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Suppose i try to derive the most generic Dirac-like equation (that is, as factors of first-order expression in momenta and mass operator where we allow coefficients that are associative, don't necessarily commute to each other, but they commute with the space-time degrees of freedom)

Usually the Dirac equation is obtained by taking this expression:

$E^2 = P^2 + m^2$ with $E=i \hbar \partial_{t}$, $P=-i \hbar \partial_{x}$

and getting $E^2 = ( \beta m + \alpha_{i} P_{i} )^2$

and obtaining a satisfying algebra:

  • $\beta \alpha_{i} + \alpha_{i} \beta = 0$

  • $\alpha_{i} \alpha_{j} + \alpha_{j} \alpha_{i} = 0$

  • $\beta^2 = \alpha_{i}^2 = 1$

However what happens if i don't assume a squared expression and just try to factorize the operator; I note that the most generic expression satisfying the above would be this:

$$E^2 = ( \hat{\beta} m + \hat{\alpha_{i}} P_{i} ) ( \hat{\beta^{-1}} m + \hat{\alpha^{-1}_{i}} P_{i} )$$

where i use the hat to denote that these are more generic that the usual $\beta, \alpha_{i}$ that we deal with traditionally

This system satisfies the original expression if we demand the following algebra;

  • $\hat{\beta} \hat{ \alpha^{-1}_{i}} + \hat{\alpha_{i} } \hat{\beta^{-1}} = 0$

  • $\hat{\alpha_{i}} \hat{ \alpha^{-1}_{j}} + \hat{\alpha_{j} } \hat{\alpha^{-1}_{i}} = 0$

The question here is: It seems to me that there is a gauge freedom in specifying these quantities; what is the meaning of these gauge symmetries and what, if any, interesting invariants we should expected to get from them?

Note 1: if i assume the following representation of these hatted cliffords in terms of the well-known ones;

  • $\hat{\beta} = \boldsymbol{L} \beta \boldsymbol{M} $

  • $\hat{\alpha_{i}} = \boldsymbol{L} \alpha_{i} \boldsymbol{M}$

i notice that this satisfies the required algebra (and makes the extra degrees of freedom explicit!)

Note 2; it seems to me that the most generic solutions look like:

  • $( \hat{\beta} m + \hat{\alpha_{i}} P_{i} ) \boldsymbol{u} = \lambda_{0} \boldsymbol{v} $
  • $( \hat{\beta^{-1}} m + \hat{\alpha^{-1}_{i}} P_{i} ) \boldsymbol{v} = \lambda_{1} \boldsymbol{u} $

is this just a rewrite of the left-hand and right-hand Weyl spinors, or is something different?

I have purposely not assumed any space-time or spinorial dimensionality in the above, but feel free to make your answers about concrete dimensions

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What is it going to take to fix the display of "\hat" on this site? –  user346 Mar 13 '11 at 8:58

1 Answer 1

There is no "gauge symmetry" in any of these manipulations and you haven't offered any evidence for such a non-existent gauge symmetry.

If I understand you well, you have just realized that the exact numerical form of the Dirac matrices depends on conventions. In particular, the matrix entries of the Dirac gamma (and therefore also alpha, beta) matrices depend on the choice of the basis of the complex four-dimensional space of the Dirac spinors.

A "solution" - a set of matrices $\alpha^k$ which contains the $\beta$ as well (or similarly for $\gamma$) - may also be conjugated by an arbitrary matrix $$\{ \alpha^k \} \to \{ U \alpha^k U^{-1} \} $$ with a universal, $k$-independent matrix $U$. We often want to preserve the hermiticity or antihermiticity of the matrices so we only allow orthonormal bases of the 4D space of Dirac matrices - which means that $U$ is unitary and $U^{-1}$ is the same thing as $U^\dagger$.

It is a well-known thing that one can choose the bases of such spaces differently. For example, one may do it for vectors as well: the coordinates of a vector depend on the choice of the coordinate axes. To some extent, the choice for the spinors is independent of the choice of axes for a vector. But once we choose the gamma matrices (or, equivalently, the beta and alpha matrices), all the conventions for spinors are settled and there is no freedom left.

The ambiguity in choosing the numerical form of the gamma (or alpha, beta) matrices is not a symmetry of any theory. It's not a gauge symmetry and it's not a global symmetry, either.

It's because a particular theory describing spinorial degrees of freedom has to depend on the gamma matrices - so the Lagrangian (or the Hamiltonian or other defining quantities) explicitly depends on the form of the Dirac matrices which must be chosen in a particular way. So any well-defined Lagrangian for spinors actually breaks the would-be symmetry and removes the ambiguity. (Of course, we're still allowed to define any theory with any convention, but the precise form of the theory will be different - the theories will be equivalent, but not identical, which is why the change of the gamma matrices is not a symmetry.) Once it says which form of the gamma matrices are used, one must use the same form for all other manipulations with the spinors, otherwise the key identities etc. wouldn't hold.

But I may have completely misunderstand what you're asking. The individual sentences of your question don't really make sense. For example, you say that you find a general "solution", but the "solution" is actually another unsolved equation. It seems you are confusing the gamma matrices with the spinors themselves, solutions with equations, and so on. Consequently, you also seem to confuse the Dirac equation with the identities that allow one to prove that the Dirac equation implies the Klein-Gordon equation, and so on and so on. It's very hard to read this kind of stuff.

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Thanks for your answer. About your comments; there is a difference - in the normal derivation, there is one and only one first-order equation that is satisfied, you can use a similarity transformation in the Clifford space and see how the solutions will transform accordingly - However, here we have two first-order equations at any given time, please see Note 2. And is not evident (to me) that they can be made equivalent to the original system. Isn't the variation in the similarity transformation on each system (which might vary) not affecting any physics, hence a gauge symmetry? –  lurscher Mar 13 '11 at 13:41
    
Apologies, lurscher, I don't understand what you're saying at all. If something is not "affecting any physics", then it is called a "symmetry", not a "gauge symmetry". Gauge symmetry is a kind of symmetry that one requires the physical states to be invariant under; that's not true for every symmetry. Also, if a change of a system makes the system inequivalent to the original one, then the change surely wasn't a symmetry, was it? What do all those things have to do with the Clifford space and/or first-order equation? What you write just makes no sense to me. –  Luboš Motl Mar 13 '11 at 20:34

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