Lorentz transformation: unitary $U(\Lambda)$ on the state of the Hilbert space, but the boost $\Lambda_{1/2}$ is not unitary on $\psi$

In p.59 of Peskin and Schroeder QFT book, he mentioned that

the operator $$U(\Lambda)$$ that implements the Lorentz transformations on the state of the Hilbert space is unitary, even though the boost $$\Lambda_{1/2}$$ is not unitary.

I interpret this sentence as follows ---

• The unitary transformation should be $$U(\Lambda)$$ on a state $$|q \rangle$$ as $$U(\Lambda) |q \rangle$$ for states with quantum observables $$q$$.

• The non unitary transformation should be $$\Lambda_{1/2}$$ on a field $$\psi$$ as $$\Lambda_{1/2}\psi$$.

questions

1. What is the significance of unitary $$U(\Lambda) |q\rangle$$ and non unitary $$\Lambda_{1/2}\psi$$?

2. But there is a relation Peskin's book eq 3.122 between $$U(\Lambda) \psi(x) U(\Lambda)^{-1} =\Lambda_{1/2}^{-1}\psi(\Lambda x) \tag{3.122}.$$ If $$U(\Lambda)$$ is unitary transformation, we have $$U(\Lambda)^{\dagger}U(\Lambda)=U(\Lambda)^{-1}U(\Lambda)=1$$. Then we also have: $$U(\Lambda) \psi(x) U(\Lambda)^{\dagger} =\Lambda_{1/2}^{-1}\psi(\Lambda x) \tag{3.122'}.$$

If we sandwitch this operator by states on two sides: $$\langle q| U(\Lambda) \psi(x) U(\Lambda)^{\dagger} |p\rangle =\langle q| \Lambda_{1/2}^{-1}\psi(\Lambda x) |p\rangle\tag{3.122''} .$$

1. How come the $$U(\Lambda)^{\dagger} |p\rangle$$ is unitary on $$|p\rangle$$ , and $$\langle q| U(\Lambda)$$ is unitary on $$\langle q|$$, but the $$\Lambda_{1/2}^{-1}\psi(\Lambda x)$$ is non unitary?
• To 1.: $U(\Lambda)$ should better be unitary, since you want quantum amplitudes $\langle a|b \rangle$ (i.e. observables) to be independent of the frame of reference (follows immediately). This is ensured if $U$ is a unitary representation. On the other hand such a necessity is not there for the representation as transformations on fields. To 2.: Is this a question?. To 3.: I do not understand the question. In particular what do you mean by "unitary on" something? Commented Oct 4, 2021 at 14:45
• Commented Oct 4, 2021 at 15:36

The necessity of having unitary representations of a symmetry group on the Hilbert space of states is well understood as a consuqeunce of Wigner's theorem; you can look at Weinberg Vol 1 chapter 2.

Instead the representation of the group acting on the space target of the fields, need not to be unitary. There is no such a necessity, since Wigner theorem concerns the predictive aspects of a quantum theory, hence it's transition amplitudes.

The fact that the spinorial representation of the lorentz group $$SO(1,3)^+$$ (we consider the proper ortochronus part) is non unitary, is directly related to the fact that the Lorentz group is non compact. (There are no non trivial finite dimensional unitary representations for a non compact Lie group, the unitary rep on the target space of fields should be finite dim, hence there aren't any). This group is generated by the angular momentum operators, which gives the rotations, and by boosts, which gives the hyperbolic rotations. The latter are responsible for the non compactness of the group, as their parameters are not limited in value. Being non unitary you can check that it's generators are not all hermitian.

In particular considering the representation $$\Lambda_{\frac{1}{2}} = \exp[-i/2 \omega_{\mu \nu}\sigma^{\mu \nu}]$$ with $$\sigma^{\mu \nu}=\frac{i}{4}[\gamma^{\mu}, \gamma^{\nu}]$$, by looking for the generators $$J^i=1/2\epsilon^{ijk}\sigma^{jk}$$, related to rotations, and $$K^i=\sigma^{0i}$$, related to the boosts, you can check directly, in the Weyl-chiral base, that $$J^{i\dagger}=J^i$$ and $$K^{i\dagger}=-K^i$$. This is necessary to recover the correct transformations of the group from the algebra. If the boosts were hermitian in fact I would recover from the exponentation some rotations between time and coordinates, and not the hyperbolic rotations of the Lorentz group.

But this shouldn't surprise, since the whole spinorial representation is built through the gamma matrices algebra (Clifford algebra) for this purpose. In fact from $$\{\gamma^{\mu}, \gamma^{\nu}\}=2g^{\mu\nu}$$ you get $$\gamma^0\gamma^0=\mathbb{1}$$ and $$\gamma^i\gamma^i=-\mathbb{1}$$, hence the $$\gamma^0$$ has eigenvalues $$+1, -1$$, while $$\gamma^i$$ has eigenvealues $$+i, -i$$. So the former is hermitian and the latter isn't. In this way, recollecting that $$\sigma^{\mu \nu}=\frac{i}{4}[\gamma^{\mu},\gamma^{\nu} ]$$, we can see that:

1. $$\sigma^{ij}=\frac{i}{4}[\gamma^{i},\gamma^{j}]$$ is $$i$$ times a commutator of two anti hermitian operators (and so $$i$$ times an antihermitian operator), hence $$\sigma^{ij}$$ is hermitian.

2. While for boosts, $$\sigma^{0i}=\frac{i}{4}[\gamma^{0},\gamma^{i}]$$ is $$i$$ times the commutator of an hermitian and an anti hermitian operators (and so $$i$$ times an hermitian operator). This shows that $$K^i\sigma^{0i}$$ is anti hermitian. As it should be, boosts generators are represented by anti hermitian operators.

To sum up this should ( I hope ) answer 1,3. About 2, i don't remember what Peskin does, but $$U(\Lambda)$$ for the reason above in the third paragraph shouldn't be unitary.