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Consider a vector field $A^\mu(x)$. Its Lorentz transformation is given by

\begin{equation}\tag{1} A'^{\mu}(x') = \Lambda^\mu_{\:\nu}\,A^\nu(x), \end{equation}

where $x'\equiv x'^\mu = \Lambda^\mu_{\:\nu}\, x^\nu$. After quantization, the vector field is now an operator acting on the Hilbert space of states ($\equiv \mathcal{H}$), and its transformations under Lorentz is no longer given by (1), but by

\begin{equation}\tag{2} U^\dagger(\Lambda) \widehat{A}^\mu(x') U(\Lambda) = \Lambda^\mu_{\:\nu}\,\widehat{A}^\nu(x), \end{equation}

where $U(\Lambda)$ is a unitary representation of the Lorentz transformation $\Lambda$ on $\mathcal{H}$. Here is my question; the RHS mixes $\Lambda^\mu_{\:\nu}$ and $A^\nu$. Respectively, the first object is a Lorentz transformation but not represented on $\mathcal{H}$, and the second one is an operator acting on elements of $\mathcal{H}$. How to reconcile the two pictures?

Indeed, when writing down the exponential forms of the transformations, one has; $\Lambda = \exp(-i\omega_{\mu\nu}\mathcal{J}^{\mu\nu}/2)$, the defining representation of the Lorentz group, and $U(\Lambda) = \exp(-i\omega_{\mu\nu}J^{\mu\nu}/2)$, where $J$ now is an operator on $\mathcal{H}$. So it seems we have two different representations of the Lorentz group, which I don't see how it makes sense.

Qmechanic
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QuantizedObject
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1 Answers1

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TL;DR: OP's classical and quantum field is similar to a $G$-equivariant map $A:M\to V$ and $\widehat{A}:M\to V\otimes L({\cal H},{\cal H})$, respectively.

Here:

  • $G$ is the Lorentz group;
  • $M$ is Minkowski spacetime;
  • $.:G\times M\to M$ is a left group action.
  • ${\cal H}$ is a complex Hilbert space;
  • $\rho:G\to GL(V)$ and $R:G\to B({\cal H})$ are $G$-representations;
  • $L({\cal H},{\cal H})$ is the set of $\mathbb{C}$-linear operators: ${\cal H}\to{\cal H}$.

Then $$ \forall g\in G, m\in M: \quad A(g.m)~=~\rho(g)(A(m)), $$ $$ \forall g\in G, m\in M: \quad \widehat{A}(g.m)~=~\left(\rho(g)\otimes {\rm Ad}(R(g)\right)(\widehat{A}(m)), $$ where ${\rm Ad}$ denotes the adjoint representation, i.e. a similarity transformation.

(This answer is inspired by my Phys.SE answer here.)

Qmechanic
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