Maybe ##E_p## is the Energie in the momentum which replaces ##\hbar w_k## in 2.41.There is no more explanation in the book for as much as I can see.
In my eyes it must be the notation for the integral version of the discrete case of eq 2.41.I must of course convince myself that it is...
How to derive the relevant equation ?
By comparing it to eq 2.41 in the book:
##\hat H = \sum_{k=1}^N\hbar\omega_k (\hat a\dagger_k\hat a_k + 1/2)##
I think 3.34 is the integral version of 2.41
Homework Statement
How to derive eq.3.34 in Lancaster's book QFT for the gifted amateur
Homework Equations
##\hat H= \int d^3 p E_p\hat a\dagger_p\hat a_p##
The Attempt at a Solution
Comparing
I thank you for guiding me further to "homework PF"
By the way i am p 35.Thanks to your legitimate critics i made an extra effort and understood by my self the mathematics of the previous pages.
I found it relatively easy.So i am surely not as you say "going to keep on like this." I hope to...
I am just a fascinated selfstudent.
I understand your time is precious and I fully respect it!
It's also a chalenge to understand things by myself.
These are two excellent reasons to do my very best to understand before asking a question.
Anyhow i am gratefull to you for all your answers.They...
I am really doing my best to understand before I ask a question!
May I ask if it is necessary to mention each time "QFT for the gifted amateur"
Thanks again for helping!
In eq's 3.17 and 3.18 p 33 QFT for the gifted amateur
Are the constants of proportionality ##\sqrt (n_i+1) and \sqrt n_i## explained somewhere in the book?
Indeed I see I can trivialy plug in.But then I have a product of ##\rm{p_k}## with ##\rm{p_k}## .
Thats why I said "I square 2.63 and plug it in 2.59".And the same for 2.62.
In my humble opinion this yields the same result.
As for the fact that it is going to get much harder,I don't intend to...
I want to ask if i understand it well:
In the first equality the - sign becomes + because it is ##p\tilde\dagger##. In the second equality ##p\dagger## becomes ##p## because ##p## is hermitian and the third equality is definition again.
Is it correct?