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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7$\begingroup$ 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? $\endgroup$– PraharCommented Jul 8, 2013 at 18:49
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2$\begingroup$ @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. $\endgroup$– Alfred CentauriCommented Jul 8, 2013 at 20:58
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1$\begingroup$ @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. $\endgroup$– user1504Commented Jul 8, 2013 at 21:57
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$\begingroup$ @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). $\endgroup$– Alfred CentauriCommented Jul 8, 2013 at 22:48
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$\begingroup$ @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. $\endgroup$– user1504Commented Jul 8, 2013 at 22:54
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1 Answer
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There are many ways to answer this question with varying levels of sophistication but here's an attempt at a short and relatively non-sophisticated answer.
Assume the classical field obeys a wave equation such that each mode of the field obeys the equation of motion of an independent harmonic oscillator.
It's straightforward to show that promoting the classical equation of motion of the field to an operator equation of motion is equivalent to quantizing each mode of the classical field as an independent quantum harmonic oscillator.
This allows the quanta of each mode, which are created and destroyed by associated ladder operators for each mode, to be interpreted as "particles" with definite energy and momentum.
answered Jul 8, 2013 at 21:13