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A classical field $$\phi(x) =\int\frac{d^3p}{\sqrt{E_p}} (a_pe^{-ipx} + a^*_pe^{ipx})$$ is not a unique thing. We could "choose" $a_p$ and $a^*_p$ conveniently so that it represent field of our specific problem.

But when we quantize it we get

$$\phi(x) =\int\frac{d^3p}{\sqrt{E_p}} (a_pe^{-ipx} + a^{\dagger}_pe^{ipx})$$ then $a_p$ and $a^{\dagger}_p$ are unique thing (ladder operators of a SHO). So who this field represent our specific quantum system?! We are in Heisenberg picture so we expect dynamics in operators but rigidity of $a_p$ and $a^{\dagger}_p$ don't let any dynamic even for interacting fields. What is the overall picture of QFT in this manner?

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  • $\begingroup$ You guys keep asking this. Are you in a troubled class? Your classical field is the real solution to K-G, and the quantum the hermitean one. What does your text say? $\endgroup$ – Cosmas Zachos Apr 28 '20 at 20:10
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After the quantization the fields are not describing the state of the system anymore, but they are linear operators that act on it. Your question is about something we already observe in standard quantum mechanics, take for example the harmonic oscillator. The classical state variables $x,p$ are associated with Operators $X$ and $P$ composed in a very similar fashion to the fields you mentioned, see Wikipedia. The don't wiggle around as they do in a classical harmonical oscillator, instead they are a fixed objects and so are the fields in QFT. Instead the dynamic happens in the Hilbert space of our system. For a free QFT the Hilbert space is given by a Fock space and instead of saying my (classical) system is in the state $\phi(x)$ you say my (quantum) system is in the state $|\psi\rangle\in\mathcal{H}$. The be precise here, the classical states are given by solutions of your equations of motion, thus fields, while quantum states are given by a density operator $\rho$ on the Hilbert space of your system (or an element of your Hilbert space, depending on how you precise you want to be).

I hope this can answer your question.

Cheers!

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  • $\begingroup$ Thanks for your attention but at this moment I don't get my answer. I think we do QFT in Heisenberg picture and in it $X$ and $P$ are time dependent (by formulas 2.3.45 at Sakurai's text) @Johnny $\endgroup$ – moshtaba Apr 28 '20 at 20:37
  • $\begingroup$ True that, but this is just a question of where you put the time evolution. Let's put it the following way. Consider the Klein-Gordon field, a solution is fully specified once given the data $(\phi(t_o,\bullet),\partial_t \phi(t_0,\bullet))$ for some time $t_0$. In quantum theory, the state at time $t_0$ is instead discribed by an element in the Fock space and the evolution of this state is specified by the Hamiltonian. We can use the Heisenberg picture and move the time evolution into the operators, but it won't change the expectation values and thus the physical quantities. $\endgroup$ – Johnny Longsom Apr 28 '20 at 21:24
  • $\begingroup$ Anyway, the field operator as you wrote it in your question is just the most general hermitian (because it's a real field you quantize) field operater that satisfies the Klein-Gordon equation and the leader operators are then fixed by the canonical comutation relation. $\endgroup$ – Johnny Longsom Apr 28 '20 at 21:25
  • $\begingroup$ By the way, concerning the title of this question, quantization is far from unique, since commuting variables are raised to non-commuting ones. You have to decide how to manage the ordering your variables for the quantization. $\endgroup$ – Johnny Longsom Apr 28 '20 at 21:31

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