# Some confusion about understanding the relativistic quantum mechanics

S. Weinberg in his book "The quantum theory of fields" chapter 2 introduced the notion of symmetry in quantum mechanic as follows: Physical states are represented by rays in Hilbert space. To be more precise, consider a quantum system whose space of states is a Hilbert space $$\mathcal{H}$$. A pure quantum state of the system is a ray in $$\mathcal{H}$$, that is, a one dimensional subspace $$[\Psi]=\{ z\Psi : z\in \mathbb{C}^*\}$$ where $$\Psi$$ is some non-zero state vector. The set of mutually inequivalent pure states is the projective space $$\mathbb{P}\mathcal{H}\equiv \frac{\mathcal{H}\setminus \{ 0\}}{\mathbb{C}^*}.$$ Is a system is in a state represented by a ray $$R$$ and an expriment is done to test whether it is in any one of the different states represented by mutually orthogonal rays $$R_1 ,R_2 ,\ldots$$, then the probobilty of finding it in the state represented by $$R_n$$ is $$P(R\to R_n)\equiv |(\Psi ,\Psi_n )|^2$$ where $$\Psi$$ and $$\Psi_n$$ are any vectors belonging to rays $$R$$ and $$R_n$$, respectively. A symmetry is a bijection $$\mathbb{P}\mathcal{H}\to \mathbb{P}\mathcal{H}$$ that preserves the probobilty. The symmetries form a group called symmetry group of the system.

He then talks about quantum Lorentz transformation in Section 2.2. In fact, he introduces the Poincare group (inhomogeneous Lorentz group) as the set of transformations that preserves the Minkowski metric.

My question: What does Weinberg mean by quantum Lorentz transformations? I don't understand the relationship between ordinary Loerntz group and quantum one. Are these ordinary Lorentz transformations that preserve the probobilty?

• I think he probably just means the unitary operators that represent the transformation as it acts on the quantum state. But, I will let someone who has the book handy answer...
– hft
Commented Jul 17 at 16:11
• He just means the action of a Lorentz transformation on a Hilbert space. Also, note that he said "Quantum Lorentz Transformation", NOT "Quantum Lorentz Group" which are not the same thing (so your last paragraph is very different from the rest of your question) Commented Jul 17 at 16:20
• @hft That's right. I got it. Thank you very much. Commented Jul 20 at 5:10
• @Prahar You are right. Understood. Apreciate you. Commented Jul 20 at 5:12

The group is still the Poincare group (Lorentz+ translations). The tricky thing is that we need to find a way for that group to act on the Rays in such a way that it preserves the probability. There is a Theorem by Wigner that says that any transformation on $$\mathbb{P}\mathcal{H}$$ which preserves probability can be represented as either a linear unitary, or antilinear antiunitary operator on $$\mathcal{H}$$.