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I was taught that for the Dirac-equation to "work", you need matrices of the following form:

  • $Tr(\alpha^i) = 0$.
  • Eigenvalues +1 or -1
  • 2 previous points together: equal number of negative and positive eigenvalues, thus even dimension.
  • $\alpha^i$ and $\beta$ meet the anti-commutator relation.

My Advanced QM professor told me convincingly that there are only 3 linearly independent 2D matrices, the Pauli matrices, which can be used in 2D. That's one short, so you'll need at least 4D matrices.

Now my QFT professor, who excells at confusing me (and not only me, mind you), stated that it is perfectly possible to use only 2D matrices, by extending the Pauli matrix set with another one. I think he's wrong, but he seems too old (and "wise") to be wrong, and I didn't remember all conditions above at the time, so I didn't want to argue without arguments.

Is it possible to 2D-ify the Dirac equation? (and not only in a 2D system like graphene, but in general)


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up vote 9 down vote accepted

Dear rubenb, yes, what your professor says is surely based on solid maths. The reason is that the 4-component Dirac spinor is actually composed of two separate 2-component pieces.

The elementary "spinors" for 3+1 dimensions have two complex components. That results from the isomorphism between groups $$SL(2,C) \sim Spin (3,1).$$ Note that both groups have 6 real generators. In particular, in a properly chosen basis, the $\alpha^i$ and $\beta$ matrices may be brought into a block-diagonal form with $2\times 2$ blocks. It follows that the $2\times 2$ blocks themselves satisfy the same algebra.

In particular, the four matrices may be written as $$(\beta, \alpha^i) = (1_{2\times 2}, \sigma^i)\equiv \sigma^\mu$$ i.e. as the Pauli matrices supplemented with the identity matrix. Note that the $\alpha^i$ i.e. $\sigma^i$ matrices anticommute with each other while they commute with $\beta$ i.e. $\sigma^0$ and all the matrices square to the identity much like $\beta$, $\alpha^i$ do.

The isomorphism above may be viewed as a "noncompact extension" of the usual isomorphism $$SU(2) \sim Spin(3).$$ Note that the group $SU(2)$ is a subgroup of $SL(2,C)$ - it's the same pair as $Spin(3)$ which is a subgroup of $Spin(3,1)$.

The two-component spinors are directly relevant for the description of the neutrinos. They only describe a left-handed massless particle (and right-handed massless antiparticle). That's different from the 4-component Dirac spinor that describes a particle that can either left-handed or right-handed. The neutrino is given by a Weyl spinor and the free equation is simply $$\sigma^\mu \partial_\mu \chi = 0$$ which is Lorentz-covariant. However, one must realize that the 4-vector of $2\times 2$ matrices, $\sigma^\mu$, don't transform the 2-dimensional complex space (left-handed Weyl spinors) onto itself but onto another 2-dimensional complex space (of right-handed Weyl spinors) which is the complex conjugate of the first one.

Massive charged particles such as the electron require a 4-component spinor - i.e. a pair of two 2-component spinors - but for neutrinos, the minimum amount to describe a single particle is given by one 2-component spinor.

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Thanks for the prompt answer. So it is not correct to use 2x2 matrices for massive particles? But the isomorphism between the two groups should hold regardless of the value of m, no? – rubenvb Mar 8 '11 at 14:10
@rubenvb: you can write down mass terms fine--the mass term in the action will simply mix one of the spinors with the other one. It's having both a mass and a charge term that meses things up – Jerry Schirmer Mar 8 '11 at 14:45
Dear @rubenvb, exactly as Jerry says. In the equation $\sigma^\mu \partial_\mu\chi = 0$, you can't just add the mass term $m\chi$ because the derivative term transforms as a right-handed spinor while $m\chi$ transforms as the opposite left-handed spinor. However, you may add $m\bar\chi$, the complex conjugate. However, $\bar\chi$ carries the opposite charges than $\chi$, so the equation will only preserve the charges if the charges of $\chi$ vanish. In that case, the $m$ parameter is called the "Majorana mass". However, it's not possible for charged electrons etc. You need two 2-spinors then. – Luboš Motl Mar 8 '11 at 15:06
@Luboš Motl: Thank you for the extended and understandable explanation! – rubenvb Mar 8 '11 at 15:19
It was a pleasure, @rubenvb. – Luboš Motl Mar 8 '11 at 15:40

Luboš' Answer is just fine, and he's adopted a level of presentation that looks just right for you, but I prefer to put this somewhat differently, using the manifestly Lorentz covariant $\gamma^\mu$ operators instead of using the operators $(\beta,\alpha^i)$. For the massive Dirac equation, you need four objects that satisfy the relationships $\gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu=2g^{\mu\nu}$ to make the presentation manifestly covariant. Using these, you can construct 6 linearly independent objects of the form $\gamma^0\gamma^1$, 4 linearly independent objects of the form $\gamma^0\gamma^1\gamma^2$, and just one linearly independent object of the form $\gamma^0\gamma^1\gamma^2\gamma^3$, 15 in all, which together with the identity gives the same number of dimensions as the algebra of 4x4 matrices. Indeed we can prove that the algebra generated by the $\gamma^\mu$ over the complex field with the relationships above is isomorphic to the algebra of 4x4 complex matrices.

To get to 2x2 matrices without breaking Lorentz invariance (almost), we use the object $\gamma^0\gamma^1\gamma^2\gamma^3$, which is invariant under Lorentz transformations except reflections and which commutes with an 8-dimensional part of the algebra and anti-commutes with a distinct 8-dimensional part of the algebra. We can use this to construct an almost Lorentz-invariant projection $\frac{1}{2}(1+i\gamma^0\gamma^1\gamma^2\gamma^3)$. This projection splits up the 4-dimensional vectors that a 4x4 matrix acts on into two 2-dimensional parts, with eigenvalues 1 and 0. In the chiral representation, this projection operator is $\left(\array{I\ 0\\ 0\ 0}\right)$, where $I$ is a 2x2 identity matrix.

This is only a tiny piece of what there is to pick up about Dirac matrices. You will pick up lots of pieces along the way. Some people prefer to work with matrices, some people prefer to work in the slightly more abstract way I've used above, others (mathematicians!) prefer still higher levels of abstraction than you see above. The above is not entirely explicit, but it's much better to work through these things for yourself as far as possible.

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Thanks for the alternative explanation (although it's largely the identical except for the expliciteness with the $\gamma$'s. I feel I'm more at home in groups than algebra's, although both seem to have been largely skipped in all my courses thus far. Two questions: 1) what do you mean exactly by "almost Lorentz invariant"? 2) what do you mean by 8-dimensional algebra? Is it something like 4 real and 4 imaginary "dimensions" or is it something else? – rubenvb Mar 8 '11 at 16:40
@rubenvb The manifestness of the covariance is (very) worthwhile. It keeps the algebra and geometry under much better mathematical control, so that harder calculations can be done more easily. By "almost", I mean yes under rotations and boosts, no under reflections. That's why 2d left-handed spinors transform into 2d right-handed spinors under reflection. There is an 8-dimensional Lorentz-invariant subalgebra of the Dirac algebra containing the 6 $\gamma^\mu\gamma^\nu$, the identity, and $\gamma^0\gamma^1\gamma^2\gamma^3$. In the chiral rep, these are 2x2 block matrices on the diagonal. – Peter Morgan Mar 9 '11 at 12:45

There is a recent book by Jean Hladik "Spinors in Physics" that looks good. And there are websites on Clifford Algebra, and articles by Pertti Lounesto and yet more by David Hestenes. Dirac Algebra is a Clifford Algebra, and that is very straightforward stuff. You take any number of anticommuting letters, with any signature, and form a 'number' consisting of all combinations of letters taken r at a time, say: 1 + t + x + y + z + tx + ty + tz + xy + yz + zx + txy + tyz + tzx + xyz + txyz (+ - - -) and now you have a 'spacetime algebra' with all the subspaces of spacetime explicitly listed, and the 'even subalgebra' 1 + tx + ty + tz + xy + yz + zx + txyz can be used to rotate a fourVector - ie, a Lorentz transformation. Dirac uses complex coefficients. The gamma matrices are just the t,x,y,z above, in a matrix representation, with all the 'bilinear covariants' listed. The groups and matrices make it look more complicated than it is.

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What has not been explicitly pointed out in the answers above is that the Dirac equation can be expressed in 2-spinor form. Because it is really a 4 spinor object two equations are required.

$\nabla^A_{B'} \alpha_A = 2^{-1/2}M\beta_{B'}$

$\nabla^{B'}_{A} \beta_{B'} = 2^{-1/2}M\alpha_{A}$

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