Does the Renormalization of QFT Contradict Canonical Quantization?

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?

• 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