Recent content by StephvsEinst

How the hell did I not think for once of doing the Fourier transform... Thank you, I will work on that!

I was thinking that psi could be described by two different relativistic fields, with each one having a (1 0) or (0 1) (those two states must be read in a COLUMN vector). Wouldn't this give us the four states: 2 different states of spin (up and down) of the particle and 2 different states of...

EDIT: In the quote I meant:

I saw this somewhere but I think it is wrong... I already read Griffiths' "Introduction to Particle Physics" (the 1st edition) from the page 216 to the page 222 (chapter of Quantum Electrodynamics - section "Solution to the Dirac Equation") and I didn't understood why was there the imaginary...

Thank you! :)

Is this the correct way? $$ \overline{P_L \psi } = \left[ \frac{ ( 1 - \gamma_5 ) }{2} \psi \right]^{ \dagger } \gamma_0, $$ We have $$ ( \gamma_5 )^{ \dagger } = \gamma_5 $$ So: $$ = \frac{ 1 }{2} \psi^{ \dagger } \gamma_0 - \frac{ 1 }{2} \psi^{ \dagger } \gamma_5 \gamma_0 \\ =...

Thank you for the reply! It isn't homework, just need to calculate that because I want to understand the following: $$ J_L^{ \mu + } = \bar{ \psi_L } \gamma^{ \mu } \psi_L \equiv V - A $$ Thanks for the hint, I'll try to solve it that way!

Hey! I wanted to prove that: $$ P_L \bar{ \psi } = \bar{ \psi } P_R $$ And I want to know if I did it correctly. $$ --- $$ Here is what I did: $$ P_L \bar{ \psi } = \frac{ ( 1 - \gamma_5 ) }{2} \psi^{ \dagger } \gamma_0, $$ $$ = \frac{ 1 }{2} \psi^{ \dagger } \gamma_0 - \frac{ 1 }{2} \gamma_5...

Thanks for the answer. So I did it correctly?

Hey! I was working with Dirac's equation: $$ ( i \hbar \gamma^\mu \partial_\mu - m ) \psi = 0, $$ and I found that if you work with a function that depends on the momentum, $$ \psi ( \mathbf{p} ), $$ you obtain: $$ ( i \gamma \cdot \mathbf{p} + m ) \psi ( \mathbf{p} ) = 0. $$ The problem is...

It's true that ## \bar{\psi }_L \gamma^{ \mu } \psi_L = \bar{\psi }P_R \gamma^{\mu } P_L \psi ## but I can't prove mathematically this step. I see now that [ ## \gamma_{\mu }, \gamma_5 ] = 0 ## so it's true that ## \gamma_5 \gamma_{\mu } = \gamma_{\mu } \gamma_5 ## . I am still not...

Hi! Can anyone explain to me the math behind this simple step: $$ P_L \psi = \psi P_R $$ where $$ P_L = \frac{1}{2} ( 1 + \gamma_5 ) $$ and $$P_R = \frac{1}{2} ( 1 - \gamma_5 )$$. And why is $$ \bar{\psi }P_R \gamma^{\mu } \psi = \bar{\psi } \gamma^{\mu } P_L \psi ,$$ where $$ \gamma_5$$ and...

Summed up looks easy xD I will work on that, thank you for the reply.

When I said deduction I was thinking of the mathematics behind the SM Lagrangian. Probably deduction wasn't the best word to use, instead I should have said "mathematical background".

Does anyone know where can I find the deduction of all terms of the updated SM lagrangian? Although I have already looked at some lagrangians and theories like local gauge invariance, Yang-Mills theory, feynman rules, spontaneous symmetry-breaking and others, I wanted to see the deduction and...