A bit of everything

# Very low dimensional quantum field theory

Let's focus on quantum field theories of dimensions lower than $(1+1)$ dimensions. As we'll see, despite their simplicity, there is quite a lot of interesting things going on there.

## $0$-dimensional quantum field theory

Despite being this low-dimensional, there are still lessons to be learned from $0$-dimensional quantum field theory, mostly pertaining to path integrals and perturbation theory, as it serves as an example of an exactly solvable theory.

First, a bit of overview of the spacetime (this should be quick). Up to homeomorphism, there is only one connected zero-dimensional manifold, which is $\mathbb{R}^0$. More generally we can define it as the set $\{ p \}$ for some object $p$, with the discrete topology $\{ \varnothing, \{p \} \}$. It possesses a unique chart

$$\{ (\{ p \}, \phi_0) \}$$

with $\phi_0$ a homeomorphism between $p$ and $0 \in \mathbb{R}^0$. Every two $0$-manifold are homeomorphic, as can be shown easily by considering another manifold defined similarly by the singleton $\{ q \}$, in which case the homeomorphism will be

$$f(p) = q,\ f^{-1}(q) = p$$

We can define all the metrics we want on this manifold, as all the metric axioms will be satisfied vacuously, and we only have that $d(p,p) = 0$. Its orthogonal group, Lorentz group and any other group we can wish to define will be the identity group $I$, as every group defined on a singleton. Its tangent bundle is equal to itself, via the projection map $\pi(p) = p$

Fields on such a spacetime are fairly simple : any scalar field is simply a map $\phi : \{ p \} \ to \mathbb{R}$, which is roughly saying that a scalar field is just a real number (or a complex number, for a complex scalar field), since as is known, there is always an isomorphism between $\text{Fun}(\{ p \}, V)$ and $V$ itself for a vector space $V$. Vector fields are not particularly more interesting as only the zero vector exists. We can add arbitrarily many fields to our points if necessary, making our configuration space $\mathbb{R}^n$.

With that out of the way, let's consider our (Euclidian) path integral.

$$\mathscr{Z} = \int d\phi e^{- S[\phi_i] }$$

Our action is a linear functional from a tuple of real numbers to real numbers, in other words,

$$S : \mathbb{R}^n \to \mathbb{R}$$

This is simply a traditional function of $n$ variables. To integrate our path integral, similarly to the infinite-dimensional case, we can use the Gaussian measure, although the Lebesgue measure $d\mu$ can be turned into the Gaussian measure $d\gamma$ using

$$d\gamma(\phi^i) = \frac{1}{(2\pi)^\frac{n}{2}} e^{-\frac{1}{2} \| \phi^i \|^2}$$

Given the form of our path integral, we'll just have to require a "kinetic" term $S_0[\phi^i] = -\frac{1}{2} \| \phi^i \|^2$. Assuming our action analytic, we can write it as

$$S[\phi^i] = c_0 + c_{i} \phi^i + c_{ij} \phi^i \phi^j + \sum_{n = 3}^\infty c_{ijk\ldots} \phi^i \phi^j \phi^k \ldots$$

For simplicity, let's consider for now the "free" case of a single field,

$$S[\phi] = c_0 + c_1 \phi + c_{2} \phi^2$$

Here we can note that $c_2$ is roughly equivalent, in a more reasonable QFT, to a mass term, while $c_1$ is similar to a source term. We can rewrite it as

$$S[\phi] = c_0 + J \phi + m^2 \phi^2$$

Our path integral is

$$\mathscr{Z} = \int d\phi e^{- c_0 - J \phi - m^2 \phi^2 }$$

Our first term can be immediatly taken out of the integral, the rest is a Gaussian integral

$$\mathscr{Z} = e^{-c_0} \sqrt{\frac{\pi}{m^2}} e^{\frac{J^2}{4m^2}}$$

## $(0+1)$-dimensional quantum field theory

The $(0+1)$-dimensional case is already a lot more complex. We consider here the case of one time dimension and zero spatial dimensions. If everything is connected, Hausdorff and paracompact, this means that our spacetime will be either $\mathbb{R}$ or $S^1$, foliated by timelike hypersurfaces equal to $\Sigma_t = \{ t \}$. To simplify things for now, let's only consider $\mathbb{R}$, as it is the only causal one.

Here our manifold does have a tangent bundle, which is $\mathbb{R}^2$. Its Lorentz group will be $\operatorname{O}(\mathbb{R}) = \mathbb{Z}_2$, comprising only the identity element $I$ and time reversal $T$, with the special Lorentz group $\operatorname{SO}(\mathbb{R}) = \{ 1 \}$. The Poincaré group is just our Lorentz group supplemented with the time translation $\mathbb{R}$.

The geometry is defined by some einbein $e$, or a metric $g_{tt}$. It can be shown by picking the proper coordinates that this metric is always diffeomorphic to the canonical metric on $\mathbb{R}$, we therefore do not have to worry about any curvature.

Unlike previously, our fields will actually be proper functions, although only of time.

$$\phi : \mathbb{R}_t \to \mathbb{R}$$

Therefore our configuration space is simply a function space over $\mathbb{R}$. We don't know quite yet what will be a good function space, so let's keep it broad and call it $\mathcal{C}$. Our action is a linear functional on this space

$$S[\phi^i(t), t] = \int_{t_a}^{t_b} dt\ L(\phi^i(t), \dot{\phi}^i(t), t)$$

There may be some familiarity in this form. You may recall that in string theory, we could entirely describe the theory as a field theory on a $(1+1)$-dimensional manifold, completely disregarding the target manifold. The same thing is true here indeed, and a $(0+1)$-dimensional quantum field theory can be used to describe the quantum theory of point particles in a space of arbitrarily many dimensions, simply by considering the values of our field $\phi^i(t)$ to represent the position in the target manifold. The theory we'll get will depend on a variety of things, including the invariance of our fields with respect to some group.

Our path integral is then something of the form

$$\mathcal{Z}[t_a, t_b] = \int$$

We must now come to decide what our configuration space is. This will decide both the form of our action and the form of our path integral.

### Non-relativistic quantum mechanics as $(0+1)$-dimensional quantum field theory

A rather simple theory we can pick is just to have our fields obey similar equations to non-relativistic point mechanics,

$$S[\phi^i(t), t] = \int_{t_a}^{t_b} dt\ \left[ \frac{m}{2} \dot{\phi}^i(t) \dot{\phi}_i(t) + V(\phi^i(t)) \right]$$

Given an appropriate potential (depending only on $|\phi^i|$), this action is invariant under the rotation group $\operatorname{O}(n)$.

## Bibliography

0D QFT
Last updated : 2020-12-06 08:56:23