# Cut-off Regularisation and Renormalisation in Scalar Field Theory, Deriving the Cutoff Independent Physical Mass

I'm having trouble reproducing Equation 42:

$$\tag{1} m^{2}_{\text{phys}}= m^{2}_{r} + m^{2}_{r} \tilde{\lambda} \text{log} \left( \dfrac{m^{2}_{r}}{\mu^{2}} \right)$$

that appears in http://www.physics.mcgill.ca/~jcline/qft1b.pdf, page 14 (Prof. James M. Cline's 'Physics 198-730B' lecture notes, McGill University, Canada).

If the link no longer works, and you want a copy of the original text, message me via the email address on my profile and mention the reference number 1093 (it's how I organise my resources). I'll send you an electronic copy of the notes.

Here is my attempt:

Equation 40 of the source material:

$$\tag{2} \delta m^{2} = \tilde{\lambda} \left( \Lambda^{2} - m^{2}_{0}\text{log} \left( 1+ \dfrac{\Lambda^{2}}{m^{2}_{0}} \right) \right)$$

Note that the author originally uses $m^{2}$ instead of $m^{2}_{0}$ but then replaces $m^{2}$ with $m^{2}_{0}$ as a change of notation.

Equation 41 of the source material:

$$\tag{3} m^{2}_{0} = m^{2}_{r} \left( 1+\tilde{\lambda} \text{log} \left( \dfrac{\Lambda^{2}}{\mu^{2}} \right) \right) - \tilde{\lambda}\Lambda^{2}$$

Attempt to prove Equation 42 of the source material:

$$\begin{split} m^{2}_{\text{phys}} &= m^{2}_{0} + \delta m^{2}\\ &=\text{Equation 41} + \text{Equation 40}\\ &= m^{2}_{r} \left( 1+\tilde{\lambda} \text{log} \left( \dfrac{\Lambda^{2}}{\mu^{2}} \right) \right) - \tilde{\lambda}\Lambda^{2} \\ &\,\,\,\,\, + \tilde{\lambda} \left( \Lambda^{2} - m^{2}_{0}\text{log} \left( 1+ \dfrac{\Lambda^{2}}{m^{2}_{0}} \right) \right)\\ &= m^{2}_{r} + m^{2}_{r} \tilde{\lambda} \text{log} \left( \dfrac{\Lambda^{2}}{\mu^{2}} \right) - \tilde{\lambda}\Lambda^{2} \\ &\,\,\,\,\, + \tilde{\lambda} \Lambda^{2} - m^{2}_{0} \tilde{\lambda} \text{log} \left( 1+ \dfrac{\Lambda^{2}}{m^{2}_{0}} \right)\\ &= m^{2}_{r} + m^{2}_{r} \tilde{\lambda} \text{log} \left( \dfrac{\Lambda^{2}}{\mu^{2}} \right) - m^{2}_{0} \tilde{\lambda} \text{log} \left( 1+ \dfrac{\Lambda^{2}}{m^{2}_{0}} \right)\\ \end{split}$$

Plugging in Equation 41 of the source material (Equation 3 in this text) in place of the $m^{2}_{0}$ coefficient in front of the second logarithm, then ignoring all terms of order $\tilde{\lambda}^{2}$, means that only the $m_{r}$ term survives in the resulting coefficient:

$$\begin{split} &= m^{2}_{r} + m^{2}_{r} \tilde{\lambda} \text{log} \left( \dfrac{\Lambda^{2}}{\mu^{2}} \right) - m^{2}_{r} \tilde{\lambda} \text{log} \left( 1+ \dfrac{\Lambda^{2}}{m^{2}_{0}} \right)\\ \end{split}$$

Plugging in Equation 41 (Equation 3 in this text) in place of the $m^{2}_{0}$ term in the denominator of the fraction in the second logarithm gives:

$$\begin{split} &= m^{2}_{r} + m^{2}_{r} \tilde{\lambda} \text{log} \left( \dfrac{\Lambda^{2}}{\mu^{2}} \right) \\ &\,\,\,\,\, - m^{2}_{r} \tilde{\lambda} \text{log} \left( 1+ \dfrac{\Lambda^{2}}{m^{2}_{r} + m^{2}_{r} \tilde{\lambda} \text{log} \left( \dfrac{\Lambda^{2}}{\mu^{2}} \right) - \tilde{\lambda}\Lambda^{2}} \right)\\ &= m^{2}_{r} + m^{2}_{r} \tilde{\lambda} \text{log} \left( \dfrac{\Lambda^{2}}{\mu^{2}} \right) \\ &\,\,\,\,\, - m^{2}_{r} \tilde{\lambda} \text{log} \left( \dfrac{m^{2}_{r} + m^{2}_{r} \tilde{\lambda} \text{log} \left( \dfrac{\Lambda^{2}}{\mu^{2}} \right)- \tilde{\lambda}\Lambda^{2} + \Lambda^{2}}{m^{2}_{r} + m^{2}_{r} \tilde{\lambda} \text{log} \left( \dfrac{\Lambda^{2}}{\mu^{2}} \right) - \tilde{\lambda}\Lambda^{2}} \right)\\ \end{split}$$

Problematic Step: Now the only way to get the correct answer is if the complicated fraction in the second logarithm some how approximates to $\Lambda^{2}/m^{2}_{r}$. In order to do this I first use the inequality $\Lambda^{2} \gg m^{2}_{r}$. This seems okay as the cutoff is supposed to be very large. This ensures that only the $\Lambda^{2}$ contributes to the numerator of the logarithm. I also impose $\tilde{\lambda} \ll (m_{r}/\Lambda)^{2} \rightarrow \tilde{\lambda} \Lambda^{2} \ll m_{r}$, which ensures that only the $m^{2}_{r}$ contributes to the denominator of the logarithm. Unfortunately I'm not sure how to justify the second inequality that I impose. Since $m_{r}/\Lambda$ is very small (the cutoff is very large), this means that $(m_{r}/\Lambda)^{2}$ is very, very small. This means that $\tilde{\lambda}$ must be much smaller still, in order for the second inequality to hold. I'm not sure this makes sense. Continuing would give:

$$\begin{split} &= m^{2}_{r} + m^{2}_{r} \tilde{\lambda} \text{log} \left( \dfrac{\Lambda^{2}}{\mu^{2}} \right) - m^{2}_{r} \tilde{\lambda} \text{log} \left( \dfrac{\Lambda^{2}}{m^{2}_{r}} \right)\\ &= m^{2}_{r} + m^{2}_{r} \tilde{\lambda} \text{log} \left(\Lambda^{2} \right) - m^{2}_{r} \tilde{\lambda} \text{log} \left( \mu^{2} \right)\\ &\,\,\,\,\, - m^{2}_{r} \tilde{\lambda} \text{log} \left( \Lambda^{2} \right) + m^{2}_{r} \tilde{\lambda} \text{log} \left( m^{2}_{r} \right)\\ &= m^{2}_{r} - m^{2}_{r} \tilde{\lambda} \text{log} \left( \mu^{2} \right) + m^{2}_{r} \tilde{\lambda} \text{log} \left( m^{2}_{r} \right)\\ &= m^{2}_{r} + m^{2}_{r} \tilde{\lambda} \text{log} \left( \dfrac{m^{2}_{r}}{\mu^{2}} \right)\\ &= \text{Equation 42} \end{split}$$

Your way of thinking is, essentially, correct. When it comes to this $\tilde\lambda\Lambda^2$, there is this famous quote (citing from memory, don't remember which book it's from, but it's famous), "Even though it is infinitely large, we will assume that it is finite, and that is furthermore infinitely small." In most QFTs the perturbative series diverges for any finite $|\lambda|>0$ anyway, so most of what we do are formal series.
Basically you can start from $$m^{2}_{r} + m^{2}_{r} \tilde{\lambda} \text{log} \left( \dfrac{\Lambda^{2}}{\mu^{2}} \right) - m^{2}_{r} \tilde{\lambda} \text{log} \left( 1+ \dfrac{\Lambda^{2}}{m^{2}_{0}} \right) \;,$$ say that $\Lambda/m_0\gg1$ to get $$m^{2}_{r} + m^{2}_{r} \tilde{\lambda} \text{log} \left( \dfrac{\Lambda^{2}}{\mu^{2}} \right) - m^{2}_{r} \tilde{\lambda} \text{log} \left( \dfrac{\Lambda^{2}}{m^{2}_{0}} \right) + \mathcal O(\Lambda^{-2}) \;,$$ and then put $m_0=m_r + \mathcal O(\tilde\lambda)$ to finally get $$m^{2}_{r} + m^{2}_{r} \tilde{\lambda} \text{log} \left( \dfrac{m_0^{2}}{\mu^{2}} \right) + \mathcal O(\tilde\lambda^2) \;.$$ In the last step we did assume that $\tilde\lambda^2 \ln\Lambda^2$ is infinitesimally small compared to terms of order $\tilde\lambda$, which we can given the argument above.