Hello World..

1drv.ms/u/s!AozWlUoG8z4tiGAZoAck26PdtI_R

Pls see link in example-2. I don't understand the solution of equation which is 3.11

Pls pls answer

@Charlie s have a look

@123 it's the formula here

$\sin\theta + \cos\theta = \sqrt{2}\sin(\theta+\frac\pi4)$, a useful trick to know

Guys I am using frixion pens for my notes

With these pens , you can erase what you have written

but on YouTube and internet I found that if I keep my notes on the car dash or somewhere out where it is hot . My notes will be erased

If you know about these pens , then pls tell me if I should use these pens or not

I am preparing for jee

@NiharKarve thanks. How we find this equivalence relation.

Hi @JohnRennie sir

@123 did you see the link in my previous message? There's a derivation on Maths SE.

@NiharKarve thanks I see the link

I have so many PDFs entitled "An introduction to quantum field theory"

@Slereah would you rather they all had quirky names where you can't tell they're intro to QFT texts? :P

@ACuriousMind You can do both, hopefully

Maybe give it a youthful vibe

The Fresh Rap about Quantum Field Theory

Also if they are so generic that they don't deserve a more specific name, do they deserve to exist

We're not lacking in generic QFT introductions

I think a QFT intro is maybe what many theorists write before writing anything more advanced?

look at it as training wheels

Do a specific angle at least, I don't know

The Marxist introduction to quantum field theory

or even just something that's not QFT

Do a weirdly intense theoretic intro to thermodynamics

Those are less common

A bundle interpretation of rollercoaster theory : amazon.com/Coasters-101-Engineers-Roller-Coaster/dp/1468013556

I have just stumbled across this gem. It needs only one more delete vote if any 20k users are feeling in the mood.

:O he is dangerously close to 10k

To write a Marxist QFT book one would first have to seize the means of pair production

@ACuriousMind dear moderator , I want some of my question to removed. I contacted stack exchange team many a times but no one responded. Can you help me with that and what is the way to do that

Also , if anyone else’s knows.Please do help me

You can delete your own questions, there's a button for it under your question, under the tags

It says you can’t sometimes

When a lot of people have put their answers in it

@Charlie

Oh, perhaps after a certain amount of time you can't

But it also says if you contact the team , they will help

Why do you want to remove the questions anyway? If they're off-topic and/or bad questions they'll get closed/removed anyway

In that case I don't know, you'd have to talk to the team

:p

Closed doesn’t mean people can’t see it

I think closed posts get removed after some time but I can't be sure

@Charlie You can't delete questions that have upvoted answers, among other restrictions, see meta.stackexchange.com/a/5222/263383.

Ah, I see

How come some sources have the Dirac Lagrangian with the derivative going both ways and some don't? The Lagrangian $\mathcal L=\bar\psi(i\gamma^\mu\partial_\mu-m)\psi $ gives the correct EOM's for $\psi^\dagger$ but not for $\psi$ because there's no derivative acting on $\psi^\dagger$?

I don't know what you mean by the "correct e.o.m. for $\psi^\dagger$ but not for $\psi$"

There is only one dynamical field here, $\psi$. When you vary the action w.r.t. it, you get an e.o.m. What's the problem?

as in if I calculate $\partial\mathcal L/\partial(\partial_\mu\phi^\dagger)$ I get zero

Some sources will write the partial derivative with a double ended arrow over it to indicate that we also take the derivative of the $\bar\psi$, but a lot don't

@Charlie sure, but that's not the e.o.m.

From the EL equation you'd then get $-m\psi=0$ no?

hmm maybe I am misinterpreting the role of the conjugate eom

no, you get $(\mathrm{i}\gamma^\mu\partial_\mu - m)\psi = 0$ - why has the derivative disappeared in your version?

oh

woops

lol

ty, I see what I did and I am ashamed

I think there's a "symmetric" version of the action

$$\mathcal{L} = \frac{1}{2} \left[ \bar{\psi} (i \gamma^\mu \partial_\mu - m) \psi + \psi (i \gamma^\mu \partial_\mu - m) \bar{\psi} \right]$$

here for instance they have the lagrangian with the derivative going both ways

But it's not fundamentally important I think?

Because whether your EOM applies to $\psi$ or $\bar{\psi}$, you get the same theory

I've never seen it like that actually, that just looks more complicated without gaining anything

Yeah I'm not 100% sure why some people use it

Maybe it has some pleasant properties

Actually... Wouldn't the two terms be related by partial integration?

I was going to guess something in the integration by parts makes the distinction unnecessary if I hadn't made a high-school level algebra booboo

There may be some slight complications if it's a Berezin integrals, maybe, but I think it's just equivalent

@Slereah there's no Berezin integrals here

you only get them in a path integral context where you want to integrate over the $\psi$s

Also I think that a variable and its conjugate commute in the Grassmann algebra

So even in that context it's not an issue

Without the bar it really is possible to get confused/make a mistake and is ambiguous in a sense because of the total derivatives right

Does the Hermitian conjugate operation commute with the partial derivative? I.e. $(\partial_\mu\psi)^\dagger=\partial^\mu(\psi^\dagger)$?

actually hmm

I have to think in this case of the derivative being basically a matrix right

$(\partial_{\mu} \psi)^{\dagger} = \psi^{\dagger} \overleftarrow{\partial}^{\dagger}$?

yeah i was just in the middle of writing that down

the derivative is just a spacetime derivative, conjugation does nothing to it

$\psi$ is a four-component vector and the adjoint acts on this as a vector, the derivative doesn't live in this vector space so I'd say it's irrelevant and $\partial_{\mu}$ commutes with it

Yo

?!? nature.com/subjects/astronomy-and-astrophysics