A complex classical field, $\phi(t,\vec{x})$, is represented by an operator $\hat{\phi}(t,\vec{x})$ in QFT. Why is its complex conjugate, $\phi^*(t,\vec{x})$, gets replaced by the adjoint operator, $\hat{\phi}^\dagger(t,\vec{x})$? Is it obvious?
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$\begingroup$ We're generalizing $|\phi|^2\ge0$ to $\left\Vert\hat{\phi}|\psi\rangle\right\Vert^2\ge0$ for any state ket $|\psi\rangle$. $\endgroup$J.G.– J.G.2026-01-31 15:10:01 +00:00Commented Jan 31 at 15:10
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1 Answer
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It's obvious if you consider a complex scalar field $\phi$ simply as the combination of two real-valued scalar fields $\phi_1,\phi_2$. Real-valued scalar fields are self-adjoint: $\phi_i^\dagger = \phi_i$, and so writing $$\phi = \phi_1 + \mathrm{i}\phi_2 $$ immediately implies $$ \phi^\dagger = (\phi_1+\mathrm{i}\phi_2)^\dagger = \phi_1 - \mathrm{i}\phi_2 = \phi^\ast.$$
