Finiteness of quantum mechanics

Question

In most standard undergraduate treatments of quantum mechanics, there is rarely any need to treat divergences in perturbation theory, other than the Casimir energy perhaps? The subtleties of renormalization are only discussed once we start examining quantum field theory.

However, in hindsight, quantum mechanics is nothing but QFT in 1 spacetime dimension, and perturbation theory can be performed in the path integral formulation exactly as any higher dimensional QFT (evaluate diagrams using Feynman rules). This begs the question: is there a direct power counting argument which shows that loop divergences will never occur in 1d QFT? Because it's not at all obvious to me. Thanks!

Let's keep things simple and restrict to Lagrangians which are smooth functions of the coordinates (1d fields) $x^i$, fermions $\psi^i$, and their derivatives.

Answer 1

The most `standard' loop diagrams involve products of bosonic propagators, and these don't diverge in $d=1$ $$\int_{\Bbb R} \frac{dk}{2\pi}\frac{1}{k^2+m^2}=\frac{1}{2m},$$ $$\int_{\Bbb R} \frac{dk}{2\pi}\frac{1}{(k^2+m^2)((k+p)^2+m^2)}=\frac{1}{m}\frac{1}{p^2+4m^2},$$ and so on. By counting powers of $k$ even without doing the integral you would expect no divergence.

But if you take a one-loop diagram of fermion in $d=1$ this is logarithmically divergent, and looks something like (it depends a bit on your conventions) $$\int \frac{dk}{2\pi}\frac{1}{-ik+m}.$$ You can get different answers depending on how you regularize this, and the ambiguity corresponds to the operator ordering ambiguity of a product of fermionic creation and annihilation operators $\psi^\dagger \psi$ in quantum mechanics. For $d=1$ quantum field theory with fermions see arXiv:2110.04681 or Boozer, (2007) .

Also if you have a quantum mechanical particle trapped on a curved surface (a non-linear sigma model), you will generically run into divergences in perturbation theory which may be considered to arise from the derivatives in the interaction. For examples of this see arXiv:2109.06597 or arXiv:9208059.

Answer 2

1. In QM = 0+1 QFT with spacetime dimension $d=1$, and with a Lagrangian of the form $$ L(\phi,\psi,\bar{\psi})~=~\frac{1}{2}\dot{\phi}^2 -\frac{1}{2}M^2 \phi^2 + \bar{\psi}(i\partial_t-m)\psi -V(\phi,\psi,\bar{\psi}), $$ we have $[L]=1=[M]=[m]$, $$[\phi]~=~-\frac{1}{2}\quad\text{and}\quad [\psi]~=~0. $$ Let us exclude derivative couplings and singular potentials$^1$, and assume that the potential $V(\phi,\psi,\bar{\psi})$ is a power series. A coupling constant $\lambda_{n_bn_f}$ for a vertex with $n_b$ bosons and $n_f$ fermions has mass dimension $[\lambda_{n_bn_f}]=1+\frac{n_b}{2}~>~0$, which is positive/relevant/super-renormalizable.

2. For a connected Feynman diagram the superficial degree of (UV) divergence is defined as$^2$ $$\begin{align} D~:=~& \#\{\text{$\mathrm{d}p$ in int. measure}\} ~+~ \#\{\text{$p$ in numerator}\}\cr &~-~ \#\{\text{$p$ in denominator}\}\cr\cr ~=~& \underbrace{\#\{\text{loops}\}}_{\leq I_b+I_f} +0 -2I_b-I_f\cr ~\leq~&-I_b, \end{align} $$ where $I_b$ and $I_f$ are the numbers of internal boson and fermion propagators, respectively. It follows that superficially divergent diagrams $D\geq 0$ must have no internal boson propagators $I_b=0$, i.e. all boson lines are external.

3. We can therefore treat the boson $\phi$ as a background field. Then the Lagrangian becomes of the form $$ L(\psi,\bar{\psi})~=~L_0(\phi,\dot{\phi}) + \bar{\psi}(i\partial_t-m)\psi -\sum_n\lambda_{n}(\phi)\psi^n\bar{\psi}^n, $$ where the $\phi$-dependent coupling constants $\lambda_{n}(\phi)$ all have mass dimension $[\lambda_{n}(\phi)]=1$. One may show that the superficial degree of (UV) divergence is $$ D~=~ d - \sum_n [\lambda_n(\phi)] V_n =~1- \#\{\text{vertices}\}, $$ where $V_n$ is the number of vertices of type $n$, cf. e.g. this Phys.SE post.

   It follows that a Feynman diagram with at least 1 vertex and which is superficially divergent $D\geq 0$ consists of a single fermion vertex $\lambda_{n}(\phi)$. The attached fermionic lines are either external legs or self-loops.

   A fermionic self-loop integral is finite since we will assume that integral over the odd part of the integrand vanishes. So the diagram is actually finite.

   Hence there are no actual divergencies and the theory is a finite, cf. OP's title question.

References:

1. M.E. Peskin & D.V. Schroeder, An Intro to QFT, 1995; p. 316.

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$^1$ The coupling constants for (some higher) time-derivative interactions, singular power law potentials $\phi^{\leq -2}$, higher-dimensional Dirac delta distributions $\delta^{\geq 2}(\phi)=\delta(\phi_1)\delta(\phi_2)\ldots$, derivatives $\delta^{(\geq 1)}(\phi)$ of Dirac delta distributions, etc., have non-positive mass dimension. Also note that the standard perturbative technique of working with Fourier transformed fields $\widetilde{\phi}$ is not useful for singular interactions.

$^2$ If the diagram contains a divergent subdiagram, its actual divergence may be worse than indicated by $D$, cf. Ref. 1. Let us exclude divergent subdiagrams.