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1 vote
0 answers
87 views

How does the two index spinor $(v_{a\dot{b}})$ transforms?

Using the Van der Waerden Notation, we define the four-vector as: $$v_{a\dot{b}}=v_\nu \sigma^\nu_{a\dot{b}}$$ I'm trying to see how this transforms. Defining: $$\Lambda \equiv e^{i\vec{\theta}\cdot \...
Young Kindaichi's user avatar
1 vote
0 answers
81 views

Derivation of the transformation law for spinors

I'm reading the book Quantum Field Theory: An Integrated Approach by Eduardo Fradkin, and I got stuck where the transformation law for spinors $$ \psi'(x') = S(\Lambda) \psi(x) $$ is derived. In ...
SrJaimito's user avatar
  • 601
3 votes
1 answer
187 views

What is the relationship between spinors and rotating motion geometrically?

Spinors are famously like spinning tops, but not actually like spinning tops since they are point particles and thus cannot rotate around their axis. It is easy to show algebraically how spinors must ...
user438236's user avatar
1 vote
0 answers
66 views

Doubt on transformation laws of tensors and spinors using standard tensor calculus and group theory

1) Introduction From standard tensor calculus, here restricted to Minkowski spacetime, we learned that: A scalar field is a object that transforms as: $$\phi'(x^{\mu'}) = \phi(x^{\mu})\tag{1}$$ A ...
M.N.Raia's user avatar
  • 3,085
1 vote
1 answer
219 views

How to contract spinor indices?

In normal vector representation, vectors can be contracted as follows: $$v^\mu v_\mu$$ with one covariant and one contravariant index. But in spinor representation, there are 4 possible type of ...
Habouz's user avatar
  • 1,324
0 votes
1 answer
90 views

Lorentz boost property of gamma matrices

I was watching this video where he boosted the Dirac equation. He reached this equation: $$S^{-1}(\Lambda)\gamma^\mu S(\Lambda)=\Lambda^\mu{}_\nu \gamma^\nu$$ My question is since $\gamma^\mu$ is a ...
Habouz's user avatar
  • 1,324
0 votes
0 answers
38 views

Can one express the evolution of a particle with a one-parameter group of $SO(3,1)$?

Can one express the evolution of a particles using a sequence of $SO(3,1)$ transformations? If yes, how? Is it sufficient to apply $SO(3,1)$ transformations to a spinor? $$ \psi(t) = e^{t\mathfrak{so}(...
Anon21's user avatar
  • 1,548
2 votes
0 answers
146 views

Angular-momentum of the Dirac spinor theory

The standard Dirac action $$ S = \int d^4 x \bar \psi (i \gamma^\mu \partial_\mu - m) \psi $$ is invariant under Lorentz transformation. In David Tong's lecture note, eq (4.96) lists that the ...
Lelouch's user avatar
  • 669
5 votes
2 answers
451 views

What is the idea behind 2-spinor calculus?

In the book by Penrose & Rindler of "Spinors and Space-Time", the preface says that there is an alternative to differential geometry and tensor calculus techniques known as 2-spinor ...
Cathartic Encephalopathy's user avatar
1 vote
1 answer
212 views

Does the Dirac Spinor live in the complexification of the Lorentz group?

In this question I learned that when working with quaternionic representations of (the double cover of) our relevant orthogonal group, we cannot avoid working in complex vector spaces. However it is ...
Craig's user avatar
  • 1,117
3 votes
0 answers
147 views

Is the real spinor representation of the Lorentz group irreducible?

Specifically the $(\frac{1}{2},0)\oplus(0,\frac{1}{2})$ representation. Given that we label representations by the corresponding representations of the complexified Lie group, the direct sum can be ...
Craig's user avatar
  • 1,117
0 votes
1 answer
146 views

Quantum Field Theory Unitary Transformations

I am currently reading through Itzyskon and Zuber for my quantum field theory class, and I came across this regarding the unitary transformations of the Dirac bispinors in chapter 2. They show that ...
user132849's user avatar
1 vote
1 answer
250 views

Linearization of the Klein-Gordon equation and decoupling of ''spinors''! [closed]

We know that the K-G equation is deduced from the Einstein relation: $E^{2}=m^{2} +\vec{p}^{2} \;\;\;\;$ (with $c=1$) It is known that :$E^{2}=\frac{m^{2}}{1-\beta^{2}}=\left(\frac{m}{1-\beta}\...
The Tiler's user avatar
  • 1,400
1 vote
2 answers
160 views

To construct a Lorentz scalar we use $\psi^{\dagger}\gamma^{0}\psi$. Could we use $\gamma^{5}$ instead of $\gamma^{0}$ seen as both are Hermitian?

Both $\gamma^{0}$ and $\gamma^{5}$ are Hermitian, so could we replace $\gamma^{0}$ with $\gamma^{5}$ to construct a Lorentz scalar with the same properties as $\bar{\psi}\psi$?
sputnik44's user avatar
  • 115
0 votes
0 answers
155 views

Unitarity and boost

I wonder if anyone could shed some light on the representation theory of the Lorentz group. In particular, I would like to understand unitary and spinorial representations of boosts better. To my ...
egj's user avatar
  • 9

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