Observables and Commutation (newbie questions?)

[JDoolin]

My point was that since the dot product for vectors in $\mathbb R^n$ is defined by $\mathbf{x}\cdot\mathbf{y}=\sum_i\, x_i y_i$, your first guess about what $\mathbf{\sigma}\cdot\mathbf a$ means should (or at least could) have been $\sum_i\sigma_i a_i$. Of course, if the author hasn't even defined the sigmas, I can see how it would be confusing.

Ah. Empathy. Thank you.
:)

You got the determinant wrong by the way.

In fact, I'm glad you told me, because I've been scratching my head since yesterday morning wondering why I got that answer. Now I see I just screwed up a couple of minus-signs.

 

[Ben Niehoff]

No sense in using spoiler tags for this stuff, we already know the answers.

Tr[X] = 2 a_0
Tr[σ_1 X] = 2 a_1
Tr[σ_2 X] = 2 a_2
Tr[σ_3 X] = 2 a_3
Det[X]= a_0^2-a_1^2-a_2^2 -a_3^2
(Edit: Determinant modified since Fredrik's correction.)

At this point, I just see a really neat mathematical pattern, but I don't yet have any particular insight as to how that pattern matches anything in physical reality. I hope to gain more insight as I do more problems. (I started #3 in the book, but forgot to finish #2b!)

OK, so if I tell you det(X) = 0, what does that tell you about the 4-vector (a_0, \vec a)? Similarly for det(X) < 0, and det(X) > 0.

What would you say is the relationship between the space of 2x2 Hermitian matrices, and 3+1-dimensional Minkowski space?

 

[JDoolin]

No sense in using spoiler tags for this stuff, we already know the answers.



OK, so if I tell you det(X) = 0, what does that tell you about the 4-vector (a_0, \vec a)? Similarly for det(X) < 0, and det(X) > 0.

If det[X]=0 then
a_0^2=a_1^2+a_2^2+a_3^2

and

If det[X]<0 then
a_0^2&lt;a_1^2+a_2^2+a_3^2

If det[X]>0 then
a_0^2&gt;a_1^2+a_2^2+a_3^2

(a_0, \vec a) = \left ( \pm\sqrt{a_1^2+a_2^2+a_3^3},\vec a \right )

What would you say is the relationship between the space of 2x2 Hermitian matrices, and 3+1-dimensional Minkowski space?

Have we generated the complete space of 2x2 Hermitian matrices by defining X? I suppose we have!

A 2x2 Hermitian matrix, in general form, would be written,

\begin{pmatrix} x &amp; u+vi\\ u-vi &amp; y \end{pmatrix}

Now the determinant of this thing is:


xy+u^2+v^2

However, with insight provided by the preceding problem,

Let \begin{align*} a_0 &amp;=\frac{x+y}{2} \\ a_1 &amp;= u \\ a_2 &amp;= v\\ a_3 &amp;= \frac{y-x}{2} \end{align*}

We can further break down the determinant into:

Det\begin{pmatrix} x &amp; u+iv \\ u-iv &amp; y \end{pmatrix}= -\left ( \frac{x+y}{2} \right )^2+\left ( \frac{y-x}{2} \right )^2+u^2+v^2

Now, what would I say is the relationship between the space of 2x2 Hermitian matrices, and 3+1-dimensional Minkowski space? It would be a rather complicated statement to put into words, involving the midpoint and half-the-difference of the real terms of the Hermitian Matrix, and the real and imaginary parts of the complex terms. Once I specified those four terms, if I were to decide to give those four variables names like cΔt, Δx, Δy, Δz, then you would see that the determinant turned out to be identical to the definition of the space-time-interval in Minkowski space-time.

Hmmm, so could we define the variables at the beginning, in such a way that this mathematical identicalness actually means something more than a superficial similarity?

 

[Fredrik]

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.

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).

 

[Ben Niehoff]

If det[X]=0 then
a_0^2=a_1^2+a_2^2+a_3^2

and

If det[X]<0 then
a_0^2&lt;a_1^2+a_2^2+a_3^2

If det[X]>0 then
a_0^2&gt;a_1^2+a_2^2+a_3^2

(a_0, \vec a) = \left ( \pm\sqrt{a_1^2+a_2^2+a_3^3},\vec a \right )

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.

 

[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 &amp; \Delta x+i \Delta y \\ \Delta x - i \Delta y &amp; 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.