Subtraction point not physical mass

[geoduck]

In conventional renormalization, for the self-energy, is it possible to make a subtraction from a point not equal to the physical mass?

$$\frac{1}{p^2-m_o^2-\Sigma(\mu^2)-\Sigma'(\mu^2)(p^2-\mu^2)-...}$$

Now define $m_o^2+\Sigma(\mu^2)\equiv m(\mu^2)$

Then:

$$\frac{1}{p^2-m(\mu)^2-\Sigma'(\mu^2)(p^2-\mu^2)-...}$$

But you can't seem to write this in the form $\frac{Z}{p^2-m(\mu)^2-\text{finite}}$

unless you choose $\mu^2=m(\mu^2)$. But this choice corresponds to the physical mass.

But in BPZ renormalization, you have no problems working with a mass that depends on scale μ, and a scale is like a subtraction point is it not?

 

[king vitamin]

It is possible to choose a subtraction point besides the physical mass. I'm not familiar with how you're writing the self-energy (what's the utility in expanding about $p^2 = \mu^2$?), but writing the bare propagator as

$$ \frac{1}{p^2-m_o^2-\Sigma_{0}(p^2)} $$

we can choose any subtraction scheme to obtain a finite self energy $\Sigma_R$, and then simply write the propagator as

$$ \frac{1}{p^2-m_R^2-\Sigma_{R}(p^2)} .$$

Now $m_R$ is a parameter dependent on your subtraction scheme, and thus dependent on $\mu$, and it will take the value $m_R^2 = m_P^2 - \Sigma_R(m_P^2)$. The dependence of renormalized mass on the energy scale $\mu$ is physically important, especially for studying critical exponents.

EDIT: And of course, the $\mu$ dependence of $Z = (1-\Sigma'_R(m_P^2))^{-1}$ is also very important

 

[geoduck]

It is possible to choose a subtraction point besides the physical mass. I'm not familiar with how you're writing the self-energy (what's the utility in expanding about $p^2 = \mu^2$?), but writing the bare propagator as

$$ \frac{1}{p^2-m_o^2-\Sigma_{0}(p^2)} $$

we can choose any subtraction scheme to obtain a finite self energy $\Sigma_R$, and then simply write the propagator as

$$ \frac{1}{p^2-m_R^2-\Sigma_{R}(p^2)} .$$

Textbooks start with the bare propagator as you have, but then make an expansion about the physical mass:

$$ \frac{1}{p^2-m_o^2-\Sigma_{0}(p^2)}=\frac{1}{p^2-m_o^2-\Sigma_{0}(m_p^2)-\Sigma'_{0}(m_p^2)(p^2-m_p^2)-...} $$

They then say the definition of the physical mass is that $m_o^2+\Sigma_{0}(m_p^2)=m_p^2$, so that

$$ \frac{1}{p^2-m_o^2-\Sigma_{0}(p^2)}=\frac{1}{p^2-m_o^2-\Sigma_{0}(m_p^2)-\Sigma'_{0}(m_p^2)(p^2-m_p^2)-...}=
\frac{1}{(p^2-m_p^2)-\Sigma'_{0}(m_p^2)(p^2-m_p^2)-...}
$$

Then the (1-Ʃ')^-1 can be factored out, which is Z:

$$
=\frac{(1-\Sigma'(m_p))^{-1}}{(p^2-m_p^2)-\Sigma_{R}(p^2)}
$$

That is how they got the relation: Z=(1-Ʃ')^-1.

But this doesn't seem to work if Taylor expand Ʃ about a different point than the physical mass.

So textbooks basically manipulate Ʃ₀ directly and define the physical mass from it, whereas you define the physical mass from the renormalized Ʃ_R and the finite m²(μ).

 

[andrien]

If you will calculate the self energy correction by taking into account the inifinities of diagram and use some identity like 1/A+B=1/A-(1/A)B(1/A)+... you will see the result getting around the physical mass.This idea is just translated into a much convenient way without doing it always .They make an expansion around physical mass because that is the way it comes around.