# The doubling effect

Fermion doubling is a strange effect, which has been observed when people have tried to make computations with fermions using lattice approximations. It happens if one tries to approximate the Dirac equation (the basic equation for fermions) on a lattice. The naive way to do this is quite straightforward: The partial derivatives in the point $$x = n \Delta$$ (where $$\Delta$$ is the distance between the lattice nodes) are replaced by the corresponding central differences. In the one-dimensional case, the central difference is $\frac{d}{dx} f(x) \to \frac{f(n+1)-f(n-1)}{2\Delta}$

The generalization to higher dimensions is straightforward: $\frac{\partial}{\partial x} f(x,y,z) \to \frac{f(n_1+1,n_2,n_3)-f(n_1-1,n_2,n_3) }{2\Delta}$ $\frac{\partial}{\partial y} f(x,y,z) \to \frac{f(n_1,n_2+1,n_3)-f(n_1,n_2-1,n_3) }{2\Delta}$ $\frac{\partial}{\partial z} f(x,y,z) \to \frac{f(n_1,n_2,n_3+1)-f(n_1,n_2,n_3-1) }{2\Delta}$

and so on for all spacetime dimensions.

Unfortunately, the resulting lattice equation does not describe what it was intended to do — the Dirac equation for a single fermion. Instead, it describes much more, namely, 16 Dirac particles. We have a "doubling factor" two in each of the four spacetime dimensions.

What happens is not that difficult to understand. Especially there is no need to know the Dirac equation. The interesting, qualitative effects can be reproduced in simple one- or two-dimensional examples.

## One-dimensional example

Let's try to describe what happens in a simple example. The simplest case is the quite trivial equation $\frac{d}{dx} f(x) = 0$

which gives a constant f(x) = c as the solution. With our naive lattice equation, we obtain $f(n+1)-f(n-1) = 0.$

But this equation has not only the constant solution f(n) = c. The solution may be different for even and odd nodes: $f(2n) = c_0; \qquad f(2n+1) = c_1$

## Two-dimensional example

This example can be easily generalized to the two-dimensional case. Here we have two equations: $\frac{\partial}{\partial x} f(x,y) = 0; \qquad \frac{\partial}{\partial y} f(x,y) =0$

Again, the general solution of the continuous equation is a single constant f(x,y) = c.

The lattice is two-dimensional, thus, we have a lattice function f(m,n) of two integer variables, and the lattice equations are $f(m+1,n) - f(m-1,n) = 0; \qquad f(m,n+1) - f(m,n-1) = 0.$

and obtain now a general solution which depends on four independent constants: $f(2m,2n) = c_{00}; \qquad f(2m+1,2n) = c_{10};\qquad f(2m,2n+1) = c_{01};\qquad f(2m+1,2n+1) = c_{11};$

It is now easy to generalize this to arbitrary dimension. Each additional dimension gives an additional factor two for the number of doublers. Especially, in our lattice model, we have a three-dimensional lattice. This gives a doubling factor $$2^3=8$$. If we consider a four-dimensional spacetime lattice, we obtain a doubling factor $$2^4=16$$.

## An example without decoupling

In these two minimal examples, the equations for the different "doublers" simply decouple: We have independent lattice equations on odd and even (in each direction) lattice nodes. Thus, there is a simple way to get rid of the additional fields: Simply throw away the equation on the odd nodes, leaving only one of the equations.

For more complicate equations this is not possible. Let's consider, as an example, the equation $\frac{d}{dx} f(x)=f(x),$

with $$f(x) = c e^{x}$$ as the solution. The equation no longer splits into two independent equations: $f(n+1) - f(n-1) = 2\Delta f(n).$

Nonetheless, we can give two independent initial values f(0) and f(1), instead of only a single initial value f(0) of the continuous equation. Thus, the decoupling disappears, but the doubling effect remains.

## An example with a staggered sublattice

The case of the Dirac equation in four dimensions is an intermediate one. From the sixteen Dirac particles, which we obtain by naive discretization of the Dirac equation, four groups of four Dirac fermions decouple. So we can restrict the lattice equation to a subset of the equations. The equations on this subset are known as "staggered fermions". The reason it is named "staggered" is that different components of the Dirac fermion are located on different sublattices.

How such a staggered sublattice appears can also be demonstrated in a simple case. The equation is $\frac{d}{dx} f(x)=g(x),\qquad \frac{d}{dx} g(x)=f(x),$

The corresponding naive lattice equation $\frac{f(n+1) - f(n-1)}{2\Delta} = g(n); \qquad \frac{g(n+1) - g(n-1)}{2\Delta} = f(n).$

decomposes into two independent equations: One part connects the f(2n) with the g(2n+1), the other one connects f(2n+1) with g(2n). The solution of this "doubling problem" consists of taking one of these sub-equations (say the first): $\frac{f(2n+2) - f(2n)}{2\Delta} = g(2n+1);\qquad \frac{ g(2n+1) - g(2n-1)}{ 2\Delta} = f(2n).$

This gives not simply a subset of even or odd nodes, but a "staggered lattice", where we have the function f(x) defined on even nodes f(2n) and the function g(x) defined on odd nodes g(2n+1).

## Almost continuous and heavily oscillating solutions

To understand the nature of the additional solution - the doublers - it is useful to identify the original solution among the set of lattice solutions. In the first cases, the original solution is given by $c=c_0 = c_1$

resp. $c = c_{00} = c_{01} = c_{10} = c_{11}$

In the last case, the original solution may be characterized by the initial conditions $f(1) = f(0) e^{\Delta} \sim f(0)$

for a fine lattice with small $$\Delta$$. We see that the "expected" solutions are almost continuous — the values of neighbour nodes are close to each other. Instead, the "unexpected" doublers behave very different — they heavily oscillate.

Note also that the "oscillating" modes not necessarily follow the same equations as the almost continuous solutions. Especially in the last example, the "oscillating" solution with the initial values $f(1) = - f(0)$

gives a solution which is approximately $f(n) \sim (-1)^n f(0) e^{-\Delta n}$

Its "smooth part" $$f(0) e^{-x}$$ is the solution of the equation $\frac{d}{dx} f(x) = -f(x)$

Thus, in the last example the two doublers follow different, but independent, equations.

## A way to take a square root of quadratic equations

But note that they also fulfill a common equation: $\frac{d^2}{dx^2} f(x) = f(x)$

This may be interesting if we do not really want to solve the original first order equation, but this quadratic equation. In this case, the "doubling effect" may appear as a useful tool: The continuous limit has to be described by two solutions — one almost continuous, the other heavily oscillating. The smooth parts of above solutions fulfill the quadratic equation, and the general solution of this quadratic equation is a linear combination of them.

It is also interesting to consider the connection between some of the previous examples:

• Taking the square root of $$\frac{d^2}{dx^2} f(x) = f(x)$$ in the continuous case gives the set of equations $$\frac{d}{dx} f(x)=g(x),\frac{d}{dx} g(x)=f(x)$$.
• The naive discretization of this equation gives $$f(n+1) - f(n-1) = 2\Delta g(n); g(n+1) - g(n-1) = 2\Delta f(n)$$, which decouples into two equations on two staggered lattices.
• The equation on the staggered sublattice containing f(2n) and g(2n+1) becomes, after renaming g(2n+1) into f(2n+1), the lattice equation $$f(n+1) - f(n-1) = 2\Delta f(n)$$, which naively approximates $$\frac{d}{dx} f(x)=f(x)$$. Note that $$\frac{d}{dx} f(x)=f(x)$$ is one of the ways to take the square root of $$\frac{d^2}{dx^2} f(x) = f(x)$$.
• The "naive" discretization $$f(n+1) - f(n-1) = 2\Delta f(n).$$ of the "naive" square root $$\frac{d}{dx} f(x)=f(x)$$ of $$\frac{d^2}{dx^2} f(x) = f(x)$$ gives, via the doubling effect, the other solutions of $$\frac{d^2}{dx^2} f(x) = f(x)$$ as doublers.

## The case of the Dirac equation

Now, taking the square root of the Klein-Gordon equation was the aim of Dirac when he developed the Dirac equation. And what happens in the case of the Dirac equation is quite similar:

• Taking the square root of the Klein-Gordon equation in the continuous case gives the Dirac equation — a set of equations with four complex components.
• The naive discretization of the Dirac equation gives a set of lattice equations, which decouples into four equations on staggered sublattices.
• The equation on the staggered sublattice contains one complex field in each lattice node. The components may be renamed, forgetting about their staggered nature. This gives a lattice equation for a single complex field on the lattice. It may be interpreted as the "naive" discretization for one of the ways to take a square root out of the Klein-Gordon equation.
• Via the doubling effect, this lattice equation gives, as doublers, other solutions, which describe other ways to take the square root out of the Klein-Gordon equation.

Thus, some of the qualitative properties of the "fermion doubling" effect may be understood by considering our simple one-dimensional analog.

But there remains a difference: What we obtain on the staggered lattice — resp. as an equation for a single complex field on the lattice — is not a single Dirac fermion.

• In the case of a four-dimensional spacetime lattice, we obtain an equation for four Dirac fermions.
• In our case, we have only a lattice in space. As a consequence, we do not obtain a doubling effect in the direction of time. Thus, we have a factor two less doublers as in the case of the spacetime lattice. Thus, we obtain only two Dirac fermions.

But in the SM, all Dirac fermions appear in pairs, known as electroweak doublets. Thus, there is a natural solution for the "doubling problem" in our case, where doubling gives only two Dirac fermions — we interpret the resulting lattice equation for a single complex field on the lattice as the lattice Dirac equation for a whole electroweak doublet.

## The "almost continuous" solution and translations

As we have seen, the doublers look qualitatively different: One of them is an almost continuous solution, the others oscillate heavily.

In our approach, the almost continuous solution plays a special role too. The point is Euclidean symmetry, especially translational symmetry. Translations $$t^i$$ act on the lattice fields which describe lepton doublets as $a^i_0 \to a^i_0 + t^i$

Obviously, the translation shifts only the almost continuous parts. The oscillating parts oscillate as before.

We identify the almost continuous solution with some linear combination of the right-handed neutrino and the corresponding antineutrino. This identification has an important consequence: Any gauge field, which does not interact with the right-handed neutrino and its antiparticle, commutes with translations, thus, preserves translational symmetry.