# Dirac equation in 1+1D spacetime compared to “standard” 3+1D Dirac equation

In the past couple of weeks I've been studying the Dirac equation and its solutions. During a discussion with a tutor it was pointed out to me that one could formulate something similar to the Dirac equation in (1,1)-spacetime. I got interested in the topic, although it doesn't really seem to be relevant for the course I'm taking (so I can't really ask the tutor to explain things here in length...), and looked it up. My main resource here is this document, which gives an nice introduction. I've reproduced most of the calculations up to page 8, where the author seems to shift the focus on the Clifford algebra in different dimensions.

After reading  I've been wondering if there are some conclusions that we could draw from studying this ''toy Dirac equation'' that relate to the Dirac equation, but I'm having a hard time identifying them.

Similar porperties to the Dirac equation:

• $$\gamma$$-matrices satisfying the Clifford algebra,
• transformation properties of the fields under appropriate Lorentz transformations (eq. (18.2) is true for both equations),
• method to find generators of the Field transformation for this specific representation (see page 5 with the result in eq. (21) compared to for example this derivation on page 318/319)

Differences between the "toy" and Dirac equation:

• $$SO^+(1,1)$$ only admits boosts compared to $$SO^+(1,3)$$ with rotations and boosts,

• the spinors in the "toy Dirac equation" are two-component vectors in contrast to the four-component vectors in the Dirac equation,

• ''In the present context, however, the two branches of the spinor representation of the Lorentz group $$O(1, 1)$$ are — uncharacteristically — disjoint.'' page 7 of .

To my question: The first question would be if there are other noteworthy differences that I missed in the above list. The second question is what kind of consequences do these differences entail for the two dimensional spacetime. For example, does the fact that we don't have rotations imply that the notion of spin doesn't exist in this space? What kind of "particles" would these two component vectors in (1,1) spacetime describe? What does it actually imply that the "two branches of spinor representations" are disjoint?

The author, as far as I can see, didn't really write anything on that matter and I really couldn't find any understandable answer on my level (physics undergrad) so I was hoping that someone had maybe a reference on the topic or could answer the question.

Some differences are that "properly speaking", "in (1 + 1) dimensions, there is no such thing as spin" which would explain the "I had hoped to be able to assign meaning to the “spin” of the ψ-field" quote in your document, and comes from the fact that "The little group $$G(0)$$ can be taken as the definition of the spin of a particle" (P.307) , but despite this "there are Majorana or Weyl spinors in two dimensions for any choice of signature. In addition, in two-dimensional Minkowki spacetime there are Majorana-Weyl fermions", as can be seen from the table (From Polchinski Vol. 2 Appendix A) below, where the size of the Dirac gamma matrices depends on the dimension of space and the possibility of reducing the Dirac representation to Weyl or Majorana representations depends on the dimensions, where e.g. in 4D we can only have Weyl or Majorana but not both, while in 1-1 we can have both. These are some basic differences. Note the $$SO(2,1)$$ dimensional case also has bizarre properties allowing for "anyons", analogues of bosons and fermions. Regarding the "branches" comment on P7 of your notes - just like the Lorentz group $$SO(3,1)$$ is disconnected, the group $$SO(1,1)$$ is also disconnected, as is any $$SO(p,q)$$ group when both $$p,q > 1$$. The author is comparing this disconnected group to the connected group $$SO(3)$$ in trying to understand what's going on, and so refers to $$SO(1,1)$$ being disconnected as "uncharacteristically disjoint" when compared with $$SO(3)$$ however this is very natural since the Lorentz group $$SO(3,1)$$ is also disconnected.

• Thank you very much for the answer! There are still some points in your answer that are a bit unclear to me... For example: "in two dimensions we do not have 'two branches' we now have an infinite number of 'branches'." What exactly do you mean here? It seems I don't really have the right idea about these "branches".. Could you maybe explain this in a bit more detail? And about the anyons, do I understand correct that these particles would only exist in (1,1)-spacetime? – Sito Apr 13 at 16:36
• Edited to address issues like "spin" and other differences directly, also addressed the branches topic which is hopefully natural (proof for the $SO(3,1)$ analogue on page 3 and mentioned here) and added the $SO(1,d-1)$ table in place of the $SO(d)$ table. – bolbteppa Jul 13 at 0:52

On the notion of "does spin exist in 1+1 dimensions", I think this depends on how one defines it. The gamma matrices can be split into time plus the Pauli spin matrices; what you are doing when going from 3 dimensions in the Pauli spin matrices to just 1 dimension is that you are restricting to $$\sigma_z$$ and leaving the other $$\sigma$$ matrices off. So spin still exists as eigenvectors to $$\sigma_z$$.

What "the two branches of the spinor are disjoint" means: A difference between vectors (which follow the fundamental representation of SO(3)) and spinors (which follow the fundamental representation of SU(2)) is that spinors are "double valued". Rotating a spinor by $$2\pi$$ changes its sign, i.e. multiplies it by -1, no such change occurs to a vector.

Now the phrase "rotating a spinor by $$2\pi$$'' needs to be more carefully defined. To get the complex phase $$-1 = \exp(2i\pi\;\;(2\pi/4\pi))$$ you need for the rotation to carve out $$2\pi$$ ster-radians of the Bloch sphere whose total surface area is $$4\pi$$. You could do this by rotating along a great circle arc from +z to +x to -z to -x and then to +z. But this requires the use of the x dimension and so is impossible in 1+1 dimensions. Hence the double cover is disjoint in 1+1 dimensions while you can use rotations to show it is not in 2+1 or 3+1 or 4+1 or bigger.

As an excuse for doing calculations, you can do the rotation by multiplying bras and kets for spin in various directions. The minus sign from rotating a spinor through the great circle route going from +z to +x and back around through to +z again is given by a product of projection operators (pure density matrices) which will give a minus sign times the projection operator for spin in the +z direction. That is:

$$|+z\rangle\langle+z||-x\rangle\langle-x||-z\rangle\langle-z||+x\rangle\langle+x||+z\rangle\langle+z| = -|+z\rangle\langle+z|/4$$

where the - sign comes from the geometric phase $$\exp(2i\pi\;(2\pi/4\pi)) = \exp(i\pi) = -1$$ and the 1/4 comes from the four losses of each $$\sqrt{1/2}$$ in amplitude at each spin change.

Now switch from spinors to vectors (say the fundamental rep of SO(3) or spin-1 rep of SU(2) and do the same calculation as above the minus sign goes away as the representations are not double covers.