Understanding spinors, double cover and professor's expanation

Question

I'm following an introductory course in QFT, and we are facing the spin group part. I think that most of the details are left apart because it would take too much time to be developd, and my profesor doesn't really look like "an expert" in the sector to say it all. I think he just gives us facts, but me I want to understand those facts.

Here is what he wrote:

Representation of Spin$(1, 3)$ which is the double vocer of $SO(1, 3)$. It turns out that Spin$(1, 3) = SL(2, \mathbb{C})$, hence $SO(1, 3) \equiv SL(2, \mathbb{C})/\mathbb{Z}_2$.

$SO(1, 3)$ is homomorphic to $SL(2, \mathbb{C})$ but not isomorphic since it is a $1-2$ map, that is $A, B \in SL(2, \mathbb{C})$, then $\Lambda(A)\Lambda(B) = \Lambda(AB)$ where $\Lambda$ are Lorentz's matrices.

Also to "show" the double cover question, he drew a segment, from point $a$ to point $b$, and then he drew a circle around the segment in order to make the segment a chord of this circle saying "it passes through two points of the segment, so it double covers it".

EDIT: I got answers of two of the questions here: SL$(2, \mathbb{C})$ double cover of SO${}^{+}$(1, 3)

My remaining doubts:

• Finally, $SO^+(1, 3)$ is homomorphic to $SL(2, \mathbb{C})$ but not isomorphic since it is a $1-2$ map. Why the example with Lorentz's matrices?

• to "show" the double cover question, he drew a segment, from point $a$ to point $b$, and then he drew a circle around the segment in order to make the segment a chord of this circle saying "it passes through two points of the segment, so it double covers it". WHAT?

I recognise it's a lot to ask for, but trust me when I tell you that I have been searching for two days in books, notes and so on, and my professor is just rude and not availabel to explanations.

Answer 1

Why is $Spin(1,3)$ the double cover of $SO(1,3)$? You have first to realize that $Spin(1,3)$ is related to how spinor/fermion $\psi$ transforms, whereas $SO(1,3)$ is related to how vector $V$ transforms.

The essence of "double cover" can be elucidated pretty clearly, when vector $V$ is expressed as $V=\Sigma_\mu V_\mu \gamma^\mu$, where $\gamma^\mu$ is the Dirac Gamma operator. Let's look at the different ways spinor/fermion $\psi$ and vector $V$ transform under a rotation: $$ Spin(1,3) \space rotation: \quad \psi \rightarrow e^{iT\theta/2}\psi, \\ SO(1,3) \space rotation: \quad V \rightarrow e^{iT\theta/2}Ve^{-iT\theta/2} , $$ where $T$ is a rotation generator of the Lie algebra.

Because of the double-sided nature of $e^{iT\theta/2}Ve^{-iT\theta/2}$, a $\theta/2$ rotation of a fermion is translated to a $\theta$ rotation of a vector. For example, when the $\theta/2 = \pi$ $$ \psi \rightarrow e^{iT\pi}\psi =e^{i\pi}\psi = -\psi\\ V \rightarrow e^{iT\pi}Ve^{-iT\pi}= e^{2iT\pi}V= e^{2i\pi}V = V $$ for any rotation generator $T$ anti-commuting with $V$ (we also used the property that $T$ squares to unit operator $I$: $T^2 = I$).

So when a fermion is half way around a circle, a vector has already finished the whole circle rotation. And when a fermion finishes the full circle, a vector has already went around the circle twice. That is what we mean by "$Spin(1,3)$ is a double cover of $SO(1,3)$".

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Added note:

If you are curious enough, you might want to ask why we demand that vector $V$ should transform as $V \rightarrow e^{iT\theta/2}Ve^{-iT\theta/2}$ and why $V$ is expressed as $V=\Sigma_\mu V_\mu \gamma^\mu $.

Well, it has to do with how we build a typical vector from spinor-biliear in Quantum Field Theory (QFT): $$ V_\mu = \bar{\psi}\gamma_\mu \psi $$ which are just the coefficients of the vector $\gamma^\mu$ components of $$ \psi\bar{\psi} $$ How does $\psi\bar{\psi}$ transform? Based on the spinor transform property $\psi \rightarrow e^{iT\theta/2}\psi$, you can easy verify that: $$ \psi\bar{\psi} \rightarrow e^{iT\theta/2}\psi\bar{\psi}e^{-iT\theta/2} $$ So the transformation property is dictated by how spinor $\psi$ and vector $V_\mu$ are related in QFT.

In other words, spinor $\psi$ transforms single-sidedly as $Spin(1,3)$ and (the vector portion of) spinor-bilinear $\psi\bar{\psi}$ transforms double-sidedly as $SO(1,3)$. That's the whole story. Pretty neat, huh?