$\varphi^4$ via renormalization group with hard cut-off I am studying the application of the renormalization group to the $\varphi^4$ theory:
$$\mathcal{L} = -\frac{1}{2} \varphi (\Box + m^2)\varphi -\frac{\lambda}{4!}\varphi^4.$$
In particular I wanted to follow two different regularization methods, and verify that the resulting critical exponents $\nu$ did not differ, as I expect.
If you want to calculate the contributions up to one loop, you encounter two diagrams which diverge: the tadpole for the 2-points correlation function, and the loop for the 4-points correlation function. The first goes like $\int \frac{d^4k}{k^2-m^2}$, the second like $\int \frac{d^4k}{(k^2-m^2)^2}$.
In every textbook I know (e.g. Schwartz, Quantum Field Theory and the Standard Model, 23.5.1), you see the treamtment of the $\varphi^4$ and of the renormalization group with the dimensional regularization, which leads to the following equations:
$$\beta = -\varepsilon\lambda + \frac{3\lambda^2}{16\pi^2}$$
$$\mu\frac{d}{d\mu}m^2 = \frac{\lambda}{16\pi^2}m^2$$
where $\mu$ is the renormalization scale, $\lambda$ and $m^2$ are the renormalized constant of interaction and mass, and $d=4-\varepsilon$ is the dimension.
Solving for the Wilson-Fisher fixed point, we find $\lambda^* = \frac{16\pi^2\varepsilon}{3}$ and $m^{2*} =0$.
At the fixed point, the anomalous dimension of the mass is then $\gamma_m=\frac{\lambda^{*}}{16\pi^2}=\frac{\varepsilon}{3}$ giving $\nu=\frac{1}{2-\gamma_m}=\frac{3}{6-\varepsilon}$.
Fine. Now, if I try to introduce an explicit cut-off $\Lambda$ in the divergent integrals, in four dimensions I get something like:
$\int \frac{d^4k}{k^2-m^2} \propto \frac{\Lambda^2}{m^2} + \log(1+\frac{\Lambda^2}{m^2})$
$$\int \frac{d^4k}{(k^2-m^2)^2} \propto \log(1+\frac{\Lambda^2}{m^2}).$$
This is something worrisome, since we have two very different behaviours between the two integrals, and in the critical exponent they enter as a ratio.
But let's get into the calculation. I switch to the Euclidean integrals, and I define:
$f(\Lambda,m^2,d) = \int_\Lambda \frac{d^dk}{k^2+m^2}$
and
$g(\Lambda,m^2,d) = \int_\Lambda \frac{d^dk}{(k^2+m^2)^2}$
Taking into account that
$\Lambda \frac{d}{d\Lambda} f(\Lambda,m^2,d) = S_d \frac{\Lambda^d}{\Lambda^2+m^2} + \frac{\partial f(\Lambda,m^2,d)}{\partial m^2}\Lambda \frac{d}{d\Lambda}m^2$
and
$\Lambda \frac{d}{d\Lambda} g(\Lambda,m^2,d) = S_d \frac{\Lambda^d}{(\Lambda^2+m^2)^2} + \frac{\partial g(\Lambda,m^2,d)}{\partial m^2}\Lambda \frac{d}{d\Lambda}m^2$,
where $S_d$ is the area of the d-dimensional unit sphere,
I obtain the following equations for d-dimensions:
$\beta = -(4-d)\lambda - \frac{3}{2^{d+1}\pi^d}\lambda^2(-(4-d)f + \frac{\Lambda^d}{\Lambda^2+m^2}S_d)$
$\Lambda\frac{d}{d\Lambda}m^2 = -\frac{\lambda}{2^{d+1}\pi^d}m^2 (-(4-d)g + \frac{\Lambda^d}{(\Lambda^2+m^2)^2}S_d)$
If I now try to evaluate $\gamma_m$ at the fixed point, I find something like:
$\gamma_m = \frac{4-d}{3}\frac{-(4-d)g + \frac{\Lambda^d}{(\Lambda^2+m^2)^2}S_d}{-(4-d)f + \frac{\Lambda^d}{\Lambda^2+m^2}S_d}$
If I did not make any error, which is an assumption, that $\gamma_m$ is not equivalent to the one obtained via dimensional regularization. I am probably missing something.
Any suggestions?
EDIT: I realized (thanks @TehMeh) that I defined the functions $f$ and $g$ differently from my pen and paper calculation, and came up with a mixed notation and a lot of mess, which ended up in a lot of errors. Sorry to everyone. Let me now correct.
$f(\Lambda,m^2,d) = \int_\Lambda \frac{d^dk}{(k^2+m^2)^2}$
and
$g(\Lambda,m^2,d) = \frac{1}{m^2}\int_\Lambda \frac{d^dk}{k^2+m^2}$
Taking into account that
$\Lambda \frac{d}{d\Lambda} f(\Lambda,m^2,d) = S_d \frac{\Lambda^d}{(\Lambda^2+m^2)^2} + \frac{\partial f(\Lambda,m^2,d)}{\partial m^2}\Lambda \frac{d}{d\Lambda}m^2$,
and
$\Lambda \frac{d}{d\Lambda} g(\Lambda,m^2,d) = \frac{S_d}{m^2} \frac{\Lambda^d}{\Lambda^2+m^2} + \frac{\partial g(\Lambda,m^2,d)}{\partial m^2}\Lambda \frac{d}{d\Lambda}m^2$
where $S_d$ is the area of the d-dimensional unit sphere,
I obtain the following equations for d-dimensions:
$\beta = -(4-d)\lambda - \frac{3}{2^{d+1}\pi^d}\lambda^2(-(4-d)f + \frac{\Lambda^d}{(\Lambda^2+m^2)^2}S_d)$
$\Lambda\frac{d}{d\Lambda}m^2 = -\frac{\lambda}{2^{d+1}\pi^d}m^2 (-(4-d)g + \frac{\Lambda^d}{\Lambda^2+m^2}\frac{S_d}{m^2})$
If I now try to evaluate $\gamma_m$ at the fixed point, I find something like:
$\gamma_m = \frac{4-d}{3}\frac{-(4-d)g + \frac{\Lambda^d}{\Lambda^2+m^2}\frac{S_d}{m^2}}{-(4-d)f + \frac{\Lambda^d}{(\Lambda^2+m^2)^2}S_d}$
$f$ and $g$ are representable with the hypergeometric function, but if we take the limit for small $4-d$ it should not anyway matter their expression.
 A: I've been doing the very same problem (23.6 - right?), hopefully, this is still helpful.
First of all, I notice that your $\beta$ function is probably incorrect. In $d=4$ it should be dimensionless and yet one of the terms is of mass dimension 2, which is also different from the two other terms.
If you fix that, maybe in $d=4-\epsilon$ dimensions your $\gamma_m$ is correct once expanded in $\epsilon$ as you have the seemingly correct prefactor of $\frac{4-d}{3}$, which would immediately give the correct answer.
I myself did the problem in a slightly different manner.
We have to work in $d=4-\epsilon$ dimensions and I introduced the usual subtraction point $\mu$, also there were no $\epsilon$ poles as these are regulated by the cut off $\Lambda$.
To get $\beta$ functions I differentiated bare parameters with respect to $\mu$, for example,
$\mu \frac{d\lambda_0}{d\mu}=\mu \frac{d(\lambda_R \mu^{\epsilon}(\mu)Z_\lambda)}{d\mu}$
Where $\lambda_R(\mu)$ is the renormalized coupling and $Z_\lambda=1+\delta_\lambda$ is the renormalization constant - all as in Schwartz. I got the counterterms by expanding the integrands in powers of $m^2$ and keeping only divergent (after integration) terms.
The counterterms are quite nasty due to $\epsilon$ and the regulator $\lambda$, hence, I used Mathematica to do expansions and solve for $\beta$ functions. In the end, the result matches the one from dimensional regularization.
A: Taking a derivative (with regard to a parameter) to a quadratically divergent integral
$$
\Lambda \frac{d}{d\Lambda} f(\Lambda,m^2,d) = S_d \frac{\Lambda^d}{\Lambda^2+m^2} + \frac{\partial f(\Lambda,m^2,d)}{\partial m^2}\Lambda \frac{d}{d\Lambda}m^2
$$
will open a can of worms (e.g. the order of $\partial m^2$ and $d^4k$ is not interchangeable) when hard cutoff is involved, albeit
$$
\Lambda \frac{d}{d\Lambda} g(\Lambda,m^2,d)
$$
is OK, since $g(\Lambda,m^2,d)$ is only logarithmically divergent.
The cutoff and boundary conditions are very tricky for divergent Feynman integrals beyond logarithmic divergence. A typical example is the triangular diagram (linearly divergent) of the ABJ anomaly, where seemingly innocuous shift of integrals are prohibited.
