Observables and Commutation (newbie questions?)

[JDoolin]

I was hoping maybe you would interpret those formulas, maybe with words like "timelike", "spacelike", or "lightlike". It helps to step back from the math and think about what you're doing.

I may be nitpicking, but as given, the problem is unitless, so I don't see how it has any intrinsic reference to time or space.

But if by "interpret" you mean I can modify the question so the numbers actually refer to physical measurements of time and space, I would suggest the following possible interpretation.

$$X = \begin{pmatrix} c \Delta t+\Delta z & \Delta x+i \Delta y \\ \Delta x - i \Delta y & c \Delta t -\Delta z \end{pmatrix}$$

Then if my clumsy calculations are right, a positive determinant would be timelike, a negative determinant would be spacelike, and a zero determinant would be lightlike.

A further note: Does this math work out in some way to "pick out" one dimension ; the z-dimension in particular? I'm wondering whether some yet more careful interpretation of the math (i.e. definition of variables) might yield some expression of Heisenberg's uncertainty principle.

 

[JDoolin]

You're on the right track. The set of complex self-adjoint (=hermitian) 2×2 matrices is a 4-dimensional vector space over ℝ, so it's isomorphic to the vector space ℝ⁴. ℝ⁴ is of course also the underlying set of Minkowski spacetime, so any map that takes complex self-adjoint 2×2 matrices to complex self-adjoint 2×2 matrices can be used to define a map from ℝ⁴ into ℝ⁴. The maps of the form
$$X\mapsto AXA^\dagger$$ where A is a complex 2×2 matrix with determinant 1 (i.e. A is a member of SL(2,ℂ)) are especially interesting, because they are linear and preserve determinants, i.e. $\det(AXA^\dagger)=\det X$. This means that they correspond to Lorentz transformations. Note that if you replace A by -A, you get the same map. So there are two members of SL(2,ℂ) for each Lorentz transformation.

This relationship between the Lorentz group SO(3,1) and SL(2,ℂ) is the main part of the reason why SL(2,ℂ) is used instead of SO(3,1) in relativistic QM.

I'm curious about your statement that each Lorentz Transformation corresponds to two members of SL(2,C). So, are you saying that taking any matrix A in SL(2,C), and Hermitian X, composed of (cΔt, Δx, Δy, Δt) then $AXA^\dagger$ will yield X' composed of (cΔt', Δx', Δy', Δt'); i.e. a Lorentz transformed version of the original four-vector?

If you had started with complex traceless self-adjoint 2×2 matrices, you could have made essentially the same argument with ℝ³ and SU(2) instead of ℝ⁴ and SL(2,ℂ).

And that, I'll have to look at later.

 

[Fredrik]

I'm curious about your statement that each Lorentz Transformation corresponds to two members of SL(2,C). So, are you saying that taking any matrix A in SL(2,C), and Hermitian X, composed of (cΔt, Δx, Δy, Δt) then $AXA^\dagger$ will yield X' composed of (cΔt', Δx', Δy', Δt'); i.e. a Lorentz transformed version of the original four-vector?

Yes, it's a linear transformation from ℝ⁴ to ℝ⁴ that preserves the Minkowski "norm" (which as you know by now isn't really a norm). Such a transformation is called a Lorentz transformation.