# Reason behind canonical quantization in QFT?

Reason behind canonical quantization in QFT?

In the scalar field theory we simply promote the scalar field, $\phi(x)$ to a set of operators: $\hat{\phi}(x)$. What is the reason behind this?

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Well, we are doing Quantum Field theory, so we want to quantize the fields $\phi(x)$. Thus, the initially classical field $\phi(x)$ is promoted to a quantum field ${\hat \phi}(x)$ (in the same way that the classical position $x$ is promoted to a quantum operator ${\hat x}$. Maybe you want to ask why a quantum field theory is the right thing to study in the first place? –  Prahar Jul 8 '13 at 18:49
@Prahar, I think that's somewhat misleading since $\hat x$ is the operator corresponding to the classical observable $x$. This contrasts with the quantum field operators which create and/or destroy particles at an event. –  Alfred Centauri Jul 8 '13 at 20:58
@AlfredCentauri: You're repeating a misconception. The field operators $\phi(x)$ (more precisely, the smeared field operators $\int\phi(x)f(x)dx$) really do measure the values of fields. They also, in some cases, create and destroy particles (with wave functions derived from $f$). However, this interpretation isn't necessary or universal; some QFTs don't have particle excitations. The field interpretation isn't dispensable; even the Standard Model requires it, e.g., for the Higgs mechanism. Prahar is correct; the classical observable $\phi \mapsto \phi(x)$ is promoted to an operator. –  user1504 Jul 8 '13 at 21:57
@user1504, I did not claim that Prahar was incorrect. In every introductory text on QFT that I have, there is no mention of interpreting the field operator as an observable as the path to QFT. For that reason, I wrote that his statement was somewhat misleading (to the OP). –  Alfred Centauri Jul 8 '13 at 22:48
@AlfredCentauri: I agree that most textbooks approach the subject this way. I suspect it's a bad way to do it, since so many students have difficulty learning the subject. –  user1504 Jul 8 '13 at 22:54