# Observables and Commutation (newbie questions?)

#### JDoolin

Gold Member
In this case, I think it was reasonable to expect the students to see the dot, and recall the definition of the dot product. But if I had been in his shoes, I would have explained it, just in case.
You seem to be under the impression that recalling the definition of the dot-product will help you solve this problem. I strongly disagree. I was perfectly aware of the definition of the dot-product, but that did not help me in any way.

You are given:

$$X=a_0 + \sigma \cdot a$$

You are told that a0 is a number

You are told that a is a three-by-one vector

You are told that X is a 2x2 matrix

You are not told anything at all about the nature of σ.

The key to this problem is knowing the definition of σ. There's no way to do it if you don't know.

#### Ben Niehoff

Gold Member
Well, now that you do understand the statement of the problem, what do you learn from it?

#### JDoolin

Gold Member
Well, now that you do understand the statement of the problem, what do you learn from it? Well, although I don't know if this gives any particular insight, I did come up with these results for part a (and your additional question about the determinant.

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!)

Last edited:

#### Fredrik

Staff Emeritus
Gold Member
You seem to be under the impression that recalling the definition of the dot-product will help you solve this problem. I strongly disagree. I was perfectly aware of the definition of the dot-product, but that did not help me in any way.

You are given:

$$X=a_0 + \sigma \cdot a$$

You are told that a0 is a number

You are told that a is a three-by-one vector

You are told that X is a 2x2 matrix

You are not told anything at all about the nature of σ.

The key to this problem is knowing the definition of σ. There's no way to do it if you don't know.
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.

You got the determinant wrong by the way. Last edited:

#### JDoolin

Gold Member
Ah, but how much time and frustration could be saved by would-be students of quantum-mechanics if the practice were changed? \begin{align*} X &=a_0 + \sigma \cdot a \\ &= a_0 + \sigma^k a_k\\ &= \begin{pmatrix} a_0 & 0\\ 0 & a_0 \end{pmatrix} + \begin{pmatrix} \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} & \begin{pmatrix} 0 & -i\\ i & 0 \end{pmatrix} & \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} \end{pmatrix} \begin{pmatrix} a_1\\ a_2\\ a_3 \end{pmatrix} \\ &= \begin{pmatrix} a_0+a_3 & a_2-i a_2 \\ a_1 + i a_2 & a_0-a_3 \end{pmatrix} \end{align*}

Do I have it right, now?
Just a teeny mistake in X12

It should have been:

$$X = \begin{pmatrix} a_0+a_3 & a_1-i a_2 \\ a_1 + i a_2 & a_0-a_3 \end{pmatrix}$$

Then for part (b):

$$a_0=\frac{X_{11}+X_{22}}{2}$$
$$a_1=\frac{X_{12}+X_{21}}{2}$$
$$a_2=\frac{X_{12}-X_{21}}{2i}$$
$$a_3=\frac{X_{11}-X_{22}}{2}$$

#### JDoolin

Gold Member
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

Gold Member
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

Gold Member
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
$$a_0^2<a_1^2+a_2^2+a_3^2$$

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

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

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,

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

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 &=\frac{x+y}{2} \\ a_1 &= u \\ a_2 &= v\\ a_3 &= \frac{y-x}{2} \end{align*}

We can further break down the determinant into:

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

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

Staff Emeritus
Gold Member
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

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

and

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

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

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

Gold Member
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.

Last edited:

#### JDoolin

Gold Member
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

Staff Emeritus
Gold Member
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.

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving