jonjacson said:
WHy is that each one of you gives me a different version?
I will search info about wave functionals.
I appreciate any comment really, thank you so much.
As mentioned above there are many ways of formulating Quantum Field Theory.
The Wavefunctional approach is the form most closely related to the way people normally do non-relativistic QM. However solving for the Wave Functional or Wave Functional eigenvalue problems are virtually impossible outside of free field theories, so people usually use the method of calculating the correlation functions I mentioned. Non-relativistic QM can also be expressed via correlation functions just to be clear.
One thing I didn't mention in my post is renormalization. When we write down QFT we often make use of expressions like ##\phi^{3}(x)##. However because the object ##\phi(x)## is so singular, it turns out that ##\phi^{3}(x)## isn't a well defined mathematical expression. So one has to attempt to define a quantity similar to ##\phi^{3}(x)## that is well defined: ##[\phi^{3}(x)]_R##.
I'll give an example using plain old functions. Say you have the Heaviside function ##\Theta(x)## and the function ##\frac{1}{x}##. We have their multiple ##\frac{\Theta(x)}{x}##. However consider the following integral:
$$\int_{\mathbb{R}}{\frac{\Theta(x)}{x}g(x)dx} = \int^{\infty}_{0}{\frac{g(x)}{x}dx}$$
with ##g## some function. The ##\Theta## function has been used to restrict the integral in the second expression.
In general this will diverge if ##g## is non-zero about the point ##x = 0##.
However the following modified version of the integral is well-defined:
$$\int^{a}_{0}{\frac{g(x) - g(0)}{x}dx} + \int^{\infty}_{a}{\frac{g(x)}{x}dx}$$
So we isolate the problematic region and add a term that removes the divergence.
This can be rewritten as:
$$\int^{\infty}_{0}{\frac{g(x)}{x}dx} - \int^{a}_{0}{\frac{g(0)}{x}dx}$$
And since ##g(0)## can be obtained from a delta function, we can further rewrite it as:
$$\int^{\infty}_{0}{\frac{g(x)}{x}dx} - \left(\int^{a}_{0}{\frac{1}{x}dx}\right)\int{\delta(x)g(x)dx}$$
Finally if we put back in the Heaviside function, this can be written as:
$$\int_{\mathbb{R}}{\left[\frac{\Theta(x)}{x} - \left(\int^{a}_{0}{\frac{1}{x}dx}\right)\delta(x)\right]g(x)dx}$$
So we see that:
$$\frac{\Theta(x)}{x} - \left(\int^{a}_{0}{\frac{1}{x}dx}\right)\delta(x)$$
is a modified version of:
$$\frac{\Theta(x)}{x}$$
that produces well defined integrals.
The physicist way of doing this is to call
$$\left(\int^{a}_{0}{\frac{1}{x}dx}\right)\delta(x)$$
a counterterm and to consider
$$\left(\int^{a}_{0}{\frac{1}{x}dx}\right)$$
to define an "infinite constant" ##C_a##. So they would write:
$$\left[\frac{\Theta(x)}{x}\right]_R = \frac{\Theta(x)}{x} - C_a\delta(x)$$
In physics language you obtain a version of the product of the Heaviside step function and ##\frac{1}{x}## that behaves well under integrals by subtracting off a counterterm which has an infinite constant. This "well behaved" version is called "renormalized".
For some Quantum Field Theories the terms you have to add are no problem. For instance ##\phi^{4}## theory in three dimensions requires:
$$\left[\phi^{4}\right]_R = \phi^{4} - \delta m^{2}\phi^{2}$$
That is making the ##\phi^{4}## interaction well-defined requires adding a counterterm with the ##\phi^{2}## term. However since the Lagrangian already has a ##\phi^{2}## this is no problem, it doesn't change the character of the theory.
However some theories require the addition of new terms to their Lagrangian to be well defined. If this process doesn't stop, i.e. the new terms then need further new terms, the theory is called non-renormalizable.
