quantum one, and have the same value. Now, in renormalization the constants that appear in the (bare) Lagrangian are not the physical ones, and so are different from the constants that appear in the "classical" Lagrangian (and therefore Hamiltonian). So the "bare" quantum Hamiltonian will have a similar form to the classical Hamiltonian, but the constants will be totally (and perhaps infinitely) different!

So does the quantization scheme need to be reformulated?

share|improve this question
1  
Well, the correct commutation relations for the bare fields must be extracted using the standard method (velocities, momenta - derivatives of L with respect to velocities) from the full Lagrangian for the bare fields (which includes counterterms). The full commutation relations for any kind of renormalized fields must be extracted from the methods applied to the full Lagrangian as a function of renormalized fields. What's the problem? It's the same Lagrangian, just in different variables so the two methods (and results for the commutators) are related by a field redefinition only. –  Luboš Motl Jun 23 '13 at 5:30
    
Take a classical field. It is defined with a Lagrangian that involves physical constants. Compute the Hamiltonian (via Noether or Legendre). It too will contain physical constants. When you quantize the theory, you turn the fields to operators, and replace the operators in H. So now you have a quantum H with the same physical constants and form as the classical one. On the other hand, if you take the L expressed with bare fields and constants (therefore no counterterms) and find the H, you get an identical expression but with bare rather than physical constants. –  Lior Jun 23 '13 at 14:17
    
When I quantize a classical field, does the field turn into the operator of the bare field or into the operator of the physical field (with unit field strength Z)? –  Lior Jun 23 '13 at 14:20
    
@Lior: you might want to check out Tacciati's Quantum Field Theory for Mathematicians...it walks through renormalization in the canonical setting quite beautifully. –  Alex Nelson Jun 24 '13 at 15:55
add comment

1 Answer

up vote 1 down vote accepted

(After thinking about the above issue, I came up with this explanation which I think is correct. I'm aware that the question itself hasn't been 100% well stated).

There is no contradiction.

Although "pedagogically" it is helpful to think about the procedure of quantization, in which the quantum theory is obtained from the classical one, it is more "natural" to think the other way around - that the classical theory emerges from the quantum one via the process of "classicization" (This is because the quantum theory is a more fundamental description of reality). Now, quantum field theory and renormalization group flow tell us that the effective Lagrangian of a theory depends on the momentum scale of the interacting fields. In other words, coupling constants and mass depend on the momentum scale. When we do classical physics we usually use Lagrangians that correspond to low momentum scales. These Lagrangians should be thought of as obtained from the quantum theory in the process of "classicization", and the effective Lagrangian that was used in the process corresponded to the low momentum scale. If we would work in a higher momentum scale, we would "flow" to a different Lagrangian in the quantum theory, and then when "classicizating" the theory we would get a different Lagrangian. (Of course, there is no difference between the classical and quantum Lagrangians; but the quantum physics and the classical physics that relate to these identical Lagrangians are different). So it's true that if you use only a specific Lagrangian of a classical theory and you quantize it, then you get an Hamiltonian which has a different dependence on the physical constants than the classical one. This is because the quantum Hamiltonian has additional counterterms that "absorb" the fact that the theory behaves differently at different momentum scales. This different behaviour was forgotten in the classical theory since it was "classicized" from a specific effective Lagrangian at a specific momentum scale.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.