I was thinking about the following thing: we know that if the Lagrangian in field theory doesn't depend on the spacetime position, the Noether's theorem says that the stress-energy tensor is conserved, and that T^00 is the energy density at spacetime point x.
Then if one integrates this h(x)...
Ok, but I tried that example because I was able to treat it nonperturbatively (after all we agreed that it is a free theory!). We just started feynman diagrams in our course and didn't get to renormalization theory yet (I don't even knopw if we will do it in this course) so my knowledge of these...
Question about this "interaction" in QFT
Hi, I started two months ago my course in QFT, and since I heard about the fact that the bare mass appearing in the Lagrangian of a theory isn't the physical mass of a particle (due to self interaction, I guess), I tried to find an example explicitly...
My knowledge about this is very limited.. So i'd be glad if you could post some references. I just started a course on quantum field theory and since (but it also happens in the mathematical formalism of quantum mechanics) we're dealing with operator valued functions, I wanted to know more about...
In the original post, things should look like this:
f'(s_0)
is an operator such that:
lim_{\Delta s\rightarrow 0}\frac{|f(s_0+\Delta s) - f(s_0) - \Delta s\,f'(s_0)|_L}{\Delta s} = 0
The oroginal post was wrong since it compared a number to an operator :smile:
Ok, one doesn't need to put || in THAT definition, and I know that f'(s_0) is an operator, what I meant was that when i use the usual \epsilon,\delta
definition of limit, the norm on L will surely pop out: I mean that f'(s_0) is the operator that does this thing...
If I have a function
f:R\rightarrow L
where L is the space of linear operators from an hilbert space to itself, how can i define the derivative of f at a particular point of R? I mean, it is "obvious" that one should try:
f'(s_0)=lim_{\Delta s\rightarrow0}\frac{|f(s_0 + \Delta s)...
Yes, I know these mathematics behind that, but my question is why do we do so? I mean, why one says the things you wrote in bold? Is it somehow related to active vs passive transformations?
My question is the following: when in quantum mechanics one introduces symmetry, says that a states and observables transform both, in order to mantain mean values intact (kind of like a change of coordinate system), i.e.:
|\psi>\rightarrow U|\psi>
and
O\rightarrow UOU^\dagger...
No the example is ok! I did something similar for the free particle.. The thing is that it doesn't answer my problem.. I am looking for a vector field whose flow I can use to solve EXACTLY a PDE evolution equation, as I can do in classical mechanics.
The main point of my last post was, in...
Hi spocchio!
I didn't get what you were trying to say in your last post, but I tried to solve that problem.
First off I took as initial state ket the function \psi (p) = p^2 and made it evolve: i took terms not higher than order t and got (I set \hbar = 1 )
(\hat{U}_t\psi)(p)...
I know.. But in classical mechanics an observable is an element of a infinite dimensional space too! The space of functions of position and momentum. Maybe I didn't made myself clear enough in my last post: from what i understood about CM and its equations, one can find the hamiltonian flow by...
But how can I find the vector field that generates this flow? I tried (naively) to get the three "equations of motion" by chosing as \psi(x,y,z) the functions \psi(x,y,z) = x , \psi(x,y,z) = y , \psi(x,y,z) = z . By doing this I got that:
\frac{d}{dt}\vec{r} =...
Here's my question: as soon as I learned Quantum Mechanics and Schrodinger equation, I saw a "similarity" with the equation one gets in classical mechanics for the evolution of a function in phase space. In QM one has:
i\hbar\frac{d}{dt}\psi = \hat{H}\psi
and this is a evolution...