# Scalar fields and spinors in 0+1D

As part of learning about SUSY quantum mechanics, I am trying to get a grasp on the following Lagrnagians in 1 (temporal dimension):

But since these early times the treatment and methods of field theory have changed drastically, and not all beginners have the solid background which is required to understand the introductions to SUSY and SUGRA in d=4 and higher dimensions. We shall descend from d=3+1 to d=0+1 dimensions: quantum mechanics (QM).

No details of QFT are required, since we shall only deal with real scalar “fields” $$\phi(t)$$ and real one-component anti-commuting spinors $$\psi(t)$$. Their free field actions are the time integral of

$$L = \frac{1}{2} \dot{\phi}\dot{\phi}$$

which one can view as the one-dimensional limit of the Klein-Gordon action for Higgs scalars; and

$$L = \frac{i}{2} \psi \dot{\psi}$$

which one can view as the one-dimensional limit of the Dirac action for quarks or leptons.

My understanding of the Klein Gordan and Dirac Lagrangian (densities) respectively are:

$$\mathcal{L} = \frac{1}{2} \partial^\mu \phi \partial_\nu \phi -\frac{1}{2}m^2 \phi^2 \\ \mathcal{L} = \bar{\psi}(i\gamma^\mu \partial_\mu -m)\psi$$

I am very close to understanding the 1 (temporal) dimensional limit of the KG action mentioned above, splitting time and space up we have

\begin{align} \mathcal{L} &= \frac{1}{2} \eta^{\mu\nu} \partial_\mu \phi \partial_\nu \phi -\frac{1}{2}m^2 \phi^2 \\ &= \frac{1}{2} \eta^{00} \partial_0 \phi \partial_0 \phi + \frac{1}{2} \eta^{ii} \partial_i \phi \partial_i \phi -\frac{1}{2}m^2 \phi^2 \\ &= \frac{1}{2} (1) \partial_0 \phi \partial_0 \phi + \frac{1}{2} (-1) \partial_i \phi \partial_i \phi -\frac{1}{2}m^2 \phi^2 \\ &\rightarrow \frac{1}{2} \dot{\phi}\dot{\phi} -\frac{1}{2}m^2 \phi^2 \end{align}

where the $$\rightarrow$$ indicates we have taken the limit $$\phi(t,\vec{x}) \rightarrow \phi(t)$$.

My only remaining confusion is:

why are we allowed to call $$L = \frac{1}{2} \dot{\phi}\dot{\phi}$$ a Klein Gordon action when it disregards the potential term? Surely the kinetic term makes since for a 1-dimensional $$\phi=\phi(t)$$, but even the generalized KG equation with some unspecified $$V(\phi)$$ potential, $$\partial^2 \phi +\frac{\partial V}{\partial \phi} =0$$ has a potential. Thus how is a purely kinetic Lagrangian a KG one?

I am much more confused with the 1-d limit of the Dirac Lagrangian, the gamma matrices are indeed matrices, not components of a matrix like $$\eta^{00}$$ from before. This is characteristic of the Dirac equation, the gamma matrices are necessary to maintain Lorentz invariance. If we look at the temporal part of (the kinetic part of) the Dirac Lagrangian, we have

$$i\bar{\psi}\gamma^0\partial_0 \psi$$

I've never experienced Lorentz symmetry in a dimension less that 4, I don't even know if this is possible. Thus how can we write a "Dirac" equation in 1 dimension?

• 1. When you are quoting things, please give a proper citation so readers know where the quote is from. 2. Are you sure your source does not simply assume the fields are massless? 3. You don't have gamma matrices or Lorentz symmetry in 0+1 dimensions - that's why your source didn't write any in their "Dirac equation". What exactly is your question about that? Commented Aug 9, 2020 at 17:45

In the scalar case, the text is probably just considering the (time-only version of) a massless scalar field (setting $$m^2=0$$), because the goal is to illustrate SUSY in the simplest possible way. That's probably also why they didn't include a potential. We don't stop calling it Klein-Gordon just because we're considering a limiting case ($$m^2\to 0$$ and $$V\to 0$$).
The Dirac side is more interesting. The Dirac equation can be formulated in any number of dimensions, and the case of one time and no space dimensions might be easier to appreciate if we consider the generalization first. In $$1+D$$ dimensions for any $$D\in\{0,1,2,...\}$$, the massless version of the Dirac equation is $$\gamma^a\partial_a\psi=0 \tag{1}$$ where the $$\gamma$$s are the smallest possible set of matrices (with complex components) satisfying the Clifford algebra relation $$\gamma^a\gamma^b+\gamma^b\gamma^a = 2\eta^{ab}, \tag{2}$$ where $$\eta^{ab}$$ is the Minkowski metric. How small can these matrices be? This is an exercise in linear algebra (cf Dimension of Dirac $\gamma$ matrices). The matrices need to have size $$2^n\times 2^n$$ where $$n$$ is the integer part of $$(1+D)/2$$. Here's a table: $$\begin{matrix} 1+D & n & 2^n \\ \hline 1 & 0 & 1 \\ 2 & 1 & 2 \\ 3 & 1 & 2 \\ 4 & 2 & 4 \\ 5 & 2 & 4 \\ 6 & 3 & 8 \\ 7 & 3 & 8 \\ \end{matrix}$$ The pattern should be clear. The case described in the question has $$1+D=1$$, so the $$\gamma$$-matrices have size $$1\times 1$$, and therefore the spinor $$\psi$$ only needs one component.
For some values of $$1+D$$, we can have a Majorana representation of the Clifford algebra, in which they have only real components but still have the same size $$2^n\times 2^n$$. We can also consider pseudo-Majorana representations (often also called Majorana representations), in which the components are purely imaginary. In either case, we can take $$\psi$$ to satisfy a reality condition of the form $$\psi^*=C\psi$$ for some matrix $$C$$. This is possible, in particular, when $$1+D=1$$. Based on the notation in the question, the text is considering a Majorana spinor, so its one-and-only component is self-adjoint (the Grassmann version of "real").
• @LopeyTall -- Weyl (chiral) spinors exist only in even-dimensional spacetimes, because only then does the product of all $1+D$ Dirac matrices have two distinct eigenvalues (whose eigenspinors are the left- and right-handed parts of a Dirac spinor). The existence of Majorana (real) spinors depends in a more complicated way on the number of dimensions (and on the signature). Equation (1) is equally valid whether $\psi$ is Dirac, Weyl, or Majorana, and I omitted the $i$ only because the right-hand side is zero, so the $i$ doesn't matter. Commented Aug 9, 2020 at 22:10
• @LopeyTall The representation of the Dirac matrices is not unique, because any similarity transform preserves equation (2). Still, in the single-component case, that component needs to have magnitude $1$ in order to be consistent with (2). On the other hand, we are free to adjust the coefficients of the different terms in the lagrangian however we want to. Those coefficients aren't dictated by the Dirac algebra. Changing the coefficients may change the model's properties, and in this case, a judicious choice of coefficients can make the model supersymmetric. Commented Aug 9, 2020 at 22:32