Lorentz transformation in QFT

Question

A Lorentz transformation can be seen as a change in reference frame. So, after apply a Lorentz transformation to a system (or change the reference frame), how should the state and field operator change? I can't find a book which introduces those things very well. I am considering three kinds of possible explanation. Here is my consideration.

1. The field operator $\phi(x)$ doesn't change. But the state vector will change. A unitary transformation $U(\Lambda)$ will be applied to the state: $|\psi\rangle\rightarrow U(\Lambda)|\psi\rangle$.
2. The state vector $|\psi\rangle$ doesn't change. But the operator will change. $\phi(x)\rightarrow U(\Lambda)\phi(x)U^{-1}(\Lambda)$? or $\phi(x)\rightarrow U^{-1}(\Lambda)\phi(x)U(\Lambda)$? Which one is true?
3. The state vector and the operator both change.

I know 1 and 2 may be the same. If they are true, they are just different forms.

But I don't know which is the correct one, 1,2 or 3? Can someone explain it in details? Thanks.

Answer 1

Lets look at a Lorentz transformation (LT) from the unprimed to the primed system (I will look at transformations on Hilbert space, Fock space is analogous). The state that was described in the unprimed system as $|\psi\rangle$, will now be described in the new coordinate system as $U|\psi\rangle=|\psi^{\prime}\rangle$. All we have done is to change the description of the state, the state itself is unchanged under a passive transformation. And the same operator $O$ (after a passive transformation) will now be described in the new reference frame as $UOU^{-1}$. So $3$ is correct, the description of both operators and states change under a change of coordinate system. This is consistent with requiring that measurements are independent of the coordinate system used:

$$\langle\psi|O|\psi\rangle = \langle U\psi|UOU^{-1}|U\psi\rangle$$

Beware of a possible source of confusion though. Sometimes the notation $O^{\prime}=UOU^{-1}$ is in use. This is to be understood as the operator $O$ is described/interpreted physically in the primed system as $UOU^{-1}$. It does not mean that $O$ transforms into an analogous operator, but defined relative to the primed coordinate system. For example, let the operator $O$ be $P^{x}$, the $x$ - momentum operator in the unprimed system. Note that (as one would expect):

$$UP^{x}U^{-1}=\Lambda^{x}_{\mu^{\prime}}P^{\mu^{\prime}} \ .$$

In other words, the operator is interpreted in the unprimed system as the $x$ - momentum operator, while in the primed system the same operator is interpreted as the sum of primed momentum - operator components $\Lambda^{x}_{\mu^{\prime}}P^{\mu^{\prime}}$. Certainly,

$$UP^{x}U^{-1}\neq P^{x^{\prime}} \ .$$

The exception is when the operator transforms as a relativistic world scalar - then there is equality. As an example, under a LT:

$$P_{\mu}P^{\mu}\to UP_{\mu}P^{\mu}U^{-1}=P_{\mu^{\prime}}P^{\mu^{\prime}} \ .$$

The form of the operator is the same in the primed system (in terms of primed operators), as in the unprimed system - i.e. the physical interpretation is the same in both systems.