I agree with Brian Moths that $p \rightarrow 0$ means going to small energies.

However, I do not understand the following statement.

as you integrate out the wavenumbers between $𝑀$ and $p$, the coupling constant $\bar{\lambda}$ gets multiplied by a large factor to give the renormalized coupling $\lambda$.

In the example given in the book, we have a positive $\beta = \frac{3\lambda^2}{16\pi^2}$. Therefore the factor is in fact smaller than 1 as we move to smaller energies ($p \rightarrow 0$).

This means that if we fix $\bar{\lambda}$ at $M$ and move to smaller energies ($p \rightarrow 0$), the effective $\lambda$ should become smaller and smaller. Now if we instead fix $\lambda$ as we move to smaller energies ($p \rightarrow 0$), we would have to make $\bar{\lambda}$ bigger and bigger to make the equation hold.

For reference, see the sentence immediately following (12.28) on Page 404.

[Update]

After dreaming about this thing the whole night, I believe I have a more satisfactory answer.

The bacteriological analogy is a formal analogy, in the sense that Equation (12.68) as the same form as Equation (12.66) with the replacements in (12.69). It does not mean that we should interpret $\bar{\lambda}$, which is something that we just make up to make it look like $\bar{x}$ defined by (12.70), in the same way as we interpret $\bar{x}$.

In fact, the definition of $\lambda$ has never changed since (12.30). It is simply the effective coupling at $p^2=-M^2$ (or equivalently at $p=M$).

Later we "figure out" that $\bar{\lambda}$ is in fact the effective coupling at any $p$ different to $M$.

I agree with Brian Moths that $p \rightarrow 0$ means going to small energies.

However, I do not understand the following statement.

as you integrate out the wavenumbers between $𝑀$ and $p$, the coupling constant $\bar{\lambda}$ gets multiplied by a large factor to give the renormalized coupling $\lambda$.

In the example given in the book, we have a positive $\beta = \frac{3\lambda^2}{16\pi^2}$. Therefore the factor is in fact smaller than 1 as we move to smaller energies ($p \rightarrow 0$).

This means that if we fix $\bar{\lambda}$ at $M$ and move to smaller energies ($p \rightarrow 0$), the effective $\lambda$ should become smaller and smaller. Now if we instead fix $\lambda$ as we move to smaller energies ($p \rightarrow 0$), we would have to make $\bar{\lambda}$ bigger and bigger to make the equation hold.

For reference, see the sentence immediately following (12.28) on Page 404.

[Update]

After dreaming about this thing the whole night, I believe I have a more satisfactory answer.

I am afraid that Brian Moths was wrong with the following statement, for the objection that I posted originally.

My interpretation is that you are right when you say...

Below is what I believe should be the correct interpretation.

The bacteriological analogy is a formal analogy, in the sense that Equation (12.68) has the same form as Equation (12.66) with the replacements in (12.69). It does not mean that we should interpret $\bar{\lambda}$, which is something that we just make up to make it formally look like $\bar{x}$ defined by (12.70), in the same way as we interpret $\bar{x}$.

In fact, the definition of $\lambda$ has never changed since (12.30). It is simply the effective coupling at $p^2=-M^2$ (or equivalently at $p=M$).

Later we "figure out" that $\bar{\lambda}$ is in fact the effective coupling at any $p$, which can be different to $M$.

I agree with Brian Moths that $p \rightarrow 0$ means going to small energies.

However, I do not understand the following statement.

as you integrate out the wavenumbers between $𝑀$ and $p$, the coupling constant $\bar{\lambda}$ gets multiplied by a large factor to give the renormalized coupling $\lambda$.

In the example given in the book, we have a positive $\beta = \frac{3\lambda^2}{16\pi^2}$. Therefore the factor is in fact smaller than 1 as we move to smaller energies ($p \rightarrow 0$).

This means that if we fix $\bar{\lambda}$ at $M$ and move to smaller energies ($p \rightarrow 0$), the effective $\lambda$ should become smaller and smaller. Now if we instead fix $\lambda$ as we move to smaller energies ($p \rightarrow 0$), we would have to make $\bar{\lambda}$ bigger and bigger to make the equation hold.

For reference, see the sentence immediately following (12.28) on Page 404.

I agree with Brian Moths that $p \rightarrow 0$ means going to small energies.

However, I do not understand the following statement.

as you integrate out the wavenumbers between $𝑀$ and $p$, the coupling constant $\bar{\lambda}$ gets multiplied by a large factor to give the renormalized coupling $\lambda$.

In the example given in the book, we have a positive $\beta = \frac{3\lambda^2}{16\pi^2}$. Therefore the factor is in fact smaller than 1 as we move to smaller energies ($p \rightarrow 0$).

This means that if we fix $\bar{\lambda}$ at $M$ and move to smaller energies ($p \rightarrow 0$), the effective $\lambda$ should become smaller and smaller. Now if we instead fix $\lambda$ as we move to smaller energies ($p \rightarrow 0$), we would have to make $\bar{\lambda}$ bigger and bigger to make the equation hold.

For reference, see the sentence immediately following (12.28) on Page 404.

[Update]

After dreaming about this thing the whole night, I believe I have a more satisfactory answer.

The bacteriological analogy is a formal analogy, in the sense that Equation (12.68) as the same form as Equation (12.66) with the replacements in (12.69). It does not mean that we should interpret $\bar{\lambda}$, which is something that we just make up to make it look like $\bar{x}$ defined by (12.70), in the same way as we interpret $\bar{x}$.

In fact, the definition of $\lambda$ has never changed since (12.30). It is simply the effective coupling at $p^2=-M^2$ (or equivalently at $p=M$).

Later we "figure out" that $\bar{\lambda}$ is in fact the effective coupling at any $p$ different to $M$.