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Under the Lorentz transformation, classical Dirac fields transform like: $$ \psi(x)\rightarrow\psi^\prime(x^\prime)=\Lambda_{1/2}\psi(\Lambda^{-1}x) $$ where $\Lambda_{1/2}=e^{-i\omega_{\mu\nu}S^{\mu\nu}/2}$ and $\Lambda$ is an element of Lorentz group, while quantum Dirac fields transform like: $$ \psi(x)\rightarrow U(\Lambda)\psi U^{-1}(\Lambda)=\Lambda_{1/2}^{-1}\psi(\Lambda x) $$ where $U(\Lambda)$ is a representation of $\Lambda$ in Hilbert space.

In Peskin & Schroesder, An Introduction to quantum field theory, page 60, it said when we deal with classical field, "we transformed a pre-existing field distribution that was measured by $\phi(x)$" and when we deal with quantum field "we are transforming the action of $\phi(x)$ in creating or destroying particles". Does it make any difference to these 2 equations? If so, how? I can't see it.

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  • $\begingroup$ In the classical theory, the Dirac field is just a spinor-valued function of space-time. In the quantum theory, the Dirac field is an operator-valued function that carries spinor degrees of freedom. In the latter case, the Lorentz transformations can either operate on the spinor degrees of freedom (left-hand side) or they can operate as quantum operators in the Hilbert space in which the Dirac field operator is defined (right-hand side). $\endgroup$ – flippiefanus Oct 23 '16 at 4:25
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In fact, the quantum Dirac field - like every quantum field - transforms both in a finite-dimensional representation $\rho_\text{fin}$ of the Lorentz group and in an infinite-dimensional unitary representation $U$ of the Poincaré group, such that $$ \rho_\text{fin}(\Lambda)\phi(\Lambda^{-1} x + a) = U(\Lambda,a)\phi(x)U(\Lambda,a)^\dagger$$ holds for every Lorentz transformation $\Lambda$ and every translation vector $a\in\mathbb{R}^4$. This property - a sort of compatibility between the classical transformation of the field as a field and the quantum transformation of the field as an operator - is one of the Wightman axioms.

It's not really that the quantum field transforms "differently" from the classical field, it's that it has two ways of being transformed. We can apply the classical $\rho_\text{fin}$ to transform the field, but it's now operator-valued. And we know from linear algebra that if we let a unitary transformation $U$ act on a vector space as $x\mapsto Ux$, then all operators on the space transform as $O\mapsto UOU^\dagger$. So the second way of transforming the quantum field is an inevitable consequence of us having declared that the quantum field is operator-valued.

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